Question

diagonal of rectangle is 5cm and one. side of rectangle is. 4 cm find other. side. of. rectangle
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Answer 1

Here's your answer

hope it help you

Answer 2

• The other side of the rectangle (breadth) = 3 cm.

Given :

• The diagonal of the rectangle = 5 cm.
• One side of the rectangle (length) = 4 cm.

To Find :

• The other side of the rectangle (breadth).

Diagram :

$\setlength{\unitlength}{15mm}\begin{picture}(2,2)\put(2 , 2){\line(0,2){2}}\put(2,4){\line(2,0){3}}\put(5,2){\line(0,2){2}}\put(2,2){\line(2,0){3}}\put(2,2){\line(3,2){3}}\put(1.8,4.1){\bf{A}}\put(1.8,1.7){\bf{B}}\put(5,4.1){\bf{D}}\put(4.9,1.7){\bf{C}}\put(3.2,1.7){\bf{4 cm}}\put(2.9,3.1){\bf{5 cm}}\put(5.2,2.9){\bf{x}}\end{picture}$

Solution :

Let,

The other side of the rectangle (breadth) be x.

According to the question,

A rectangle have a diagonal that means, a rectangle divided into two right angled triangles.

Let, take one right angled triangle.

By pythagoras theorem.

$\huge\blue{\star} \: \: \: \large{ \boxed{ \bf{ {h}^{2} = {b}^{2} + {p}^{2} }}}$

In ∆DBC,

Where,

• BC = B = base (length of the rectangle)
• CD = P = perpendicular (breadth of the rectangle)
• BD = H = hypotenuse (diagonal of the rectangle)

We have,

• H = 5 cm.
• B = 4 cm.

We have to find P (Perpendicular OR breadth of the rectangle).

$\bf \implies {(5 \: cm)}^{2} = {(4 \: cm)}^{2} + {(x)}^{2} \\ \\ \\ \bf \implies 25 \: {cm}^{2} = 16 \: {cm \: }^{2} + {x}^{2} \\ \\ \\ \bf \implies 25 \: {cm}^{2} - 16 \: {cm}^{2} = {x}^{2} \\ \\ \\ \bf \implies 9 \: {cm}^{2} = {x}^{2} \\ \\ \\ \bf \implies \sqrt{9 \: {cm}^{2} } = x \\ \\ \\ \bf \implies 3 \: cm = x \\ \\ \\ \: \: \bf \therefore \: \: Perpendicular = 3 \: cm$

Hence,

The other side of the rectangle (breadth) is 3 cm.