Question

Base of a triangle is 9 and height is 5. Base of another triangle is 10 and height is 6. Find the ratio of areas of these triangles.

Answer 1

Let ABC and PQR be two right triangles with AD ⊥ BC and PS ⊥ QR.

BC = 9, AD = 5, PS = 6 and QR = 10.

The ratio of areas of two triangles is equal to the ratio of the products of their bases and corresponding heights.

`("A"(Δ"ABC"))/("A"(Δ"PQR")) = ("AD" × "BC")/("PS" × "QR")`

∴ `("A"(Δ"ABC"))/("A"(Δ"PQR")) = (5 × 9)/(6 × 10)`

∴ `("A"(Δ"ABC"))/("A"(Δ"PQR")) = 3/4`

Hence, Ratio of Area of △ABC : Area of △PQR = 3 : 4.

Answer 2

Let the base, height, and area of the first triangle be b₁, h₁, and A₁ respectively. Let the base, height and area of the second triangle be b₂, h₂, and A₂ respectively.

b₁ = 9, h₁ = 5, b₂ = 10 and h₂ = 6.

The ratio of areas of two triangles is equal to the ratio of the products of their bases and corresponding heights.

`("A"_1)/("A"_2) = ("b"_1 × "h"_1)/("b"_2 × "h"_2)`

∴ `("A"_1)/("A"_2) = (9 × 5)/(10 × 6)`

∴ `("A"_1)/("A"_2) = 3/4`

The ratio of the areas of the triangles is 3 : 4.