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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.
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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 |
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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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