Question

Ramesh has a kite. the figure shows his kite with the corners marked as ABCD . where angle ABD=60° angle DAB = 90° AB =BC and AD=CD given that the area of the kite is 25
$\sqrt{}$
3 sq (1) the length of ab and ad (2) the perimeter of the kite ​

Answer 1

1) the length of ab and ad (2) the perimeter of the kite ​ is solved in the picture attached.

Answer 2

$\mathbf{AB=5,AD =5\sqrt{3}, \ Perimeter =10(1+\sqrt{3})}$

Step-by-step explanation:

• Given data

        ABCD is a kite

       AB=BC

       AD=CD

      $\mathbf{\angle DAB=90^{\circ}}$

      $\mathbf{\angle DCB=90^{\circ}}$

      $\mathbf{\textrm{Area of kite ABCD}=25\sqrt{3}}$

• Here from figure, it is clear that $\mathbf{\Delta ABD \ and \Delta BDC}$ is a right angle triangle.

        We know area of right angle triangle

        $\mathbf{Area =\frac{1}{2}\times (Base\times Altitude)}$

       So

       $\mathbf{Area \ of \Delta ABD =\frac{1}{2}\times (AD\times AB)}$       ...1)

       Similarly

       $\mathbf{Area \ of \Delta BDC =\frac{1}{2}\times (CD\times BC)}$

       But DC= AB and CB = AB

       So we can write

      $\mathbf{Area \ of \Delta BDC =\frac{1}{2}\times (AD\times AB)}$   ...2)

• Now

      $\mathbf{Area \ of \Delta ABD +Area \ of \Delta BDC =Area \ of \ ABCD}$

      $\mathbf{\frac{1}{2}\times (AD\times AB)+\frac{1}{2}\times (AD\times AB)=25\sqrt{3}}$

      Means

      $\mathbf{AD\times AB=25\sqrt{3}}$       ...3)

• Here BD bisect the $\mathbf{\angle ADC}$

       So

       $\mathbf{\angle ADB=30^{\circ}}$      ...4)

       From

       $\mathbf{\tan 30=\frac{AB}{AD}}$

       $\mathbf{\frac{1}{\sqrt{3}}=\frac{AB}{AD}}$

       So

      $\mathbf{AB=\frac{AD}{\sqrt{3}}}$          ...5)

• From equation 3) and equation 5), We get

        $\mathbf{\frac{AD}{\sqrt{3}}\times AD=25\sqrt{3}}$

       On solving ,we get

       $\mathbf{AD=5\sqrt{3}}$       ...6)

       From equation 5) and equation 6)

      $\mathbf{AB=\frac{AD}{\sqrt{3}}=\frac{5\sqrt{3}}{\sqrt{3}}=5}$    ...7)

• Now Perimeter of ABCD

       Perimeter = AB +BC+ CD+ DA

       $\mathbf{Perimeter=5+5\sqrt{3}+5+5\sqrt{3}=10+10\sqrt{3}=10(1+\sqrt{3})}$