{"id":140327,"date":"2021-02-20T11:54:43","date_gmt":"2021-02-20T11:54:43","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=140327"},"modified":"2023-09-15T05:28:27","modified_gmt":"2023-09-15T05:28:27","slug":"ncert-solutions-for-7th-class-maths-chapter-6-the-triangle-and-its-properties","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-7th-class-maths-chapter-6-the-triangle-and-its-properties\/","title":{"rendered":"NCERT Solutions for 7th Class Maths: Chapter 6-The Triangle and its Properties"},"content":{"rendered":"\n

Class 7: Maths Chapter 6 solutions. Complete Class 7 Maths Chapter 6 Notes.<\/p>\n\n\n\n

NCERT Solutions for 7th Class Maths: Chapter 6-The Triangle and its Properties<\/strong><\/h2>\n\n\n\n

NCERT 7th Maths Chapter 6, class 7 Maths Chapter 6 solutions<\/p>\n\n\n\n

1. In APQR, D is the mid-point of <\/strong><\/p>\n\n\n\n

 is ________<\/p>\n\n\n\n

PD is_______<\/p>\n\n\n\n

Is QM = MR?<\/strong><\/p>\n\n\n\n

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

Answer<\/strong><\/p>\n\n\n\n

QD = DR<\/p>\n\n\n\n

\u2234  is altitude.<\/p>\n\n\n\n

PD is median.<\/p>\n\n\n\n

No, QM (eq) MR as D is the mid-point of QR.<\/p>\n\n\n\n

2. Draw rough sketches for the following: <\/strong><\/p>\n\n\n\n

(a) In \u0394ABC, BE is a median. <\/strong><\/p>\n\n\n\n

(b) In \u0394PQR, PQ and PR are altitudes of the triangle. <\/strong><\/p>\n\n\n\n

(c) In \u0394XYZ, YL is an altitude in the exterior of the triangle.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(a) Here, BE is a median in \u0394ABC and AE = EC.<\/p>\n\n\n\n

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

(b) Here, PQ and PR are the altitudes of the \u0394PQR and RP \u22a5 QP.<\/p>\n\n\n\n

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

(c) YL is an altitude in the exterior of \u0394XYZ.<\/p>\n\n\n\n

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

3. Verify by drawing a diagram if the median and altitude of a isosceles triangle can be same. <\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Isosceles triangle means any two sides are same.<\/p>\n\n\n\n

Take \u0394ABC and draw the median when AB = AC.<\/p>\n\n\n\n

AL is the median and altitude of the given triangle.<\/p>\n\n\n\n

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

NCERT 7th Maths Chapter 6, class 7 Maths Chapter 6 solutions<\/p>\n\n\n\n

Exercise 6.2<\/h4>\n\n\n\n

1. Find the value of the unknown exterior angle x in the following diagrams:<\/strong><\/p>\n\n\n\n

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

Answer<\/strong><\/p>\n\n\n\n

(i) x = 50\u00b0 + 70\u00b0 = 120\u00b0<\/p>\n\n\n\n

(ii) x = 65\u00b0 + 45\u00b0 = 110\u00b0<\/p>\n\n\n\n

(iii) x = 30\u00b0 + 40\u00b0 = 70\u00b0<\/p>\n\n\n\n

(iv) x = 60\u00b0+60\u00b0 = 120\u00b0<\/p>\n\n\n\n

(v) x = 50\u00b0 + 50\u00b0 = 100\u00b0<\/p>\n\n\n\n

(vi) x = 60\u00b0 + 30\u00b0 = 90\u00b0<\/p>\n\n\n\n

2. Find the value of the unknown interior angle x in the following figures:<\/strong><\/p>\n\n\n\n

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

Answer<\/strong><\/p>\n\n\n\n

(i)  x + 50\u00b0 = 115\u00b0   \u21d2    x = 115\u00b0 – 50\u00b0 = 65\u00b0<\/p>\n\n\n\n

(ii)  70\u00b0+ x = 100\u00b0    \u21d2   x = 100\u00b0- 70\u00b0 =  30\u00b0<\/p>\n\n\n\n

(iii)  x + 90\u00b0 = 125\u00b0  \u21d2  x = 120\u00b0- 90\u00b0 =  35\u00b0<\/p>\n\n\n\n

(iv)  60\u00b0+ x = 120\u00b0   \u21d2  x = 120\u00b0- 60\u00b0 = 60\u00b0<\/p>\n\n\n\n

(v)   30\u00b0 + x = 80\u00b0   \u21d2   x = 80\u00b0- 30\u00b0 = 50\u00b0<\/p>\n\n\n\n

(vi)  x + 35\u00b0= 75\u00b0   \u21d2 x = 75\u00b0- 35\u00b0 = 40\u00b0<\/p>\n\n\n\n

NCERT 7th Maths Chapter 6, class 7 Maths Chapter 6 solutions<\/p>\n\n\n\n

Exercise 6.3<\/h4>\n\n\n\n

1. Find the value of unknown x in the following diagrams:<\/strong><\/p>\n\n\n\n

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

Answer<\/strong><\/p>\n\n\n\n

(i) In \u0394ABC,<\/p>\n\n\n\n

\u2220 BAC + \u2220 ACB + \u2220 ABC = 180\u00b0  [ By angle sum property of a triangle]<\/p>\n\n\n\n

\u21d2 x + 50\u00b0+ 60\u00b0 = 180\u00b0<\/p>\n\n\n\n

\u21d2  x  + 110\u00b0 = 180\u00b0<\/p>\n\n\n\n

\u21d2  x = 180\u00b0-110\u00b0 = 70\u00b0<\/p>\n\n\n\n

(ii) In \u0394PQR,<\/p>\n\n\n\n

\u2220 RPQ + \u2220 PQR + \u2220 RPQ = 180\u00b0 [By angle sum property of a triangle]<\/p>\n\n\n\n

\u21d2  90\u00b0+30\u00b0+ x = 180\u00b0<\/p>\n\n\n\n

\u21d2  x+ 120\u00b0 = 180\u00b0<\/p>\n\n\n\n

\u21d2   x= 180\u00b0-120\u00b0= 60\u00b0<\/p>\n\n\n\n

(iii) In \u0394XYZ,<\/p>\n\n\n\n

\u2220 ZXY + \u2220 XYZ + \u2220 YZX = 180\u00b0 [By angle sum property of a triangle]<\/p>\n\n\n\n

\u21d2   30\u00b0 + 110\u00b0 + x = 180\u00b0<\/p>\n\n\n\n

\u21d2  x + 140\u00b0 = 180\u00b0<\/p>\n\n\n\n

\u21d2 x = 180\u00b0-140\u00b0 = 40\u00b0<\/p>\n\n\n\n

(iv)  In the given isosceles triangle,<\/p>\n\n\n\n

x+x + 50\u00b0 = 180\u00b0   [By angle sum property of a triangle]<\/p>\n\n\n\n

\u21d2  2x+50\u00b0= 180\u00b0<\/p>\n\n\n\n

\u21d2  2x = 180\u00b0- 50\u00b0<\/p>\n\n\n\n

\u21d2  2x = 130\u00b0<\/p>\n\n\n\n

\u21d2 x = 130\u00b0\/2 = 65\u00b0<\/p>\n\n\n\n

(v) In the given equilateral triangle,<\/p>\n\n\n\n

x +x+x = 180\u00b0 [By angle sum property of a triangle]<\/p>\n\n\n\n

\u21d2 3x = 180\u00b0<\/p>\n\n\n\n

\u21d2 x = 180\u00b0\/3 = 60\u00b0<\/p>\n\n\n\n

(vi) In the given right angled triangle,<\/p>\n\n\n\n

x + 2x+90\u00b0 = 180\u00b0    [By angle sum property of a triangle)<\/p>\n\n\n\n

\u21d2  3x+90\u00b0 = 180\u00b0<\/p>\n\n\n\n

\u21d2  3* = 180\u00b0- 90\u00b0<\/p>\n\n\n\n

\u21d2 3x = 90\u00b0<\/p>\n\n\n\n

\u21d2 x = 90\u00b0\/3 = 30\u00b0<\/p>\n\n\n\n

2. Find the values of the unknowns x and y in the following diagrams:<\/strong><\/p>\n\n\n\n

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

Answer<\/strong><\/p>\n\n\n\n

(i) 50\u00b0 + x = 120\u00b0    [Exterior angle property of a \u0394]<\/p>\n\n\n\n

\u21d2   x = 120\u00b0-50\u00b0 = 70\u00b0<\/p>\n\n\n\n

Now, 50\u00b0 + x + y =180\u00b0    [Angle sum property of a \u0394]<\/p>\n\n\n\n

\u21d2  50\u00b0 + 70\u00b0 + y = 180\u00b0<\/p>\n\n\n\n

\u21d2  120\u00b0 + y= 180\u00b0<\/p>\n\n\n\n

\u21d2  y = 180\u00b0 -120\u00b0 = 60\u00b0<\/p>\n\n\n\n

(ii) y = 80\u00b0 ….. (i)    [Vertically opposite angle]<\/p>\n\n\n\n

Now, 50\u00b0 + x + y =180\u00b0    [Angle sum property of a \u0394]<\/p>\n\n\n\n

\u21d2  50\u00b0 + 80\u00b0+y = 180\u00b0    [From equation (i)]<\/p>\n\n\n\n

\u21d2 130\u00b0 + y = 180\u00b0<\/p>\n\n\n\n

\u21d2 y = 180\u00b0 -130\u00b0 = 50\u00b0<\/p>\n\n\n\n

(iii) 50\u00b0+ 60\u00b0 = x    (Exterior angle property of a \u0394]<\/p>\n\n\n\n

x = 110\u00b0<\/p>\n\n\n\n

Now 50\u00b0 + 60\u00b0+ y = 180\u00b0    [Angle sum property of a \u0394]<\/p>\n\n\n\n

\u21d2  110\u00b0 + y = 180\u00b0<\/p>\n\n\n\n

\u21d2  y = 180\u00b0 – 110\u00b0<\/p>\n\n\n\n

\u21d2  y = 70\u00b0<\/p>\n\n\n\n

(iv) x = 60\u00b0 ….. (i) [Vertically opposite angle]<\/p>\n\n\n\n

Now, 30\u00b0 + x + y = 180\u00b0    [Angle sum property of a \u0394 ]<\/p>\n\n\n\n

\u21d2  50\u00b0 + 60\u00b0 + y = 180\u00b0    [From equation (i)]<\/p>\n\n\n\n

\u21d2 90\u00b0 + y = 180\u00b0<\/p>\n\n\n\n

\u21d2  y = 180\u00b0 – 90\u00b0 = 90\u00b0<\/p>\n\n\n\n

(v) y = 90\u00b0   ….(i)    [Vertically opposite angle]<\/p>\n\n\n\n

Now, y + x + x = 180\u00b0    [Angle sum property of a \u0394]<\/p>\n\n\n\n

\u21d2  90\u00b0 + 2x = 180\u00b0    [From equation (i)]<\/p>\n\n\n\n

\u21d2 2x = 180\u00b0- 90\u00b0<\/p>\n\n\n\n

\u21d2  2x = 90\u00b0<\/p>\n\n\n\n

\u21d2 x = 90\u00b0\/2 = 45\u00b0<\/p>\n\n\n\n

(vi) x = y  …(i)   [Vertically opposite angle]<\/p>\n\n\n\n

Now, x + x + y = 180\u00b0    [Angle sum property of a \u0394]<\/p>\n\n\n\n

\u21d2 2x+x = 180\u00b0    [From equation (i)]<\/p>\n\n\n\n

\u21d2 3x = 180\u00b0<\/p>\n\n\n\n

\u21d2 x = 180\u00b0\/3 = 60\u00b0<\/p>\n\n\n\n

NCERT 7th Maths Chapter 6, class 7 Maths Chapter 6 solutions<\/p>\n\n\n\n

Exercise 6.4<\/h4>\n\n\n\n

1. Is it possible to have a triangle with the following sides? <\/strong><\/p>\n\n\n\n

(i) 2 cm, 3 cm, 5 cm <\/strong><\/p>\n\n\n\n

(ii) 3 cm, 6 cm, 7 cm <\/strong><\/p>\n\n\n\n

(iii) 6 cm, 3 cm, 2 cm <\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) 2 cm, 3 cm, 5 cm<\/p>\n\n\n\n

2 + 3 > 5 No<\/p>\n\n\n\n

2 + 5 > 3 Yes<\/p>\n\n\n\n

3 + 5 > 2 Yes<\/p>\n\n\n\n

This triangle is not possible.<\/p>\n\n\n\n

(ii) 3 cm, 6 cm, 7 cm<\/p>\n\n\n\n

3 + 6 > 7 Yes<\/p>\n\n\n\n

6 + 7 > 3 Yes<\/p>\n\n\n\n

3 + 7 > 6 Yes<\/p>\n\n\n\n

This triangle is possible.<\/p>\n\n\n\n

(iii) 6 cm, 3 cm, 2 cm<\/p>\n\n\n\n

6 + 3 > 2 Yes<\/p>\n\n\n\n

6 + 2 > 3 Yes<\/p>\n\n\n\n

2 + 3 > 6 No<\/p>\n\n\n\n

This triangle is not possible.<\/p>\n\n\n\n

2. Take any point O in the interior of a triangle PQR. Is:<\/strong><\/p>\n\n\n\n

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

(i) OP + OQ > PQ ?<\/strong><\/p>\n\n\n\n

(ii) OQ + OR > QR?<\/strong><\/p>\n\n\n\n

(iii) OR + OP > RP?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

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

(i) Is OP + OQ > PQ ?<\/p>\n\n\n\n

Yes, POQ form a triangle.<\/p>\n\n\n\n

(ii) Is OQ + OR > QR ?<\/p>\n\n\n\n

Yes, RQO form a triangle.<\/p>\n\n\n\n

(iii) Is OR + OP > RP ?<\/p>\n\n\n\n

Yes, ROP form a triangle.<\/p>\n\n\n\n

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

3. AM is a median of a triangle ABC. Is AB + BC + CA > 2 AM? (Consider the sides of triangles \u0394ABM and \u0394AMC.)<\/strong><\/p>\n\n\n\n

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

Answer<\/strong><\/p>\n\n\n\n

Since, the sum of lengths of any two sides in a triangle should be greater titan the length of third side.<\/p>\n\n\n\n

Therefore, In \u0394ABM, AB + BM > AM … (i)<\/p>\n\n\n\n

In \u0394AMC, AC + MC > AM … (ii)<\/p>\n\n\n\n

Adding eq. (i) and (ii),<\/p>\n\n\n\n

AB + BM + AC + MC > AM + AM <\/p>\n\n\n\n

\u21d2 AB + AC + (BM + MC) > 2AM <\/p>\n\n\n\n

\u21d2 AB + AC + BC > 2AM<\/p>\n\n\n\n

Hence, it is true.<\/p>\n\n\n\n

4. ABCD is a quadrilateral. Is AB + BC + CD + DA > AC + BD?<\/strong><\/p>\n\n\n\n

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

Answer<\/strong><\/p>\n\n\n\n

Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side.<\/p>\n\n\n\n

Therefore, In \u0394 ABC, AB + BC > AC ………(i)<\/p>\n\n\n\n

In \u0394 ADC, AD + DC > AC  (ii)<\/p>\n\n\n\n

In \u0394DCB, DC + CB > DB  (iii)<\/p>\n\n\n\n

In \u0394ADB, AD + AB > DB   (iv)<\/p>\n\n\n\n

Adding equations (i), (ii), (iii) and (iv), we get<\/p>\n\n\n\n

AB + BC + AD + DC + DC + CB + AD + AB > AC + AC + DB + DB <\/p>\n\n\n\n

\u21d2 (AB + AB) + (BC + BC) + (AD + AD) + (DC + DC) > 2AC + 2DB <\/p>\n\n\n\n

\u21d2 2AB + 2BC +2AD + 2DC > 2(AC+DB)<\/p>\n\n\n\n

\u21d2 2(AB +BC + AD +DC) > 2(AC +DB)<\/p>\n\n\n\n

\u21d2 AB + BC + AD + DC > AC + DB<\/p>\n\n\n\n

\u21d2 AB + BC + CD + DA > AC + DB<\/p>\n\n\n\n

5. ABCD is quadrilateral. Is AB + BC + CD + DA < 2 (AC + BD)?<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side.<\/p>\n\n\n\n

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

Therefore, In A AOB, AB < OA + OB ……… (i)<\/p>\n\n\n\n

In A BOC, BC < OB + OC  (ii)<\/p>\n\n\n\n

In A COD, CD<OC + OD  (iii)<\/p>\n\n\n\n

InAAOD, DA < OD + OA  ….(iv)<\/p>\n\n\n\n

Adding equations (i), (ii), (iii) and (iv), we get<\/p>\n\n\n\n

AB + BC + CD + DA < OA + OB + OB + OC + OC + OD + OD + OA <\/p>\n\n\n\n

\u21d2 AB + BC + CD + DA < 2OA + 20B + 2OC + 2OD<\/p>\n\n\n\n

\u21d2 AB + BC + CD + DA < 2[(AO + OC) + (DO + OB)]<\/p>\n\n\n\n

\u21d2 AB + BC + CD + DA < 2(AC + BD)<\/p>\n\n\n\n

Hence, it is proved.<\/p>\n\n\n\n

6. The lengths of two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall? <\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side.<\/p>\n\n\n\n

It is given that two sides of triangle are 12 cm and 15 cm.<\/p>\n\n\n\n

Therefore, the third side should be less than 12 + 15 = 27 cm.<\/p>\n\n\n\n

Also the third side cannot be less than the difference of the two sides.<\/p>\n\n\n\n

Therefore, the third side has to be more than 15 \u2013 12 = 3 cm. Hence, the third side could be the length more than 3 cm and less than 27 cm.<\/p>\n\n\n\n

NCERT 7th Maths Chapter 6, class 7 Maths Chapter 6 solutions<\/p>\n\n\n\n

Exercise 6.5<\/h4>\n\n\n\n

1. PQR is a triangle, right angled at P. If PQ = 10 cm and PR = 24 cm, find QR.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

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

Given: PQ = 10 cm, PR = 24 cm<\/p>\n\n\n\n

Let QR be x cm.<\/p>\n\n\n\n

In right angled triangle QPR,<\/p>\n\n\n\n

[Hypotenuse)2<\/sup> = (Base)2<\/sup> + (Perpendicular)2<\/sup>   [By Pythagoras theorem]<\/p>\n\n\n\n

\u21d2 (QR)2<\/sup> = (PQ)2<\/sup> + (PR]2<\/sup><\/p>\n\n\n\n

\u21d2 x2<\/sup> = (10)2<\/sup> +(24)2<\/sup> <\/p>\n\n\n\n

\u21d2 x2<\/sup> = 100 + 576 = 676<\/p>\n\n\n\n

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

= 26 cm<\/p>\n\n\n\n

Thus, the length of QR is 26 cm.<\/p>\n\n\n\n

2. ABC is a triangle, right angled at C. If AB = 25 cm and AC = 7 cm, find BC.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Given: AB = 25 cm, AC = 7 cm<\/p>\n\n\n\n

 Let BC be x cm.<\/p>\n\n\n\n

 In right angled triangle ACB,<\/p>\n\n\n\n

 (Hypotenuse)2 = (Base)2 + (Perpendicular)2      [By Pythagoras theorem]<\/p>\n\n\n\n

 \u21d2 (AB)2 = (AC)2 + (BC)2<\/p>\n\n\n\n

 \u21d2 (25)2 = (7)2+x2<\/p>\n\n\n\n

 \u21d2 625 = 49 + x2<\/p>\n\n\n\n

 \u21d2 x2 = 625 – 49 = 576<\/p>\n\n\n\n

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

= 24 cm<\/p>\n\n\n\n

Thus, the length of BC is 24 cm.<\/p>\n\n\n\n

3. A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.<\/strong><\/p>\n\n\n\n

<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

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

Let AC be the ladder and A be the window.<\/p>\n\n\n\n

Given: AC = 15 m, AB = 12 m, CB = a m<\/p>\n\n\n\n

In right angled triangle ACB,<\/p>\n\n\n\n

(Hypotenuse)2<\/sup> = (Base)2<\/sup> + (Perpendicular)2<\/sup>    [By Pythagoras theorem]<\/p>\n\n\n\n

\u21d2 (AC)2<\/sup> = (CB)2<\/sup> + (AB)2<\/sup><\/p>\n\n\n\n

\u21d2 (15)2<\/sup> + (a)2<\/sup> = (12)2<\/sup><\/p>\n\n\n\n

\u21d2 225 = a2<\/sup> + 144<\/p>\n\n\n\n

\u21d2 a2<\/sup> = 225 – 144 = 81<\/p>\n\n\n\n

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

= 9 m<\/p>\n\n\n\n

Thus, the distance of the foot of the ladder from the wall is 9 m.<\/p>\n\n\n\n

4. Which of the following can be the sides of a right triangle? <\/strong><\/p>\n\n\n\n

(i) 2.5 cm, 6.5 cm, 6 cm <\/strong><\/p>\n\n\n\n

(ii) 2 cm, 2 cm, 5 cm <\/strong><\/p>\n\n\n\n

(iii) 1.5 cm, 2 cm, 2.5 cm <\/strong><\/p>\n\n\n\n

In the case of right angled triangles, identify the right angles.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Let us consider, the larger side be the hypotenuse and also using Pythagoras theorem,<\/p>\n\n\n\n

(Hypotenuse)2<\/sup> = (Base)2<\/sup> + (Perpendicular)2<\/sup><\/p>\n\n\n\n

(i) 2.5 cm, 6.5 cm, 6 cm<\/p>\n\n\n\n

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

In AABC, (AC)2<\/sup> = (AB)2<\/sup> + (BC)2<\/sup><\/p>\n\n\n\n

L.H.S. = (6.5)2<\/sup> = 42.25 cm<\/p>\n\n\n\n

R.H.S. = (6)2<\/sup> + (2.5)2<\/sup> = 36 + 6.25 = 42.25 cm<\/p>\n\n\n\n

Since, L.H.S. = R.H.S.<\/p>\n\n\n\n

Therefore, the given sides are of the right angled triangle.<\/p>\n\n\n\n

Right angle lies on the opposite to the greater side 6.5 cm, i.e., at B.<\/p>\n\n\n\n

(ii) 2 cm, 2 cm, 5 cm<\/p>\n\n\n\n

In the given triangle, (5)2<\/sup> = (2)2<\/sup> + (2)2<\/sup> <\/p>\n\n\n\n

L.H.S. = (5)2<\/sup> = 25<\/p>\n\n\n\n

R.H.S. = (2)2<\/sup> + (2)2<\/sup> = 4 + 4 = 8 <\/p>\n\n\n\n

Since, L.H.S. \u2260 R.H.S.<\/p>\n\n\n\n

Therefore, the given sides are not of the right angled triangle.<\/p>\n\n\n\n

(iii) 1.5 cm, 2 cm, 2.5 cm<\/p>\n\n\n\n

In \u0394PQR, (PR)2<\/sup> = (PQ)2<\/sup> + (RQ)2<\/sup><\/p>\n\n\n\n

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

L.H.S. = (2 5)2<\/sup> = 6.25 cm<\/p>\n\n\n\n

R.H.S. = (1.5)2<\/sup> + (2)2<\/sup> = 2.25 + 4 = 6.25 cm<\/p>\n\n\n\n

Since, L.H.S. = R.H.S.<\/p>\n\n\n\n

Therefore, the given sides are of the right angled triangle.<\/p>\n\n\n\n

Right angle lies on the opposite to the greater side 2.5 cm, i.e., at Q.<\/p>\n\n\n\n

5. A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree. <\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Let A’CB represents the tree before it broken at the point C and let the top A\u2019 touches the ground at A after it broke. Then \u0394ABC is a right angled triangle, right angled at B.<\/p>\n\n\n\n

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

AB = 12 m and BC = 5 m<\/p>\n\n\n\n

Using Pythagoras theorem, In \u0394ABC<\/p>\n\n\n\n

(AC)2<\/sup> =(AB)2<\/sup> + (BC)2<\/sup> <\/p>\n\n\n\n

\u21d2 (AC)2<\/sup> = (12)2<\/sup> + (5)2<\/sup><\/p>\n\n\n\n

\u21d2 (AC)2<\/sup> = 144 + 25<\/p>\n\n\n\n

\u21d2 (AC)2<\/sup> = 169<\/p>\n\n\n\n

\u21d2 AC = 13 m<\/p>\n\n\n\n

Hence, the total height o f the tree = AC + CB 13 + 5 = 18 m.<\/p>\n\n\n\n

6. Angles Q and R of a \u0394PQR are 25\u00b0 and 65\u00b0<\/strong><\/p>\n\n\n\n

Write which of the following is true:<\/strong><\/p>\n\n\n\n

(i) PQ2<\/sup> + QR2<\/sup> = RP2<\/sup><\/strong><\/p>\n\n\n\n

(ii) PQ2<\/sup> + RP2<\/sup> = QR2<\/sup> <\/strong><\/p>\n\n\n\n

(iii) RP2<\/sup> + QR2<\/sup> = PQ2<\/sup><\/strong><\/p>\n\n\n\n

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

Answer<\/strong><\/p>\n\n\n\n

In \u0394PQR, \u2220PQR + \u2220QRP + \u2220RPQ = 180\u00b0      [By Angle sum property of a \u0394] <\/p>\n\n\n\n

\u21d2 25\u00b0 + 65\u00b0 + \u2220RPQ = 180\u00b0<\/p>\n\n\n\n

\u21d2 90\u00b0 + \u2220RPQ = 180\u00b0<\/p>\n\n\n\n

\u21d2 \u2220RPQ = 180\u00b0- 90\u00b0 = 90\u00b0<\/p>\n\n\n\n

Thus, \u0394PQR is a right angled triangle, right angled at P.<\/p>\n\n\n\n

\u2234 (Hypotenuse)2<\/sup> = (Base)2<\/sup> + (Perpendicular)2<\/sup>    [By Pythagoras theorem]<\/p>\n\n\n\n

\u21d2 (QR)2<\/sup> = (PR)2<\/sup> + (QP)2<\/sup><\/p>\n\n\n\n

Hence, Option (ii) is correct.<\/p>\n\n\n\n

7. Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm. <\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Given diagonal (PR) = 41 cm, length (PQ) = 40 cm<\/p>\n\n\n\n

Let breadth (QR) be x cm.<\/p>\n\n\n\n

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

Now, in right angled triangle PQR,<\/p>\n\n\n\n

(PR)2<\/sup> = (RQ)2<\/sup> +(PQ)2<\/sup>      [By Pythagoras theorem]<\/p>\n\n\n\n

\u21d2 (41)2<\/sup> = x2<\/sup> + (40)2<\/sup><\/p>\n\n\n\n

\u21d2 1681 = x2<\/sup> + 1600<\/p>\n\n\n\n

\u21d2 x2<\/sup> = 1681 – 1600<\/p>\n\n\n\n

\u21d2 x2<\/sup> = 81<\/p>\n\n\n\n

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

= 9 cm<\/p>\n\n\n\n

Therefore the breadth of the rectangle is 9 cm,<\/p>\n\n\n\n

Perimeter of rectangle – 2(length + breadth)<\/p>\n\n\n\n

= 2 [9 + 49)<\/p>\n\n\n\n

= 2 \u00d7 49 = 98 cm<\/p>\n\n\n\n

Hence, the perimeter of the rectangle is 98 cm.<\/p>\n\n\n\n

8. The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter. <\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Given: Diagonals AC = 30 cm and DB = 16 cm.<\/p>\n\n\n\n

Since the diagonals of the rhombus bisect at right angle to each other.<\/p>\n\n\n\n

Therefore, OD = DB\/2 = 16\/2 = 8 cm<\/p>\n\n\n\n

And OC = AC\/2 = 30\/2 = 15 cm<\/p>\n\n\n\n

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

Now , In right angle triangle DOC,<\/p>\n\n\n\n

(DC)2 =(OD)2 + (OC)2    [By Pythagoras dieorem]<\/p>\n\n\n\n

\u21d2  (DC)2 = (8)2+(15)2<\/p>\n\n\n\n

\u21d2 (DC)2 = 64 + 225 = 289<\/p>\n\n\n\n

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

= 17 cm<\/p>\n\n\n\n

Perimeter of rhombus = 4 x side = 4 x 17 = 68 cm<\/p>\n\n\n\n

Thus, die perimeter of rhombus is 68 cm.<\/p>\n\n\n\n

NCERT 7th Maths Chapter 6, class 7 Maths Chapter 6 solutions<\/p>\n\n\n\n

NCERT Solutions for 7th Class Maths: Chapter 6: Download PDF<\/strong><\/h2>\n\n\n\n

NCERT Solutions for 7th Class Maths: Chapter 6-The Triangle and its Properties<\/p>\n\n\n\n

Download PDF<\/strong>: NCERT Solutions for 7th Class Maths: Chapter 6-The Triangle and its Properties PDF<\/a><\/p>\n\n\n\n

Chapter-wise NCERT Solutions Class 7 Maths<\/h2>\n\n\n\n
Chapter 1 Integers<\/a><\/td><\/tr>
Chapter 2 Fractions and Decimals<\/a><\/td><\/tr>
Chapter 3 Data Handling<\/a><\/td><\/tr>
Chapter 4 Simple Equations<\/a><\/td><\/tr>
Chapter 5 Lines and Angles<\/a><\/td><\/tr>
Chapter 6 The Triangle and its Properties<\/a><\/td><\/tr>
Chapter 7 Congruence of Triangles<\/a><\/td><\/tr>
Chapter 8 Comparing Quantities<\/a><\/td><\/tr>
Chapter 9 Rational Numbers<\/a><\/td><\/tr>
Chapter 10 Practical Geometry<\/a><\/td><\/tr>
Chapter 11 Perimeter and Area<\/a><\/td><\/tr>
Chapter 12 Algebraic Expressions<\/a><\/td><\/tr>
Chapter 13 Exponents and Powers<\/a><\/td><\/tr>
Chapter 14 Symmetry<\/a><\/td><\/tr>
Chapter 15 Visualising Solid Shapes<\/a><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n


<\/strong>About NCERT<\/h2>\n\n\n\n

The National Council of Educational Research and Training is an autonomous organization of the Government of India which was established in 1961 as a literary, scientific, and charitable Society under the Societies Registration Act. Its headquarters are located at Sri Aurbindo Marg in New Delhi. Visit the Official NCERT website<\/a> to learn more. <\/p>\n\n\n\n

\n
NCERT Solutions<\/a><\/div>\n\n\n\n
NCERT Solutions for Class 7<\/a><\/div>\n\n\n\n
NCERT Solutions for Class 7 Maths<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Class 7: Maths Chapter 6 solutions. Complete Class 7 Maths Chapter 6 Notes. NCERT Solutions for 7th Class Maths: Chapter 6-The Triangle and its Properties NCERT 7th Maths Chapter 6, class 7 Maths Chapter 6 solutions 1. In APQR, D is the mid-point of   is ________ PD is_______ Is QM = MR? Answer QD = DR […]<\/p>\n","protected":false},"author":302,"featured_media":627316,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[1411,907],"tags":[1550],"boards":[1180],"class_list":["post-140327","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-book-solutions","category-class-7","tag-ncert-maths-class-7","boards-ncert","entry"],"yoast_head":"\nNCERT Solutions Class 7, Maths Chapter 6 - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"NCERT Solutions for 7th Class Maths: Chapter 6-The Triangle and its Properties | Browse all Class 7 Maths Solutions- IndCareer Schools\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-7th-class-maths-chapter-6-the-triangle-and-its-properties\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for 7th Class Maths: Chapter 6-The Triangle and its Properties\" \/>\n<meta property=\"og:description\" content=\"Class 7: Maths Chapter 6 solutions. 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