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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`
संबंधित प्रश्न
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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 ∆ABC, B – D – C and BD = 7, BC = 20 then Find following ratio.
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NR = `square` cm
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- Draw two triangles, give the names of all points, and show heights.
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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`
