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प्रश्न
The area of the parallelogram ABCD is 90 cm2 (see figure). Find
- ar (ΔABEF)
- ar (ΔABD)
- ar (ΔBEF)

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उत्तर
Given, area of parallelogram, ABCD = 90 cm2.
i. We know that, parallelograms on the same base and between the same parallel are equal in areas.
Here, parallelograms ABCD and ABEF are on same base AB and between the same parallels AB and CF.
So, ar (ΔBEF) = ar (ABCD) = 90 cm2
ii. We know that, if a triangle and a parallelogram are on the same base and between the same parallels, then area of triangle is equal to half of the area of the parallelogram.
Here, ΔABD and parallelogram ABCD are on the same base AB and between the same parallels AB and CD.
So, ar (ΔABD) = `1/2` ar (ABCD)
= `1/2 xx 90` ...[∴ ar (ABCD) = 90 cm2]
= 45 cm2
iii. Here, ABEF and parallelogram ABEF are on the same base EF and between the same parallels AB and EF.
ar (ΔBEF) = `1/2` ar (ABEF)
= `1/2 xx 90` ...[∴ ar (ABEF) = 90 cm2, from part (i)]
= 45 cm2
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संबंधित प्रश्न
In the given figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:
(i) ar (DOC) = ar (AOB)
(ii) ar (DCB) = ar (ACB)
(iii) DA || CB or ABCD is a parallelogram.
[Hint: From D and B, draw perpendiculars to AC.]

Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).
In the given figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).

In the following figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that:-

(i) ΔMBC ≅ ΔABD
(ii) ar (BYXD) = 2 ar(MBC)
(iii) ar (BYXD) = ar(ABMN)
(iv) ΔFCB ≅ ΔACE
(v) ar(CYXE) = 2 ar(FCB)
(vi) ar (CYXE) = ar(ACFG)
(vii) ar (BCED) = ar(ABMN) + ar(ACFG)
Note : Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in Class X.
In the below fig. D and E are two points on BC such that BD = DE = EC. Show that ar
(ΔABD) = ar (ΔADE) = ar (ΔAEC).

PQRS is a parallelogram whose area is 180 cm2 and A is any point on the diagonal QS. The area of ∆ASR = 90 cm2.
In the following figure, ABCD and EFGD are two parallelograms and G is the mid-point of CD. Then ar (DPC) = `1/2` ar (EFGD).

The area of the parallelogram ABCD is 90 cm2 (see figure). Find ar (ΔABD)
The area of the parallelogram ABCD is 90 cm2 (see figure). Find ar (ΔBEF)
In the following figure, X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. Prove that ar (ABP) = ar (ACQ).

