RD Sharma Class 10 Solutions Chapter 4 Exercise 4.5 (Updated for 2021-22)

RD Sharma Class 10 Solutions Chapter 4 Exercise 4.5: The resemblance of triangles can be demonstrated using a variety of criteria. In this exercise, all of these criteria are explored. Students who wish to learn more about these criteria can use the RD Sharma Class 10 Solutions to get high or full marks in their exams. For additional help, check the RD Sharma Solutions for Class 10 Maths Chapter 4 Exercise 4.5 PDF provided below.

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RD Sharma Class 10 Solutions Chapter 4 Exercise 4.5

Access answers to RD Sharma Solutions Class 10 Maths Chapter 4 Exercise 4.5- Important Question with Answers

Question 1.
In the figure, ∆ACB ~ ∆APQ. If BC = 8 cm, PQ = 4 cm, BA = 6.5 cm and AP = 2.8 cm, find CA and AQ. (C.B.S.E. 1991)
Solution:
In the figure,
∆ACB ~ ∆APQ
BC = 8 cm, PQ = 4 cm, BA = 6.5 cm and AP = 2.8 cm

Question 2.
In the figure, AB || QR. Find the length of PB. (C.B.S.E. 1994)

Solution:
In the figure,
In ∆PQR, AB || QR
AB = 3 cm, QR =9 cm, PR = 6 cm
In ∆PAB and ∆PQR
∠P = ∠P (common)
∠PAB = ∠PQR (corresponding angles)
∠PBA = ∠PRQ (corresponding angles)
∠PAB = ∠PQR (AAA axiom)

Question 3.
In the figure, XY || BC. Find the length of XY. (C.B.S.E. 1994C)
Solution:
In the figure
In ∆ABC XY || BC
AX = 1 cm, BC = 6 cm

Question 4.
In a right-angled triangle with sides a and b and hypotenuse c, the altitude drawn on the hypotenuse is x. Prove that ab = cx.
Solution:
Given : In right ∆ABC, ∠B is right angle BD ⊥ AC
Now AB = a, BC = b, AC = c and BD = x

Question 5.
In the figure, ∠ABC = 90° and BD ⊥ AC. If BD = 8 cm and AD = 4 cm, find CD.
Solution:

Question 6.
In the figure, ∠ABC = 90° and BD ⊥ AC. If AB = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm, find BC.
Solution:
In right ∆ABC, ∠B = 90°
BD ⊥ AC
AB = 5.7 cm, BD = 3.8 and CD = 5.4 cm

Question 7.
In the figure, DE || BC such that AE = 14 AC. If AB = 6 cm, find AD.
Solution:
In the figure, in ∆ABC, DE || BC
AE = 14 AC, AB = 6 cm

Question 8.
In the figure, if AB ⊥ BC, DC ⊥ BC, and DE ⊥ AC, prove that ∆CED ~ ∆ABC.

Solution:
Given : In the figure AB ⊥ BC, DC ⊥ BC and DE ⊥ AC
To prove : ∆CED ~ ∆ABC.
Proof: AB ⊥ BC
∠B = 90°
and ∠A + ∠ACB = 90° ….(i)
DC ⊥ BC
∠DCB = 90°
=> ∠ACB + ∠DCA = 90° ….(ii)
From (i) and (ii)
∠A = ∠DCA
Now in ∆CED and ∆ABC,
∠E = ∠B (each 90°)
∠DEA or ∠DCE = ∠A (proved)
∆CED ~ ∆ABC (AA axiom)
Hence proved.

Question 9.
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using similarity criterion
for two triangles, show that OAOC = OBOD
Solution:
Given: ABCD is a trapezium in which AB || DC and diagonals AC and BD intersect each other at O
To Prove : OAOC = OBOD

Question 10.
In ∆ABC and ∆AMP are two right triangles, right-angled at B and M respectively such that ∠MAP = ∠BAC. Prove that
(i) ∆ABC ~ ∆AMP
(ii) CAPA = BCMP.
Solution:
Given : In ∆ABC and ∆AMP,
∠B = ∠M = 90°
∠MAP = ∠BAC

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)
Solution:
The shadows are cast by a vertical stick and a tower at the same time
Their angles will be equal

Question 12.
The figure, A = CED, proves that ∠CAB ~ ∠CED. Also, find the value of x.

Solution:
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 the first triangle is 9 cm, what is the corresponding side of the other triangle? (C.B.S.E. 2002C)
Solution:
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.
Solution:
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 = 52 DE.
Solution:
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


Hence proved

Question 16.
D is the mid-point of side BC of an ∆ABC. AD is bisected at point E and BE produced cuts AC at point X. Prove that BE: EX = 3: 1.
Solution:
Given: In ∆ABC, D is the midpoint of BC, and E is the midpoint 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


Hence proved.

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.
Solution:
Given: ABCD is a parallelogram.
APQ is a straight line that 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) OAOC = OMOL
Solution:
Given: In ∆ABC, AL ⊥ BC, CM ⊥ AB which intersect each other at O

(ii) OAOC = OMOL
Hence proved.

Question 19.
ABCD is a quadrilateral in which AD = BC. If P, Q, R, S be the mid-points of AB, AC, CD, and BD respectively, show that PQRS is a rhombus.
Solution:
Given : In quadrilateral ABCD, AD = BC
P, Q, R, and S are the midpoints of AB, AC, CD, and AD respectively
PQ, QR, RS, SP are joined

Question 20.
In an isosceles ∆ABC, the base AB has produced both the ways to P and Q such that AP x BQ = AC². Prove that ∆APC ~ ∆BCQ.
Solution:
Given : In ∆ABC, AC = BC
Base AB is produced to both sides and points P and Q are taken in such a way that
AP x BQ = AC²
CP and CQ are joined
To prove: ∆APC ~ ∆BCQ

Question 21.
A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/sec. If the lamp is 3.6 m above ground, find the length of her shadow after 4 seconds.
Solution:
Let AB be the lamp post and CD be the girl and AB = 3.6 m, CD = 90100 = 9 m
Distance covered in 4 seconds = 1.2 m x 4 = 4.8 m
BD = 4.8 m

x = 1.6
Length of her shadow = 1.6 m

Question 22.
A vertical stick of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Solution:
Let AB be stick and DE be a tower.
A stick 6 m long casts a shadow of 4 m i.e., AB = 6 m and BC = 4 m
Let DE casts a shadow at the same time which is EF = 28 m
Let the height of tower DE = x
Now in ∆ABC and ∆DEF,
∠B = ∠E (each 90°)
∠C = ∠F (shadows at the same time)
∆ABC ~ ∆DEF (AA criterion)

Question 23.
In the figure, ∆ABC is right-angled at C and DE ⊥ AB. Prove that ∆ABC ~ ∆ADE and hence find the lengths of AE and DE.

Solution:
Given: In the figure, ∆ABC is a right-angled triangle right angle at C.
DE ⊥ AB
To prove:
(i) ∆ABC ~ ∆ADE
(ii) Find the length of AE and DE
Proof: In ∆ABC and ∆ADE,
∠ACB = ∠AED (each 90°)
∠BAC = ∠DAE (common)
∆ABC ~ ∆ADE (AA axiom)

Question 24.
In the figure, PA, QB, and RC are each perpendicular to AC. Prove that

Solution:
Given: In the figure, PA, QB, and RC are perpendicular on AC and PA = x, QB = z, and RC = y



Hence proved.

Question 25.
In the figure, we have AB || CD || EF. If AB = 6 cm, CD = x cm, EF = 10 cm, BD = 4 cm and DE = y cm, calculate the values of x and y.

Solution:
In the figure, AB || CD || EF
AB = 6 cm, EF = 10 cm, BD = 4 cm, CD = x cm and DE = y cm
In ∆ABE, CE || AB
∆CED ~ ∆AEB

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