Question

The ratio between the altitudes of two similar triangles is 3 : 5; write the ratio between their :

1. corresponding medians.
2. perimeters.
3. areas.

Answer 1

In ΔABC ∼ ΔPQR and AL : PM = 3 : 5

Now, we have to find the ratio between their.

1. corresponding medians
2. perimeters and
3.  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