Question

If ABCD is a rectangle which length is double of his breadth and perimeter of the rectangle 300 cm then find his l, b and area​

Answer 1

Solution :

Suppose we have rectangle ABCD which length is double of his breadth & perimeter of the rectangle is 300 cm.

$\underline{\bf{Explanation\::}}$

Let the breadth of the rectangle be r cm & length of the rectangle be 2r cm respectively.

As we know that formula of the perimeter of rectangle;

$\boxed{\bf{Perimeter = 2(length + breadth)}}$

A/q

$\mapsto\sf{2(2r + r ) =300}\\\\\mapsto\sf{2(3r) = 300}\\\\\mapsto\sf{6r = 300}\\\\\mapsto\sf{r=\cancel{300/6}}\\\\\mapsto\bf{r=50\:cm}$

Thus;

• Length of the rectangle, (L) = 2r = 2 × 50 = 100 cm
• Breadth of the rectangle, (B) = r = 50 cm

Now;

As we know that formula of the area of rectangle;

$\mapsto\sf{Area\:of\:the\:rectangle=Length \times breadth }\\\\\mapsto\sf{Area\:of\:the\:rectangle=100 cm \times 50 cm }\\\\\mapsto\bf{Area\:of\:the\:rectangle=5000\:cm^{2} }$