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Mathematics

The length and breadth of a rectangular piece of paper are 9 cm and 102310\dfrac{2}{3} cm respectively. Find:

(i) its area

(ii) its perimeter

Rational Numbers

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Answer

Length = 9 cm

Breadth = 102310\dfrac{2}{3} cm

Area = length x breadth

=9×1023=9×323=9×321×3=2883=96 cm2= 9 \times 10\dfrac{2}{3}\\[1em] = 9 \times \dfrac{32}{3}\\[1em] = \dfrac{9 \times 32}{1 \times 3}\\[1em] = \dfrac{288}{3}\\[1em] = 96 \text{ cm}^2

Area = 96 cm2

Perimeter = 2 x (length + breadth)

=2×(9+1023)=2×(91+323)= 2 \times \Big(9 + 10\dfrac{2}{3}\Big)\\[1em] = 2 \times \Big(\dfrac{9}{1} + \dfrac{32}{3}\Big)

LCM of 1 and 3 is 3.

=2×(9×31×3+32×13×1)=2×(273+323)=2×(27+323)=2×(593)=(59×23×1)=(1183)=39(13)= 2 \times \Big(\dfrac{9 \times 3}{1 \times 3} + \dfrac{32 \times 1}{3 \times 1}\Big)\\[1em] = 2 \times \Big(\dfrac{27}{3} + \dfrac{32}{3}\Big)\\[1em] = 2 \times \Big(\dfrac{27 + 32}{3}\Big)\\[1em] = 2 \times \Big(\dfrac{59}{3}\Big)\\[1em] = \Big(\dfrac{59 \times 2}{3 \times 1}\Big)\\[1em] = \Big(\dfrac{118}{3}\Big)\\[1em] = 39\Big(\dfrac{1}{3}\Big)

Perimeter = 391339\dfrac{1}{3} cm

Hence, area of the rectangular piece of paper is 96 cm2 and its perimeter is 391339\dfrac{1}{3} cm.

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