Question

A rectangle has diagonal $2$ and length $\sqrt{3}$. Find its width.

Short answer

The width of rectangle is $1$.

Step-by-step solution

Step 1. Given information.

The rectangle has length $\sqrt{3}$ and diagonal $2$.

Step 2. Concept used.

Pythagoras Theorem state that in a right angle triangle, the square of the hypotenuse is equal to the sum of squares of other sides.

Step 3. Now find the width.

First we draw a rectangle of length $\sqrt{3}$and diagonal $2$and labeled $\square ABCD$.

Let be width be $x$.

Inright angle triangle $\Delta ADC$.

So, we use Pythagoras Theorem.

$\begin{array}{c}{x}^2+{\sqrt{3}}^2=2^2\\ x^2+3=4\\ x^2=1\end{array}$

Taking square root we get.

$\begin{array}{l}x=\pm \sqrt{1}\\ x=\pm 1\end{array}$

Since, distance is always a positive quantity. So, we neglect negative sign.

Therefore, the width of rectangle is $1$.