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Constructions

Question
CBSEENMA9002884
Wired Faculty App

Show that the area of a rhombus is half the product of the lengths of its diagonals.

Or

Prove that the area of a rhombus is equal to half the rectangle contained by its diagonals.

Solution

Let ABCD be a rhombus whose diagonals are AC and BD.
Then,
Area of rhombus ABCD
= Area of ΔABD + Area of ΔCBD
equals space fraction numerator left parenthesis BD right parenthesis left parenthesis AO right parenthesis over denominator 2 end fraction plus fraction numerator left parenthesis BD right parenthesis left parenthesis OC right parenthesis over denominator 2 end fraction
∵ Diagonals of a rhombus are perpendiculars to each other

equals fraction numerator left parenthesis BD right parenthesis over denominator 2 end fraction left parenthesis AO plus OC right parenthesis equals fraction numerator left parenthesis BD right parenthesis left parenthesis AC right parenthesis over denominator 2 end fraction equals space 1 half
Product of the lengths of its diagonals.

Some More Questions From Constructions Chapter

Prove that of all parallelograms of which the sides are given, the parallelogram which is a rectangle, has the greatest area.

In the figure, diagonals AC and BD of a trapezium ABCD with AB || CD intersect each other at O. Show that ar(Δ AOD) = ar(Δ BOC).

Parallelograms on the same base and between same parallels are equal in area. Prove this.

ABCD is a quadrilateral and BD is one of its diagonals as shown in figure. Show that ABCD is a parallelogram and find its area.

In the given figure, AB | | DC. Show that ar(BDE) = ar(ACED).

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