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प्रश्न
In ∆ABC, B – D – C and BD = 7, BC = 20, then find the following ratio.
`(A(triangleABD))/(A(triangleABC))`
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उत्तर
In ∆ABC, B – D – C and BD = 7, BC = 20
As ΔABD and ∆ABC have the same height.
∴ Areas of triangles with the same height are proportional to their corresponding bases.
`(A(triangle ABD))/(A(triangle ABC)) = (BD)/(BC)`
= `7/20`
संबंधित प्रश्न
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In the given figure, BC ⊥ AB, AD ⊥ AB, BC = 4, AD = 8, then find `("A"(∆"ABC"))/("A"(∆"ADB"))`
In adjoining figure, PQ ⊥ BC, AD ⊥ BC then find following ratios.
- `("A"(∆"PQB"))/("A"(∆"PBC"))`
- `("A"(∆"PBC"))/("A"(∆"ABC"))`
- `("A"(∆"ABC"))/("A"(∆"ADC"))`
- `("A"(∆"ADC"))/("A"(∆"PQC"))`
In ∆ABC, B - D - C and BD = 7, BC = 20 then find following ratio.
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The ratio of the areas of two triangles with the common base is 4 : 3. Height of the larger triangle is 2 cm, then find the corresponding height of the smaller triangle.
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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In ΔABC, B-D-C and BD = 7, BC = 20, then find the following ratio.
(i) `(A(ΔABD))/(A(ΔADC))`
(ii) `(A(ΔABD))/(A(ΔABC))`
(iii) `(A(ΔADC))/(A(ΔABC))`
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`
