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Saurabh sharma

· started a discussion

· 1 Months ago

IN FIGURE TUPE QUESTION (Question: 114078.0 )

IN THESE Q PLEASE ADD FIGURE AT LEAST SOLUTION TIME

Saurabh sharma

· started a discussion

· 1 Months ago

IN FIGURE TUPE QUESTION (Question: 114078.0 )

IN THESE Q PLEASE ADD FIGURE AT LEAST SOLUTION TIME

Question:
In the figure, ABCD is a parallelogram with AD = a units, DC = 2a units and DE : EC = 1 : 2. CEFG is a rectangle with FE = 3AE. What is the ratio of the area of a parallelogram and the rectangle?

Options:
A) 1 : 1
B)  1 : 2
C) \(\cfrac{a^2}{5}:1\)
D) 2:1
Solution:

Ans:(b)


Area of parallelogram ABCD

= 2(Area of \(\Delta\)ADE)+(Area of rectangle AECH)

= \(\left (2\times \cfrac {1}{2}\times DE\times AE \right )+(AE \times EC)\)

=DE × AE + AE × EC

=AE × DC

Area of rectangle EFGC = FE × EC

∴ Required ratio = \(\cfrac{AE\times DC}{FE\times EC}\) 

$$=\cfrac{AE(EC + DE)}{3AE \times EC}$$

=AE(EC+DE)3AE×EC

\(=\cfrac{EC+\cfrac{1}{2}EC}{3EC}\)

= \(\cfrac{1}{2}\)

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