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Question
ABCD is a trapezium with parallel sides AB = a cm and DC = b cm (Figure). E and F are the mid-points of the non-parallel sides. The ratio of ar (ABFE) and ar (EFCD) is ______.
Options
a : b
(3a + b) : (a + 3b)
(a + 3b) : (3a + b)
(2a + b) : (3a + b)
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Solution
ABCD is a trapezium with parallel sides AB = a cm and DC = b cm (Figure). E and F are the mid-points of the non-parallel sides. The ratio of ar (ABFE) and ar (EFCD) is (3a + b) : (a + 3b).
Explanation:
Given, AB = a cm, DC = b cm and AB || DC.
Also, E and F are mid-points of AD and BC, respectively.
So, distance between CD, EF and AB, EF will be same say h.
Join BD which intersect EF at M.
Now, in ΔABD, E is the mid-point of AD and EM || AB
So, M is the mid-point of BD
And EM = `1/2`AB [By mid-point theorem] ...(i)
Similarly in ΔCBD, MF = `1/2`CD ...(ii)
On adding equations (i) and (ii), we get
EM + MF = `1/2` AB + `1/2` CD
⇒ EF = `1/2`(AB + CD) = `1/2`(a + b)
Now, area of trapezium ABFE
= `1/2`(sum of parallel sides) × (distance between parallel sides)
= `1/2(a + 1/2(a + b)) xx h`
= `1/4(3a + b)h`
Now, area of trapezium EFCD
= `1/2[b + 1/2(a + b)] xx h`
= `1/4(3b + a)h`
∴ Required ratio = `"Area of ABFE"/"Area of EFCD"`
= `(1/4(3a + b)h)/(1/4(3b + a)h)`
= `((3a + b))/((a + 3b))` or (3a + b) : (a + 3b)
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