Solve the following mathematical problems:

Twice the width of a rectangle is 10 meters more the its length. If the area of the region enclosed by the rectangle is 600 square meters , then find its perimeter.

Created: 4 months ago | Updated: 6 days ago

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Let, the length of the rectangle is x meter

And the width of the rectangle is y meter

According to question,

$2 y = x + 10 . . . . . . . . . . . . (i) A n d x y = 600 . . . . . . . (i i) F r o m e q u a t i o n (i i) x y = 600 x = \frac{600}{y} . . . . . . . . . . . (i i i) P u t t i n g t h e v a l u e o f x i n e q u a t i o n (i) 2 y = x + 10 \Rightarrow 2 y = \frac{600}{y} + 10 \Rightarrow 2 y^{2} - 10 y - 600 = 0 \Rightarrow y^{2} - 5 y - 300 = 0 [D i v i d i n g b o t h s i d e s b y 2] \Rightarrow y^{2} - 20 y + 15 y - 300 = 0 \Rightarrow y (y - 20) + 15 (y - 20) = 0 \Rightarrow (y - 20) (y + 15) = 0 ∴ (y - 20) (y + 15) = 0$

Since width cannot be (-)ve. So we find y=20

That means the width of the rectangle is 20 meters

The length of the rectangle is $x = \frac{600}{y}$

$∴ x = \frac{600}{20} = 30$

The length of the rectangle is 30 meters.

The perimeter of the rectangle = $2 \times (30 + 20) = 2 \times 50 = 100$ meters.

ans. The perimeter is 100m.

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Solve the following mathematical problems:

Twice the width of a rectangle is 10 meters more the its length. If the area of the region enclosed by the rectangle is 600 square meters , then find its perimeter.

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