Question

In the given figure,O is the centre of larger circle.
BC=9cm and ED=5cm. Find the area of the shaded region

Answer 1

Answer -

Area of the shaded region is 643.5 cm²

$\rule{200}2$

Let the radius of the smaller circle be r and radius of the bigger circle be R.

Given -

• BC = 9 cm and ED = 5 cm
• O is the centre of larger circle

Find -

Area of the shaded region.

Solution -

Radius of bigger circle is R. So, OE = OA = OF = OB = R

→ OE = OD + ED

→ OD = R - 5

Similarly,

→ OB = OC + BC

→ OC = R - 9

Now, join AD and CD.

In ∆ADC, which is a right angled triangle.

$\sf{\angle{ADC}}$ = 90°

In ∆AOD and ∆DOC

→ OD = OD (common)

→ $\sf{\angle{O}}$ = $\sf{\angle{O}}$ = 90°

→ $\sf{\angle{ADO}}$ = $\sf{\angle{DCO}}$

So, ∆AOD ~ ∆DOC

$\therefore\:\sf{\dfrac{OA}{OD}}$ = $\sf{\dfrac{OD}{OC}}$

$\implies\:\sf{OD^2}$ = $\sf{OA\:\times\:OC}$

$\implies\:\sf{(R\:-\:5)^2}$ = $\sf{R(R\:-\:9)}$

$\implies\:\sf{R^2\:+\:25\:-\:10R}$ = $\sf{R^2\:-\:9R}$

$\implies\:\sf{10R\:-\:9R}$ = $\sf{25}$

$\implies\:\sf{R}$ = $\sf{25}$

Now,

→ AB = OA + OB

→ AB = 25 + 25

→ AB = 50

So,

→ AC = 50 - 9

→ AC = 41

So, radius = $\sf{\dfrac{41}{2}}$

•°• Area of shaded region = Area of bigger circle - Area of smaller circle.

→ πR² - πr²

→ π(R² - r²)

→ $\sf{\dfrac{22}{7}\bigg[(25)^2\:-\:\bigg(\dfrac{41}{2}\bigg)^2\bigg]}$

→ $\sf{\dfrac{22}{7}\bigg(625\:-\:\dfrac{1681}{4}\bigg)}$

→ $\sf{\dfrac{22}{7}\bigg(\dfrac{2500\:-\:1681}{4}\bigg)}$

→ $\sf{\dfrac{1287}{2}}$

→ $\sf{643.5\:cm^2}$

Answer 2

$\huge\tt\blue{Answer}$

Let R be the bigger radius.

________________________________________________

→OE = OD + ED

→OD = R - 5

________________________________________________

→OB = OC + BC

→OC = R - 9

________________________________________________

Joining AD with CD

∆ ADC = 90°

________________________________________________

→∆ AOD, ∆ DOE

→OD = OD

→∠O = ∠O = 90°

→∠ADO = ∠DCO

________________________________________________

∆ AOD = ∠ DCO

→OA / OD = OD / OC

________________________________________________

→OD² = OA × DC

→(R - 5)² = R (R - 9)

→R² + 25 - 10R = R² - 9R

10R - 9R = 25

→

→R = 25

________________________________________________

→AB = OA + OB

→AB = 25 + 25

→AB = 50

________________________________________________

→AC = 50 - 9

→AC = 41

→41 / 2 = R

________________________________________________

→π R² - π r²

→π (R² / r²)

→22 / 7 { ( 25)² - ( 41 / 2 )² }

→22 / 7 ( 625 - 1681 / 4 )

→1287 / 2

→643.5 cm²