Question

The breadth of a rectangle is 8 cm less than its length. If the length is increased by 4 cm and the breadth is increased by 9 cm, the area of the rectangle increases by 147 sq. cm. Find the length and breadth of the rectangle.


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Answer 1

Given : The breadth of a rectangle is 8 cm less than its length. If the length is increased by 4 cm and the breadth is increased by 9 cm, the area of the rectangle increases by 147 sq. cm.

To Find : Find the Length and Breadth of the Rectangle ?

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Solution : Let the length be x.

$~$

• Given that the breadth of a rectangle is 8 cm less than its length.

$~$

So,

$\qquad{\sf:\implies{Breadth~=~x~-~8}}$

$\qquad{\sf:\implies{Area~of~Rectangle~=~x(x~-~8)}}$

$\qquad{\sf:\implies{x^2-~8x}}$

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Also,

• If the length is increased by 4 cm and the breadth is increased by 9 cm, the area of the rectangle increases by 147 cm² .

$~$

$\qquad{\sf:\implies{Length~=~x~+~4}}$

$\qquad{\sf:\implies{Breadth~=~x~-~8~+~9~=~x~+~1}}$

$\qquad{\sf:\implies{Area~=~{x}^{2}~-~8x~+~147}}$

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$\underline{\frak{\pmb{According~ to ~the ~given~ question~:}}}$

$~$

$\qquad{\sf:\implies{(x~+~4)~(x~+~1)~=~{x}^{2}~-~8x~+~147}}$

$\qquad{\sf:\implies{{x}^{2}~+~x~+~4x~+~4~=~{x}^{2}~-~8x~+~147}}$

$\qquad{\sf:\implies{{x}^{2}~+~5x~+~4~=~{x}^{2}~-~8x~+~147}}$

$\qquad{\sf:\implies{{x}^{2}~-~{x}^{2}~+~5x+~8x~+~4~-~147~=~0}}$

$\qquad{\sf:\implies{13x~-~143~=~0}}$

$\qquad{\sf:\implies{13x~=~143}}$

$\qquad{\sf:\implies{x=\cancel\dfrac{143}{13}}}$

$\qquad{:\implies{\underline{\boxed{\pmb{\frak{\pink{x~=~11 }}}}}}}$★

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Hence,

• ${\sf{Length\:of\:Rectangle=x}}$

$\qquad{\sf:\implies{\underline{\pmb{11\:cm}}}}$

• ${\sf{Breadth\:of\:Rectangle=11-8}}$

$\qquad{\sf:\implies{\underline{\pmb{3\:cm}}}}$

Answer 2

Answer:

The length of a rectangle is 4 cm more than its breadth. If the length is increased by 4 cm and breadth is decreased by 2 cm, the area remains the same as that of the original rectangle. What is the length and breadth of the rectangle?

W = breadth, W+4 = length

Area = LxB = W^2 + 4W

New area is (W+4+4)(W-2) = W^2+6W-16.

Set the two equal and solve for W to get breadth = 8 and length = 12.

$oln:

Let the length of rectangle be l and breadth be b

In 1st condition

l = b+4

A= l*b = (b+4)*b = b^2 + 4b

In 2nd condition

l = b+4+4 =b+8

b = b — 2

A = l*b = (b+8) (b—2) = b^2 + 6b — 16

$ince, Area of rectangles in both condition are same

b^2 +4b = b^2 +6b — 16

b^2— b^2 + 16 = 6b— 4b

16 = 2b

Therefore , b = 8

l = (b+4) = (8+4) = 12

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Step-by-step explanation:

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