Question

Rectangle ABCD is formed in a circle as shown in Figure. If AE = 8 cm and AD = 5 cm, find the perimeter of the rectangle. ​

Answer 1

⠀Question -

Rectangle ABCD is formed in a circle as shown in Figure. If AE = 8 cm and AD = 5 cm, find the perimeter of the rectangle.

Stated -

• ABCD is a rectangle that formed in a circle.
• AE measures 8cm
• AD measures 5cm

To Attain -

• Perimeter of Rectangle

Where -

• L stands length
• B stands breadth

Formulae to be used -

• Perimeter of Rectangle = 2(l + b)
• Pythagoras Theorm = a² + b² = c²

Explanation -

⠀⠀Given, ABCD makes a rectangle inside of the circle whereas AE measures 8cm and AD measures 5cm. After all, we are asked to find the perimeter of rectangle. So, first we will find DE which is radius as DB which diagonal of rectangle are equal. In order to find the right angled triangle ADC, we will put formula of pythagoras theorm then we will find the perimeter of rectangle by putting its formula.

Solution -

First, we will find the radius DE.

$\dashrightarrow\sf{DE (Radius) = AE + AD}$

$\dashrightarrow\sf{DE = 8cm + 5cm}$

$\dashrightarrow\underline{\sf{\red{DE = 13cm}}}$

Since, -

DB is also a radius So, DB = AC = 13cm [Diagonal of Rectangle are equal]

Now, -

we will find the right ∆ADC by using Pythagoras Theorm for required Breadth.

$\:\dashrightarrow\sf{{(AD)}^{2}} + {(DC)}^{2} = {(AC)}^{2}$

$\:\dashrightarrow\sf{{(5)}^{2}} + {(DC)}^{2} = {(13)}^{2}$

$\:\dashrightarrow\sf{25} + {(DC)}^{2} = {169}$

$\:\dashrightarrow\sf{DC}^{2} = {169 - 25} = 144$

$\:\dashrightarrow\sf{DC = \sqrt{144} }$

$\:\dashrightarrow{\sf{\red{DC = 12 }}}$

Now, -

Now, we have our required Breadth, So, we will now gonna find the perimeter of Rectangle by using its appropriate formula

$\dashrightarrow\sf{Perimeter_{\:(Rectangle)} = 2(l+b)}$

$\dashrightarrow\sf{Perimeter = 2(5cm +12cm)}$

$\dashrightarrow\sf{Perimeter = 2 \times 17cm}$

$\dashrightarrow{\sf{\red{Perimeter = 34cm}}}$

Final Answer -

$\:\;\;\;\;\;\leadsto$ Hence, Perimeter of given ABCD rectangle is 34cm.

_____________________________________________

Additional Information:-

• Perimeter of Rectangle = 2(l+b)
• Perimeter of Square = 4× Side
• Area of Rectangle = lb sq.units
• Area of Square = (side)²
• Area of Parallelogram = b×h
• Area of Triangle = 1/2 × b × h
• Pythagoras Theorm = a² + b² = c²
• Area of trapezium = 1/2 × (a+b) × h

Answer 2

Answer:

Perimeter the rectangle = 34 cm.

Step-by-step explanation:

GIVEN :

• ABCD is a rectangle.

• AE = 8 cm.

• AD (breadth) = 5 cm.

TO FIND :

• The perimeter of the rectangle.

SOLUTION :

First, we need to find the length of the rectangle (DC).

DE is the radius of the circle.

DE = EA + AD

DE = 8 cm + 5 cm.

DE = 13 cm

• The radius of the circle is 13 cm.

Also, DB is the radius of the circle.

So, DE = DB = 13 cm.

And,

DB = AC [Diagonals of the rectangle are equal in length.]

Therefore, AC = 13 cm.

Now,

From ∆ADC,

=> DC² = AC² - AD²

=> DC² = (13cm)² - (5cm)²

=> DC² = 169cm² - 25cm²

=> DC² = 144cm²

=> DC = √144cm²

=> DC = 12 cm

So, the length of the rectangle is 12 cm.

Now,

Perimeter of the rectangle = 2(length + breadth)

• Length = DC = 12 cm.

• Breadth = AD = 5 cm.

=> 2(12cm + 5cm)

=> 2(17cm)

=> 34 cm

Hence, the perimeter of the rectangle is 34 cm.