In the given figure,ABCD is a trapezium in which AB||DC such that AB=a cm and DC= b cm.If E and F are the midpoints of AD an BC respectively then ar(ABFE):ar(EFCD)=?

(a) a:b

(b) (a+3b):(3a+b)

(c) (3a+b):(a+3b)

(d) (2a+b):(3a+b)

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Solution

(c) (3 a + b) : ( a +3b)
Clearly, EF = 12 (a + b)
[Mid point theorem]
Let d be the distance between AB and EF .
Then d is the distance between DC and EF .

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Q. In the given figure ABCD is a trapezium in which AB || DC such that AB = a cm and DC = b cm. If E and F are the midpoints of AD and BC respectively. Then, ar(ABFE) : ar(EFCD) = ?
(a) a : b
(b) (a + 3b) : (3a + b)
(c) (3a + b) : (a + 3b)
(d) (2a + b) : (3a + b)

Q. ABCD is a trapezium with parallel sides AB =a and DC = b. If E and F are mid-points of non-parallel sides AD and BC respectively, then the ratio of areas of quadrilaterals ABFE and EFCD is

(a) a : b

(b) (a + 3b): (3a + b)

(c) (3a + b) : (a + 3b)

(d) (2a + b) : (3a + b)

ABCD is a tapezium with parallel sides AB = a and DC = b. If E and F are mid points of non parallel sides AD and BC respectively, then the ratio of areas of quadrialaterals ABFE and EFCD is

In a trapezium ABCD,AB||DC,AB=a cm,and DC=b cm.If M and N are the midpoints of the nonparallel lines,AD and BC respectively then find the ratio of ar(DCNM) and ar(MNBA).

ABCD is a trapezium in which AB||DC,AB=16 cm and DC=24 cm.If E and Fa re respectively the midpoints of Ad andf BC,prove that $a r (A B F E) = \frac{9}{11} a r (E F C D)$ .

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In the given figure,ABCD is a trapezium in which AB||DC such that AB=a cm and DC= b cm.If E and F are the midpoints of AD an BC respectively then ar(ABFE):ar(EFCD)=? (a) a:b (b) (a+3b):(3a+b) (c) (3a+b):(a+3b) (d) (2a+b):(3a+b)

(c) (3 a + b) : ( a +3b) Clearly, EF = 12 (a + b) [Mid point theorem] Let d be the distance between AB and EF . Then d is the distance between DC and EF .

In the given figure,ABCD is a trapezium in which AB||DC such that AB=a cm and DC= b cm.If E and F are the midpoints of AD an BC respectively then ar(ABFE):ar(EFCD)=?

(a) a:b

(b) (a+3b):(3a+b)

(c) (3a+b):(a+3b)

(d) (2a+b):(3a+b)

(c) (3 a + b) : ( a +3b)
Clearly, EF = 12 (a + b)
[Mid point theorem]
Let d be the distance between AB and EF .
Then d is the distance between DC and EF .