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# Find the diagonal of a rectangle whose length is 16 cm and area is 192 sq. cm?

Last updated date: 02nd Aug 2024
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Hint: Here we will first find the breadth of the rectangle by using the area of rectangle formula. Then we will draw our rectangle and form a triangle in it. Finally we will use Pythagora's theorem and substitute the values of length and breadth as perpendicular and base to find the diagonal and get the required answer.

Formula Used:
We will use the following formulas:
1. Area of rectangle $= l \times b$, where $l$ is the length of the rectangle and $b$ is the breadth of the rectangle
2. Pythagoras Theorem: ${{P}^{2}}+{{B}^{2}}={{H}^{2}}$, where $P$ is the perpendicular, $B$ is the base and $H$ is the hypotenuse of the right angled triangle.

It is given that:
The length of the rectangle $= 16$ cm
The Area of the rectangle $= 192$ sq. cm
We know that Area of rectangle $= l \times b$
Substituting these values in the above formula, we get
$\Rightarrow 192 =16 \times b$
Dividing both sides by 16, we get
$\Rightarrow b = \dfrac{{192}}{{16}}$
$\Rightarrow b = 12$ cm….$\left( 1 \right)$
Now, we will draw the rectangle by using the above data.

Now we will use Pythagoras theorem to find the length of the diagonal.
In $\vartriangle ABD$,
$A{B^2} + A{D^2} = B{D^2}$
Substituting the value from the diagram in above formula, we get
$\Rightarrow {\left( {16} \right)^2} + {\left( {12} \right)^2} = B{D^2}$
Applying the exponent on the terms, we get
$\Rightarrow 256 + 144 = B{D^2}$
$\Rightarrow B{D^2} = 400$
$\Rightarrow BD = \sqrt{400}$
$\Rightarrow BD = 20$ cm