Question

class 10 step by step solution ​

Answer 1

Answer:

length of AB and AD are 5 and 5√3

and perimeter = 10(1+√3)

Step-by-step explanation:

ABCD is a kite

AB = BC

AD = CD

area of the kite = 25√3

solution :

triangle ABD and BDC are the righht angled triangles

area of triangle ABD = \frac{1}{2}\times base \times height

2

1

×base×height

= \frac{1}{2} \times AD \times AB ..(1)

2

1

×AD×AB..(1)

Similarly

area of triangle BDC = \frac{1}{2} \times CD \times BC ..(2)

2

1

×CD×BC..(2)

but DC = AB and BC = AB

so we can write

area of triangle BDC = \frac{1}{2} \times AD \times AB ..(2)

2

1

×AD×AB..(2)

NOW

area of triangle ABD + area of triangle BDC = area of ABCD

\frac{1}{2} \times AD \times AB +\frac{1}{2} \times AD \times AB = 25\sqrt{3}

2

1

×AD×AB+

2

1

×AD×AB=25

3

\begin{lgathered}AD \times AB = 25\sqrt{3}...(3)\\\end{lgathered}

AD×AB=25

3

...(3)

here BD bisects \angle ADC∠ADC

SO, \begin{lgathered}\angle ADB = 30..(4)\\\end{lgathered}

∠ADB=30..(4)

using trigonometry

\begin{lgathered}tan 30^\circ = \frac{AB}{AD} \\\frac{1}{\sqrt{3} } = \frac{AB}{AD}\\AB =\frac{AD}{\sqrt{3}}\end{lgathered}

tan30

∘

=

AD

AB

3

1

=

AD

AB

AB=

3

AD

On solving we get

AD = 5√3

then

AB = \frac{AD}{\sqrt{3} } = \frac{5\sqrt{3} }{\sqrt{3} } = 5AB=

3

AD

=

3

5

3

=5

NOW

perimeter = AB + BC +CD +DA

= 5+ 5√3+5+5√3

= 10 +10√3

= 10(1+√3)

hence ,

length of AB and AD are 5 and 5√3

and perimeter = 10(1+√3)

Answer 2

REFER TO THE ATTACHMENT