Question

5.a) Length of a rectangle is 25 + x metres and its breadth is 25 -x metres. What is the
perimeter of the rectangle?
b) Find the length and breadth of a rectangle having perimeter 100 metres and area
525 square metres.​

Answer 1

Step-by-step explanation:

(a)

length = 25+x m

breadth = 25-x m

so, the perimeter = 2(length+breadth)

= 2(50)

= 100 m

b)

let length is x m and breadth is y m

2(x+y) = 100

so, x+y = 50

so, x = 50-y

xy = 525 m²

(50-y)(y) = 525

50y-y² = 525

so, 0= y²-50y+525

y²-35y-15y + 525 = 0

y(y-35) -15(y-35) = 0

So, y = 35 m or y = 15 m

if y = 35 m , x = 50-(35) = 15 m

y = 15 m , x = 50-(15). = 35 m

Answer 2

Answer:

a. Perimeter of the rectangle = 100m

b. length = 35m, breadth =15 m

Step-by-step explanation:

a. Given length = 25+x

Given breadth = 25-x

Now we find the Perimeter of the rectangle.

We know that, the perimeter of the rectangle= 2(l+b)

  = 2(25+x+25-x)

= 2(25+25)

=2×50=100m

The perimeter of the rectangle is 100m.

b.

Given the perimeter of the rectangle= 100m

Area of rectangle = 525 $m^{2}$

Let us assume length = x

breadth = y

We know that, the perimeter of the rectangle= 2(l+b)

100=2(x+y)

50=x+y

x=50-y--------( 1)

Area of rectangle = length × breadth

525 =xy

Value of x is put in equation 2.

So, 525=(50-y)y

525=50y-$y^2$

$y^2-50y+525=0\\y^2-(35+15)y+525=0\\y^2-35y-15y+525=0\\y(y-5=35)-15(y-35)=0\\y=35, y=15$

If the value of y=35 is put in equation 1

x=50-35=15

If the value of y=15 put in equation 1.

x=50-15=35

So, length= 35m

So, breadth= 15m

Hence, the answer is length 35 and breadth is 15 m.

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