RD Chapter 4 Triangles Ex 4.7 Solutions
Question - 11 : - ABCD is a square, F is the mid point of AB. BE is one third of BC. If the area of ∆FBE = 108 cm² find the length of AC. (C.B.S.E. 1995)
Answer - 11 : - In square ABCD, F is mid point of AB i.e.,
Question - 12 : - In an isosceles triangle ABC, if AB = AC = 13 cm and the altitude from A on BC is 5 cm, find BC. (C.B.S.E. 2000)
Answer - 12 : -
In ∆ABC, AB = AC
AD ⊥ BC
AB = AC = 13 cm, AD = 5 cm
AD ⊥ BC
AD bisects BC at D
BD = (1/2) BC
=> BC = 2BD
Question - 13 : - In a ∆ABC, AB = BC = CA = 2a and AD ⊥ BC. Prove that
(i) AD = a √3
(ii) area (∆ABC) = √3 a² (C.B.S.E. 1991)
Answer - 13 : -
In ∆ABC, AB = BC = AC = 2a
AD ⊥ BC
AD bisects BC at D
BD = DC = (1/2) BC = a
Question - 14 : - The lengths of the diagonals of a rhombus are 24 cm and 10 cm. Find each side of the rhombus. (C.B.S.E. 1993C)
Answer - 14 : -
ABCD is a rhombus whose diagonals AC = 24 cm and BD = 10 cm
The diagonals of a rhombus bisect eachother at right angles
AO = OC = 242 = 12 cm
and BO = OD = 102 = 5 cm
Now in right ∆AOB,
AB² = AO² + BO² (Pythagoras Theorem)
(12)² + (5)² = 144 + 25 = 169 = (13)²
AB = 13
Each side of rhombus = 13 cm
Question - 15 : - Each side of a rhombus is 10 cm. If one of its diagonals is 16 cm, find the length of the other diagonal.
Answer - 15 : -
In rhombus ABCD, diagonals AC and BD bisect eachother at O at right angles
Each side = 10 cm and one diagonal AC = 16 cm
AO = OC = 162 = 8 cm
Now in right angled triangle AOB,
AB² = AO² + OB² (Pythagoras Theorem)
(10)² = (8)² + (BO)²
=> 100 = 64 + BO²
=> BO² = 100 – 64 = 36 = (6)²
BO = 6
BD = 2BO = 2 x 6 = 12 cm
Question - 16 : - Calculate the height of an equilateral triangle each of whose sides measures 12 cm.
Answer - 16 : - Each side of the equilateral ∆ABC = 12 cm
AD ⊥ BC which bisects BC at D
BD = DC = 122 = 6 cm
Question - 17 : - In the figure, ∠B < 90° and segment AD ⊥ BC. Show that:
(i) b² = h² + a² + x² – 2ax
(ii) b² = a² + c² – 2ax
Answer - 17 : -
Given : In ∆ABC, ∠B < 90°
AD ⊥ BC
AD = c, BC = a, CA = b AD = h, BD = x, DC = a – x
To prove: (i) b² = h² + a² + x² – 2ax
(ii) b² = a² + c² – 2ax
Proof: (i) In right ∆ADC, AC² = AD² + DC² (Pythagoras Theorem)
=> b² = h² + (a – x)² = h² + a² + x² – 2ax
(ii) Similarly in right ∆ADB
AB² = AD² + BD²
c² = h² + x² ….(i)
b² = h² + a² + x² – 2ax = h² + x² + a² – 2ax
= c² + a² – 2ax {From (i)}
= a² + c² – 2ax
Hence proved.
Question - 18 : - In an equilateral ∆ABC, AD ⊥ BC, prove that AD² = 3 BD². (C.B.S.E. 2002C)
Answer - 18 : -
Given : ∆ABC is an equilateral in which AB = BC = CA .
AD ⊥ BC
Question - 19 : - ∆ABD is a right triangle right-angled at A and AC ⊥ BD. Show that
(i) AB² = BC.BD
(ii) AC² = BC.DC
(iii) AD² = BD.CD
(iv) 
Answer - 19 : -
In ∆ABD, ∠A = 90°
AC ⊥ BD
Question - 20 : - A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut ?
Answer - 20 : - Let AB be the vertical pole whose height = 18 m