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RD Chapter 4 Triangles Ex 4.5 Solutions

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)

Answer - 1 : -

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)

Answer - 2 : -

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)

Answer - 3 : -

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 hypotenuse is x. Prove that ab = cx.

Answer - 4 : -

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.

Answer - 5 : -


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.

Answer - 6 : -

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 = 1/4 AC. If AB = 6 cm, find AD.

Answer - 7 : -

In the figure, in ∆ABC, DE || BC
AE = 1/4 AC, AB = 6 cm

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

Answer - 8 : -

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 OA/OC = OB/OD

Answer - 9 : -

Given : ABCD is a trapezium in which AB || DC and diagonals AC and BD intersect each other at O
To Prove : OA/OC = OB/OD

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) CA/PA = BC/MP.

Answer - 10 : -

Given : In ∆ABC and ∆AMP,
∠B = ∠M = 90°
∠MAP = ∠BAC

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