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Question
The ratio between the areas of two similar triangles is 16 : 25. State the ratio between their :
- perimeters.
- corresponding altitudes.
- corresponding medians.
Solution
ΔABC ∼ ΔDEF, AL ⊥ BC and DM ⊥ EF and AP and DQ are the medians and also area ΔABC : area ΔDEF = 16 : 25
∵ ΔABC ~ ΔDEF ...(Given)
∴ `(Area ΔABC)/(Area ΔDEF) = (AB^2)/(DE^2)`
`\implies (AB^2)/(DE^2) = 16/25 = (4)^2/(5)^2`
∴ `(AB)/(DE) = 4/5` or AB : DE = 4 : 5 ...(i)
∵ ΔABC ∼ ΔDEF
∴ ∠B = ∠E and `(AB)/(DE) = (BC)/(EF)` ...(i)
i. ∵ ΔABC ∼ ΔDEF
∴ `(AB)/(DE) = (BC)/(EF) = (CA)/(FD)`
= `(AB + BC + CA)/(DE + EF + FD)`
= `4/5` ...[From (i)]
∴ The ratio between two perimeters = 4 : 5
ii. Now, in ΔABC and ΔDEM,
∴ ∠B = ∠E, ∠L = ∠M ...(Each 90°)
∴ ΔABL ∼ ΔDEM ...(AA criterion of similarity)
∴ `(AB)/(DE) = (AL)/(DM) = 4/5` ...[From (i)]
∵ AL : DM = 4 : 5
iii. ΔABC ∼ ΔDEF, ∠B = ∠E and
`(AB)/(DE) = (BC)/(EF) = (2BP)/(2EQ) = (BP)/(EQ)`
∴ ΔABD ∼ ΔDEQ
∴ `(AB)/(DE) = (AP)/(DQ) = 4/5` ...[From (i)]
∴ AB : DE = 4 : 5
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