The length and width of a rectangle are in ratio of 3:4. If the rectangle has an area of $300c{m^2}$, what is the length of the diagonal?
A. 15 cm
B. 12 cm
C. 25 m
D. 25 cm
Answer
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Hint: We will first find the length and width of the rectangle by using the given ratio and area of the rectangle. Let the length of rectangle is $3x$ and the width of the rectangle is $4x$and then substitute these values in the formula is $l \times b$, where $l$ is the breadth of rectangle and $b$ is the width of the rectangle to find the value of $x$. Then, apply Pythagoras theorem in the triangle formed by the diagonal and sides of the rectangle to determine the length of the rectangle.
Complete step-by-step answer:
Let the common ratio of length and width of rectangle be $x$.
Then, the length of rectangle is $3x$ and the width of the rectangle is $4x$
We know that the area of the rectangle is $l \times b$, where $l$ is the breadth of the rectangle and $b$ is the width of the rectangle.
Then, the area of the given rectangle is $3x\left( {4x} \right) = 12{x^2}$
But, we are given that the area of the rectangle is $300c{m^2}$.
$12{x^2} = 300$
Divide both sides by 12
$
\Rightarrow {x^2} = \dfrac{{300}}{{12}} \\
\Rightarrow {x^2} = 25 \\
$
Take square root on both sides,
$x = 5$
Therefore, the length of the rectangle is $3\left( 5 \right) = 15cm$ and the width of the rectangle is $4\left( 5 \right) = 20cm$.
We have to find the length of the diagonal of the rectangle.
In a rectangle all angles are right angles.
Therefore, $ABC$ is a right triangle.
We want to find the length $AC$, which is the diagonal of the rectangle.
We will use Pythagoras theorem to find the value of $AC$.
Hence, ${\left( {AC} \right)^2} = {\left( {AB} \right)^2} + {\left( {BC} \right)^2}$
On substituting the values of $AB$ and $BC$ in the above equation, we will get,
$
{\left( {AC} \right)^2} = {\left( {20} \right)^2} + {\left( {15} \right)^2} \\
\Rightarrow {\left( {AC} \right)^2} = 400 + 225 \\
\Rightarrow {\left( {AC} \right)^2} = 625 \\
$
After taking square root on both sides, we will get,
$AC = 25cm$
Also, the diagonals of the rectangle are equal.
Hence, the value of the rectangle is 25 cm
Thus, option D is correct.
Note: One should know the basic properties of a rectangle, such as all the angles of a rectangle are equal to ${90^ \circ }$, opposite sides of a rectangle are equal and diagonals of a rectangle are equal.
Complete step-by-step answer:
Let the common ratio of length and width of rectangle be $x$.
Then, the length of rectangle is $3x$ and the width of the rectangle is $4x$
We know that the area of the rectangle is $l \times b$, where $l$ is the breadth of the rectangle and $b$ is the width of the rectangle.
Then, the area of the given rectangle is $3x\left( {4x} \right) = 12{x^2}$
But, we are given that the area of the rectangle is $300c{m^2}$.
$12{x^2} = 300$
Divide both sides by 12
$
\Rightarrow {x^2} = \dfrac{{300}}{{12}} \\
\Rightarrow {x^2} = 25 \\
$
Take square root on both sides,
$x = 5$
Therefore, the length of the rectangle is $3\left( 5 \right) = 15cm$ and the width of the rectangle is $4\left( 5 \right) = 20cm$.
We have to find the length of the diagonal of the rectangle.
In a rectangle all angles are right angles.
Therefore, $ABC$ is a right triangle.
We want to find the length $AC$, which is the diagonal of the rectangle.
We will use Pythagoras theorem to find the value of $AC$.
Hence, ${\left( {AC} \right)^2} = {\left( {AB} \right)^2} + {\left( {BC} \right)^2}$
On substituting the values of $AB$ and $BC$ in the above equation, we will get,
$
{\left( {AC} \right)^2} = {\left( {20} \right)^2} + {\left( {15} \right)^2} \\
\Rightarrow {\left( {AC} \right)^2} = 400 + 225 \\
\Rightarrow {\left( {AC} \right)^2} = 625 \\
$
After taking square root on both sides, we will get,
$AC = 25cm$
Also, the diagonals of the rectangle are equal.
Hence, the value of the rectangle is 25 cm
Thus, option D is correct.
Note: One should know the basic properties of a rectangle, such as all the angles of a rectangle are equal to ${90^ \circ }$, opposite sides of a rectangle are equal and diagonals of a rectangle are equal.
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