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प्रश्न
In the given figure, BC ⊥ AB, AD ⊥ AB, BC = 4, AD = 8, then find `("A"(∆"ABC"))/("A"(∆"ADB"))`
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उत्तर
In ∆ABC and ∆ADB,
BC ⊥ AB, AD ⊥ AB, BC = 4, AD = 8 ...(Given)
∆ABC and ∆ADB have same base AB. ...(Given)
∴ Areas of triangles with equal bases are proportional to their corresponding heights.
`("A"(∆"ABC"))/("A"(∆"ADB")) = "BC"/"AD"`
∴ `("A"(∆"ABC"))/("A"(∆"ADB")) = 4/8`
∴ `("A"(∆"ABC"))/("A"(∆"ADB")) = 1/2`.
संबंधित प्रश्न
In the given figure, AD is the bisector of the exterior ∠A of ∆ABC. Seg AD intersects the side BC produced in D. Prove that:
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From adjoining figure, ∠ABC = 90°, ∠DCB = 90°, AB = 6, DC = 8, then `("A"(Δ"ABC"))/("A"(Δ"BCD"))` = ?
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(i) `"A(ΔABD)"/"A(ΔADC)"`
(ii) `"A(ΔABD)"/"A(ΔABC)"`
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If ΔABC ∼ ΔDEF, length of side AB is 9 cm and length of side DE is 12 cm, then find the ratio of their corresponding areas.
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,
`(A(ΔPQB))/(A(ΔPBC)) = (1/2 xx square xx square)/(1/2 xx square xx square)`
= `square/square`
