Question

What is the length of the diagonal of a rectangle that is 8 feet long and 15 feet wide?
A. 13 feet
B. 11 feet
C. 17 feet
D. 289 feet

Answer 1

Hint: this question, we have given the length and width of the rectangle. We need to find the diagonal of the rectangle. First draw the figure of the rectangle and mention its diagonal. By figure we can see that the diagonal of the rectangle is also the hypotenuse of the triangle with sides equal to length and breadth of the given rectangle. Pythagoras Theorem to find the diagonal of the rectangle.

Complete step-by-step answer:
Let ABCD be the rectangle with in which length AB = 8 feet and BC = 15 feet.

Given,
Length of the rectangle (AB) = 8 feet
Width of the rectangle (BC) = 15 feet
We need to find the diagonal of a rectangle (AC) which is also a hypotenuse of a rectangle.
We know that all angles of the rectangle is 90°. So, in triangle ABC, ∠B = 90°.
Using Pythagoras Theorem, we get;
  $ A{C^2} = A{B^2} + B{C^2} $
  $ \Rightarrow A{C^2} = {8^2} + {15^2} $
  $ \Rightarrow A{C^2} $ = 64 + 225 = 289
  $ \Rightarrow AC = \sqrt {289} $ = 17 feet
Thus, the diagonal of a rectangle is 17 feet.

So, the correct answer is “Option C”.

Note: In these types of questions, draw figures and draw the line which is asked in question. Here, use basic properties of rectangle like, all opposite sides of the rectangle are equal and all angles of a rectangle is 90°. Also the diagonal of a rectangle divides the rectangle in two triangles which are similar to each other. Whenever asked about diagonal of a rectangle use Pythagoras Theorem, we have $ {H^2} = {P^2} + {B^2} $ , where H is the Hypotenuse, P is Perpendicular, and B is the Base.