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Question
In ∆ABC, B - D - C and BD = 7, BC = 20 then find following ratio.
`"A(∆ ABD)"/"A(∆ ADC)"`
Solution
Construction: Draw a perpendicular from vertex A to line BC.
BC = BD + DC ...[B - D - C]
DC = BC − BD
DC = 20 − 7
DC = 13
Ratio of areas of two triangles is equal to the ratio of the products of their bases and corresponding heights.
∴ `"A(∆ ABD)"/"A(∆ ADC)" = (1/2 xx "AX" xx "BD")/(1/2 xx "AX" xx "DC")`
∴ `"A(∆ ABD)"/"A(∆ ADC)" = "BD"/"DC"`
∴ `"A(∆ ABD)"/"A(∆ ADC)" = 7/13`
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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`
