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Maharashtra State BoardSSC (English Medium) 10th Standard

ΔPQR ~ ΔABC. In ΔPQR, PQ = 3.6 cm, QR = 4 cm, PR = 4.2 cm. Ratio of the corresponding sides of triangle is 3 : 4, then construct ΔPQR and ΔABC.

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Question

ΔPQR ~ ΔABC. In ΔPQR, PQ = 3.6 cm, QR = 4 cm, PR = 4.2 cm. Ratio of the corresponding sides of triangle is 3 : 4, then construct ΔPQR and ΔABC.

Geometric Constructions
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Solution


Analysis:

∆PQR ∼ ∆ABC

∴ `(PQ)/(AB) = (QR)/(BC) = (PR)/(AC)`   ...[Corresponding sides of similar triangles]

∴ `3.6/(AB) = 4/(BC) = 4.2/(AC) = 3/4`   ...[Given]

∴ `3.6/(AB) = 3/4`

∴ `AB = (3.6 xx 4)/3`

∴ AB = 1.2 × 4

∴ AB = 4.8 cm 

`4/(BC) = 3/4`

∴ `BC = (4 xx 4)/3`

∴ `BC = 16/3`

∴ BC = 5.3 cm (approx)

`4.2/(AC) = 3/4`

∴ `AC = (4.2 xx 4)/3`

∴ AC = 1.4 × 4

∴ AC = 5.6 cm

Steps of construction:

  ∆PQR ∆ABC
i. Draw seg QR of 4 cm Draw seg BC of 5.3 cm
ii. Taking 3.6 cm and 4.2 cm distances on compass draw two arcs from Q and R respectively. Taking 4.8 cm and 5.6 cm distance on compass draw two arcs from point B and C respectively.
iii. Name the point of intersection as P. Name the point of intersection as A.
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Chapter 4: Geometric Constructions - Q.3 (B)

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