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Question
Two triangles are similar. Smaller triangle’s sides are 4 cm, 5 cm, 6 cm. Perimeter of larger triangle is 90 cm then find the sides of larger triangle.
Solution
Given: ΔABC ~ ΔPQR
In ΔABC, AB = 4 cm, BC = 5 cm, AC = 6 cm
In ΔPQR, PQ + QR + PR = 90 cm
To find: PQ, QR and PR
ΔABC ~ ΔPQR .....[Given]
∴ `"AB"/"PQ" = "BC"/"QR" = "AC"/"PR"` .....[Corresponding sides of similar triangles]
Let `"AB"/"PQ" = "BC"/"QR" = "AC"/"PR"` = k
∴ `4/"PQ" = 5/"QR" = 6/"PR"` = k .....[Given]
∴ `4/"PQ"` = k, `5/"QR"` = k and `6/"PR"` = k
∴ PQ = `4/"k"`, QR = `5/"k"` and PR = `6/"k"` ......(i)
∴ PQ + QR + PR = `4/"k" + 5/"k" + 6/"k"`
∴ 90 =`15/"k"` .....[Given]
∴ k = `15/90`
= `1/6`
∴ PQ = `4/((1/6))` = 4 × 6 = 24 cm ...[From (i)]
QR = `5/((1/6))` = 5 × 6 = 30 cm ...[From (i)]
PR = `6/((1/6))` = 6 × 6 = 36 cm ...[From (i)]
∴ The sides of the larger triangle are 24 cm, 30 cm and 36 cm.
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