{"id":598462,"date":"2022-05-03T06:04:30","date_gmt":"2022-05-03T06:04:30","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=598462"},"modified":"2022-05-10T07:49:01","modified_gmt":"2022-05-10T07:49:01","slug":"selina-class-7-icse-solutions-mathematics-chapter-15-triangles","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-15-triangles\/","title":{"rendered":"Selina Class 7 ICSE Solutions Mathematics : Chapter 15-\u00a0Triangles"},"content":{"rendered":"\n

Class 7: Maths Chapter 15 solutions. Complete Class 7 Maths Chapter 15 Notes.<\/p>\n\n\n\n

Selina Class 7 ICSE Solutions Mathematics : Chapter 15- Triangles<\/h2>\n\n\n\n

Selina 7th Maths Chapter 15, Class 7 Maths Chapter 15 solutions<\/p>\n\n\n\n

Exercise 15A page: 176<\/h4>\n\n\n\n

1. State, if the triangles are possible with the following angles :
(i) 20\u00b0, 70\u00b0 and 90\u00b0
(ii) 40\u00b0, 130\u00b0 and 20\u00b0
(iii) 60\u00b0, 60\u00b0 and 50\u00b0
(iv) 125\u00b0, 40\u00b0 and 15\u00b0<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

In a triangle, the sum of three angles is 1800<\/sup><\/p>\n\n\n\n

(i) 20\u00b0, 70\u00b0 and 90\u00b0<\/p>\n\n\n\n

Sum = 200<\/sup> + 700<\/sup> + 900<\/sup> = 1800<\/sup><\/p>\n\n\n\n

Here the sum is 1800<\/sup> and therefore it is possible.<\/p>\n\n\n\n

(ii) 40\u00b0, 130\u00b0 and 20\u00b0<\/p>\n\n\n\n

Sum = 40\u00b0 + 130\u00b0 + 20\u00b0 = 1900<\/sup><\/p>\n\n\n\n

Here the sum is not 1800<\/sup> and therefore it is not possible.<\/p>\n\n\n\n

(iii) 60\u00b0, 60\u00b0 and 50\u00b0<\/p>\n\n\n\n

Sum = 60\u00b0 + 60\u00b0 + 50\u00b0 = 1700<\/sup><\/p>\n\n\n\n

Here the sum is not 1800<\/sup> and therefore it is not possible.<\/p>\n\n\n\n

(iv) 125\u00b0, 40\u00b0 and 15\u00b0<\/p>\n\n\n\n

Sum = 125\u00b0 + 40\u00b0 + 15\u00b0 = 1800<\/sup><\/p>\n\n\n\n

Here the sum is 1800<\/sup> and therefore it is possible.<\/p>\n\n\n\n

2. If the angles of a triangle are equal, find its angles.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

In a triangle, the sum of three angles is 1800<\/sup><\/p>\n\n\n\n

So each angle = 1800<\/sup>\/3 = 600<\/sup><\/p>\n\n\n\n

3. In a triangle ABC, \u2220A = 45\u00b0 and \u2220B = 75\u00b0, find \u2220C.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

In a triangle, the sum of three angles is 1800<\/sup><\/p>\n\n\n\n

\u2220A + \u2220B + \u2220C = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

450<\/sup> + 750<\/sup> + \u2220C = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

1200<\/sup> + \u2220C = 1800<\/sup><\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

\u2220C = 1800<\/sup> \u2013 1200<\/sup> = 600<\/sup><\/p>\n\n\n\n

4. In a triangle PQR, \u2220P = 60\u00b0 and \u2220Q = \u2220R, find \u2220R.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider \u2220Q = \u2220R = x<\/p>\n\n\n\n

\u2220P = 60\u00b0<\/p>\n\n\n\n

We can write it as<\/p>\n\n\n\n

\u2220P + \u2220Q + \u2220R = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

600<\/sup> + x + x = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

600<\/sup> + 2x = 1800<\/sup><\/p>\n\n\n\n

2x = 1800<\/sup> \u2013 600<\/sup> = 1200<\/sup><\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

x = 1200<\/sup>\/2 = 600<\/sup><\/p>\n\n\n\n

\u2220Q = \u2220R = 600<\/sup><\/p>\n\n\n\n

Therefore, \u2220R = 600<\/sup>.<\/p>\n\n\n\n

5. Calculate the unknown marked angles in each figure:<\/strong><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

In a triangle, the sum of three angles is 1800<\/sup><\/p>\n\n\n\n

(i) From figure (i)<\/p>\n\n\n\n

900<\/sup> + 300<\/sup> + x = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

1200<\/sup> + x = 1800<\/sup><\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

x = 1800<\/sup> \u2013 1200<\/sup> = 600<\/sup><\/p>\n\n\n\n

Therefore, x = 600<\/sup>.<\/p>\n\n\n\n

(ii) From figure (ii)<\/p>\n\n\n\n

y + 800<\/sup> + 200<\/sup> = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

y + 1000<\/sup> = 1800<\/sup><\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

y = 1800<\/sup> \u2013 1000<\/sup> = 800<\/sup><\/p>\n\n\n\n

Therefore, y = 800<\/sup>.<\/p>\n\n\n\n

(iii) From figure (iii)<\/p>\n\n\n\n

a + 900<\/sup> + 400<\/sup> = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

a + 1300<\/sup> = 1800<\/sup><\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

a = 1800<\/sup> \u2013 1300<\/sup> = 500<\/sup><\/p>\n\n\n\n

Therefore, a = 500<\/sup>.<\/p>\n\n\n\n

6. Find the value of each angle in the given figures:<\/strong><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) From the figure (i)<\/p>\n\n\n\n

\u2220A + \u2220B + \u2220C = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

5x0<\/sup> + 4x0<\/sup> + x0<\/sup> = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

10x0<\/sup> = 1800<\/sup><\/p>\n\n\n\n

x = 180\/10 = 180<\/sup><\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

\u2220A = 5x0<\/sup> = 5 \u00d7 180<\/sup> = 900<\/sup><\/p>\n\n\n\n

\u2220B = 4x0<\/sup> = 4 \u00d7 180<\/sup> = 720<\/sup><\/p>\n\n\n\n

\u2220C = x = 180<\/sup><\/p>\n\n\n\n

(ii) From the figure (ii)<\/p>\n\n\n\n

\u2220A + \u2220B + \u2220C = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

x0<\/sup> + 2x0<\/sup> + 2x0<\/sup> = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

5x0<\/sup> = 1800<\/sup><\/p>\n\n\n\n

x0<\/sup> = 1800<\/sup>\/5 = 360<\/sup><\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

\u2220A = x0<\/sup> = 360<\/sup><\/p>\n\n\n\n

\u2220B = 2x0<\/sup> = 2 \u00d7 360<\/sup> = 720<\/sup><\/p>\n\n\n\n

\u2220C = 2x0<\/sup> = 2 \u00d7 360<\/sup> = 720<\/sup><\/p>\n\n\n\n

7. Find the unknown marked angles in the given figure:<\/strong><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) From the figure (i)<\/p>\n\n\n\n

\u2220A + \u2220B + \u2220C = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

b0<\/sup> + 500<\/sup> + b0<\/sup> = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

2b0<\/sup> = 1800<\/sup> \u2013 500<\/sup> = 1300<\/sup><\/p>\n\n\n\n

b0<\/sup> = 1300<\/sup>\/2 = 650<\/sup><\/p>\n\n\n\n

Therefore, \u2220A = \u2220C = b0<\/sup> = 650<\/sup>.<\/p>\n\n\n\n

(ii) From the figure (ii)<\/p>\n\n\n\n

\u2220A + \u2220B + \u2220C = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

x0<\/sup> + 900<\/sup> + x0<\/sup> = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

2x0<\/sup> = 1800<\/sup> \u2013 900<\/sup> = 900<\/sup><\/p>\n\n\n\n

x0<\/sup> = 900<\/sup>\/2 = 450<\/sup><\/p>\n\n\n\n

Therefore, \u2220A = \u2220C = x0<\/sup> = 450<\/sup>.<\/p>\n\n\n\n

(iii) From the figure (iii)<\/p>\n\n\n\n

\u2220A + \u2220B + \u2220C = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

k0<\/sup> + k0<\/sup> + k0<\/sup> = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

3k0<\/sup> = 1800<\/sup><\/p>\n\n\n\n

k0<\/sup> = 1800<\/sup>\/3 = 600<\/sup><\/p>\n\n\n\n

Therefore, \u2220A = \u2220B = \u2220C = 600<\/sup>.<\/p>\n\n\n\n

(iv) From the figure (iv)<\/p>\n\n\n\n

\u2220A + \u2220B + \u2220C = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

(m0<\/sup> \u2013 50<\/sup>) + 600<\/sup> + (m0<\/sup> + 50<\/sup>) = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

m0<\/sup> \u2013 50<\/sup> + 600<\/sup> + m0<\/sup> + 50<\/sup> = 1800<\/sup><\/p>\n\n\n\n

2m0<\/sup> = 1800<\/sup> \u2013 600<\/sup> = 1200<\/sup><\/p>\n\n\n\n

m0<\/sup> = 1200<\/sup>\/2 = 600<\/sup><\/p>\n\n\n\n

Therefore, \u2220A = m0<\/sup> \u2013 50<\/sup> = 600<\/sup> \u2013 50<\/sup> = 550<\/sup><\/p>\n\n\n\n

\u2220C = m0<\/sup> + 50<\/sup> = 600<\/sup> + 50<\/sup> = 650<\/sup><\/p>\n\n\n\n

8. In the given figure, show that: \u2220a = \u2220b + \u2220c<\/strong><\/p>\n\n\n\n

(i) If \u2220b = 60\u00b0 and \u2220c = 50\u00b0; find \u2220a.
(ii) If \u2220a = 100\u00b0 and \u2220b = 55\u00b0; find \u2220c.
(iii) If \u2220a = 108\u00b0 and \u2220c = 48\u00b0; find \u2220b.<\/strong><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

From the figure<\/p>\n\n\n\n

AB || CD<\/p>\n\n\n\n

b = \u2220C and \u2220A = c are alternate angles<\/p>\n\n\n\n

In triangle PCD<\/p>\n\n\n\n

Exterior \u2220APC = \u2220C + \u2220D<\/p>\n\n\n\n

a = b + c<\/p>\n\n\n\n

(i) If \u2220b = 60\u00b0 and \u2220c = 50\u00b0<\/p>\n\n\n\n

\u2220a = \u2220b + \u2220c<\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

\u2220a = 60 + 50 = 1100<\/sup><\/p>\n\n\n\n

(ii) If \u2220a = 100\u00b0 and \u2220b = 55\u00b0<\/p>\n\n\n\n

\u2220a = \u2220b + \u2220c<\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

\u2220c = 100 \u2013 55 = 450<\/sup><\/p>\n\n\n\n

(iii) If \u2220a = 108\u00b0 and \u2220c = 48\u00b0<\/p>\n\n\n\n

\u2220a = \u2220b + \u2220c<\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

\u2220b = 108 \u2013 48 = 600<\/sup><\/p>\n\n\n\n

9. Calculate the angles of a triangle if they are in the ratio 4 : 5 : 6.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

In a triangle, the sum of angles of a triangle is 1800<\/sup><\/p>\n\n\n\n

\u2220A + \u2220B + \u2220C = 1800<\/sup><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

It is given that<\/p>\n\n\n\n

\u2220A: \u2220B: \u2220C = 4: 5: 6<\/p>\n\n\n\n

Consider \u2220A = 4x, \u2220B = 5x and \u2220C = 6x<\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

4x + 5x + 6x = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

15x = 1800<\/sup><\/p>\n\n\n\n

x = 1800<\/sup>\/15 = 120<\/sup><\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

\u2220A = 4x = 4 \u00d7 120<\/sup> = 480<\/sup><\/p>\n\n\n\n

\u2220B = 5x = 5 \u00d7 120<\/sup> = 600<\/sup><\/p>\n\n\n\n

\u2220C = 6x = 6 \u00d7 120<\/sup> = 720<\/sup><\/p>\n\n\n\n

10. One angle of a triangle is 60\u00b0. The, other two angles are in the ratio of 5 : 7. Find the two angles.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

From the triangle ABC<\/p>\n\n\n\n

Consider \u2220A = 600<\/sup>, \u2220B: \u2220C = 5:7<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

In a triangle<\/p>\n\n\n\n

\u2220A + \u2220B + \u2220C = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

600<\/sup> + \u2220B + \u2220C = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

\u2220B + \u2220C = 1800<\/sup> \u2013 600<\/sup> = 1200<\/sup><\/p>\n\n\n\n

Take \u2220B = 5x and \u2220C = 7x<\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

5x + 7x = 1200<\/sup><\/p>\n\n\n\n

12x = 1200<\/sup><\/p>\n\n\n\n

x = 1200<\/sup>\/12 = 100<\/sup><\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

\u2220B = 5x = 5 \u00d7 100<\/sup> = 500<\/sup><\/p>\n\n\n\n

\u2220C = 7x = 7 \u00d7 100<\/sup> = 700<\/sup><\/p>\n\n\n\n

11. One angle of a triangle is 61\u00b0 and the other two angles are in the ratio 1 \u00bd : 1 1\/3. Find these angles.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

From the triangle ABC<\/p>\n\n\n\n

Consider \u2220A = 610<\/sup><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

In a triangle<\/p>\n\n\n\n

\u2220A + \u2220B + \u2220C = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

610<\/sup> + \u2220B + \u2220C = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

\u2220B + \u2220C = 1800<\/sup> \u2013 610<\/sup> = 1190<\/sup><\/p>\n\n\n\n

\u2220B: \u2220C = 1 \u00bd: 1 1\/3 = 3\/2: 4\/3<\/p>\n\n\n\n

Taking LCM<\/p>\n\n\n\n

\u2220B: \u2220C = 9\/6: 8\/ 6<\/p>\n\n\n\n

\u2220B: \u2220C = 9: 8<\/p>\n\n\n\n

Consider \u2220B = 9x and \u2220C = 8x<\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

9x + 8x = 1190<\/sup><\/p>\n\n\n\n

17x = 1190<\/sup><\/p>\n\n\n\n

x = 1190<\/sup>\/ 17 = 70<\/sup><\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

\u2220B = 9x = 9 \u00d7 70<\/sup> = 630<\/sup><\/p>\n\n\n\n

\u2220C = 8x = 8 \u00d7 70<\/sup> = 560<\/sup><\/p>\n\n\n\n

12. Find the unknown marked angles in the given figures:<\/strong><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

In a triangle, if one side is produced<\/p>\n\n\n\n

Exterior angle is the sum of opposite interior angles<\/p>\n\n\n\n

(i) From the figure (i)<\/p>\n\n\n\n

1100<\/sup> = x0<\/sup> + 300<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

x0<\/sup> = 1100<\/sup> \u2013 300<\/sup> = 800<\/sup><\/p>\n\n\n\n

(ii) From the figure (ii)<\/p>\n\n\n\n

1200<\/sup> = y0<\/sup> + 600<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

y0<\/sup> = 1200<\/sup> \u2013 600<\/sup> = 600<\/sup><\/p>\n\n\n\n

(iii) From the figure (iii)<\/p>\n\n\n\n

1220<\/sup> = k0<\/sup> + 350<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

k0<\/sup> = 1220<\/sup> \u2013 350<\/sup> = 870<\/sup><\/p>\n\n\n\n

(iv) From the figure (iv)<\/p>\n\n\n\n

1350<\/sup> = a0<\/sup> + 730<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

a0<\/sup> = 1350<\/sup> \u2013 730<\/sup> = 620<\/sup><\/p>\n\n\n\n

(v) From the figure (v)<\/p>\n\n\n\n

1250<\/sup> = a + c \u2026\u2026 (1)<\/p>\n\n\n\n

1400<\/sup> = a + b \u2026\u2026 (2)<\/p>\n\n\n\n

By adding both the equations<\/p>\n\n\n\n

a + c + a + b = 1250<\/sup> + 1400<\/sup><\/p>\n\n\n\n

On further calculation<\/p>\n\n\n\n

a + a + b + c = 2650<\/sup><\/p>\n\n\n\n

We know that a + b + c = 1800<\/sup><\/p>\n\n\n\n

Substituting it in the equation<\/p>\n\n\n\n

a + 1800<\/sup> = 2650<\/sup><\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

a = 265 \u2013 180 = 850<\/sup><\/p>\n\n\n\n

If a + b = 1400<\/sup><\/p>\n\n\n\n

Substituting it in the equation<\/p>\n\n\n\n

850<\/sup> + b = 1400<\/sup><\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

b = 140 \u2013 85 = 550<\/sup><\/p>\n\n\n\n

If a + c = 1250<\/sup><\/p>\n\n\n\n

Substituting it in the equation<\/p>\n\n\n\n

850<\/sup> + c = 1250<\/sup><\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

c = 125 \u2013 85 = 400<\/sup><\/p>\n\n\n\n

Therefore, a = 850<\/sup>, b = 550<\/sup> and c = 400<\/sup>.<\/p>\n\n\n\n

\n\n\n\n

Exercise 15B page: 180<\/h4>\n\n\n\n

1. Find the unknown angles in the given figures:<\/strong><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) From the figure (i)<\/p>\n\n\n\n

x = y as the angles opposite to equal sides<\/p>\n\n\n\n

In a triangle<\/p>\n\n\n\n

x + y + 800<\/sup> = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

x + x + 800<\/sup> = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

2x = 1800<\/sup> \u2013 800<\/sup> = 1000<\/sup><\/p>\n\n\n\n

x = 1000<\/sup>\/2 = 500<\/sup><\/p>\n\n\n\n

Therefore, x = y = 500<\/sup>.<\/p>\n\n\n\n

(ii) From the figure (ii)<\/p>\n\n\n\n

b = 400<\/sup> as the angles opposite to equal sides<\/p>\n\n\n\n

In a triangle<\/p>\n\n\n\n

a + b + 400<\/sup> = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

a + 400<\/sup> + 400<\/sup> = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

a = 180 \u2013 80 = 1000<\/sup><\/p>\n\n\n\n

Therefore, a = 1000<\/sup> and b = 400<\/sup>.<\/p>\n\n\n\n

(iii) From the figure (iii)<\/p>\n\n\n\n

x = y as the angles opposite to equal sides<\/p>\n\n\n\n

In a triangle<\/p>\n\n\n\n

x + y + 900<\/sup> = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

x + x + 900<\/sup> = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

2x = 180 \u2013 90 = 900<\/sup><\/p>\n\n\n\n

x = 90\/2 = 450<\/sup><\/p>\n\n\n\n

Therefore, x = y = 450<\/sup>.<\/p>\n\n\n\n

(iv) From the figure (iv)<\/p>\n\n\n\n

a = b as the angles opposite to equal sides are equal<\/p>\n\n\n\n

In a triangle<\/p>\n\n\n\n

a + b + 800 <\/sup>= 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

a + a + 800 <\/sup>= 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

2a = 180 \u2013 80 = 1000<\/sup><\/p>\n\n\n\n

a = 1000<\/sup>\/2 = 500<\/sup><\/p>\n\n\n\n

Here a = b = 500<\/sup><\/p>\n\n\n\n

We know that in a triangle the exterior angle is equal to sum of its opposite interior angles<\/p>\n\n\n\n

x = a + 800<\/sup><\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

x = 50 + 80 = 1300<\/sup><\/p>\n\n\n\n

Therefore, a = 500<\/sup>, b = 500<\/sup> and x = 1300<\/sup>.<\/p>\n\n\n\n

(v) From the figure (v)<\/p>\n\n\n\n

In an isosceles triangle consider each equal angle = x<\/p>\n\n\n\n

x + x = 860<\/sup><\/p>\n\n\n\n

2x = 860<\/sup><\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

x = 860<\/sup>\/2 = 430<\/sup><\/p>\n\n\n\n

For a linear pair<\/p>\n\n\n\n

p + x = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

p + 430<\/sup> = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

p = 180 \u2013 43 = 1370<\/sup><\/p>\n\n\n\n

Therefore, p = 1370<\/sup>.<\/p>\n\n\n\n

2. Apply the properties of isosceles and equilateral triangles to find the unknown angles in the given figures:<\/strong><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) a = 700<\/sup> as the angles opposite to equal sides are equal<\/p>\n\n\n\n

In a triangle<\/p>\n\n\n\n

a + 700<\/sup> + x = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

700<\/sup> + 700<\/sup> + x = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

x = 180 \u2013 140 = 400<\/sup><\/p>\n\n\n\n

y = b as the angles opposite to equal sides are equal<\/p>\n\n\n\n

Here a = y + b as the exterior angle is equal to sum of interior opposite angles<\/p>\n\n\n\n

700<\/sup> = y + y<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

2y = 700<\/sup><\/p>\n\n\n\n

y = 700<\/sup>\/2 = 350<\/sup><\/p>\n\n\n\n

Therefore, x = 400<\/sup> and y = 350<\/sup>.<\/p>\n\n\n\n

(ii) From the figure (ii)<\/p>\n\n\n\n

Each angle is 600<\/sup> in an equilateral triangle<\/p>\n\n\n\n

In an isosceles triangle<\/p>\n\n\n\n

Consider each base angle = a<\/p>\n\n\n\n

a + a + 1000<\/sup> = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

2a = 180 \u2013 100 = 800<\/sup><\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

a = 800<\/sup>\/2 = 400<\/sup><\/p>\n\n\n\n

x = 600<\/sup> + 400<\/sup> = 1000<\/sup><\/p>\n\n\n\n

y = 600<\/sup> + 400<\/sup> = 1000<\/sup><\/p>\n\n\n\n

(iii) From the figure (iii)<\/p>\n\n\n\n

1300<\/sup> = x + p as the exterior angle is equal to the sum of interior opposite angles<\/p>\n\n\n\n

It is given that the lines are parallel<\/p>\n\n\n\n

Here p = 600<\/sup> is the alternate angles and y = a<\/p>\n\n\n\n

In a linear pair<\/p>\n\n\n\n

a + 1300<\/sup> = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

a = 180 \u2013 130 = 500<\/sup><\/p>\n\n\n\n

Here x + p = 1300<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

x + 600<\/sup> = 1300<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

x = 130 \u2013 60 = 700<\/sup><\/p>\n\n\n\n

Therefore, x = 700<\/sup>, y = 500<\/sup> and p = 600<\/sup>.<\/p>\n\n\n\n

(iv) From the figure (iv)<\/p>\n\n\n\n

x = a + b<\/p>\n\n\n\n

Here b = y and a = c as the angles opposite to equal sides are equal<\/p>\n\n\n\n

a + c + 300<\/sup> = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

a + a + 300<\/sup> = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

2a = 180 \u2013 30 = 1500<\/sup><\/p>\n\n\n\n

a = 150\/2 = 750<\/sup><\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

b + y = 900<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

y + y = 900<\/sup><\/p>\n\n\n\n

2y = 900<\/sup><\/p>\n\n\n\n

y = 90\/2 = 450<\/sup><\/p>\n\n\n\n

where b = 450<\/sup><\/p>\n\n\n\n

Therefore, x = a + b = 75 + 45 = 1200<\/sup> and y = 450<\/sup>.<\/p>\n\n\n\n

(v) From the figure (v)<\/p>\n\n\n\n

a + b + 400<\/sup> = 1800<\/sup><\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

a + b = 180 \u2013 40 = 1400<\/sup><\/p>\n\n\n\n

The angles opposite to equal sides are equal<\/p>\n\n\n\n

a = b = 140\/2 = 700<\/sup><\/p>\n\n\n\n

x = b + 400<\/sup> = 700<\/sup> + 400<\/sup>= 1100<\/sup><\/p>\n\n\n\n

Here the exterior angle of a triangle is equal to the sum of its interior opposite angles<\/p>\n\n\n\n

In the same way<\/p>\n\n\n\n

y = a + 400<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

y = 700<\/sup> + 400<\/sup> = 1100<\/sup><\/p>\n\n\n\n

Therefore, x = y = 1100<\/sup>.<\/p>\n\n\n\n

3. The angle of vertex of an isosceles triangle is 100\u00b0. Find its base angles.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider \u2206 ABC<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here AB = AC and \u2220B = \u2220C<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

\u2220A = 1000<\/sup><\/p>\n\n\n\n

In a triangle<\/p>\n\n\n\n

\u2220A + \u2220B + \u2220C = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

1000<\/sup> + \u2220B + \u2220B = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

2\u2220B = 1800<\/sup> \u2013 1000<\/sup> = 800<\/sup><\/p>\n\n\n\n

\u2220B = 80\/2 = 400<\/sup><\/p>\n\n\n\n

Therefore, \u2220B = \u2220C = 400<\/sup>.<\/p>\n\n\n\n

4. One of the base angles of an isosceles triangle is 52\u00b0. Find its angle of vertex.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that the base angles of isosceles triangle ABC = 520<\/sup><\/p>\n\n\n\n

Here \u2220B = \u2220C = 520<\/sup><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

In a triangle<\/p>\n\n\n\n

\u2220A + \u2220B + \u2220C = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

\u2220A + 520<\/sup> + 520<\/sup> = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

\u2220A = 180 \u2013 104 = 760<\/sup><\/p>\n\n\n\n

Therefore, \u2220A = 760<\/sup>.<\/p>\n\n\n\n

5. In an isosceles triangle, each base angle is four times of its vertical angle. Find all the angles of the triangle.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider the vertical angle of an isosceles triangle = x<\/p>\n\n\n\n

So the base angle = 4x<\/p>\n\n\n\n

In a triangle<\/p>\n\n\n\n

x + 4x + 4x = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

9x = 1800<\/sup><\/p>\n\n\n\n

x = 180\/9 = 200<\/sup><\/p>\n\n\n\n

So the vertical angle = 200<\/sup><\/p>\n\n\n\n

Each base angle = 4x = 4 \u00d7 200<\/sup> = 800<\/sup><\/p>\n\n\n\n

6. The vertical angle of an isosceles triangle is 15\u00b0 more than each of its base angles. Find each angle of the triangle.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider the angle of the base of isosceles triangle = x0<\/sup><\/p>\n\n\n\n

So the vertical angle = x + 150<\/sup><\/p>\n\n\n\n

In a triangle<\/p>\n\n\n\n

x + x + x + 150<\/sup> = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

3x = 180 \u2013 15 = 1650<\/sup><\/p>\n\n\n\n

x = 165\/3 = 550<\/sup><\/p>\n\n\n\n

Therefore, the base angle = 550<\/sup><\/p>\n\n\n\n

Vertical angle = 55 + 15 = 700<\/sup>.<\/p>\n\n\n\n

7. The base angle of an isosceles triangle is 15\u00b0 more than its vertical angle. Find its each angle.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider the vertical angle of the isosceles triangle = x0<\/sup><\/p>\n\n\n\n

Here each base angle = x + 150<\/sup><\/p>\n\n\n\n

In a triangle<\/p>\n\n\n\n

x + 150<\/sup> + x + 150<\/sup> + x = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

3x + 300<\/sup> = 1800<\/sup><\/p>\n\n\n\n

3x = 180 \u2013 30 = 1500<\/sup><\/p>\n\n\n\n

x = 150\/3 = 500<\/sup><\/p>\n\n\n\n

Therefore, vertical angle = 500<\/sup> and each base angle = 50 + 15 = 650<\/sup>.<\/p>\n\n\n\n

8. The vertical angle of an isosceles triangle is three times the sum of its base angles. Find each angle.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider each base angle of an isosceles triangle = x<\/p>\n\n\n\n

Vertical angle = 3 (x + x) = 3 (2x) = 6x<\/p>\n\n\n\n

In a triangle<\/p>\n\n\n\n

6x + x + x = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

8x = 1800<\/sup><\/p>\n\n\n\n

x = 180\/8 = 22.50<\/sup><\/p>\n\n\n\n

Therefore, each base angle = 22.50<\/sup> and vertical angle = 3 (22.5 + 22.5) = 3 \u00d7 45 = 1350<\/sup>.<\/p>\n\n\n\n

9. The ratio between a base angle and the vertical angle of an isosceles triangle is 1 : 4. Find each angle of the triangle.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that the ratio between a base angle and the vertical angle of an isosceles triangle = 1: 4<\/p>\n\n\n\n

Consider base angle = x<\/p>\n\n\n\n

Vertical angle = 4x<\/p>\n\n\n\n

In a triangle<\/p>\n\n\n\n

x + x + 4x = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

6x = 1800<\/sup><\/p>\n\n\n\n

x = 180\/6 = 300<\/sup><\/p>\n\n\n\n

Therefore, each base angle = x = 300<\/sup> and vertical angle = 4x = 4 \u00d7 300<\/sup> = 1200<\/sup>.<\/p>\n\n\n\n

10. In the given figure, BI is the bisector of \u2220ABC and CI is the bisector of \u2220ACB. Find \u2220BIC.<\/strong><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

In \u2206 ABC<\/p>\n\n\n\n

BI is the bisector of \u2220ABC and CI is the bisector of \u2220ACB<\/p>\n\n\n\n

Here AB = AC<\/p>\n\n\n\n

\u2220B = \u2220C as the angles opposite to equal sides are equal<\/p>\n\n\n\n

We know that \u2220A = 400<\/sup><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

In a triangle<\/p>\n\n\n\n

\u2220A + \u2220B + \u2220C = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

400<\/sup> + \u2220B + \u2220B = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

400<\/sup> + 2\u2220B = 1800<\/sup><\/p>\n\n\n\n

2\u2220B = 180 \u2013 40 = 1400<\/sup><\/p>\n\n\n\n

\u2220B = 140\/2 = 700<\/sup><\/p>\n\n\n\n

Here BI and CI are the bisectors of \u2220ABC and \u2220ACB<\/p>\n\n\n\n

\u2220IBC = \u00bd \u2220ABC = \u00bd \u00d7 700<\/sup> = 350<\/sup><\/p>\n\n\n\n

\u2220ICB = \u00bd \u2220ACB = \u00bd \u00d7 700<\/sup> = 350<\/sup><\/p>\n\n\n\n

In \u2206 IBC<\/p>\n\n\n\n

\u2220BIC + \u2220IBC + \u2220ICB = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

\u2220BIC + 350<\/sup> + 350<\/sup> = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

\u2220BIC = 180 \u2013 70 = 1100<\/sup><\/p>\n\n\n\n

Therefore, \u2220BIC = 1100<\/sup>.<\/p>\n\n\n\n

11. In the given figure, express a in terms of b.<\/strong><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

From the \u2206 ABC<\/p>\n\n\n\n

BC = BA<\/p>\n\n\n\n

\u2220BCA = \u2220BAC<\/p>\n\n\n\n

Here the exterior \u2220CBE = \u2220BCA + \u2220BAC<\/p>\n\n\n\n

a = \u2220BCA + \u2220BCA<\/p>\n\n\n\n

a = 2\u2220BCA \u2026\u2026 (1)<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here \u2220ACB = 1800<\/sup> \u2013 b<\/p>\n\n\n\n

Where \u2220ACD and \u2220ACB are linear pair<\/p>\n\n\n\n

\u2220BCA = 1800<\/sup> \u2013 b \u2026\u2026. (2)<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

a = 2 \u2220BCA<\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

a = 2 (1800<\/sup> \u2013 b)<\/p>\n\n\n\n

a = 3600<\/sup> \u2013 2b<\/p>\n\n\n\n

12. (a) In Figure (i) BP bisects \u2220ABC and AB = AC. Find x.
(b) Find x in Figure (ii) Given: DA = DB = DC, BD bisects \u2220ABC and \u2220ADB = 70\u00b0.<\/strong><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(a) From the figure (i)<\/p>\n\n\n\n

AB = AC and BP bisects \u2220ABC<\/p>\n\n\n\n

AP is drawn parallel to BC<\/p>\n\n\n\n

Here PB is the bisector of \u2220ABC<\/p>\n\n\n\n

\u2220PBC = \u2220PBA<\/p>\n\n\n\n

\u2220APB = \u2220PBC are alternate angles<\/p>\n\n\n\n

x = \u2220PBC \u2026.. (1)<\/p>\n\n\n\n

In \u2206 ABC<\/p>\n\n\n\n

\u2220A = 600<\/sup><\/p>\n\n\n\n

Since AB = AC we get \u2220B = \u2220C<\/p>\n\n\n\n

In a triangle<\/p>\n\n\n\n

\u2220A + \u2220B + \u2220C = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

600<\/sup> + \u2220B + \u2220C = 1800<\/sup><\/p>\n\n\n\n

We get<\/p>\n\n\n\n

600<\/sup> + \u2220B + \u2220B = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

2\u2220B = 180 \u2013 60 = 1200<\/sup><\/p>\n\n\n\n

\u2220B = 120\/2 = 600<\/sup><\/p>\n\n\n\n

\u00bd \u2220B = 60\/2 = 300<\/sup><\/p>\n\n\n\n

\u2220PBC = 300<\/sup><\/p>\n\n\n\n

So from figure (i) x = 300<\/sup><\/p>\n\n\n\n

(b) From the figure (ii)<\/p>\n\n\n\n

DA = DB = DC<\/p>\n\n\n\n

Here BD bisects \u2220ABC and \u2220ADB = 700<\/sup><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

In a triangle<\/p>\n\n\n\n

\u2220ADB + \u2220DAB + \u2220DBA = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

700<\/sup> + \u2220DBA + \u2220DBA = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

700<\/sup> + 2\u2220DBA = 1800<\/sup><\/p>\n\n\n\n

2\u2220DBA = 180 \u2013 70 = 1100<\/sup><\/p>\n\n\n\n

\u2220DBA = 110\/2 = 550<\/sup><\/p>\n\n\n\n

Here BD is the bisector of \u2220ABC<\/p>\n\n\n\n

So \u2220DBA = \u2220DBC = 550<\/sup><\/p>\n\n\n\n

In \u2206 DBC<\/p>\n\n\n\n

DB = DC<\/p>\n\n\n\n

\u2220DCB = \u2220DBC<\/p>\n\n\n\n

Hence, x = 550<\/sup>.<\/p>\n\n\n\n

13. In each figure, given below, ABCD is a square and \u2206 BEC is an equilateral triangle.
<\/strong>
\"Selina<\/p>\n\n\n\n

Find, in each case: (i) \u2220ABE (ii) \u2220BAE<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The sides of a square are equal and each angle is 900<\/sup><\/p>\n\n\n\n

In an equilateral triangle all three sides are equal and all angles are 600<\/sup><\/p>\n\n\n\n

In figure (i) ABCD is a square and \u2206 BEC is an equilateral triangle<\/p>\n\n\n\n

(i) \u2220ABE = \u2220ABC + \u2220CBE<\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

\u2220ABE = 900<\/sup> + 600<\/sup>= 1500<\/sup><\/p>\n\n\n\n

(ii) In \u2206 ABE<\/p>\n\n\n\n

\u2220ABE + \u2220BEA + \u2220BAE = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

1500<\/sup> + \u2220BAE + \u2220BAE = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

2\u2220BAE = 180 \u2013 150 = 300<\/sup><\/p>\n\n\n\n

\u2220BAE = 30\/2 = 150<\/sup><\/p>\n\n\n\n

In figure (ii) ABCD is a square and \u2206 BEC is an equilateral triangle<\/p>\n\n\n\n

(i) \u2220ABE = \u2220ABC \u2013 \u2220CBE<\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

\u2220ABE = 900<\/sup> \u2013 600<\/sup> = 300<\/sup><\/p>\n\n\n\n

(ii) In \u2206 ABE<\/p>\n\n\n\n

\u2220ABE + \u2220BEA + \u2220BAE = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

300<\/sup> + \u2220BAE + \u2220BAE = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

2\u2220BAE = 180 \u2013 30 = 1500<\/sup><\/p>\n\n\n\n

\u2220BAE = 150\/2 = 750<\/sup><\/p>\n\n\n\n

14. In \u2206 ABC, BA and BC are produced. Find the angles a and h. if AB = BC.<\/strong><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

In \u2206 ABC, BA and BC are produced<\/p>\n\n\n\n

\u2220ABC = 540<\/sup> and AB = BC<\/p>\n\n\n\n

In \u2206 ABC<\/p>\n\n\n\n

\u2220BAC + \u2220BCA + \u2220ABC = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

\u2220BAC + \u2220BAC + 540 <\/sup>= 1800<\/sup><\/p>\n\n\n\n

2\u2220BAC = 180 \u2013 54 = 1260<\/sup><\/p>\n\n\n\n

\u2220BAC = 126\/2 = 630<\/sup><\/p>\n\n\n\n

\u2220BCA = 630<\/sup><\/p>\n\n\n\n

In a linear pair<\/p>\n\n\n\n

\u2220BAC + b = 1800<\/sup><\/p>\n\n\n\n

Substituting the value<\/p>\n\n\n\n

630<\/sup> + b = 1800<\/sup><\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

b = 180 \u2013 63 = 1170<\/sup><\/p>\n\n\n\n

In a linear pair<\/p>\n\n\n\n

\u2220BCA + a = 1800<\/sup><\/p>\n\n\n\n

Substituting the value<\/p>\n\n\n\n

630<\/sup> + a = 1800<\/sup><\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

a = 180 \u2013 63 = 1170<\/sup><\/p>\n\n\n\n

Therefore, a = b = 1170<\/sup>.<\/p>\n\n\n\n

\n\n\n\n

Exercise 15C page: 185<\/h4>\n\n\n\n

1. Construct a \u2206ABC such that:
(i) AB = 6 cm, BC = 4 cm and CA = 5.5 cm
(ii) CB = 6.5 cm, CA = 4.2 cm and BA = 51 cm
(iii) BC = 4 cm, AC = 5 cm and AB = 3.5 cm<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment BC = 4 cm.<\/p>\n\n\n\n

2. Taking B as centre and 6 cm as radius construct an arc.<\/p>\n\n\n\n

3. Taking C as centre and 5.5 cm as radius construct another arc which intersects the first arc at the point A.<\/p>\n\n\n\n

4. Now join AB and AC.<\/p>\n\n\n\n

Therefore, \u2206ABC is the required triangle.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(ii) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment CB = 6.5 cm<\/p>\n\n\n\n

2. Taking C as centre and 4.2 cm as radius construct an arc.<\/p>\n\n\n\n

3. Taking B as centre and 5.1 cm as radius construct another arc which intersects the first arc at the point A.<\/p>\n\n\n\n

4. Now join AC and AB.<\/p>\n\n\n\n

Therefore, \u2206ABC is the required triangle.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(iii) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment BC = 4 cm<\/p>\n\n\n\n

2. Taking B as centre and 3.5 cm as radius construct an arc.<\/p>\n\n\n\n

3. Taking C as centre and 5 cm as radius construct another arc which intersects the first arc at the point A.<\/p>\n\n\n\n

4. Now join AB and AC.<\/p>\n\n\n\n

Therefore, \u2206ABC is the required triangle.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

2. Construct a <\/strong>\u2206 ABC such that:
(i) AB = 7 cm, BC = 5 cm and \u2220ABC = 60\u00b0
(ii) BC = 6 cm, AC = 5.7 cm and \u2220ACB = 75\u00b0
(iii) AB = 6.5 cm, AC = 5.8 cm and \u2220A = 45\u00b0<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment AB = 7 cm.<\/p>\n\n\n\n

2. At the point B construct a ray which makes an angle 600<\/sup> and cut off BC = 5cm.<\/p>\n\n\n\n

3. Now join AC.<\/p>\n\n\n\n

Therefore, \u2206ABC is the required triangle.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(ii) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment BC = 6 cm.<\/p>\n\n\n\n

2. At the point C construct a ray which makes an angle 750<\/sup> and cut off CA = 5.7 cm.<\/p>\n\n\n\n

3. Now join AB.<\/p>\n\n\n\n

Therefore, \u2206ABC is the required triangle.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(iii) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment AB = 6.5 cm.<\/p>\n\n\n\n

2. At the point A construct a ray which makes an angle 450<\/sup> and cut off AC = 5.8 cm.<\/p>\n\n\n\n

3. Now join CB.<\/p>\n\n\n\n

Therefore, \u2206ABC is the required triangle.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

3. Construct a \u2206 PQR such that :
(i) PQ = 6 cm, \u2220Q = 60\u00b0 and \u2220P = 45\u00b0. Measure \u2220R.
(ii) QR = 4.4 cm, \u2220R = 30\u00b0 and \u2220Q = 75\u00b0. Measure PQ and PR.
(iii) PR = 5.8 cm, \u2220P = 60\u00b0 and \u2220R = 45\u00b0.
Measure \u2220Q and verify it by calculations<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment PQ = 6 cm.<\/p>\n\n\n\n

2. At point P construct a ray which makes an angle 450<\/sup>.<\/p>\n\n\n\n

3. At point Q construct another ray which makes an angle 600<\/sup> which intersect the first ray at point R.<\/p>\n\n\n\n

Therefore, \u2206 PQR is the required triangle.<\/p>\n\n\n\n

By measuring \u2220R = 750<\/sup>.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(ii) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment QR = 4.4 cm.<\/p>\n\n\n\n

2. At point Q construct a ray which makes an angle 750<\/sup>.<\/p>\n\n\n\n

3. At point R construct another ray which makes an angle 300<\/sup> which intersect the first ray at point R.<\/p>\n\n\n\n

Therefore, \u2206 PQR is the required triangle.<\/p>\n\n\n\n

By measuring the length, PQ = 2.1 cm and PR = 4.4 cm.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(iii) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment PR = 5.8 cm.<\/p>\n\n\n\n

2. At point P construct a ray which makes an angle 600<\/sup>.<\/p>\n\n\n\n

3. At point R construct another ray which makes an angle 450<\/sup> which intersect the first ray at point Q.<\/p>\n\n\n\n

Therefore, \u2206 PQR is the required triangle.<\/p>\n\n\n\n

By measuring \u2220Q = 750<\/sup>.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Verification \u2013<\/p>\n\n\n\n

\u2220P + \u2220Q + \u2220R = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

600<\/sup> + \u2220Q + 450<\/sup> = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

\u2220Q = 180 \u2013 105 = 750<\/sup><\/p>\n\n\n\n

4. Construct an isosceles <\/strong>\u2206 ABC such that:
(i) base BC = 4 cm and base angle = 30\u00b0
(ii) base AB = 6.2 cm and base angle = 45\u00b0
(iii) base AC = 5 cm and base angle = 75\u00b0.
Measure the other two sides of the triangle.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Steps of Construction<\/p>\n\n\n\n

In an isosceles triangle the base angles are equal<\/p>\n\n\n\n

1. Construct a line segment BC = 4 cm.<\/p>\n\n\n\n

2. At the points B and C construct rays which makes an angle 300<\/sup> intersecting each other at the point A.<\/p>\n\n\n\n

Therefore, \u2206 ABC is the required triangle.<\/p>\n\n\n\n

By measuring the equal sides, each is 2.5 cm in length approximately.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(ii) Steps of Construction<\/p>\n\n\n\n

In an isosceles triangle the base angles are equal<\/p>\n\n\n\n

1. Construct a line segment AB = 6.2 cm.<\/p>\n\n\n\n

2. At the points A and B construct rays which makes an angle 450<\/sup> intersecting each other at the point C.<\/p>\n\n\n\n

Therefore, \u2206 ABC is the required triangle.<\/p>\n\n\n\n

By measuring the equal sides, each is 4.3 cm in length approximately.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(iii) Steps of Construction<\/p>\n\n\n\n

In an isosceles triangle the base angles are equal<\/p>\n\n\n\n

1. Construct a line segment AC = 5 cm.<\/p>\n\n\n\n

2. At the points A and C construct rays which makes an angle 750<\/sup> intersecting each other at the point B.<\/p>\n\n\n\n

Therefore, \u2206 ABC is the required triangle.<\/p>\n\n\n\n

By measuring the equal sides, each is 9.3 cm in length approximately.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

5. Construct an isosceles \u2206ABC such that:
(i) AB = AC = 6.5 cm and \u2220A = 60\u00b0
(ii) One of the equal sides = 6 cm and vertex angle = 45\u00b0. Measure the base angles.
(iii) BC = AB = 5-8 cm and ZB = 30\u00b0. Measure \u2220A and \u2220C.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment AB = 6.5 cm.<\/p>\n\n\n\n

2. At point A construct a ray which makes an angle 600<\/sup>.<\/p>\n\n\n\n

3. Now cut off AC = 6.5cm<\/p>\n\n\n\n

4. Join BC.<\/p>\n\n\n\n

Therefore, \u2206 ABC is the required triangle.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(ii) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment AB = 6 cm.<\/p>\n\n\n\n

2. At point A construct a ray which makes an angle 450<\/sup>.<\/p>\n\n\n\n

3. Now cut off AC = 6cm<\/p>\n\n\n\n

4. Join BC.<\/p>\n\n\n\n

Therefore, \u2206 ABC is the required triangle.<\/p>\n\n\n\n

By measuring \u2220B and \u2220C, both are equal to 67 \u00bd 0<\/sup>.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(iii) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment BC = 5.8 cm.<\/p>\n\n\n\n

2. At point B construct a ray which makes an angle 300<\/sup>.<\/p>\n\n\n\n

3. Now cut off BA = 5.8cm<\/p>\n\n\n\n

4. Join AC.<\/p>\n\n\n\n

Therefore, \u2206 ABC is the required triangle.<\/p>\n\n\n\n

By measuring \u2220C and \u2220A is equal to 750<\/sup>.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

6. Construct an equilateral triangle ABC such that:
(i) AB = 5 cm. Draw the perpendicular bisectors of BC and AC. Let P be the point of intersection of these two bisectors. Measure PA, PB and PC.
(ii) Each side is 6 cm.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment AB = 5cm.<\/p>\n\n\n\n

2. Taking A and B as centres and 5 cm radius, construct two arcs which intersect each other at the point C.<\/p>\n\n\n\n

3. Now join AC and BC where \u2206 ABC is the required triangle.<\/p>\n\n\n\n

4. Construct perpendicular bisectors of sides AC and BC which intersect each other at the point p.<\/p>\n\n\n\n

5. Join PA, PB and PC.<\/p>\n\n\n\n

By measuring each is 2.8 cm.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(ii) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment AB = 6cm.<\/p>\n\n\n\n

2. Taking A and B as centres and 6 cm radius, construct two arcs which intersect each other at the point C.<\/p>\n\n\n\n

3. Now join AC and BC<\/p>\n\n\n\n

Therefore, \u2206 ABC is the required triangle.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

7. (i) Construct a \u2206 ABC such that AB = 6 cm, BC = 4.5 cm and AC = 5.5 cm. Construct a circumcircle of this triangle.
(ii) Construct an isosceles \u2206PQR such that PQ = PR = 6.5 cm and \u2220PQR = 75\u00b0. Using ruler and compasses only construct a circumcircle to this triangle.
(iii) Construct an equilateral triangle ABC such that its one side = 5.5 cm.
Construct a circumcircle to this triangle.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment BC = 4.5 cm.<\/p>\n\n\n\n

2. Taking B as centre and 6 cm radius construct an arc.<\/p>\n\n\n\n

3. Taking C as centre and 5.5 cm radius construct another arc which intersects the first arc at point A.<\/p>\n\n\n\n

4. Now join AB and AC<\/p>\n\n\n\n

Therefore, \u2206 ABC is the required triangle.<\/p>\n\n\n\n

5. Construct a perpendicular bisector of AB and AC which intersect each other at the point O.<\/p>\n\n\n\n

6. Now join OB, OC and OA.<\/p>\n\n\n\n

7. Taking O as centre and radius OA construct a cirlce which passes through the points A, B and C.<\/p>\n\n\n\n

This is the required circumcircle of \u2206 ABC.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(ii) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment PQ = 6.5 cm.<\/p>\n\n\n\n

2. At point Q, construct an arc which makes an angle 750<\/sup>.<\/p>\n\n\n\n

3. Taking P as centre and radius 6.5 cm construct an arc which intersects the angle at point R.<\/p>\n\n\n\n

4. Join PR.<\/p>\n\n\n\n

\u2206 PQR is the required triangle.<\/p>\n\n\n\n

4. Construct the perpendicular bisector of sides PQ and PR which intersects each other at the point O.<\/p>\n\n\n\n

5. Join OP, OQ and OR.<\/p>\n\n\n\n

6. Taking O as centre and radius equal to OP or OQ or OR construct a circle which passes through P, Q and R.<\/p>\n\n\n\n

This is the required circumcircle of \u2206 PQR.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(iii) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment AB = 5.5 cm.<\/p>\n\n\n\n

2. Taking A and B as centres and radius 5.5 cm construct two arcs which intersect each other at point C.<\/p>\n\n\n\n

3. Now join AC and BC.<\/p>\n\n\n\n

\u2206 ABC is the required triangle.<\/p>\n\n\n\n

4. Construct perpendicular bisectors of sides AC and BC which intersect each other at the point O.<\/p>\n\n\n\n

5. Now join OA, OB and OC.<\/p>\n\n\n\n

6. Taking O as centre and OA or OB or OC as radius construct a circle which passes through A, B and C.<\/p>\n\n\n\n

This is the required circumcircle.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

8. (i) Construct a \u2206ABC such that AB = 6 cm, BC = 5.6 cm and CA = 6.5 cm. Inscribe a circle to this triangle and measure its radius.
(ii) Construct an isosceles \u2206 MNP such that base MN = 5.8 cm, base angle MNP = 30\u00b0. Construct an incircle to this triangle and measure its radius.
(iii) Construct an equilateral \u2206DEF whose one side is 5.5 cm. Construct an incircle to this triangle.
(iv) Construct a \u2206 PQR such that PQ = 6 cm, \u2220QPR = 45\u00b0 and angle PQR = 60\u00b0. Locate its incentre and then draw its incircle.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment AB = 6 cm.<\/p>\n\n\n\n

2. Taking A as centre and 6.5 cm as radius and B as centre and 5.6 cm as radius construct arcs which intersect each other at point C.<\/p>\n\n\n\n

3. Now join AC and BC.<\/p>\n\n\n\n

4. Construct the angle bisector of \u2220A and \u2220B which intersect each other at point I.<\/p>\n\n\n\n

5. From the point I construct IL which is perpendicular to AB.<\/p>\n\n\n\n

6. Taking I as centre and IL as radius construct a circle which touches the sides of \u2206ABC internally.<\/p>\n\n\n\n

By measuring the required incircle the radius is 1.6 cm.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(ii) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment MN = 5.8 cm.<\/p>\n\n\n\n

2. At points M and N construct two rays which make an angle 300<\/sup> each intersecting each other at point P.<\/p>\n\n\n\n

3. Construct the angle bisectors of \u2220M and \u2220N which intersect each other at point I.<\/p>\n\n\n\n

4. From the point I draw perpendicular IL on MN.<\/p>\n\n\n\n

5. Taking I as centre and IL as radius construct a circle which touches the sides of \u2206 PMB internally.<\/p>\n\n\n\n

By measuring the required incircle the radius is 0.6 cm.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(iii) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment BC = 5.5 cm.<\/p>\n\n\n\n

2. Taking B and C as centres and 5.5 cm radius construct two arcs which intersect each other at point A.<\/p>\n\n\n\n

3. Now join AB and AC.<\/p>\n\n\n\n

4. Construct the perpendicular bisectors of \u2220B and \u2220C which intersect each other at the point I.<\/p>\n\n\n\n

5. From the point I construct IL which is perpendicular to BC.<\/p>\n\n\n\n

6. Taking I as centre and IL as radius construct a circle which touches the sides of \u2206ABC internally.<\/p>\n\n\n\n

This is the required incircle.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(iv) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment PQ = 6 cm.<\/p>\n\n\n\n

2. At the point P construct rays which make an angle of 450<\/sup> and at point Q which makes an angle 600<\/sup> thats intersects each other at point R.<\/p>\n\n\n\n

3. Construct the bisectors of \u2220P and \u2220Q which intersect each other at point I.<\/p>\n\n\n\n

4. From the point I construct IL which is perpendicular to PQ.<\/p>\n\n\n\n

5. Taking I as centre and IL as radius construct a circle which touches the sides of \u2206PQR internally.<\/p>\n\n\n\n

This is the required incircle where the point I is incentre.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Exercise 15B<\/h3>\n\n\n\n

Exercise 15B page: 180<\/h4>\n\n\n\n

1. Find the unknown angles in the given figures:<\/strong><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) From the figure (i)<\/p>\n\n\n\n

x = y as the angles opposite to equal sides<\/p>\n\n\n\n

In a triangle<\/p>\n\n\n\n

x + y + 800<\/sup> = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

x + x + 800<\/sup> = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

2x = 1800<\/sup> \u2013 800<\/sup> = 1000<\/sup><\/p>\n\n\n\n

x = 1000<\/sup>\/2 = 500<\/sup><\/p>\n\n\n\n

Therefore, x = y = 500<\/sup>.<\/p>\n\n\n\n

(ii) From the figure (ii)<\/p>\n\n\n\n

b = 400<\/sup> as the angles opposite to equal sides<\/p>\n\n\n\n

In a triangle<\/p>\n\n\n\n

a + b + 400<\/sup> = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

a + 400<\/sup> + 400<\/sup> = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

a = 180 \u2013 80 = 1000<\/sup><\/p>\n\n\n\n

Therefore, a = 1000<\/sup> and b = 400<\/sup>.<\/p>\n\n\n\n

(iii) From the figure (iii)<\/p>\n\n\n\n

x = y as the angles opposite to equal sides<\/p>\n\n\n\n

In a triangle<\/p>\n\n\n\n

x + y + 900<\/sup> = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

x + x + 900<\/sup> = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

2x = 180 \u2013 90 = 900<\/sup><\/p>\n\n\n\n

x = 90\/2 = 450<\/sup><\/p>\n\n\n\n

Therefore, x = y = 450<\/sup>.<\/p>\n\n\n\n

(iv) From the figure (iv)<\/p>\n\n\n\n

a = b as the angles opposite to equal sides are equal<\/p>\n\n\n\n

In a triangle<\/p>\n\n\n\n

a + b + 800 <\/sup>= 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

a + a + 800 <\/sup>= 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

2a = 180 \u2013 80 = 1000<\/sup><\/p>\n\n\n\n

a = 1000<\/sup>\/2 = 500<\/sup><\/p>\n\n\n\n

Here a = b = 500<\/sup><\/p>\n\n\n\n

We know that in a triangle the exterior angle is equal to sum of its opposite interior angles<\/p>\n\n\n\n

x = a + 800<\/sup><\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

x = 50 + 80 = 1300<\/sup><\/p>\n\n\n\n

Therefore, a = 500<\/sup>, b = 500<\/sup> and x = 1300<\/sup>.<\/p>\n\n\n\n

(v) From the figure (v)<\/p>\n\n\n\n

In an isosceles triangle consider each equal angle = x<\/p>\n\n\n\n

x + x = 860<\/sup><\/p>\n\n\n\n

2x = 860<\/sup><\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

x = 860<\/sup>\/2 = 430<\/sup><\/p>\n\n\n\n

For a linear pair<\/p>\n\n\n\n

p + x = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

p + 430<\/sup> = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

p = 180 \u2013 43 = 1370<\/sup><\/p>\n\n\n\n

Therefore, p = 1370<\/sup>.<\/p>\n\n\n\n

2. Apply the properties of isosceles and equilateral triangles to find the unknown angles in the given figures:<\/strong><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) a = 700<\/sup> as the angles opposite to equal sides are equal<\/p>\n\n\n\n

In a triangle<\/p>\n\n\n\n

a + 700<\/sup> + x = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

700<\/sup> + 700<\/sup> + x = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

x = 180 \u2013 140 = 400<\/sup><\/p>\n\n\n\n

y = b as the angles opposite to equal sides are equal<\/p>\n\n\n\n

Here a = y + b as the exterior angle is equal to sum of interior opposite angles<\/p>\n\n\n\n

700<\/sup> = y + y<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

2y = 700<\/sup><\/p>\n\n\n\n

y = 700<\/sup>\/2 = 350<\/sup><\/p>\n\n\n\n

Therefore, x = 400<\/sup> and y = 350<\/sup>.<\/p>\n\n\n\n

(ii) From the figure (ii)<\/p>\n\n\n\n

Each angle is 600<\/sup> in an equilateral triangle<\/p>\n\n\n\n

In an isosceles triangle<\/p>\n\n\n\n

Consider each base angle = a<\/p>\n\n\n\n

a + a + 1000<\/sup> = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

2a = 180 \u2013 100 = 800<\/sup><\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

a = 800<\/sup>\/2 = 400<\/sup><\/p>\n\n\n\n

x = 600<\/sup> + 400<\/sup> = 1000<\/sup><\/p>\n\n\n\n

y = 600<\/sup> + 400<\/sup> = 1000<\/sup><\/p>\n\n\n\n

(iii) From the figure (iii)<\/p>\n\n\n\n

1300<\/sup> = x + p as the exterior angle is equal to the sum of interior opposite angles<\/p>\n\n\n\n

It is given that the lines are parallel<\/p>\n\n\n\n

Here p = 600<\/sup> is the alternate angles and y = a<\/p>\n\n\n\n

In a linear pair<\/p>\n\n\n\n

a + 1300<\/sup> = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

a = 180 \u2013 130 = 500<\/sup><\/p>\n\n\n\n

Here x + p = 1300<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

x + 600<\/sup> = 1300<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

x = 130 \u2013 60 = 700<\/sup><\/p>\n\n\n\n

Therefore, x = 700<\/sup>, y = 500<\/sup> and p = 600<\/sup>.<\/p>\n\n\n\n

(iv) From the figure (iv)<\/p>\n\n\n\n

x = a + b<\/p>\n\n\n\n

Here b = y and a = c as the angles opposite to equal sides are equal<\/p>\n\n\n\n

a + c + 300<\/sup> = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

a + a + 300<\/sup> = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

2a = 180 \u2013 30 = 1500<\/sup><\/p>\n\n\n\n

a = 150\/2 = 750<\/sup><\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

b + y = 900<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

y + y = 900<\/sup><\/p>\n\n\n\n

2y = 900<\/sup><\/p>\n\n\n\n

y = 90\/2 = 450<\/sup><\/p>\n\n\n\n

where b = 450<\/sup><\/p>\n\n\n\n

Therefore, x = a + b = 75 + 45 = 1200<\/sup> and y = 450<\/sup>.<\/p>\n\n\n\n

(v) From the figure (v)<\/p>\n\n\n\n

a + b + 400<\/sup> = 1800<\/sup><\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

a + b = 180 \u2013 40 = 1400<\/sup><\/p>\n\n\n\n

The angles opposite to equal sides are equal<\/p>\n\n\n\n

a = b = 140\/2 = 700<\/sup><\/p>\n\n\n\n

x = b + 400<\/sup> = 700<\/sup> + 400<\/sup>= 1100<\/sup><\/p>\n\n\n\n

Here the exterior angle of a triangle is equal to the sum of its interior opposite angles<\/p>\n\n\n\n

In the same way<\/p>\n\n\n\n

y = a + 400<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

y = 700<\/sup> + 400<\/sup> = 1100<\/sup><\/p>\n\n\n\n

Therefore, x = y = 1100<\/sup>.<\/p>\n\n\n\n

3. The angle of vertex of an isosceles triangle is 100\u00b0. Find its base angles.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider \u2206 ABC<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here AB = AC and \u2220B = \u2220C<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

\u2220A = 1000<\/sup><\/p>\n\n\n\n

In a triangle<\/p>\n\n\n\n

\u2220A + \u2220B + \u2220C = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

1000<\/sup> + \u2220B + \u2220B = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

2\u2220B = 1800<\/sup> \u2013 1000<\/sup> = 800<\/sup><\/p>\n\n\n\n

\u2220B = 80\/2 = 400<\/sup><\/p>\n\n\n\n

Therefore, \u2220B = \u2220C = 400<\/sup>.<\/p>\n\n\n\n

4. One of the base angles of an isosceles triangle is 52\u00b0. Find its angle of vertex.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that the base angles of isosceles triangle ABC = 520<\/sup><\/p>\n\n\n\n

Here \u2220B = \u2220C = 520<\/sup><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

In a triangle<\/p>\n\n\n\n

\u2220A + \u2220B + \u2220C = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

\u2220A + 520<\/sup> + 520<\/sup> = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

\u2220A = 180 \u2013 104 = 760<\/sup><\/p>\n\n\n\n

Therefore, \u2220A = 760<\/sup>.<\/p>\n\n\n\n

5. In an isosceles triangle, each base angle is four times of its vertical angle. Find all the angles of the triangle.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider the vertical angle of an isosceles triangle = x<\/p>\n\n\n\n

So the base angle = 4x<\/p>\n\n\n\n

In a triangle<\/p>\n\n\n\n

x + 4x + 4x = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

9x = 1800<\/sup><\/p>\n\n\n\n

x = 180\/9 = 200<\/sup><\/p>\n\n\n\n

So the vertical angle = 200<\/sup><\/p>\n\n\n\n

Each base angle = 4x = 4 \u00d7 200<\/sup> = 800<\/sup><\/p>\n\n\n\n

6. The vertical angle of an isosceles triangle is 15\u00b0 more than each of its base angles. Find each angle of the triangle.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider the angle of the base of isosceles triangle = x0<\/sup><\/p>\n\n\n\n

So the vertical angle = x + 150<\/sup><\/p>\n\n\n\n

In a triangle<\/p>\n\n\n\n

x + x + x + 150<\/sup> = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

3x = 180 \u2013 15 = 1650<\/sup><\/p>\n\n\n\n

x = 165\/3 = 550<\/sup><\/p>\n\n\n\n

Therefore, the base angle = 550<\/sup><\/p>\n\n\n\n

Vertical angle = 55 + 15 = 700<\/sup>.<\/p>\n\n\n\n

7. The base angle of an isosceles triangle is 15\u00b0 more than its vertical angle. Find its each angle.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider the vertical angle of the isosceles triangle = x0<\/sup><\/p>\n\n\n\n

Here each base angle = x + 150<\/sup><\/p>\n\n\n\n

In a triangle<\/p>\n\n\n\n

x + 150<\/sup> + x + 150<\/sup> + x = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

3x + 300<\/sup> = 1800<\/sup><\/p>\n\n\n\n

3x = 180 \u2013 30 = 1500<\/sup><\/p>\n\n\n\n

x = 150\/3 = 500<\/sup><\/p>\n\n\n\n

Therefore, vertical angle = 500<\/sup> and each base angle = 50 + 15 = 650<\/sup>.<\/p>\n\n\n\n

8. The vertical angle of an isosceles triangle is three times the sum of its base angles. Find each angle.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Consider each base angle of an isosceles triangle = x<\/p>\n\n\n\n

Vertical angle = 3 (x + x) = 3 (2x) = 6x<\/p>\n\n\n\n

In a triangle<\/p>\n\n\n\n

6x + x + x = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

8x = 1800<\/sup><\/p>\n\n\n\n

x = 180\/8 = 22.50<\/sup><\/p>\n\n\n\n

Therefore, each base angle = 22.50<\/sup> and vertical angle = 3 (22.5 + 22.5) = 3 \u00d7 45 = 1350<\/sup>.<\/p>\n\n\n\n

9. The ratio between a base angle and the vertical angle of an isosceles triangle is 1 : 4. Find each angle of the triangle.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that the ratio between a base angle and the vertical angle of an isosceles triangle = 1: 4<\/p>\n\n\n\n

Consider base angle = x<\/p>\n\n\n\n

Vertical angle = 4x<\/p>\n\n\n\n

In a triangle<\/p>\n\n\n\n

x + x + 4x = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

6x = 1800<\/sup><\/p>\n\n\n\n

x = 180\/6 = 300<\/sup><\/p>\n\n\n\n

Therefore, each base angle = x = 300<\/sup> and vertical angle = 4x = 4 \u00d7 300<\/sup> = 1200<\/sup>.<\/p>\n\n\n\n

10. In the given figure, BI is the bisector of \u2220ABC and CI is the bisector of \u2220ACB. Find \u2220BIC.<\/strong><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

In \u2206 ABC<\/p>\n\n\n\n

BI is the bisector of \u2220ABC and CI is the bisector of \u2220ACB<\/p>\n\n\n\n

Here AB = AC<\/p>\n\n\n\n

\u2220B = \u2220C as the angles opposite to equal sides are equal<\/p>\n\n\n\n

We know that \u2220A = 400<\/sup><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

In a triangle<\/p>\n\n\n\n

\u2220A + \u2220B + \u2220C = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

400<\/sup> + \u2220B + \u2220B = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

400<\/sup> + 2\u2220B = 1800<\/sup><\/p>\n\n\n\n

2\u2220B = 180 \u2013 40 = 1400<\/sup><\/p>\n\n\n\n

\u2220B = 140\/2 = 700<\/sup><\/p>\n\n\n\n

Here BI and CI are the bisectors of \u2220ABC and \u2220ACB<\/p>\n\n\n\n

\u2220IBC = \u00bd \u2220ABC = \u00bd \u00d7 700<\/sup> = 350<\/sup><\/p>\n\n\n\n

\u2220ICB = \u00bd \u2220ACB = \u00bd \u00d7 700<\/sup> = 350<\/sup><\/p>\n\n\n\n

In \u2206 IBC<\/p>\n\n\n\n

\u2220BIC + \u2220IBC + \u2220ICB = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

\u2220BIC + 350<\/sup> + 350<\/sup> = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

\u2220BIC = 180 \u2013 70 = 1100<\/sup><\/p>\n\n\n\n

Therefore, \u2220BIC = 1100<\/sup>.<\/p>\n\n\n\n

11. In the given figure, express a in terms of b.<\/strong><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

From the \u2206 ABC<\/p>\n\n\n\n

BC = BA<\/p>\n\n\n\n

\u2220BCA = \u2220BAC<\/p>\n\n\n\n

Here the exterior \u2220CBE = \u2220BCA + \u2220BAC<\/p>\n\n\n\n

a = \u2220BCA + \u2220BCA<\/p>\n\n\n\n

a = 2\u2220BCA \u2026\u2026 (1)<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Here \u2220ACB = 1800<\/sup> \u2013 b<\/p>\n\n\n\n

Where \u2220ACD and \u2220ACB are linear pair<\/p>\n\n\n\n

\u2220BCA = 1800<\/sup> \u2013 b \u2026\u2026. (2)<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

a = 2 \u2220BCA<\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

a = 2 (1800<\/sup> \u2013 b)<\/p>\n\n\n\n

a = 3600<\/sup> \u2013 2b<\/p>\n\n\n\n

12. (a) In Figure (i) BP bisects \u2220ABC and AB = AC. Find x.
(b) Find x in Figure (ii) Given: DA = DB = DC, BD bisects \u2220ABC and \u2220ADB = 70\u00b0.<\/strong><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(a) From the figure (i)<\/p>\n\n\n\n

AB = AC and BP bisects \u2220ABC<\/p>\n\n\n\n

AP is drawn parallel to BC<\/p>\n\n\n\n

Here PB is the bisector of \u2220ABC<\/p>\n\n\n\n

\u2220PBC = \u2220PBA<\/p>\n\n\n\n

\u2220APB = \u2220PBC are alternate angles<\/p>\n\n\n\n

x = \u2220PBC \u2026.. (1)<\/p>\n\n\n\n

In \u2206 ABC<\/p>\n\n\n\n

\u2220A = 600<\/sup><\/p>\n\n\n\n

Since AB = AC we get \u2220B = \u2220C<\/p>\n\n\n\n

In a triangle<\/p>\n\n\n\n

\u2220A + \u2220B + \u2220C = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

600<\/sup> + \u2220B + \u2220C = 1800<\/sup><\/p>\n\n\n\n

We get<\/p>\n\n\n\n

600<\/sup> + \u2220B + \u2220B = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

2\u2220B = 180 \u2013 60 = 1200<\/sup><\/p>\n\n\n\n

\u2220B = 120\/2 = 600<\/sup><\/p>\n\n\n\n

\u00bd \u2220B = 60\/2 = 300<\/sup><\/p>\n\n\n\n

\u2220PBC = 300<\/sup><\/p>\n\n\n\n

So from figure (i) x = 300<\/sup><\/p>\n\n\n\n

(b) From the figure (ii)<\/p>\n\n\n\n

DA = DB = DC<\/p>\n\n\n\n

Here BD bisects \u2220ABC and \u2220ADB = 700<\/sup><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

In a triangle<\/p>\n\n\n\n

\u2220ADB + \u2220DAB + \u2220DBA = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

700<\/sup> + \u2220DBA + \u2220DBA = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

700<\/sup> + 2\u2220DBA = 1800<\/sup><\/p>\n\n\n\n

2\u2220DBA = 180 \u2013 70 = 1100<\/sup><\/p>\n\n\n\n

\u2220DBA = 110\/2 = 550<\/sup><\/p>\n\n\n\n

Here BD is the bisector of \u2220ABC<\/p>\n\n\n\n

So \u2220DBA = \u2220DBC = 550<\/sup><\/p>\n\n\n\n

In \u2206 DBC<\/p>\n\n\n\n

DB = DC<\/p>\n\n\n\n

\u2220DCB = \u2220DBC<\/p>\n\n\n\n

Hence, x = 550<\/sup>.<\/p>\n\n\n\n

13. In each figure, given below, ABCD is a square and \u2206 BEC is an equilateral triangle.
<\/strong>
\"Selina<\/p>\n\n\n\n

Find, in each case: (i) \u2220ABE (ii) \u2220BAE<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The sides of a square are equal and each angle is 900<\/sup><\/p>\n\n\n\n

In an equilateral triangle all three sides are equal and all angles are 600<\/sup><\/p>\n\n\n\n

In figure (i) ABCD is a square and \u2206 BEC is an equilateral triangle<\/p>\n\n\n\n

(i) \u2220ABE = \u2220ABC + \u2220CBE<\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

\u2220ABE = 900<\/sup> + 600<\/sup>= 1500<\/sup><\/p>\n\n\n\n

(ii) In \u2206 ABE<\/p>\n\n\n\n

\u2220ABE + \u2220BEA + \u2220BAE = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

1500<\/sup> + \u2220BAE + \u2220BAE = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

2\u2220BAE = 180 \u2013 150 = 300<\/sup><\/p>\n\n\n\n

\u2220BAE = 30\/2 = 150<\/sup><\/p>\n\n\n\n

In figure (ii) ABCD is a square and \u2206 BEC is an equilateral triangle<\/p>\n\n\n\n

(i) \u2220ABE = \u2220ABC \u2013 \u2220CBE<\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

\u2220ABE = 900<\/sup> \u2013 600<\/sup> = 300<\/sup><\/p>\n\n\n\n

(ii) In \u2206 ABE<\/p>\n\n\n\n

\u2220ABE + \u2220BEA + \u2220BAE = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

300<\/sup> + \u2220BAE + \u2220BAE = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

2\u2220BAE = 180 \u2013 30 = 1500<\/sup><\/p>\n\n\n\n

\u2220BAE = 150\/2 = 750<\/sup><\/p>\n\n\n\n

14. In \u2206 ABC, BA and BC are produced. Find the angles a and h. if AB = BC.<\/strong><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

In \u2206 ABC, BA and BC are produced<\/p>\n\n\n\n

\u2220ABC = 540<\/sup> and AB = BC<\/p>\n\n\n\n

In \u2206 ABC<\/p>\n\n\n\n

\u2220BAC + \u2220BCA + \u2220ABC = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

\u2220BAC + \u2220BAC + 540 <\/sup>= 1800<\/sup><\/p>\n\n\n\n

2\u2220BAC = 180 \u2013 54 = 1260<\/sup><\/p>\n\n\n\n

\u2220BAC = 126\/2 = 630<\/sup><\/p>\n\n\n\n

\u2220BCA = 630<\/sup><\/p>\n\n\n\n

In a linear pair<\/p>\n\n\n\n

\u2220BAC + b = 1800<\/sup><\/p>\n\n\n\n

Substituting the value<\/p>\n\n\n\n

630<\/sup> + b = 1800<\/sup><\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

b = 180 \u2013 63 = 1170<\/sup><\/p>\n\n\n\n

In a linear pair<\/p>\n\n\n\n

\u2220BCA + a = 1800<\/sup><\/p>\n\n\n\n

Substituting the value<\/p>\n\n\n\n

630<\/sup> + a = 1800<\/sup><\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

a = 180 \u2013 63 = 1170<\/sup><\/p>\n\n\n\n

Therefore, a = b = 1170<\/sup>.<\/p>\n\n\n\n

Exercise 15C page: 185<\/h4>\n\n\n\n

1. Construct a \u2206ABC such that:
(i) AB = 6 cm, BC = 4 cm and CA = 5.5 cm
(ii) CB = 6.5 cm, CA = 4.2 cm and BA = 51 cm
(iii) BC = 4 cm, AC = 5 cm and AB = 3.5 cm<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment BC = 4 cm.<\/p>\n\n\n\n

2. Taking B as centre and 6 cm as radius construct an arc.<\/p>\n\n\n\n

3. Taking C as centre and 5.5 cm as radius construct another arc which intersects the first arc at the point A.<\/p>\n\n\n\n

4. Now join AB and AC.<\/p>\n\n\n\n

Therefore, \u2206ABC is the required triangle.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(ii) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment CB = 6.5 cm<\/p>\n\n\n\n

2. Taking C as centre and 4.2 cm as radius construct an arc.<\/p>\n\n\n\n

3. Taking B as centre and 5.1 cm as radius construct another arc which intersects the first arc at the point A.<\/p>\n\n\n\n

4. Now join AC and AB.<\/p>\n\n\n\n

Therefore, \u2206ABC is the required triangle.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(iii) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment BC = 4 cm<\/p>\n\n\n\n

2. Taking B as centre and 3.5 cm as radius construct an arc.<\/p>\n\n\n\n

3. Taking C as centre and 5 cm as radius construct another arc which intersects the first arc at the point A.<\/p>\n\n\n\n

4. Now join AB and AC.<\/p>\n\n\n\n

Therefore, \u2206ABC is the required triangle.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

2. Construct a <\/strong>\u2206 ABC such that:
(i) AB = 7 cm, BC = 5 cm and \u2220ABC = 60\u00b0
(ii) BC = 6 cm, AC = 5.7 cm and \u2220ACB = 75\u00b0
(iii) AB = 6.5 cm, AC = 5.8 cm and \u2220A = 45\u00b0<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment AB = 7 cm.<\/p>\n\n\n\n

2. At the point B construct a ray which makes an angle 600<\/sup> and cut off BC = 5cm.<\/p>\n\n\n\n

3. Now join AC.<\/p>\n\n\n\n

Therefore, \u2206ABC is the required triangle.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(ii) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment BC = 6 cm.<\/p>\n\n\n\n

2. At the point C construct a ray which makes an angle 750<\/sup> and cut off CA = 5.7 cm.<\/p>\n\n\n\n

3. Now join AB.<\/p>\n\n\n\n

Therefore, \u2206ABC is the required triangle.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(iii) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment AB = 6.5 cm.<\/p>\n\n\n\n

2. At the point A construct a ray which makes an angle 450<\/sup> and cut off AC = 5.8 cm.<\/p>\n\n\n\n

3. Now join CB.<\/p>\n\n\n\n

Therefore, \u2206ABC is the required triangle.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

3. Construct a \u2206 PQR such that :
(i) PQ = 6 cm, \u2220Q = 60\u00b0 and \u2220P = 45\u00b0. Measure \u2220R.
(ii) QR = 4.4 cm, \u2220R = 30\u00b0 and \u2220Q = 75\u00b0. Measure PQ and PR.
(iii) PR = 5.8 cm, \u2220P = 60\u00b0 and \u2220R = 45\u00b0.
Measure \u2220Q and verify it by calculations<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment PQ = 6 cm.<\/p>\n\n\n\n

2. At point P construct a ray which makes an angle 450<\/sup>.<\/p>\n\n\n\n

3. At point Q construct another ray which makes an angle 600<\/sup> which intersect the first ray at point R.<\/p>\n\n\n\n

Therefore, \u2206 PQR is the required triangle.<\/p>\n\n\n\n

By measuring \u2220R = 750<\/sup>.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(ii) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment QR = 4.4 cm.<\/p>\n\n\n\n

2. At point Q construct a ray which makes an angle 750<\/sup>.<\/p>\n\n\n\n

3. At point R construct another ray which makes an angle 300<\/sup> which intersect the first ray at point R.<\/p>\n\n\n\n

Therefore, \u2206 PQR is the required triangle.<\/p>\n\n\n\n

By measuring the length, PQ = 2.1 cm and PR = 4.4 cm.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(iii) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment PR = 5.8 cm.<\/p>\n\n\n\n

2. At point P construct a ray which makes an angle 600<\/sup>.<\/p>\n\n\n\n

3. At point R construct another ray which makes an angle 450<\/sup> which intersect the first ray at point Q.<\/p>\n\n\n\n

Therefore, \u2206 PQR is the required triangle.<\/p>\n\n\n\n

By measuring \u2220Q = 750<\/sup>.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Verification \u2013<\/p>\n\n\n\n

\u2220P + \u2220Q + \u2220R = 1800<\/sup><\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

600<\/sup> + \u2220Q + 450<\/sup> = 1800<\/sup><\/p>\n\n\n\n

By further calculation<\/p>\n\n\n\n

\u2220Q = 180 \u2013 105 = 750<\/sup><\/p>\n\n\n\n

4. Construct an isosceles <\/strong>\u2206 ABC such that:
(i) base BC = 4 cm and base angle = 30\u00b0
(ii) base AB = 6.2 cm and base angle = 45\u00b0
(iii) base AC = 5 cm and base angle = 75\u00b0.
Measure the other two sides of the triangle.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Steps of Construction<\/p>\n\n\n\n

In an isosceles triangle the base angles are equal<\/p>\n\n\n\n

1. Construct a line segment BC = 4 cm.<\/p>\n\n\n\n

2. At the points B and C construct rays which makes an angle 300<\/sup> intersecting each other at the point A.<\/p>\n\n\n\n

Therefore, \u2206 ABC is the required triangle.<\/p>\n\n\n\n

By measuring the equal sides, each is 2.5 cm in length approximately.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(ii) Steps of Construction<\/p>\n\n\n\n

In an isosceles triangle the base angles are equal<\/p>\n\n\n\n

1. Construct a line segment AB = 6.2 cm.<\/p>\n\n\n\n

2. At the points A and B construct rays which makes an angle 450<\/sup> intersecting each other at the point C.<\/p>\n\n\n\n

Therefore, \u2206 ABC is the required triangle.<\/p>\n\n\n\n

By measuring the equal sides, each is 4.3 cm in length approximately.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(iii) Steps of Construction<\/p>\n\n\n\n

In an isosceles triangle the base angles are equal<\/p>\n\n\n\n

1. Construct a line segment AC = 5 cm.<\/p>\n\n\n\n

2. At the points A and C construct rays which makes an angle 750<\/sup> intersecting each other at the point B.<\/p>\n\n\n\n

Therefore, \u2206 ABC is the required triangle.<\/p>\n\n\n\n

By measuring the equal sides, each is 9.3 cm in length approximately.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

5. Construct an isosceles \u2206ABC such that:
(i) AB = AC = 6.5 cm and \u2220A = 60\u00b0
(ii) One of the equal sides = 6 cm and vertex angle = 45\u00b0. Measure the base angles.
(iii) BC = AB = 5-8 cm and ZB = 30\u00b0. Measure \u2220A and \u2220C.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment AB = 6.5 cm.<\/p>\n\n\n\n

2. At point A construct a ray which makes an angle 600<\/sup>.<\/p>\n\n\n\n

3. Now cut off AC = 6.5cm<\/p>\n\n\n\n

4. Join BC.<\/p>\n\n\n\n

Therefore, \u2206 ABC is the required triangle.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(ii) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment AB = 6 cm.<\/p>\n\n\n\n

2. At point A construct a ray which makes an angle 450<\/sup>.<\/p>\n\n\n\n

3. Now cut off AC = 6cm<\/p>\n\n\n\n

4. Join BC.<\/p>\n\n\n\n

Therefore, \u2206 ABC is the required triangle.<\/p>\n\n\n\n

By measuring \u2220B and \u2220C, both are equal to 67 \u00bd 0<\/sup>.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(iii) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment BC = 5.8 cm.<\/p>\n\n\n\n

2. At point B construct a ray which makes an angle 300<\/sup>.<\/p>\n\n\n\n

3. Now cut off BA = 5.8cm<\/p>\n\n\n\n

4. Join AC.<\/p>\n\n\n\n

Therefore, \u2206 ABC is the required triangle.<\/p>\n\n\n\n

By measuring \u2220C and \u2220A is equal to 750<\/sup>.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

6. Construct an equilateral triangle ABC such that:
(i) AB = 5 cm. Draw the perpendicular bisectors of BC and AC. Let P be the point of intersection of these two bisectors. Measure PA, PB and PC.
(ii) Each side is 6 cm.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment AB = 5cm.<\/p>\n\n\n\n

2. Taking A and B as centres and 5 cm radius, construct two arcs which intersect each other at the point C.<\/p>\n\n\n\n

3. Now join AC and BC where \u2206 ABC is the required triangle.<\/p>\n\n\n\n

4. Construct perpendicular bisectors of sides AC and BC which intersect each other at the point p.<\/p>\n\n\n\n

5. Join PA, PB and PC.<\/p>\n\n\n\n

By measuring each is 2.8 cm.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(ii) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment AB = 6cm.<\/p>\n\n\n\n

2. Taking A and B as centres and 6 cm radius, construct two arcs which intersect each other at the point C.<\/p>\n\n\n\n

3. Now join AC and BC<\/p>\n\n\n\n

Therefore, \u2206 ABC is the required triangle.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

7. (i) Construct a \u2206 ABC such that AB = 6 cm, BC = 4.5 cm and AC = 5.5 cm. Construct a circumcircle of this triangle.
(ii) Construct an isosceles \u2206PQR such that PQ = PR = 6.5 cm and \u2220PQR = 75\u00b0. Using ruler and compasses only construct a circumcircle to this triangle.
(iii) Construct an equilateral triangle ABC such that its one side = 5.5 cm.
Construct a circumcircle to this triangle.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment BC = 4.5 cm.<\/p>\n\n\n\n

2. Taking B as centre and 6 cm radius construct an arc.<\/p>\n\n\n\n

3. Taking C as centre and 5.5 cm radius construct another arc which intersects the first arc at point A.<\/p>\n\n\n\n

4. Now join AB and AC<\/p>\n\n\n\n

Therefore, \u2206 ABC is the required triangle.<\/p>\n\n\n\n

5. Construct a perpendicular bisector of AB and AC which intersect each other at the point O.<\/p>\n\n\n\n

6. Now join OB, OC and OA.<\/p>\n\n\n\n

7. Taking O as centre and radius OA construct a cirlce which passes through the points A, B and C.<\/p>\n\n\n\n

This is the required circumcircle of \u2206 ABC.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(ii) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment PQ = 6.5 cm.<\/p>\n\n\n\n

2. At point Q, construct an arc which makes an angle 750<\/sup>.<\/p>\n\n\n\n

3. Taking P as centre and radius 6.5 cm construct an arc which intersects the angle at point R.<\/p>\n\n\n\n

4. Join PR.<\/p>\n\n\n\n

\u2206 PQR is the required triangle.<\/p>\n\n\n\n

4. Construct the perpendicular bisector of sides PQ and PR which intersects each other at the point O.<\/p>\n\n\n\n

5. Join OP, OQ and OR.<\/p>\n\n\n\n

6. Taking O as centre and radius equal to OP or OQ or OR construct a circle which passes through P, Q and R.<\/p>\n\n\n\n

This is the required circumcircle of \u2206 PQR.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(iii) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment AB = 5.5 cm.<\/p>\n\n\n\n

2. Taking A and B as centres and radius 5.5 cm construct two arcs which intersect each other at point C.<\/p>\n\n\n\n

3. Now join AC and BC.<\/p>\n\n\n\n

\u2206 ABC is the required triangle.<\/p>\n\n\n\n

4. Construct perpendicular bisectors of sides AC and BC which intersect each other at the point O.<\/p>\n\n\n\n

5. Now join OA, OB and OC.<\/p>\n\n\n\n

6. Taking O as centre and OA or OB or OC as radius construct a circle which passes through A, B and C.<\/p>\n\n\n\n

This is the required circumcircle.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

8. (i) Construct a \u2206ABC such that AB = 6 cm, BC = 5.6 cm and CA = 6.5 cm. Inscribe a circle to this triangle and measure its radius.
(ii) Construct an isosceles \u2206 MNP such that base MN = 5.8 cm, base angle MNP = 30\u00b0. Construct an incircle to this triangle and measure its radius.
(iii) Construct an equilateral \u2206DEF whose one side is 5.5 cm. Construct an incircle to this triangle.
(iv) Construct a \u2206 PQR such that PQ = 6 cm, \u2220QPR = 45\u00b0 and angle PQR = 60\u00b0. Locate its incentre and then draw its incircle.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment AB = 6 cm.<\/p>\n\n\n\n

2. Taking A as centre and 6.5 cm as radius and B as centre and 5.6 cm as radius construct arcs which intersect each other at point C.<\/p>\n\n\n\n

3. Now join AC and BC.<\/p>\n\n\n\n

4. Construct the angle bisector of \u2220A and \u2220B which intersect each other at point I.<\/p>\n\n\n\n

5. From the point I construct IL which is perpendicular to AB.<\/p>\n\n\n\n

6. Taking I as centre and IL as radius construct a circle which touches the sides of \u2206ABC internally.<\/p>\n\n\n\n

By measuring the required incircle the radius is 1.6 cm.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(ii) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment MN = 5.8 cm.<\/p>\n\n\n\n

2. At points M and N construct two rays which make an angle 300<\/sup> each intersecting each other at point P.<\/p>\n\n\n\n

3. Construct the angle bisectors of \u2220M and \u2220N which intersect each other at point I.<\/p>\n\n\n\n

4. From the point I draw perpendicular IL on MN.<\/p>\n\n\n\n

5. Taking I as centre and IL as radius construct a circle which touches the sides of \u2206 PMB internally.<\/p>\n\n\n\n

By measuring the required incircle the radius is 0.6 cm.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(iii) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment BC = 5.5 cm.<\/p>\n\n\n\n

2. Taking B and C as centres and 5.5 cm radius construct two arcs which intersect each other at point A.<\/p>\n\n\n\n

3. Now join AB and AC.<\/p>\n\n\n\n

4. Construct the perpendicular bisectors of \u2220B and \u2220C which intersect each other at the point I.<\/p>\n\n\n\n

5. From the point I construct IL which is perpendicular to BC.<\/p>\n\n\n\n

6. Taking I as centre and IL as radius construct a circle which touches the sides of \u2206ABC internally.<\/p>\n\n\n\n

This is the required incircle.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

(iv) Steps of Construction<\/p>\n\n\n\n

1. Construct a line segment PQ = 6 cm.<\/p>\n\n\n\n

2. At the point P construct rays which make an angle of 450<\/sup> and at point Q which makes an angle 600<\/sup> thats intersects each other at point R.<\/p>\n\n\n\n

3. Construct the bisectors of \u2220P and \u2220Q which intersect each other at point I.<\/p>\n\n\n\n

4. From the point I construct IL which is perpendicular to PQ.<\/p>\n\n\n\n

5. Taking I as centre and IL as radius construct a circle which touches the sides of \u2206PQR internally.<\/p>\n\n\n\n

This is the required incircle where the point I is incentre.<\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Download PDF<\/strong><\/h2>\n\n\n\n

Selina Class 7 ICSE Solutions Mathematics : Chapter 15- Triangles<\/p>\n\n\n\n

Download PDF<\/strong>: Selina Class 7 ICSE Solutions Mathematics : Chapter 15-\u00a0Triangles PDF<\/a><\/p>\n\n\n\n

Chapterwise Selina Publishers ICSE Solutions for Class 7 Mathematics :<\/strong><\/h2>\n\n\n\n