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Mathematics

Directions (Q. 45 to 48): Using the given diagram answer the following questions.

In ΔPQR, AB ∥ QR, QP ∥ CB and AR intersects CB at O. Similarity of Triangles, RSA Mathematics Solutions ICSE Class 10.

In ΔPQR, AB ∥ QR, QP ∥ CB and AR intersects CB at O.

45. The triangle similar to ΔARQ is:

  1. ΔORC

  2. ΔARP

  3. ΔOBR

  4. ΔQRP

46. ΔPQR ∼ ΔBCR by axiom:

  1. SAS

  2. AAA

  3. SSS

  4. AAS

47. If QC = 6 cm, CR = 4 cm, BR = 3 cm, then the length of RP is:

  1. 4.5 cm

  2. 5 cm

  3. 7.5 cm

  4. 8 cm

48. The ratio PQ : BC is:

  1. 2 : 3

  2. 3 : 2

  3. 2 : 5

  4. 5 : 2

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Answer

45. Given,

In ΔARQ and ΔORC,

∠BRC = ∠ARQ [Common angle]

∠AQR = ∠BCR [Corresponding angles, AB ∥ QR, BC is the transversal]

ΔARQ ∼ ΔORC [By AA Similarity]

Hence, option 1 is the correct option.

46. Given,

In ΔPQR and ΔBCR,

∠PRQ = ∠BRC [Common angle]

∠PQR = ∠BCR [Corresponding angles, CB ∥ QP, BC is the transversal]

ΔPQR ∼ ΔBCR [By AA or AAA similarity]

Hence, option 2 is the correct option.

47. Given,

RQ = RC + CQ = 4 + 6 = 10 cm.

RB = 3 cm

RC = 4 cm

From 46 que we have ΔPQR ∼ ΔBCR,

Since, corresponding sides of similar triangle are proportional to each other.

⇒RPRB=RQRC⇒RP3=104⇒RP=52×3⇒RP=7.5 cm.\Rightarrow \dfrac{RP}{RB} = \dfrac{RQ}{RC} \\[1em] \Rightarrow \dfrac{RP}{3} = \dfrac{10}{4} \\[1em] \Rightarrow RP = \dfrac{5}{2} \times 3 \\[1em] \Rightarrow RP = 7.5 \text{ cm.}

Hence, option 3 is the correct option.

48. Given,

Using ΔPQR ∼ ΔBCR,

⇒PQBC=RQRC=102=52=5:2\Rightarrow \dfrac{PQ}{BC} = \dfrac{RQ}{RC} \\[1em] = \dfrac{10}{2} \\[1em] = \dfrac{5}{2} = 5:2

Hence, option 4 is the correct option.

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