1.

Diagonal of rectangle is 5cm and one. side of rectangle is. 4 cm find other. side. of. rectangle​

Answer»
  • 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.


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