Question

What is the length of the diagonal of a rectangle whose width is \[90cm\]and whose length is \[200cm\]?

Answer 1

Hint:

In rectangle we have two measurements \[length = x\] & \[width = y\], and the diagonal is a line which divide the rectangle into two equal right angled triangle
 Now according to the Pythagoras Theorem, in a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides.
Therefore, by applying the Pythagoras Theorem in rectangle, we get,
\[{(diagonal)^2} = {(length)^2} + {(width)^2}\]
By substituting the values from the above figure, we get
\[diagonal = \sqrt {{x^2} + {y^2}} \]
Formula used: \[{(diagonal)^2} = {(length)^2} + {(width)^2}\]
\[diagonal = \sqrt {{x^2} + {y^2}} \]
Where,
\[length = x\]
\[\text{width} = y\]

Complete step-by-step solution:
Given that the length of the rectangle is \[200cm\], the width of the rectangle is \[90cm\].

To find: The length of the diagonal of the rectangle.
Now, we shall get into the solution.
Step 1: According to the Pythagoras Theorem, we have
\[{(diagonal)^2} = {(length)^2} + {(width)^2}\]
Now taking square root on both sides we get
\[(diagonal) = \sqrt {{{(length)}^2} + {{(width)}^2}} \]
Step 2: On substituting the given values of length and width in the above formula, we get,
\[(diagonal) = \sqrt {{{(length)}^2} + {{(width)}^2}} \]
\[diagonal = \sqrt {{{(200)}^2} + {{(90)}^2}} \]
Step 3: On further simplification, we get,
\[diagonal = \sqrt {40000 + 8100} \]
\[diagonal = \sqrt {48100} \]
On solving, we get
\[diagonal = 219.317cm\]
Hence, the length of the diagonal of a rectangle whose width is \[90cm\] and whose length is \[200cm\] is equal to\[219.317cm\] which is the required answer.

Note: The Pythagoras Theorem can only be applied to those triangles in which the base of the triangle and one other side of the triangle is at right angles to each other and this type of triangle is known as right angled triangle. Here, in the rectangle the length and the width are at right angles to each other and therefore, we apply Pythagoras theorem.