Question

Two unequal sides of the kite are 15 cm and 20 cm and angle between these sides is a right angle. Find its diagonals and the area. ​

Answer 1

$\textbf {\underline {\underline{Step-By-Step\:Explanation:-}}}$

$\textbf {\underline {Given:}}$

• Two unequal sides are 15 cm and 20 cm
• Angle between these sides is right angle

$\textbf {\underline {To\:Find:}}$

• It's diagonals and the area

$\huge\underline\green {\tt Solution:}$

As the angle between the sides is right angle , So, Triangle ADC is a right angled triangle.

By pythagorus theorem :

${AD}^2 + {CD}^2 = {AC}^2$

➡ ${15}^2 + {20}^2 = {AC}^2$

➡ ${225} + {400} = {AC}^2$

➡ ${625} = {AC}^2$

➡ ${AC} = \sqrt{625}$

•°• AC = 25 cm

➡ Diagonal DB = Diagonal AC

Now,

Area of kite = $\dfrac{1}{2}(25 × 25 )$

= $\dfrac{625}{2}$

= 312.5 $cm^2$

Answer :

Diagonals = 25 cm and area = 312.5 $cm^2$

Answer 2

Given:

Two unequal sides are 15 cm and 20 cm

Angle between these sides is right angle

\textbf {\underline {To\:Find:}}

ToFind:

It's diagonals and the area

\huge\underline\green {\tt Solution:}

Solution:

As the angle between the sides is right angle , So, Triangle ADC is a right angled triangle.

By pythagorus theorem :

{AD}^2 + {CD}^2 = {AC}^2AD

2

+CD

2

=AC

2

➡ {15}^2 + {20}^2 = {AC}^215

2

+20

2

=AC

2

➡ {225} + {400} = {AC}^2225+400=AC

2

➡ {625} = {AC}^2625=AC

2

➡ {AC} = \sqrt{625}AC=

625

•°• AC = 25 cm

➡ Diagonal DB = Diagonal AC

Now,

Area of kite = \dfrac{1}{2}(25 × 25 )

2

1

(25×25)

= \dfrac{625}{2}

2

625

= 312.5 cm^2cm

2

Answer :

Diagonals = 25 cm and area = 312.5 cm^2cm

2