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

Advertisements
Solution
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
APPEARS IN
RELATED QUESTIONS
In the given figure, ABC and ABD are two triangles on the same base AB. If line-segment CD is bisected by AB at O, show that ar (ABC) = ar (ABD).

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.]

In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that
(i) ar (ACB) = ar (ACF)
(ii) ar (AEDF) = ar (ABCDE)

A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar (APB) × ar (CPD) = ar (APD) × ar (BPC).
[Hint : From A and C, draw perpendiculars to BD.]
P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that
(i) ar(PRQ) = 1/2 ar(ARC)
(ii) ar(RQC) = 3/8 ar(ABC)
(iii) ar(PBQ) = ar(ARC)
In a ΔABC, if L and M are points on AB and AC respectively such that LM || BC. Prove
that:
(1) ar (ΔLCM ) = ar (ΔLBM )
(2) ar (ΔLBC) = ar (ΔMBC)
(3) ar (ΔABM) ar (ΔACL)
(4) ar (ΔLOB) ar (ΔMOC)
X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. Through X, a line is drawn parallel to LM to meet MN at Z (See figure). Prove that ar (LZY) = ar (MZYX)

The medians BE and CF of a triangle ABC intersect at G. Prove that the area of ∆GBC = area of the quadrilateral AFGE.
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).

