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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.
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Solution 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.
Solution 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.
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In fig., PM = 10 cm, A(ΔPQS) = 100 sq. cm, A(ΔQRS) = 110 sq. cm, then NR?
ΔPQS and ΔQRS having seg QS common base.
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In the figure, PQ ⊥ BC, AD ⊥ BC. To find the ratio of A(ΔPQB) and A(ΔPBC), complete the following activity.
Given: PQ ⊥ BC, AD ⊥ BC
Now, A(ΔPQB) = `1/2 xx square xx square`
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Therefore,
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= `square/square`
