Question

3.2 Find the length of the diagonal of a rectangle with length 20 m and breadth 15 m.​

Answer 1

GIVEN:-

• $\rm{Length\:of\: Rectangle = 20m}$

• $\rm{Breadth\:of\: Rectangle = 15m}$.

TO FIND:-

• The Length of the Diagonal.

CONSTRUCTION:-

• Join D to B. i.e Diagonal.

Now,

$\implies\rm{ CD = 15m}$

$\implies\rm{ BC = 20m}$

$\implies\rm{ BD = ?}$.

Using Pythogoras Theorem because in rectangle the angles forms 90°.

$\implies\rm{ (BC)^2 + (CD)^2 = (BD)^2 }$

$\implies\rm{ (20)^2 + (15)^2 = (BD)^2 }$

$\implies\rm{ 400 + 225 = (BD)^2 }$

$\implies\tm{ 625 = (BD)^2 }$

$\implies\rm{\sqrt{BD} = \sqrt{625}}$

$\implies\rm{ BD = 25m}$.

Hence, The Diagonal of Rectangle is 25m

Answer 2

$\large \: \red{ \underline{ \underline{ \green{ \rm given : }}}}$

• Length = 20m

• Breadth = 15m

$\large \: \pink{ \underline{ \underline{ \orange{ \rm to \: \: find : }}}}$

• length of diagonal

$\rightarrow$ Note : We can cut a rectangle into triangle.

$\therefore$ It is a right angled triangle.

$\large \: \purple{ \underline{ \underline{ \blue{ \rm solution : }}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$ Pythagoras Theorem:-

$\large \red{ \boxed{ \boxed{ \green{ \rm {h}^{2} = p {}^{2} + {b}^{2} }}}}$

Here,

• H = diagonal (hypotenuse)

• P = length = 20m

• B = breadth ( base ) = 15m

$\implies$ H² = 20² + 15²

$\implies$ H² = 400 + 225

$\implies$ H² = 625

$\implies \large\rm h {}^{} = \sqrt{625}$

$\large \pink{ \boxed{ \boxed{ \purple{ \therefore \: diagonal = 25m}}}}$