The length of a rectangle is 5 less than twic | vedclass

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Question

(A) Let the breadth of the rectangle be $x$ and the length be $y$.According to the problem,the length is $5$ less than twice the breadth: $y = 2x - 5$

Vedclass · Accepted Answer

$700$

Vedclass · Answer

(A) Let the breadth of the rectangle be $x$ and the length be $y$.According to the problem,the length is $5$ less than twice the breadth: $y = 2x - 5$,which can be written as $2x - y - 5 = 0$.The perimeter of a rectangle is $2(l + b) = 110$,so $l + b = 55$,which means $x + y = 55$,or $x + y - 55 = 0$.We have the system of equations:$2x - y - 5 = 0$$x + y - 55 = 0$Using the cross-multiplication method:$\frac{x}{(-1)(-55) - (1)(-5)} = \frac{-y}{(2)(-55) - (1)(-5)} = \frac{1}{(2)(1) - (1)(-1)}$$\frac{x}{55 + 5} = \frac{-y}{-110 + 5} = \frac{1}{2 + 1}$$\frac{x}{60} = \frac{-y}{-105} = \frac{1}{3}$$x = \frac{60}{3} = 20$ and $y = \frac{105}{3} = 35$.The area of the rectangle is $l 	imes b = 35 	imes 20 = 700$ square units.

The length of a rectangle is $5$ less than twice its breadth. If its perimeter is $110,$ find the area of the rectangle. Use the cross-multiplication method for the solution.

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