Question

If the length of a rectangle is 16cm and its breadth is 12cm then what is the length of its each diagonal?

Answer 1

Answer-

$\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\thicklines\put(7.7,3){\large{A}}\put(9.5,1.7){\sf{\large{Diagonal}}}\put(7.7,1){\large{B}}\put(9.2,0.7){\sf{\large{16 cm}}}\put(11.1,1){\large{C}}\put(8,1){\line(1,0){3}}\put(8,1){\line(0,2){2}}\put(11,1){\line(0,3){2}}\put(8,3){\line(3,0){3}}\put(11.1,2){\sf{\large{12 cm}}}\put(8,1){\line(3,2){3}}\put(11.1,3){\large{D}}\put(10.8,1){\line(0,2){0.2}}\put(10.8,1.2){\line(2,0){0.2}}\end{picture}$

» Given :

length = 16 cm (BC)

breadth = 12 cm(CD)

» To find:

Length of Diagonal (BD)

» Solution:

Pythagoras theorem states that,

${\pink{\boxed{(Hypotenuse)^2 = (Base)^2 +(Height)^2 }}}$

The diagonal forms a right angled triangle with length and breadth.

in our case -

height = 12 cm

Base = 16 cm

(BD)² = (BC)² +(CD)²

by putting values

(Hypotenuse)² = (12)²+(16)²

= 144+256

= 400

(Hypotenuse)² = 400

Hypotenuse = √400

= 20 cm

${\purple{\boxed{\therefore \: diagonal\:of \:the\: rectangle= 20\:cm}}}$