The areas of two similar triangles are 25 cm2 and 36 cm2 respectively.

1.	The areas of two similar triangles are 25 cm2 and 36 cm2 respectively. If the altitude of the first triangle is 2.4 cm, find the corresponding altitude of the other.
Answer» We have, ΔABC ~ Δ PQR Area (ΔABC) = 25 cm² Area (PQR) = 36 cm² And AD = 2.4 cm And AD and PS are the altitudes To find: PS Proof: Since, ΔABC ~ ΔPQR Then, by area of similar triangle theorem Area of ΔABC/Area of ΔPQR = AB² /PQ² 25/36 = AB²/PQ² 5/6 = AB/PQ………………..(i) In ΔABD and Δ PQS ∠B = ∠Q (ΔABC ~ ΔPQR) ∠ADB = ∠PSQ (Each 90°) Then, ΔABD ~ Δ PQS (By AA similarity) So, AB/PS = AD/PS…………(ii) (Corresponding parts of similar Δ are proportional ) Compare (i) and (ii) AD/PS = 5/6 2.4/PS = 5/6 PS = 2.4 x 6/5 PS = 2.88cm

The areas of two similar triangles are 25 cm2 and 36 cm2 respectively. If the altitude of the first triangle is 2.4 cm, find the corresponding altitude of the other.

Answer»

We have,

ΔABC ~ Δ PQR

Area (ΔABC) = 25 cm²

Area (PQR) = 36 cm²

And AD = 2.4 cm

And AD and PS are the altitudes

To find: PS

Proof: Since, ΔABC ~ ΔPQR

Then, by area of similar triangle theorem

Area of ΔABC/Area of ΔPQR = AB² /PQ²

25/36 = AB²/PQ²

5/6 = AB/PQ………………..(i)

In ΔABD and Δ PQS

∠B = ∠Q (ΔABC ~ ΔPQR)

∠ADB = ∠PSQ (Each 90°)

Then, ΔABD ~ Δ PQS (By AA similarity)

So, AB/PS = AD/PS…………(ii) (Corresponding parts of similar Δ are proportional )

Compare (i) and (ii)

AD/PS = 5/6

2.4/PS = 5/6

PS = 2.4 x 6/5

PS = 2.88cm