RD Chapter 4 Triangles Ex 4.5 Solutions
Question - 11 : - A vertical stick 10 cm long casts a shadow of 8 cm long. At the same time a tower casts a shadow 30 m long. Determine the height of the tower. (CB.S.E. 1991)
Answer - 11 : -
The shadows are casted by a vertical stick and a tower at the same time
Their angles will be equal
Question - 12 : - In the figure, A = CED, prove that ∠CAB ~ ∠CED. Also find the value of x.
Answer - 12 : -
Given : In ∆ABC,
∠CED = ∠A
AB = 9 cm, BE = 2 cm, EC = 10 cm, AD = 7 cm and DC = 8 cm
To prove :
(i) ∆CAB ~ ∆CED
(ii) Find the value of x
Proof: BC = BE + EC = 2 + 10 = 12 cm
AC = AD + DC = 7 + 8 = 15 cm
(i) Now in ∆CAB and ∆CED,
∠A = ∠CED (given)
∠C = ∠C (common)
∆CAB ~ ∆CED (AA axiom)
Question - 13 : - The perimeters of two similar triangles are 25 cm and 15 cm respectively. If one side of first triangle is 9 cm, what is the corresponding side of the other triangle? (C.B.S.E. 2002C)
Answer - 13 : -
Let perimeter of ∆ABC = 25 cm
and perimeter of ∆DEF = 15 cm
and side BC of ∆ABC = 9 cm
Now we have to find the side EF of ∆DEF
∆ABC ~ ∆DEF (given)
Question - 14 : - In ∆ABC and ∆DEF, it is being given that: AB = 5 cm, BC = 4 cm and CA = 4.2 cm; DE = 10 cm, EF = 8 cm and FD = 8.4 cm. If AL ⊥ BC and DM ⊥ EF, find AL : DM.
Answer - 14 : -
In ∆ABC and ∆DEF,
AB = 5 cm, BC = 4 cm, CA = 4.2 cm, DE = 10 cm, EF = 8 cm and FD = 8.4 cm
AL ⊥ BC and DM ⊥ EF
Question - 15 : - D and E are the points on the sides AB and AC respectively of a ∆ABC such that: AD = 8 cm, DB = 12 cm, AE = 6 cm and CE = 9 cm. Prove that BC = (5/2) DE.
Answer - 15 : -
Given : In ∆ABC, points D and E are on the sides AB and AC respectively
and AD = 8 cm, DB = 12 cm, AE = 6 cm and CE = 9 cm
Question - 16 : - D is the mid-point of side BC of a ∆ABC. AD is bisected at the point E and BE produced cuts AC at the point X. Prove that BE : EX = 3 : 1.
Answer - 16 : -
Given : In ∆ABC, D is mid point of BC, and E is mid point of AD
BE is joined and produced to meet AC at X
To prove : BE : EX = 3 : 1
Construction : From D, draw DY || BX meeting AC at Y
Question - 17 : - ABCD is a parallelogram and APQ is a straight line meeting BC at P and DC produced at Q. Prove that the rectangle obtained by BP and DQ is equal to the rectangle contained AB and BC.
Answer - 17 : -
Given : ABCD is a parallelogram.
APQ is a straight line which meets BC at P and DC on producing at Q
Question - 18 : - In ∆ABC, AL and CM are the perpendiculars from the vertices A and C to BC and AB respectively. If AL and CM intersect at O, prove that :
(i) ∆OMA ~ ∆OLC
(ii) OA/OC = OM/OL
Answer - 18 : - Given : In ∆ABC, AL ⊥ BC, CM ⊥ AB which intersect each other at O
(ii) OAOC = OMOL
Hence proved.
Answer - 19 : -
Given : In quadrilateral ABCD, AD = BC
P, Q, R and S are the mid points of AB, AC, CD and AD respectively
PQ, QR, RS, SP are joined
Question - 20 : - In an isosceles ∆ABC, the base AB is produced both the ways to P and Q such that AP x BQ = AC². Prove that ∆APC ~ ∆BCQ.
Answer - 20 : - Given : In ∆ABC, AC = BC
Base AB is produced to both sides and points P and Q are taken in such a waythat
AP x BQ = AC²
CP and CQ are joined
To prove : ∆APC ~ ∆BCQ
x = 1.6
Length of her shadow = 1.6 m