The image shows a rectangle ABCD with length $4x+y$ cm and width $4x$ cm. We are asked to find: (a) The perimeter of the rectangle. (b) The area of the rectangle. (c) The area of triangle ABD, given that BD is a diagonal of the rectangle. (d) What must be subtracted from the length to make it a square. (e) The area of the square so formed.

1. Problem Description

The image shows a rectangle ABCD with length $4x+y$ cm and width $4x$ cm. We are asked to find:

(a) The perimeter of the rectangle.

(b) The area of the rectangle.

(c) The area of triangle ABD, given that BD is a diagonal of the rectangle.

(d) What must be subtracted from the length to make it a square.

(e) The area of the square so formed.

2. Solution Steps

(a) The perimeter of a rectangle is given by the formula:

$P = 2(length + width)$

In this case, $length = 4x + y$ and $width = 4x$.

So, $P = 2(4x + y + 4x) = 2(8x + y) = 16x + 2y$ cm.

(b) The area of a rectangle is given by the formula:

$A = length \times width$

In this case, $length = 4x + y$ and $width = 4x$.

So, $A = (4x + y)(4x) = 16x^2 + 4xy$ square cm.

(c) Since BD is a diagonal of the rectangle, it divides the rectangle into two equal parts. Therefore, the area of triangle ABD is half the area of the rectangle.

Area of triangle ABD $= \frac{1}{2} \times Area\ of\ rectangle = \frac{1}{2} (16x^2 + 4xy) = 8x^2 + 2xy$ square cm.

(d) To make the rectangle a square, the length must be equal to the width. The width is $4x$. The length is $4x+y$. Therefore, we must subtract $y$ from the length to get $4x+y-y = 4x$.

(e) If we subtract $y$ from the length, then both sides are equal to $4x$. The area of the resulting square is $(side)^2 = (4x)^2 = 16x^2$ square cm.

3. Final Answer

(a) The perimeter of the rectangle is $16x + 2y$ cm.

(b) The area of the rectangle is $16x^2 + 4xy$ square cm.

(c) The area of triangle ABD is $8x^2 + 2xy$ square cm.

(d) $y$ must be subtracted from the length to make it a square.

(e) The area of the square so formed is $16x^2$ square cm.