Class 10 RD Sharma Solutions – Chapter 4 Triangles – Exercise 4.5 | Set 1
Question 1. In fig., ΔACB ∼ ΔAPQ. If BC = 8 cm, PQ = 4 cm, BA = 6.5 cm and AP = 2.8 cm, find CA and AQ.
ΔACB ∼ ΔAPQ
BC = 8 cm, PQ = 4 cm, BA = 6.5 cm and AP = 2.8 cm
To find: CA and AQ
We know that,
ΔACB ∼ ΔAPQ [given]
BA/ AQ = CA/ AP = BC/ PQ [sides are proportional of Similar Triangles]
6.5/ AQ = 8/ 4
AQ = (6.5 x 4)/ 8
AQ = 3.25 cm
CA/ AP = BC/ PQ
CA/ 2.8 = 8/ 4
CA = 2.8 × 2
CA = 5.6 cm
Hence, CA = 5.6 cm and AQ = 3.25 cm.
Question 2. In fig., AB ∥ QR, find the length of PB.
ΔPQR, AB ∥ QR
AB = 3 cm, QR = 9 cm and PR = 6 cm
To find: length of PB
In ΔPAB and ΔPQR
∠P = ∠P [Common angle ]
∠PAB = ∠PQR [Corresponding angles]
∠PBA = ∠PRQ [Corresponding angles]
ΔPAB ∼ ΔPQR [By AAA similarity criteria]
AB/ QR = PB/ PR [sides are proportional of Similar Triangles]
⇒ 3/ 9 = PB/6
PB = 6/3
Hence the length of PB = 2 cm
Question 3. In fig. given, XY∥BC. Find the length of XY.
AX = 1 cm, XB = 3 cm and BC = 6 cm
To find: length of XY
In ΔAXY and ΔABC
∠A = ∠A [Common angle]
∠AXY = ∠ABC [Corresponding angles]
∠AYX = ∠ACB [Corresponding angles]
ΔAXY ∼ ΔABC [By AAA similarity criteria]
XY/ BC = AX/ AB [Corresponding Parts of Similar Triangles are propositional]
(AB = AX + XB = 1 + 3 = 4)
XY/6 = 1/4
XY/1 = 6/4
Hence, length of XY = 1.5 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.
Let us consider ΔABC to be a right angle triangle with sides a,b and c as hypotenuse. let BD be the altitude drawn on the hypotenuse AC.
To prove: ab = cx
In ΔABC and ΔADB
∠BAC = ∠DAB [Common]
∠ACB = ∠ABC = 90 [right angled triangle]
ΔABC ∼ ΔADB [By AA similarity criteria]
AC/ AB = BC/ EB [Corresponding Parts of Similar Triangles are propositional]
c/ a = b/ x
⇒ xc = ab
ab = cx
Question 5. In fig., ∠ABC = 90° and BD⊥AC. If BD = 8 cm, and AD = 4 cm, find CD.
ABC is a right-angled triangle and BD⊥AC.
BD = 8 cm, and AD = 4 cm
To find: CD.
Now in triangle ΔABD and ΔCBD,
∠BDC=∠BDA [each 90]
ΔDBA∼ΔDCB [By AA similarity]
BD/ CD = AD/ BD
BD2 = AD x DC
(8)2 = 4 x DC
DC = 64/4 = 16 cm
Hence, the length of side CD = 16 cm
Question 6. In fig., ∠ABC = 90o and BD ⊥ AC. If AB = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm, Find BC.
BD ⊥ AC
AB = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm
∠ABC = 90o
Required to find: BC
Now In ΔADB and ΔCDB,
∠ADB = ∠CDB = 90o [right angled triangle]
∠ABD = ∠CBD [∠ABD = ∠CBD = 90o – x]
ΔABC ∼ ΔBDC [By AA similarity]
AB/ BD = BC/ CD [Corresponding Parts of Similar Triangles are propositional]
5.7/ 3.8 = BC/ 5.4
BC = (5.7 × 5.4)/ 3.8 = 8.1
Hence, length of side BC = 8.1 cm
Question 7. In the fig. given, DE ∥ BC such that AE = (1/4)AC. If AB = 6 cm, find AD.
AE = (1/4)AC
AB = 6 cm.
To find: AD.
Now In ΔADE and ΔABC,
∠A = ∠A [Common angle]
∠ADE = ∠ABC [Corresponding angles]
ΔADE ∼ ΔABC [By AA similarity criteria]
AD/AB = AE/ AC [Corresponding Parts of Similar Triangles are propositional]
AD/6 = 1/4
4 x AD = 6
AD = 6/4
Hence length of AD = 1.5 cm
Question 8. In the fig. given, if AB ⊥ BC, DC ⊥ BC, and DE ⊥ AC, prove that ΔCED ∼ ΔABC
AB ⊥ BC,
DC ⊥ BC,
DE ⊥ AC
To prove: ΔCED∼ΔABC
Now In ΔABC and ΔCED,
∠B = ∠E = 90o [given]
∠BAC = ∠ECD [alternate angles]
ΔCED∼ΔABC [AA similarity]
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 OA/ OC = OB/ OD
Given: OC is the point of intersection of AC and BD in the trapezium ABCD, with AB ∥ DC.
To prove: OA/ OC = OB/ OD
Now In ΔAOB and ΔCOD,
∠AOB = ∠COD [Vertically Opposite Angles]
∠OAB = ∠OCD [Alternate angles]
ΔAOB ∼ ΔCOD
OA/ OC = OB/ OD [Corresponding sides are proportional]
Question 10. If Δ ABC and Δ AMP are two right triangles, right angled at B and M, respectively such that ∠MAP = ∠BAC. Prove that
(i) ΔABC ∼ ΔAMP
(ii) CA/ PA = BC/ MP
Δ ABC and Δ AMP are the two right triangles.
Now In ΔABC and ΔAMP,
∠AMP = ∠B = 90o
∠MAP = ∠BAC [Vertically Opposite Angles]
ΔABC∼ΔAMP [AA similarity]
(ii) Since, ΔABC∼ΔAMP
CA/ PA = BC/ MP [Corresponding sides are proportional]
Question 11. A vertical stick 10 cm long casts a shadow 8 cm long. At the same time, a tower casts a shadow 30 m long. Determine the height of the tower.
Length of stick = 10cm
Length of the stick’s shadow = 8cm
Length of the tower’s shadow = 30m = 3000cm
To find: the height of the tower = PQ.
Now In ΔABC and ΔPQR,
∠ABC = ∠PQR = 90o [each 90o]
∠ACB = ∠PRQ [angle are made at the same time]
ΔABC ∼ ΔPQR [By AA similarity]
AB/BC = PQ/QR [by Corresponding sides are proportional]
10/8 = PQ/ 3000
PQ = (3000×10)/ 8
PQ = 30000/8
PQ = 3750/100
Hence the length of PQ = 37.5 m
Question 12. In fig., ∠A = ∠CED, prove that ΔCAB ∼ ΔCED. Also, find the value of x.
∠A = ∠CED
To prove: ΔCAB ∼ ΔCED
In ΔCAB and ΔCED
∠C = ∠C [Common]
∠A = ∠CED [Given]
ΔCAB ∼ ΔCED [By AA similarity]
CA/ CE = AB/ ED [by Corresponding sides are proportional]
15/10 = 9/x
x = (9 × 10)/ 15
Hence, value for x = 6 cm
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