Question

plzzzzzz help me out..... ​

Answer 1

Answer:-

$\boxed{\red{\sf\:Perimeter\:of\:rectangle\:=\:\dfrac{150}{\sqrt{13}}\:cm}}$

$\boxed{\pink{\sf\:Area\:of\:rectangle\:=\:103\:\dfrac{11}{13}\:cm^2}}$

Step-by-step-explanation:

We have given that,

In $\sf\:\square\:ABCD\:,$

$\sf\:\bullet\:AC\:=\:15\:cm$

$\sf\:\bullet\:\measuredangle\:ACD\:=\:\alpha$

$\sf\:\bullet\:\cot\:\alpha\:=\:\dfrac{3}{2}$

We have to find the perimeter and area of the rectangle.

Now, we know that,

$\pink{\sf\:\cot\:\alpha\:=\:\dfrac{Adjacent\:side}{Opposite\:side}}$

$\implies\sf\:\cot\:\alpha\:=\:\dfrac{CD}{AD}$

$\implies\sf\:\dfrac{3}{2}\:=\:\dfrac{CD}{AD}$

$\implies\sf\:\dfrac{CD}{AD}\:=\:\dfrac{3}{2}$

$\implies\sf\:CD\:=\:\dfrac{3}{2}\:\times\:AD$

$\implies\boxed{\red{\sf\:CD\:=\:\dfrac{3\:AD}{2}}}$

Now, in $\sf\:\triangle\:ADC$,

$\sf\:\measuredangle\:ADC\:=\:90^{\circ}\:\:\:-\:-\:[\:Angle\:of\:rectangle\:]$

$\displaystyle{\therefore\sf\:AC^2\:=\:AD^2\:+\:CD^2\:\:\:-\:-\:[\:Pythagors\:theorem\:]}$

$\implies\sf\:\left(\:15\:\right)^2\:=\:AD^2\:+\:\left(\:\dfrac{3\:AD}{2}\:\right)^2$

$\implies\sf\:225\:=\:AD^2\:+\:\dfrac{9\:AD^2}{4}$

$\implies\sf\:\dfrac{9\:AD^2\:+\:4\:AD^2}{4}\:=\:225$

$\implies\sf\:\dfrac{13\:AD^2}{4}\:=\:225$

$\implies\sf\:13\:AD^2\:=\:225\:\times\:4$

$\implies\sf\:13\:AD^2\:=\:900$

$\implies\sf\:AD^2\:=\:\dfrac{900}{13}$

$\displaystyle{\implies\boxed{\red{\sf\:AD\:=\:\dfrac{30}{\sqrt{13}}}}\sf\:\:\:-\:-\:[\:Taking\:square\:roots\:]}$

Now,

$\pink{\sf\:CD\:=\:\dfrac{3\:AD}{2}}$

$\implies\sf\:CD\:=\:\dfrac{3\:\times\:\dfrac{30}{\sqrt{13}}}{2}$

$\implies\sf\:CD\:=\:\dfrac{\dfrac{90}{\sqrt{13}}}{2}$

$\implies\sf\:CD\:=\:\dfrac{\cancel{90}}{\sqrt{13}}\:\times\:\dfrac{1}{\cancel{2}}$

$\implies\boxed{\red{\sf\:CD\:=\:\dfrac{45}{\sqrt{13}}}}$

Now, we know that,

$\pink{\sf\:P\:(\:\square\:ABCD\:)\:=\:2\:(\:l\:+\:b\:)}$

$\implies\sf\:P\:(\:\square\:ABCD\:)\:=\:2\:(\:CD\:+\:AD\:)$

$\implies\sf\:P\:(\:\square\:ABCD\:)\:=\:2\:\left(\:\dfrac{45}{\sqrt{13}}\:+\:\dfrac{30}{\sqrt{13}}\:\right)$

$\implies\sf\:P\:(\:\square\:ABCD\:)\:=\:2\:\left(\:\dfrac{45\:+\:30}{\sqrt{13}}\:\right)$

$\implies\sf\:P\:(\:\square\:ABCD\:)\:=\:2\:\times\:\dfrac{75}{\sqrt{13}}$

$\implies\boxed{\red{\sf\:Perimeter\:of\:\square\:ABCD\:=\:\dfrac{150}{\sqrt{13}}\:cm}}$

Now,

$\pink{\sf\:A\:(\:\square\:ABCD\:)\:=\:l\:\times\:b}$

$\implies\sf\:A\:(\:\square\:ABCD\:)\:=\:CD\:\times\:AD$

$\implies\sf\:A\:(\:\square\:ABCD\:)\:=\:\dfrac{45}{\sqrt{13}}\:\times\:\dfrac{30}{\sqrt{13}}$

$\implies\sf\:A\:(\:\square\:ABCD\:)\:=\:\dfrac{45\:\times\:30}{\:(\:\sqrt{13}\:)^2}$

$\implies\sf\:A\:(\:\square\:ABCD\:)\:=\:\dfrac{1350}{13}$

$\implies\sf\:A\:(\:\square\:ABCD\:)\:=\:\dfrac{(\:103\:\times\:13\:+\:11\:)}{13}$

$\implies\sf\:A\:(\:\square\:ABCD\:)\:=\:103\:+\:\dfrac{11}{13}$

$\implies\boxed{\pink{\sf\:Area\:of\:\square\:ABCD\:=\:103\:\dfrac{11}{13}\:cm^2}}$

Answer 2

Answer:

The perimeter is 150/√13.

Step-by-step explanation:

Let's say if theta is the angle between diagonal and the side of the rectangle then

AB/BC = cot θ

If cotθ = 3/2 then sinθ = 2/√13

Sin θ = BC/AC

2/√13 = BC / 15

BC = 30/ √ 13

Similarly cos θ  = 3/√13

cos θ  = AB/AC =AB/15

3/√ 15 =AB /15 so AB = 45/√13

So perimeter = 2(AB+ BC)

                       = 2(30/√ 13  + 45/√ 13)

                      = 150/√13

Hence the perimeter is 150/√13.

hope this helps you