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Question
ΔPQR ~ ΔABC. In ΔPQR, PQ = 3.6cm, 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 |
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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