In the given figure 1.66, seg PQ || seg DE, A(∆PQF) = 20 units, PF = 2 DP,
Practice Set 1.4 | Q 7 | Page 25
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.
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
\[\therefore \frac{A\left( \bigtriangleup FDE \right)}{A\left( \bigtriangleup FPQ \right)} = \frac{{FD}^2}{{FP}^2} = \frac{\left( 3x \right)^2}{\left( 2x \right)^2} = \frac{9}{4}\]
\[A\left( \bigtriangleup FDE \right) = \frac{9}{4}A\left( \bigtriangleup FPQ \right) = \frac{9}{4} \times 20 = 45\]
\[\therefore A\left( \square DPQE \right) = A\left( \bigtriangleup FDE \right) - A\left( \bigtriangleup FPQ \right)\]
\[ = 45 - 20\]
\[ = 25\]
Solution:-
Given: seg PQ || seg DE, A(∆PQF) = 20 units, PF = 2DP
Let DP = x, then 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, A(∆FDE)/A(∆FPQ) = (FD/FP)^2 = (3x/2x)^2 = 9/4 A(∆FDE) = (9/4)A(∆FPQ) = (9/4) × 20 = 45
Therefore, A(◻DPQE) = A(∆FDE) – A(∆FPQ) = 45 – 20 = 25 square units.
Chapter 1. Similarity- Practice Set 1.4 – Page 25
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