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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))

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प्रश्न

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))`

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उत्तर


Draw AE ⊥ BC, B-E-C.

BC = BD + DC   ...[B-D-C]

∴ 20 = 7 + DC

∴ DC = 20 – 7

∴ DC = 13

(i) ΔABD and ΔADC have same height AE.

`(A(ΔABD))/(A(ΔADC)) = (BD)/(DC)`   ...[Triangles having equal height]

∴ `(A(ΔABD))/(A(ΔADC)) = 7/13`

(ii) ΔABD and ΔABC have same height AE.

`(A(ΔABD))/(A(ΔABC)) = (BD)/(BC)`   ...[Triangles having equal height]

∴ `(A(ΔABD))/(A(ΔABC)) = 7/20`

(iii) ΔADC and ΔABC have same height AE.

`(A(ΔADC))/(A(ΔABC)) = (DC)/(BC)`   ...[Triangles having equal height]

∴ `(A(ΔADC))/(A(ΔABC)) = 13/20`

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अध्याय 1: Similarity - Q.3 (B)

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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, 

`(A(ΔPQB))/(A(ΔPBC)) = (1/2 xx square xx square)/(1/2 xx square xx square)`

= `square/square`


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