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प्रश्न
In the given, seg BE ⊥ seg AB and seg BA ⊥ seg AD.
if BE = 6 and AD = 9 find `(A(Δ ABE))/(A(Δ BAD))`.

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उत्तर
`(A(Δ ABE))/(A(Δ BAD)) =(BE)/(AD)`
....[Ratio of areas of two triangles having qual base is equal to the ration of their corresponding heights.]
∴ `(A(Δ ABE))/(A(Δ BAD)) =6/9`
∴ `(A(Δ ABE))/(A(Δ BAD)) =2/3`
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संबंधित प्रश्न
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In fig. BD = 8, BC = 12, B-D-C, then `"A(ΔABC)"/"A(ΔABD)"` = ?
In ΔABC, B − D − C and BD = 7, BC = 20, then find the following ratio.

(i) `"A(ΔABD)"/"A(ΔADC)"`
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Prove that, The areas of two triangles with the same height are in proportion to their corresponding bases. To prove this theorem start as follows:
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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`
A(ΔPBC) = `1/2 xx square xx square`
Therefore,
`(A(ΔPQB))/(A(ΔPBC)) = (1/2 xx square xx square)/(1/2 xx square xx square)`
= `square/square`
