{"id":624175,"date":"2023-09-01T07:01:21","date_gmt":"2023-09-01T07:01:21","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=624175"},"modified":"2023-09-01T07:02:10","modified_gmt":"2023-09-01T07:02:10","slug":"kc-sinha-exercise-5-5-mathematics-solution-class-10-chapter-5-triangles","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-5-5-mathematics-solution-class-10-chapter-5-triangles\/","title":{"rendered":"KC Sinha: Exercise 5.5 – Mathematics Solution Class 10 Chapter 5 Triangles"},"content":{"rendered":"\n\n\n\n\n

Question 1 <\/h4>\n\n\n\n

Sides of some triangles are given below. Determine which of them are right triangles<\/strong><\/p>\n\n\n\n

(i) 8 cm, 15 cm, 17 cm<\/strong>
(ii) (2a \u20141) cm, 22\u2013\u221a <\/strong>cm and (2a + 1) cm
(iii) 7 cm, 24 cm, 25 cm<\/strong>
(iv) 1.4 cm, 4.8 cm, 5 cm<\/strong>
Sol :<\/p>\n\n\n\n

(i) Using Pythagoras theorem, i.e. if the square of the hypotenuse is equal to the sum of the other two sides. Then, the given triangle is a right angled triangle, otherwise not.
Here, (8)2<\/sup> + (15)2<\/sup> = 64 + 225 = 289 = (17)2<\/sup>
\u2234 given sides 8cm, 15cm and 17cm make a right angled triangle.
(ii) Using Pythagoras theorem, i.e. if the square of the hypotenuse is equal to the sum of the other two sides. Then, the given triangle is a right angled triangle, otherwise not.
Here, (2a \u2013 1)2<\/sup> + (2\u221a(2a))2<\/sup>
\u21d2 4a2<\/sup> + 1 \u2013 4a + 8a
\u21d2 4a2<\/sup> + 1 + 4a
= (2a + 1)2<\/sup>
\u2234 given sides (2a \u20141) cm, 22a\u2212\u2212\u221a cm and (2a + 1) cm make a right angled triangle.
(iii) Using Pythagoras theorem, i.e. if the square of the hypotenuse is equal to the sum of the other two sides. Then, the given triangle is a right angled triangle, otherwise not.
Here, (7)2<\/sup> + (24)2<\/sup> = 49 + 576 = 625 = (25)2<\/sup>
\u2234 given sides 7cm, 24cm and 25cm make a right angled triangle.
(iv) Using Pythagoras theorem, i.e. if the square of the hypotenuse is equal to the sum of the other two sides. Then, the given triangle is a right angled triangle, otherwise not.
Here, (1.4)2<\/sup> + (4.8)2<\/sup> = 1.96 + 23.04 = 25 = (5)2<\/sup>
\u2234 given sides 1.4cm, 4.8cm and 5cm make a right angled triangle.<\/p>\n\n\n\n

\n\n\n\n

Question 2 <\/h4>\n\n\n\n

A ladder 26 m long reaches a window 24 m above the ground. Find the distance of the foot of the ladder from the base of the wall.<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

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

Let AC be the position of a window from the ground and BC be the ladder, then the height of the window, AC =24m and length of the ladder, BC = 26m
Let AB = x m be the distance of the foot of the ladder from the base of the wall.
In \u2206CAB, using Pythagoras Theorm,
(Perpendicular)2<\/sup> + (Base)2<\/sup> = (Hypotenuse)2<\/sup><\/strong>
\u21d2 (AC)2<\/sup> + (AB)2<\/sup> = (BC)2<\/sup>
\u21d2 (24)2<\/sup> + (AB)2<\/sup> = (26)2<\/sup>
\u21d2 (AB)2<\/sup> = (26)2<\/sup> \u2013 (24)2<\/sup>
\u21d2 (AB)2<\/sup> = (26 \u2013 24)(26+24)
[\u2235 (a2<\/sup> \u2013 b2<\/sup>)=(a+b)(a \u2013 b)]
\u21d2 (AB)2<\/sup> = (2)(50)
\u21d2 (AB)2<\/sup> = 100
\u21d2 AB = \u00b110
\u21d2 AB = 10 [taking positive square root]
Hence, the distance of the foot of the ladder from base of the wall is 10m<\/p>\n\n\n\n

\n\n\n\n

Question 3 <\/h4>\n\n\n\n

A man goes 15 m due west and then 8 m due north. How far is he from the starting point?<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

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

Let AB = 15m and AC = 8m
In \u2206CAB, using Pythagoras Theorm,
(Perpendicular)2<\/sup> + (Base)2<\/sup> = (Hypotenuse)2<\/sup><\/strong>
\u21d2 (AC)2<\/sup> + (AB)2<\/sup> = (BC)2<\/sup>
\u21d2 (8)2<\/sup> + (15)2<\/sup> = (BC)2<\/sup>
\u21d2 (BC)2<\/sup> = 64 + 225
\u21d2 (BC)2<\/sup> = 289
\u21d2 BC = \u00b117
\u21d2 BC = 17 [taking positive square root]
Hence, the man is 17m far from the starting point.<\/p>\n\n\n\n

\n\n\n\n

Question 4 <\/h4>\n\n\n\n

A ladder 10 m long just reaches the top of a building 8 m high from the ground. Find the distance of the foot of the ladder from the building.<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

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

Let AC be the top of the building from the ground and BC be the ladder, then the height of the building, AC = 8m and length of the ladder, BC = 10m
Let AB = x m be the distance of the foot of the ladder from the building.
In \u2206CAB, using Pythagoras Theorm,
(Perpendicular)2<\/sup> + (Base)2<\/sup> = (Hypotenuse)2<\/sup><\/strong>
\u21d2 (AC)2<\/sup> + (AB)2<\/sup> = (BC)2<\/sup>
\u21d2 (8)2<\/sup> + (AB)2<\/sup> = (10)2<\/sup>
\u21d2 (AB)2<\/sup> = (10)2<\/sup> \u2013 (8)2<\/sup>
\u21d2 (AB)2<\/sup> = (10 \u2013 8)(10+8)
[\u2235 (a2<\/sup> \u2013 b2<\/sup>)=(a+b)(a \u2013 b)]
\u21d2 (AB)2<\/sup> = (2)(18)
\u21d2 (AB)2<\/sup> = 36
\u21d2 AB = \u00b16
\u21d2 AB = 6 [taking positive square root]
Hence, the distance of the foot of the ladder from building is 6m<\/p>\n\n\n\n

\n\n\n\n

Question 5<\/h4>\n\n\n\n

Find the length of a diagonal of a rectangle whose adjacent sides are 30 cm and 16 cm.<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

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

Let ABCD be a rectangle and AB and BC are the adjacent sides of length 30cm and 16cm respectively.
Let AC be the diagonal.
In \u2206CBA, using Pythagoras Theorm,
(Perpendicular)2<\/sup> + (Base)2<\/sup> = (Hypotenuse)2<\/sup><\/strong>
\u21d2 (AB)2<\/sup> + (BC)2<\/sup> = (AC)2<\/sup>
\u21d2 (30)2<\/sup> + (16)2<\/sup> = (AC)2<\/sup>
\u21d2 (AC)2<\/sup> = 900 + 256
\u21d2 (AC)2<\/sup> = 1156
\u21d2 AB = \u00b134
\u21d2 AB = 34 [taking positive square root]
Hence, the length of a diagonal of a rectangle is 34cm<\/p>\n\n\n\n

\n\n\n\n

Question 6 <\/h4>\n\n\n\n

A 13 m-long ladder reaches a window of a building 12 m above the ground. Determine the distance of the foot of the ladder from the building.<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

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

Let AC be the position of a window from the ground and BC be the ladder, then the height of the window, AC =12m and length of the ladder, BC = 13m
Let AB = x m be the distance of the foot of the ladder from the base of the wall.
In \u2206CAB, using Pythagoras Theorem,
(Perpendicular)2<\/sup> + (Base)2<\/sup> = (Hypotenuse)2<\/sup><\/strong>
\u21d2 (AC)2<\/sup> + (AB)2<\/sup> = (BC)2<\/sup>
\u21d2 (12)2<\/sup> + (AB)2<\/sup> = (13)2<\/sup>
\u21d2 (AB)2<\/sup> = (13)2<\/sup> \u2013 (12)2<\/sup>
\u21d2 (AB)2<\/sup> = (13 \u2013 12)(13+12)
[\u2235 (a2<\/sup> \u2013 b2<\/sup>)=(a+b)(a \u2013 b)]
\u21d2 (AB)2<\/sup> = (1)(25)
\u21d2 (AB)2<\/sup> = 25
\u21d2 AB = \u00b15
\u21d2 AB = 5 [taking positive square root]
Hence, the distance of the foot of the ladder from base of the wall is 5m<\/p>\n\n\n\n

\n\n\n\n

Question 7 <\/h4>\n\n\n\n

Two vertical poles of height 9 m and 14 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

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

Let BC and AD be the two poles of height 14m and 9m respectively. Again, let CD be the distance between tops of the poles.
Then, CE = BC \u2013 AD = 14 \u2013 9 = 5m [\u2235AD =BE]
Also, AB =12m
In \u2206CED, using Pythagoras theorem, we get
(Perpendicular)2<\/sup> + (Base)2<\/sup> = (Hypotenuse)2<\/sup><\/strong>
\u21d2 (CE)2<\/sup> + (DE)2<\/sup> = (CD)2<\/sup>
\u21d2 (5)2<\/sup> + (12)2<\/sup> = (CD)2<\/sup>
\u21d2 (CD)2<\/sup> = 25 + 144
\u21d2 (CD)2<\/sup> = 169
\u21d2 CD = \u221a169
\u21d2 CD = \u00b113
\u21d2 CD = 13 [taking positive square root]
Hence, the distance between the tops of the poles is 13m<\/p>\n\n\n\n

\n\n\n\n

Question 8 <\/h4>\n\n\n\n

A man goes 10 m due south and then 24 m due west. How far is he from the starting point?<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

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

Let AB = 10m and AC = 24m
In \u2206CAB, using Pythagoras Theorem,
(Perpendicular)2<\/sup> + (Base)2<\/sup> = (Hypotenuse)2<\/sup><\/strong>
\u21d2 (AC)2<\/sup> + (AB)2<\/sup> = (BC)2<\/sup>
\u21d2 (24)2<\/sup> + (10)2<\/sup> = (BC)2<\/sup>
\u21d2 (BC)2<\/sup> = 576 + 100
\u21d2 (BC)2<\/sup> = 676
\u21d2 BC = \u00b126
\u21d2 BC = 26 [taking positive square root]
Hence, the man is 26m far from the starting point.<\/p>\n\n\n\n

\n\n\n\n

Question 9 <\/h4>\n\n\n\n

A man goes 80 m due east and then 150 m due north. How far is he from the starting point?<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

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

Let AB = 80m and AC = 150m<\/p>\n\n\n\n

In \u2206CAB, using Pythagoras Theorem,
(Perpendicular)2<\/sup> + (Base)2<\/sup> = (Hypotenuse)2<\/sup><\/strong>
\u21d2 (AC)2<\/sup> + (AB)2<\/sup> = (BC)2<\/sup>
\u21d2 (150)2<\/sup> + (80)2<\/sup> = (BC)2<\/sup>
\u21d2 (BC)2<\/sup> = 22500 + 6400
\u21d2 (BC)2<\/sup> = 28900
\u21d2 BC = \u00b1170
\u21d2 BC = 170 [taking positive square root]
Hence, the man is 170 m far from the starting point.<\/p>\n\n\n\n

\n\n\n\n

Question 10 <\/h4>\n\n\n\n

\u2206ABC is an isosceles triangle with AC = BC. If AB2<\/sup> = 2AC2<\/sup>, prove that \u0394ABC is a right triangle.<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

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

Given an isosceles triangle ABC with AC = BC, and AB2<\/sup> = 2AC2<\/sup>
To Prove: \u2206ABC is a right triangle
Proof: AB2<\/sup> = 2AC2<\/sup> (given)
\u21d2 AB2<\/sup> = AC2<\/sup> +AC2<\/sup>
\u21d2 AB2<\/sup> = AC2<\/sup> +BC2<\/sup> [\u2235AC =BC]
\u21d2 \u2206ABC is a right triangle right angled at C.<\/p>\n\n\n\n

\n\n\n\n

Question 11 <\/h4>\n\n\n\n

Find the length of each side of a rhombus whose diagonals are 24 cm and 10 cm long.<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

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

Let ABCD be a rhombus where AC = 10cm and BD =24cm
Let AC and BD intersect each other at O.
Now, we know that diagonals of rhombus bisect each other at 90\u00b0
Thus, we have<\/p>\n\n\n\n

AO=12\u00d7AC<\/p>\n\n\n\n

\u21d212\u00d710=5cm<\/p>\n\n\n\n

BO=12\u00d7BD<\/p>\n\n\n\n

=12\u00d724=12cm<\/p>\n\n\n\n

Since, AOB is a right angled triangle
So, by Pythagoras theorem, we have
(Perpendicular)2<\/sup> + (Base)2<\/sup> = (Hypotenuse)2<\/sup><\/strong>
\u21d2 (AO)2<\/sup> + (BO)2<\/sup> = (AB)2<\/sup>
\u21d2 (5)2<\/sup> + (12)2<\/sup> = (AB)2<\/sup>
\u21d2 (AB)2<\/sup> = 25 + 144
\u21d2 (AB)2<\/sup> = 169
\u21d2 AB = \u221a169
\u21d2 AB = \u00b113
\u21d2 AB = 13 [taking positive square root]
Hence, AB = 13cm
Thus, length of each side of rhombus is 13cm<\/p>\n\n\n\n

\n\n\n\n

Question 12 <\/h4>\n\n\n\n

\u0394ABC is an isosceles triangle right angled at C. Prove that AB2<\/sup> = 2AC2<\/sup>.<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

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

Given: ABC is an isosceles triangle right angled at C.
Let AC = BC
In \u2206ACB, using Pythagoras theorem, we have
(Perpendicular)2<\/sup> + (Base)2<\/sup> = (Hypotenuse)2<\/sup><\/strong>
\u21d2 (AC)2<\/sup> + (BC)2<\/sup> = (AB)2<\/sup>
\u21d2 (AC)2<\/sup> + (AC)2<\/sup> = (AB)2<\/sup>
[\u2235ABC is an isosceles triangle, AC =BC]
\u21d2 2(AC)2<\/sup> = (AB)2<\/sup>
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 13 <\/h4>\n\n\n\n

\u0394ABC is an isosceles triangle with AB = AC = 13 cm. The length of altitude from A on BC is 5 cm. Find BC.<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

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

Given: \u0394ABC is an isosceles triangle with AB = AC = 13 cm
Suppose the altitude from A on Bc meets BC at M.
\u2234 M is the midpoint of BC. AM = 5 cm
In \u2206AMB, using Pythagoras theorem, we have
(Perpendicular)2<\/sup> + (Base)2<\/sup> = (Hypotenuse)2<\/sup><\/strong>
\u21d2 (AM)2<\/sup> + (BM)2<\/sup> = (AB)2<\/sup>
\u21d2 (5)2<\/sup> + (BM)2<\/sup> = (13)2<\/sup>
\u21d2 (BM)2<\/sup> = (13)2<\/sup> \u2013 (5)2<\/sup>
\u21d2 (BM)2<\/sup> = (13 \u2013 5)(13+5)
[\u2235 (a2<\/sup> \u2013 b2<\/sup>) = (a + b)(a \u2013 b)]
\u21d2 (BM)2<\/sup> = (8)(18)
\u21d2 (BM)2<\/sup> = 144
\u21d2 BM = \u00b112
\u21d2 BM = 12 [taking positive square root]
\u2234 BC = 2BM or 2MC = 2\u00d712 = 24cm<\/p>\n\n\n\n

\n\n\n\n

Question 14 <\/h4>\n\n\n\n

In an equilateral triangle ABC, AD is drawn perpendicular to BC, meeting BC in D. Prove that AD2<\/sup> = 3BD2<\/sup>.<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

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

Given: ABC is an equilateral triangle
\u2234 AB = AC = BC
and AD \u22a5 BC
Now, In \u2206ADB, using Pythagoras theorem, we have
(Perpendicular)2<\/sup> + (Base)2<\/sup> = (Hypotenuse)2<\/sup><\/strong>
\u21d2 (AD)2<\/sup> + (BD)2<\/sup> = (AB)2<\/sup>
\u21d2 (AD)2<\/sup> + (BD)2<\/sup> = (BC)2<\/sup> [\u2235 AB = BC]
\u21d2 (AD)2<\/sup> + (BD)2<\/sup> = (2BD)2<\/sup> [ as AD\u22a5BC]
\u21d2 (AD)2<\/sup> + (BD)2<\/sup> = 4BD2<\/sup>
\u21d2 AD2<\/sup> = 3BD2<\/sup><\/p>\n\n\n\n

\n\n\n\n

Question 15 <\/h4>\n\n\n\n

Find the length of altitude AD of an isosceles \u0394 ABC in which AB = AC = 2a units and BC = a units.<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

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

Given: ABC is an isosceles triangle
\u2234 AB = AC = 2a and BC = a
and AD is the altitude on BC. Therefore, BC = 2BD
Now, In \u2206ADB, using Pythagoras theorem, we have
(Perpendicular)2<\/sup> + (Base)2<\/sup> = (Hypotenuse)2<\/sup><\/strong>
\u21d2 (AD)2<\/sup> + (BD)2<\/sup> = (AB)2<\/sup>
\u21d2(AD)2+(a2)2=(2a)2
\u21d2(AD)2=(2a)2\u2212(a2)2
\u21d2(AD)2=4a2\u2212a24
\u21d2(AD)2=16a2\u2212a24
\u21d2(AD)2=15a24
\u21d2AD=15a24\u2212\u2212\u2212\u2212\u221a
\u21d2AD=15\u221a2a [taking positive square root]<\/p>\n\n\n\n

\n\n\n\n

Question 16 <\/h4>\n\n\n\n

\u0394 ABC is an equilateral triangle of side 2a units. Find each of its altitudes.<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

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

Given: ABC is an equilateral triangle
\u2234 AB = AC = BC = 2a
And let AD is an altitude on BC. <\/p>\n\n\n\n

Therefore, BD=12\u00d7BC=a
Now, In \u2206ADB, using Pythagoras theorem, we have
(Perpendicular)2<\/sup> + (Base)2<\/sup> = (Hypotenuse)2<\/sup><\/strong>
\u21d2 (AD)2<\/sup> + (BD)2<\/sup> = (AB)2<\/sup>
\u21d2 (AD)2<\/sup> + (a)2<\/sup> = (2a)2<\/sup>
\u21d2 (AD)2<\/sup> = 4a2<\/sup> \u2013 a2<\/sup>
\u21d2 (AD)2<\/sup> = 3a2<\/sup>
\u21d2 AD = a\u221a3 units<\/p>\n\n\n\n

\n\n\n\n

Question 17 <\/h4>\n\n\n\n

Find the height of an equilateral triangle of side 12 cm.<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

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

Given: ABC is an equilateral triangle
\u2234 AB = AC = BC = 12cm
And let AD is an altitude on BC. <\/p>\n\n\n\n

Therefore, BD=12\u00d7BC=6cm
Now, In \u2206ADB, using Pythagoras theorem, we have
(Perpendicular)2<\/sup> + (Base)2<\/sup> = (Hypotenuse)2<\/sup><\/strong>
\u21d2 (AD)2<\/sup> + (BD)2<\/sup> = (AB)2<\/sup>
\u21d2 (AD)2<\/sup> + (6)2<\/sup> = (12)2<\/sup>
\u21d2 (AD)2<\/sup> = 144 \u2013 36
\u21d2 (AD)2<\/sup> = 108
\u21d2 AD = \u221a108
\u21d2 AD = 6\u221a3
Hence, the height of an equilateral triangle is 6\u221a3 cm<\/p>\n\n\n\n

\n\n\n\n

Question 18 <\/h4>\n\n\n\n

L and M are the mid-points of AB and BC respectively of \u0394ABC, right-angled at B. Prove that 4LC2<\/sup> = AB2<\/sup> + 4BC2<\/sup><\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

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

Given: ABC is a right triangle right angled at B
and L and M are the mid-points of AB and BC respectively.
\u21d2 AL = LB and BM = MC
In \u2206LBC, using Pythagoras theorem we have,
(Perpendicular)2<\/sup> + (Base)2<\/sup> = (Hypotenuse)2<\/sup><\/strong>
\u21d2 (LB)2<\/sup> + (BC)2<\/sup> = (LC)2<\/sup>
\u21d2(AB2)2+(BC)2=(LC)2
\u21d2 (AB)2<\/sup> + 4(BC)2<\/sup> = 4(LC)2<\/sup>
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 19 <\/h4>\n\n\n\n

Find the length of the second diagonal of a rhombus, whose side is 5 cm and one of the diagonals is 6 cm.<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

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

Let ABCD be a rhombus having AD = 5cm and AC = 6cm
Now, we know that diagonals of rhombus bisect each other at 90\u00b0
Thus, we have
AO=12\u00d7AC\u21d212\u00d76=3cm
<\/p>\n\n\n\n

Since, AOD is a right angled triangle
So, by Pythagoras theorem, we have
(Perpendicular)2<\/sup> + (Base)2<\/sup> = (Hypotenuse)2<\/sup><\/strong>
\u21d2 (AO)2<\/sup> + (BO)2<\/sup> = (AD)2<\/sup>
\u21d2 (3)2<\/sup> + (BO)2<\/sup> = (5)2<\/sup>
\u21d2 (BO)2<\/sup> = 25 \u2013 9
\u21d2 (BO)2<\/sup> = 16
\u21d2 BO = \u221a16
\u21d2 BO = \u00b14
\u21d2 BO = 4 [taking positive square root]
Hence, BO = 4cm
\u21d2 BC = 2BO = 2 \u00d7 4 = 8cm
Thus, length of each side of rhombus is 13cm.<\/p>\n\n\n\n

\n\n\n\n

Question 20 <\/h4>\n\n\n\n

In \u0394ABC, \u2220B = 90\u00b0 and D is the midpoint of BC. Prove that AC2<\/sup> = AD2<\/sup> + 3CD2<\/sup>.<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

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

Given: \u2220B = 90\u00b0 and D is the midpoint of BC .i.e. BD = DC
To Prove: AC2<\/sup> = AD2<\/sup> + 3CD2<\/sup>
In \u2206ABC, using Pythagoras theorem we have,
(Perpendicular)2<\/sup> + (Base)2<\/sup> = (Hypotenuse)2<\/sup><\/strong>
\u21d2 (AB)2<\/sup> + (BC)2<\/sup> = (AC)2<\/sup>
\u21d2 (AB)2<\/sup> + (2CD)2<\/sup> =(AC)2<\/sup>
\u21d2 (AB)2<\/sup> + 4(CD)2<\/sup> =(AC)2<\/sup>
\u21d2 (AD2<\/sup> \u2013 BD2<\/sup>) + 4(CD2<\/sup>) = AC2<\/sup>
[\u2235 In right triangle \u2206ABD, AD2<\/sup> =AB2<\/sup> + BD2<\/sup> ]
\u21d2 AD2<\/sup> \u2013 BD2<\/sup> + 4CD2<\/sup> = AC2<\/sup>
\u21d2 AD2<\/sup> \u2013 CD2<\/sup> + 4CD2<\/sup> = AC2<\/sup>
[\u2235 D is the midpoint of BC, BD = DC]
\u21d2 AD2<\/sup> +3CD2<\/sup> = AC2<\/sup>
or AC2<\/sup> = AD2<\/sup> + 3CD2<\/sup>
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 21 <\/h4>\n\n\n\n

In \u2206ABC, \u2220C = 90\u00b0 and D is the midpoint of BC. Prove that AB2<\/sup> = 4AD2<\/sup> \u2014 3AC2<\/sup>.<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

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

Given: \u2220C = 90\u00b0 and D is the midpoint of BC .i.e. BC = 2CD
To Prove: AB2<\/sup> = 4AD2<\/sup> \u2014 3AC2<\/sup>
In \u2206ABC, using Pythagoras theorem we have,
(Perpendicular)2<\/sup> + (Base)2<\/sup> = (Hypotenuse)2<\/sup><\/strong>
\u21d2 (AC)2<\/sup> + (BC)2<\/sup> = (AB)2<\/sup>
\u21d2 (AC)2<\/sup> + (2CD)2<\/sup> =(AB)2<\/sup>
\u21d2 (AC)2<\/sup> + 4(CD)2<\/sup> =(AB)2<\/sup>
\u21d2 (AC)2<\/sup> + 4(AD2<\/sup> \u2013 AC2<\/sup>) = AB2<\/sup>
[\u2235 In right triangle \u2206ACD, AD2<\/sup> =AC2<\/sup> + CD2<\/sup> ]
\u21d2 AC2<\/sup> +4AD2<\/sup> \u2013 4AC2<\/sup> = AB2<\/sup>
\u21d2 4AD2<\/sup> \u2013 3AC2<\/sup> = AB2<\/sup>
or AB2<\/sup> = 4AD2<\/sup> \u2014 3AC2<\/sup>
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 22 <\/h4>\n\n\n\n

In an isosceles \u0394ABC, AB = AC and BD \u22a5 AC. Prove that BD2<\/sup> \u2014 CD2<\/sup> = 2CD AD.<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

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

Given: AB = AC and BD  AC
To Prove: BD2<\/sup> \u2013 CD2<\/sup> = 2CD \u00d7 AD
In \u2206BDC, using Pythagoras theorem we have,
(Perpendicular)2<\/sup> + (Base)2<\/sup> = (Hypotenuse)2<\/sup><\/strong>
\u21d2 (BD)2<\/sup> + (CD)2<\/sup> = (BC)2<\/sup> \u2026(i)
In \u2206BDA, using Pythagoras theorem we have,
(Perpendicular)2<\/sup> + (Base)2<\/sup> = (Hypotenuse)2<\/sup><\/strong>
\u21d2 (BD)2<\/sup> + (AD)2<\/sup> = (AB)2<\/sup>
\u21d2 (BD)2<\/sup> + (AD)2<\/sup> = (AC)2<\/sup> [\u2235 AB = AC]
Multiply this eq. by 2, we get
\u21d2 2(BD)2<\/sup> + 2(AD)2<\/sup> = 2(AC)2<\/sup> \u2026(ii)
Subtracting Eq. (ii) from (i), we get
\u21d2 CD2<\/sup> \u2013 BD2<\/sup> = BC2<\/sup> \u2013 2 AC2<\/sup> + 2 AD2<\/sup>
= BC2<\/sup> \u2013 2 (AD +CD)2<\/sup> + 2 AD2<\/sup>
= BC2<\/sup> \u2013 2 CD2<\/sup> \u2013 4 AD \u00d7 CD
= BD2<\/sup> + CD2<\/sup> \u2013 2 CD2<\/sup> \u2013 4 AD \u00d7 CD
= BD2<\/sup> \u2013 CD2<\/sup> \u2013 4 AD \u00d7 CD
\u21d2 CD2<\/sup> \u2013 BD2<\/sup> \u2013BD2<\/sup> +CD2<\/sup> = \u20134AD \u00d7 CD
\u21d2 \u20132(BD2<\/sup> \u2013 CD2<\/sup>) = \u20134AD \u00d7 CD
\u21d2 BD2<\/sup> \u2013 CD2<\/sup> = 2CD \u00d7 AD
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 23 <\/h4>\n\n\n\n

In a quadrilateral, \u0394BCD, \u2220B = 90\u00b0. If AD2<\/sup>= AB2<\/sup> + BC2<\/sup> + CD2<\/sup>, prove that \u2220ACD = 90\u00b0.<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

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

Given: ABCD is a quadrilateral and \u2220B = 90\u00b0
and AD2<\/sup>= AB2<\/sup> + BC2<\/sup> + CD2<\/sup>
To Prove: \u2220ACD = 90\u00b0
In right triangle \u2206ABC, using Pythagoras theorem, we have
(Perpendicular)2<\/sup> + (Base)2<\/sup> = (Hypotenuse)2<\/sup><\/strong>
\u21d2 (AB)2<\/sup> + (BC)2<\/sup> = (AC)2<\/sup> \u2026(i)
Given: AD2<\/sup>= AB2<\/sup> + BC2<\/sup> + CD2<\/sup>
\u21d2 AD2<\/sup>= AC2<\/sup> + CD2<\/sup> [from (i)]
In \u2206ACD
AD2<\/sup>= AC2<\/sup> + CD2<\/sup>
\u2234 \u2220ACD = 90\u00b0 [converse of Pythagoras theorem]
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 24 <\/h4>\n\n\n\n

In a rhombus ABCD, prove that: AB2<\/sup> + BC2<\/sup> + CD2<\/sup> + DA2<\/sup> = AC2<\/sup> + BD2<\/sup><\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

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

In rhombus ABCD, AB = BC = CD = DA
We know that diagonals bisect each other at 90\u00b0
And OA=OC=12\u00d7AC,OB=OD=12\u00d7BC
Consider right triangle AOB
(Perpendicular)2<\/sup> + (Base)2<\/sup> = (Hypotenuse)2<\/sup><\/strong>
\u21d2 (OA)2<\/sup> + (OB)2<\/sup> = (AB)2<\/sup>
\u21d2(AC2)2+(BD2)2=AB2
\u21d2 AC2<\/sup> + BD2<\/sup> = 4AB2<\/sup>
\u21d2 AC2<\/sup> + BD2<\/sup> = AB2<\/sup> + AB2<\/sup> + AB2<\/sup> + AB2<\/sup>
\u21d2 AC2<\/sup> + BD2<\/sup> = AB2<\/sup> + BC2<\/sup> + CD2<\/sup> + DA2<\/sup>
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 25 <\/h4>\n\n\n\n

In an equilateral triangle ABC, AD is the altitude drawn from A on side BC. Prove that 3AB2<\/sup>= 4AD2<\/sup>.<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

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

Given: ABC is an equilateral triangle
and AD is the altitude on side BC
To Prove: 3AB2<\/sup>= 4AD2<\/sup>
In right triangle \u2206ADB, using Pythagoras theorem
(Perpendicular)2<\/sup> + (Base)2<\/sup> = (Hypotenuse)2<\/sup><\/strong>
\u21d2 AD2<\/sup> + BD2<\/sup> = AB2<\/sup>
\u21d2AD2+(BC2)2=AB2
\u21d2 4AD2<\/sup> + BC2<\/sup> = 4AB2<\/sup>
\u21d2 4AD2<\/sup> = 4AB2<\/sup> \u2013 BC2<\/sup>
\u21d2 4AD2<\/sup> = 4AB2<\/sup> \u2013 AB2<\/sup> [\u2235ABC is an equilateral triangle]
\u21d2 4AD2<\/sup> = 3AB2<\/sup>
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 26 <\/h4>\n\n\n\n

In \u0394ABC, AB = AC. Side BC is produced to D. Prove that (AD2<\/sup> \u2014AC2<\/sup>) = BD . CD<\/strong><\/p>\n\n\n\n

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

Sol :
Construction: Draw an altitude from A on BC and named it O.<\/p>\n\n\n\n

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

Given: ABC is an isosceles triangle with AB = AC
To Prove: AD2<\/sup> \u2014AC2<\/sup> = BD \u00d7 CD
In right triangle \u2206AOD, using Pythagoras theorem, we have
(Perpendicular)2<\/sup> + (Base)2<\/sup> = (Hypotenuse)2<\/sup><\/strong>
\u21d2 AO2<\/sup> + OD2<\/sup> = AD2<\/sup> \u2026(i)
Now, in right triangle \u2206AOB, using Pythagoras theorem, we have
\u21d2 AO2<\/sup> + BO2<\/sup> = AB2<\/sup> \u2026(ii)
Subtracting eq (ii) from (i), we get
AD2<\/sup> \u2013 AB2<\/sup> = AO2<\/sup> + OD2<\/sup> \u2013 AO2<\/sup> \u2013 BO2<\/sup>
\u21d2 AD2<\/sup> \u2013 AB2<\/sup> = OD2<\/sup> \u2013 BO2<\/sup>
\u21d2 AD2<\/sup> \u2013 AB2<\/sup> = (OD + BO)(OD \u2013 OB)
[\u2235 (a2<\/sup> \u2013 b2<\/sup>)= (a + b)(a \u2013 b)]
\u21d2 AD2<\/sup> \u2013 AB2<\/sup> = (BD)(OD \u2013 OC) [\u2235OB = OC]
\u21d2 AD2<\/sup> \u2013 AB2<\/sup> = (BD)(CD)
\u21d2 AD2<\/sup> \u2013 AC2<\/sup> = (BD)(CD) [\u2235AB =AC]
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 27 <\/h4>\n\n\n\n

In \u0394ABC, D is the mid-point of BC and AE \u22a5 BC . If AC > AB, show that AB2<\/sup> = AD2<\/sup> \u2014 BC .DE + 1\/4 BC2<\/sup><\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

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

Given: In \u0394ABC, D is the mid-point of BC and AE BC
and AC > AB
In right triangle \u2206AEB, using Pythagoras theorem, we have
(Perpendicular)2<\/sup> + (Base)2<\/sup> = (Hypotenuse)2<\/sup><\/strong>
\u21d2 (AE)2<\/sup> + (BE)2<\/sup> = (AB)2<\/sup>
\u21d2 (AE)2<\/sup> + (BD \u2013 ED)2<\/sup> = (AB)2<\/sup>
\u21d2 (AE)2<\/sup> + (ED)2<\/sup> + (BD)2<\/sup> \u2013 2 (ED)(BD) = (AB)2<\/sup>
[\u2235 (a \u2013 b)2<\/sup> = a2<\/sup> + b2<\/sup> \u2013 2ab]
\u21d2 (AE2<\/sup> + ED2<\/sup>) + (BD)2<\/sup> \u2013 2 (ED)(BD) = (AB)2<\/sup>
\u21d2 (AD)2<\/sup> + (BD)2<\/sup> \u2013 2 (ED)(BD) = (AB)2<\/sup>
[\u2235 In right angled \u2206AED, AE2<\/sup> + ED2<\/sup> =AD2<\/sup>]
\u21d2(AD)2+(BC2)2\u22122ED(BC2)=(AB)2
[\u2235 D is the midpoint of BC, so 2DC = BC]
\u21d2AB2=AD2\u2212BC\u00d7ED+BC24
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 28 <\/h4>\n\n\n\n

ABC is an isosceles triangle, right angled at B. Similar triangles ACD and ABE are constructed on sides AC and AB. Find the ratio between the areas of \u0394ABE and \u0394ACD.<\/strong><\/p>\n\n\n\n

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

Sol :
Given \u2206ABC is an isosceles triangle in which \u2220B is right angled i.e. 90\u00b0
\u21d2 AB = BC
In right angled \u2206ABC, by Pythagoras theorem, we have
(Perpendicular)2<\/sup> + (Base)2<\/sup> = (Hypotenuse)2<\/sup><\/strong>
\u21d2 (AB)2<\/sup> + (BC)2<\/sup> = (AC)2<\/sup>
\u21d2 (AB)2<\/sup> + (AB)2<\/sup> = (AC)2<\/sup>
[\u2235ABC is an isosceles triangle, AB =BC]
\u21d2 2(AB)2<\/sup> = (AC)2<\/sup>
\u21d2 (AC)2<\/sup> = 2(AB)2<\/sup> \u2026(i)
It is also given that \u2206ABE ~ \u2206ADC
<\/p>\n\n\n\n

And we also know that, the ratio of similar triangles is equal to the ratio of their corresponding sides.
\u2234ar(\u0394ABE)ar(\u0394ADC)=AB2AC2
\u21d2ar(\u0394ABE)ar(\u0394ADC)=AB22AB2 [from (i)]
\u21d2ar(\u0394ABE)ar(\u0394ADC)=12
\u2234 ar(\u2206ABE) : ar(\u2206ADC) = 1 : 2<\/p>\n\n\n\n

\n\n\n\n

Question 29 <\/h4>\n\n\n\n

In the given figure, 0 is a point inside a \u2220PQR such that \u2220POR = 90\u00b0, OP = 6 cm and OR= 8 cm. If PQ = 24 cm and QR = 26 cm, prove that \u0394PQR is right angled. P<\/strong><\/p>\n\n\n\n

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

Sol :
Given: \u2220POR = 90\u00b0, OP = 6 cm and OR= 8 cm
and PQ = 24 cm and QR = 26 cm
To Prove: \u0394PQR is right angled at P
In \u2206POR, using Pythagoras theorem, we get
(Perpendicular)2<\/sup> + (Base)2<\/sup> = (Hypotenuse)2<\/sup><\/strong>
\u21d2 (PO)2<\/sup> + (OR)2<\/sup> = (PR)2<\/sup>
\u21d2 (6)2<\/sup> + (8)2<\/sup> = (PR)2<\/sup>
\u21d2 36 +64 = (PR)2<\/sup>
\u21d2 (PR)2<\/sup>= 100
\u21d2 PR =\u221a100
\u21d2 PR = 10 [taking positive square root]
In \u2206PQR
Using Pythagoras theorem, i.e. if the square of the hypotenuse is equal to the sum of the other two sides. Then, the given triangle is a right angled triangle, otherwise not.
Here, (PR)2<\/sup> + (PQ)2<\/sup>
\u21d2 (10)2<\/sup> + (24)2<\/sup>
= 100 + 576
= 676
= (26)2<\/sup> = (QR)2<\/sup>
\u2234 given sides 10cm, 24cm and 26cm make a right triangle right angled at P.
Hence Proved<\/p>\n\n\n\n

\n
KC Sinha Solutions<\/a><\/div>\n\n\n\n
KC Sinha Class 10 Solutions<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Question 1  Sides of some triangles are given below. Determine which of them are right triangles (i) 8 cm, 15 cm, 17 cm(ii) (2a \u20141) cm, 22\u2013\u221a cm and (2a + 1) cm(iii) 7 cm, 24 cm, 25 cm(iv) 1.4 cm, 4.8 cm, 5 cmSol : (i) Using Pythagoras theorem, i.e. if the square of the hypotenuse […]<\/p>\n","protected":false},"author":302,"featured_media":624114,"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":[24],"tags":[],"boards":[],"class_list":["post-624175","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-class-10","entry"],"yoast_head":"\nKC Sinha: Exercise 5.5 - Mathematics Solution Class 10 Chapter 5 Triangles - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"Question 1  Sides of some triangles are given below. 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