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Question
The ratio between the altitudes of two similar triangles is 3 : 5; write the ratio between their :
- corresponding medians.
- perimeters.
- areas.
Solution
In ΔABC ∼ ΔPQR and AL : PM = 3 : 5
Now, we have to find the ratio between their.
- corresponding medians
- perimeters and
- areas.
AD and PE are the medians of ΔABC and ΔPQR respectively.
∵ ΔABC ∼ ΔPQR
∴ ∠B = ∠Q and `(AB)/(PQ) = (BC)/(QR)`
Now in ΔABL and PQM,
∠B = ∠Q ...(Proved)
∠L = ∠M ...(Each 90°)
i. ∴ ΔABL ∼ ΔPQM
∴ `(AB)/(PQ) = (AL)/(PM) = 3/5` (Given) ...(i)
∵ ΔABC ∼ ΔPQR
∴ `(AB)/(DE) = (BC)/(QR) = (2BD)/(2QE) = (BD)/(QE)`
And ∠B = ∠Q
∴ ΔABD ∼ ΔPQE
∴ `(AB)/(PQ) = (AD)/(PE) = 3/5` ...[From (i)]
∴ `(AD)/(PE) = 3 : 5`
ii. ∵ ΔABC ∼ ΔPQR
∴ `(AB)/(PQ) = (BC)/(QR) = (CA)/(RP)`
= `(AB + BC + CA)/(PQ + QR + RP)`
= `3/5` ...[From (i)]
Hence ratio between their perimeters = 3 : 5
iii. ∵ ΔABC ∼ ΔPQR
∴ `(Area ΔABC)/(Area ΔPQR) = (AB^2)/(PQ^2) = (3)^2/(5)^2 = 9/25`
∴ The ratio between their areas = 9 : 25
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