Hushar Mulga
@Rohit
Spread the love

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.

In figure 1.66, seg PQ || seg DE, A(D PQF) = 20 units, PF = 2 DP, then find A( c DPQE) by completing the following activity.
Solution

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

Click Here for All Textbook Soutions of Chapter 1: Similarity