In the given figure, the area of ||gm ABCD is 90 cm². State giving reasons:
(i) ar (||gm ABEF)
(ii) ar (△ABD)
(iii) ar (△BEF)

Question

In the given figure, the area of ||gm ABCD is 90 cm2. State giving reasons: (i) ar (||gm ABEF) (ii) ar (△ABD) (iii) ar (△BEF)

KnowledgeBoat · Accepted Answer

(i) ar (||gm ABEF) ||gm ABCD and ||gm ABEF lie on the same base AB and between the same parallels AB and FC. ar(||gm ABEF) = ar(||gm ABCD) = 90 cm2. Hence, ar(||gm ABEF) = 90 cm2. (ii) Triangle ABD and parallelogram ABCD are on the same base AB and between the same parallels AB and CD, then area of triangle is equal to half of the area of the parallelogram. ar (△ABD) = ar(||gm ABCD) = = 45 cm2. Hence, ar (△ABD) = 45 cm2. (iii) Triangle BEF and a parallelogram ABEF are on the same base EF and between the same parallels AB and EF, then area of triangle is equal to half of the area of the parallelogram. ar (△BEF) = ar(||gm ABEF) = = 45 cm2. Hence, ar (△BEF) = 45 cm2.

Mathematics

In the given figure, the area of ||gm ABCD is 90 cm². State giving reasons:
(i) ar (||gm ABEF)
(ii) ar (△ABD)
(iii) ar (△BEF)

Theorems on Area

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Mathematics

In the given figure, the area of ||gm ABCD is 90 cm2. State giving reasons:(i) ar (||gm ABEF)(ii) ar (△ABD)(iii) ar (△BEF)

Theorems on Area

Related Questions

Find area of trapezium, whose parallel sides measure 10 cm and 8 cm respectively and the distance between these sides is 6 cm.

Show that the line segment joining the mid-points of a pair of opposite sides of a parallelogram, divides it into two equal parallelogram.

In the given figure, the area of △ABC is 64 cm2. State giving reasons:(i) ar(||gm ABCD)(ii) ar (rect. ABEF)

In the given figure, ABCD is a quadrilateral. A line through D, parallel to AC, meets BC produced in P. Prove that: ar (△ABP) = ar (quad.ABCD)

In the given figure, the area of ||gm ABCD is 90 cm². State giving reasons:
(i) ar (||gm ABEF)
(ii) ar (△ABD)
(iii) ar (△BEF)

In the given figure, the area of △ABC is 64 cm². State giving reasons:
(i) ar(||gm ABCD)
(ii) ar (rect. ABEF)