Question

The perimeter of rectangle is 68 metre and its length is 24 metre find its breadth area and diagonal

Answer 1

Answer:

The perimeter of rectangle=68

And the length of rectangle=24

=68×24

=1,632

The breadth area and diagonal of rectangle=1,632

Answer 2

Given :-

The perimeter of rectangle = 68 m

Length of the rectangle = 24 m

To Find :-

The breadth of the rectangle.

The area of the rectangle.

The diagonal of the rectangle.

Analysis :-

First we have to find the breadth. Consider the breadth as a variable and substitute the given values in the formula of perimeter of rectangle and get it's breadth.

Next substitute the values we got in the formula of area of rectangle and find it's area.

Using the Pythagoras theorem, consider the diagonal as the given in the formula and substitute the given values, then find the value accordingly.

Solution :-

We know that,

• l = Length
• b = Breadth
• a = Area
• p = Perimeter

Let the breadth be 'x' meter.

By the formula,

$\underline{\boxed{\sf Perimeter \ of \ rectangle=2(Length+Breadth)}}$

Given that,

Perimeter (p) = 68 m

Length (l) = 24 m

Substituting their values,

$\sf 68=2(24+x)$

$\sf 68=48+2x$

Transposing 48,

$\sf 2x=68-48$

$\sf 2x=20$

Finding the value of x,

$\sf x=\dfrac{20}{2}$

$\sf x=10$

Therefore, the breadth is 10 m.

Finding the area,

$\underline{\boxed{\sf Area \ of \ rectangle=Length \times Breadth}}$

Given that,

Length (l) = 24 m

Breadth (b) = 10 m

Substituting their values,

Area = 24 × 10

Area = 240 m²

Therefore, the area of the rectangle is 240 m².

Using Pythagoras theorem,

$\underline{\boxed{\sf Hypotenuse=\sqrt{a^{2}+b^{2}} }}$

Substituting their values,

$\sf c=\sqrt{24^{2}+10^{2}}$

$\sf c = \sqrt{676}$

$\sf c=28 \ m$

Therefore, the diagonal is 28 m.