Question

∆MNT ~ ∆QRS. Length of altitude drawn from point T is 5 and length of altitude drawn from point S is 9. Find the ratio `("A"(Δ"MNT"))/("A"(Δ"QRS"))`.

Answer 1

∆MNT ~ ∆QRS   ...(Given)

∴ ∠M ≅ ∠Q      ...(Corresponding angles of similar triangles)(i)

In ∆MLT and ∆QPS,

∠M ≅ ∠Q             ...[From (i)]

∠MLT ≅ ∠QPS     ...(Each angle is of measure 90°)

∴ ∆MLT ~ ∆QPS     ...(AA test of similarity)

∴ `"MT"/"QS" = "TL"/"SP"  ...("Corresponding sides of similar triangles")`

∴ `"MT"/"QS" = 5/9`      

Now, ∆MNT ~ ∆QRS        ...(ii) (Given)

By Theorem of areas of similar triangles,

∴ `("A"(∆"MNT"))/("A"(∆"QRS")) = "MT"^2/"QS"^2`

∴ `("A"(∆"MNT"))/("A"(∆"QRS")) = ("MT"/"QS")^2`

∴ `("A"(∆"MNT"))/("A"(∆"QRS")) = (5/9)^2`     ...[From (ii)]

∴ `("A"(∆"MNT"))/("A"(∆"QRS")) = 25/81`