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प्रश्न
In the given figure 1.66, seg PQ || seg DE, A(∆PQF) = 20 units, PF = 2 DP, then Find A(◻DPQE) by completing the following activity.
उत्तर
Given:
seg PQ || seg DE
A(∆PQF) = 20 units
PF = 2 DP
Let us assume DP = x
∴ PF = 2x
\[DF = DP + PF = x + 2x = 3x\]
In △FDE and △FPQ
∠FDE = ∠FPQ (Corresponding angles)
∠FED = ∠FQP (Corresponding angles)
By AA test of similarity
△FDE ∼ △FPQ
\[\therefore \frac{A\left( \bigtriangleup FDE \right)}{A\left( \bigtriangleup FPQ \right)} = \frac{{FD}^2}{{FP}^2} = \frac{\left( 3x \right)^2}{\left( 2x \right)^2} = \frac{9}{4}\]
\[A\left( \bigtriangleup FDE \right) = \frac{9}{4}A\left( \bigtriangleup FPQ \right) = \frac{9}{4} \times 20 = 45\]
\[\therefore A\left( \square DPQE \right) = A\left( \bigtriangleup FDE \right) - A\left( \bigtriangleup FPQ \right)\]
\[ = 45 - 20\]
\[ = 25\]
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