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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).
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)
In which of the following figures, you find two polygons on the same base and between the same parallels?
Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is ______.
PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ. If PS = 5 cm, then ar (PAS) = 30 cm2.
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)

If the mid-points of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram so formed will be half of the area of the given quadrilateral (Figure).
[Hint: Join BD and draw perpendicular from A on BD.]

