Question

(iii) In the fig, seg PQ || seg DE, A (A POF) = 20 units
PF = 2 DP, then find A (DPQE).​

Answer 1

Answer:

Question:-

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In the fig, seg PQ || seg DE, A (A POF) = 20 units PF = 2 DP, then find A (DPQE).

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Solution:-

Given:-

$\sf \: A(\triangle PQF)= 20 \: Units , PF=20 \\ \sf \: \: Let \: Assume \: DP=x \: \therefore PF=2x$

$\qquad \bullet\sf \: \: \: In \: \: \angle FDE \sim \angle FPQ \: \: \qquad \qquad \dots \: Corresponding \: Angle$

$\qquad \bullet \: \sf \: \: \angle FED \sim \angle FQP \qquad \qquad \dots \: Corresponding Angle$

$\sf \rightarrow \: \therefore \triangle FDE \sim \triangle FPQ$

By AA test of Similarity

$\\ \qquad \: \:\sf \longrightarrow \frac{A\big(\triangle FDE \big)}{A\big(\triangle FPQ\big)}=\frac{DF^2}{PF^2}= \frac{(3x)^2}{(2x)^2}= \frac{9}{4}$

Figure is given in Attachment,

$\sf \: \longrightarrow A\big(\triangle FDE\big)=\frac{9}{4} \: \: A\big(\triangle FPQ\big)=\frac{9}{4} ×20 =45$

$\sf \: \longrightarrow A\big( \boxed{} DPQE =A\big(\triangle FDC\big)= \: A\big(\triangle FPQ\big)\:$

So,

$\longrightarrow \qquad45-20$

$\longrightarrow \qquad \underline{ \boxed{ \purple{ \sf \: \: 25 \: Units \: }}}$

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Answer 2

Answer:

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