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प्रश्न
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 ______.

पर्याय
a : b
(3a + b) : (a + 3b)
(a + 3b) : (3a + b)
(2a + b) : (3a + b)
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उत्तर
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)
संबंधित प्रश्न
In the given figure, ABCD is parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.

P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar (BQC).
Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.
In the following figure, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).

ABCD is a parallelogram, G is the point on AB such that AG = 2 GB, E is a point of DC
such that CE = 2DE and F is the point of BC such that BF = 2FC. Prove that:
(1) ar ( ADEG) = ar (GBCD)
(2) ar (ΔEGB) = `1/6` ar (ABCD)
(3) ar (ΔEFC) = `1/2` ar (ΔEBF)
(4) ar (ΔEBG) = ar (ΔEFC)
(5)ΔFind what portion of the area of parallelogram is the area of EFG.
In the below fig. ABCD and AEFD are two parallelograms. Prove that
(1) PE = FQ
(2) ar (Δ APE) : ar (ΔPFA) = ar Δ(QFD) : ar (Δ PFD)
(3) ar (ΔPEA) = ar (ΔQFD)
ABCD is a square. E and F are respectively the mid-points of BC and CD. If R is the mid-point of EF (Figure), prove that ar (AER) = ar (AFR)

ABCD is a parallelogram in which BC is produced to E such that CE = BC (Figure). AE intersects CD at F. If ar (DFB) = 3 cm2, find the area of the parallelogram ABCD.
ABCD is a trapezium in which AB || DC, DC = 30 cm and AB = 50 cm. If X and Y are, respectively the mid-points of AD and BC, prove that ar (DCYX) = `7/9` ar (XYBA)
In the following figure, ABCD and AEFD are two parallelograms. Prove that ar (PEA) = ar (QFD). [Hint: Join PD].

