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  • ABCD is a trapezium with parallel sides AB = a cm and DC = b cm (Fig.). E and F are the mid-points of the non-parallel sides. The ratio of ar (ABFE) and ar (EFCD) is (A) a : b (B) (3a + b) : (a + 3b) (C) (a + 3b) : (3a + b) (D) (2a + b) : (3a + b)

ABCD is a trapezium with parallel sides AB = a cm and DC = b cm (Fig.). E and F are the mid-points of the non-parallel sides. The ratio of ar (ABFE) and ar (EFCD) is


 

(A) a : b
(B) (3a + b) : (a + 3b)
(C) (a + 3b) : (3a + b)
(D) (2a + b) : (3a + b)

Answers (1)

Answer: [B]

Solution.

Given: ABCD is a trapezium in which AB \parallel DC,
E and F are the mid-points of AD and BC, respectively


(Mid-segment Theorem on Trapezium: the line segment joining the midpoints of the nonparallel sides of a trapezium is half the sum of the lengths of the parallel sides and is also parallel to them.)
So,  EF=\frac{1}{2}(a+b)
Now, ABFE and EFCD are also trapeziums.
ar(ABFE)=\frac{1}{2}(EF+AB)h
={\frac{1}{2}}\left [ \frac{1}{2}(a+b)+a \right ]\times h
=\frac{h}{4}(3a+b)
ar(EFCD)=\frac{1}{2}(CD+EF)h
=\frac{1}{2}\left [ b+\frac{1}{2}(a+b) \right ]\times h
=\frac{h}{4}(a+3b)
Required ratio
\therefore \frac{ar(ABEF)}{ar(EFCD)}=\frac{\frac{h}{4}(3a+b)}{\frac{h}{4}(a+3b)}=\frac{(3a+b)}{(a+3b)}
So, the required ratio is  (3a+b):(a+3b)
Hence, (B) is the correct answer.

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