AREAS RELATED

TO CIRCLE

• Sum based on finding

Area of shaded region

ar(semicircle with

diameter PR)

+ ar(semicircle with

diameter PQ)

– ar(semicircle with

diameter RQ)

Q. In the adjoining figure,

PR = 6 units, PQ = 8 units.

Semicircles are drawn taking

sides PR, RQ and PQ as diameters.

Find the area of the shaded portion.

Diameter PR = 6 units

• Its radius (r₁) = 3 units

Diameter PQ = 8 units

• Its radius (r₂) = 4 units

In ΔPRQ,

∠RPQ = 90^o

​

∴ RQ² = PR² + PQ²

[Pythagoras theorem]

Sol.

∴ RQ²

= (6)²

+ (8)²

∴ RQ²

= 36

+ 64

∴ RQ²

[Angle inscribed

in a semicircle]

= 100

• RQ =10 units

[Taking square roots]

Diameter RQ = 10 units

∴ Its radius (r₃) = 5 units

× 6

× 8

24 sq. units

6

8

r₁ = 3

r₂ = 4

r₃ = 5

10

∴ A(ΔPRQ) =

∴ A(ΔPRQ) =

A(ΔPRQ) =

What is formula for finding Area of triangle ?

ar(shaded region) =

+ ar(ΔPRQ)

​

​

r₁ = 3, r₂ = 4, r₃ = 5

ar(ΔPRQ) = 24 sq. units

Q. Find the area of the shaded portion.

∴ Area of the shaded region is 24 sq. units

Area of the shaded region

=

Sol.

ar(semicircle with diameter PR)

+ ar(semicircle with diameter PQ)

– ar(semicircle with diameter RQ)

=

=

(r₁²

+ r₂²

– r₃²)

=

(3²

+ 4²

– 5²)

=

(9

+ 16

– 25)

= 24

(0)

= 24 sq. units

= 24

(25

– 25)

r₁ = 3

r₂ = 4

r₃ = 5

What is formula for

finding area of semicircle ?

+ ar(ΔPRQ)

+ 24

+ 24

+ 24

+ 24

AREAS RELATED

TO CIRCLE

• Sum based on finding

Area of shaded region

ar(semicircle with

diameter PR)

+ ar(semicircle with

diameter PQ)

– ar(semicircle with

diameter RQ)

Q. In the adjoining figure,

PR = 6 units, PQ = 8 units.

Semicircles are drawn taking

sides PR, RQ and PQ as diameters.

Find the area of the shaded portion.

Diameter PR = 6 units

• Its radius (r₁) = 3 units

Diameter PQ = 8 units

• Its radius (r₂) = 4 units

In ΔPRQ,

∠RPQ = 90^o

​

∴ RQ² = PR² + PQ²

[Pythagoras theorem]

Sol.

∴ RQ²

= (6)²

+ (8)²

∴ RQ²

= 36

+ 64

∴ RQ²

[Angle inscribed

in a semicircle]

= 100

• RQ =10 units

[Taking square roots]

Diameter RQ = 10 units

∴ Its radius (r₃) = 5 units

× 6

× 8

24 sq. units

6

8

r₁ = 3

r₂ = 4

r₃ = 5

10

∴ A(ΔPRQ) =

∴ A(ΔPRQ) =

A(ΔPRQ) =

What is formula for finding Area of triangle ?

ar(shaded region) =

+ ar(ΔPRQ)

​

​

r₁ = 3, r₂ = 4, r₃ = 5

ar(ΔPRQ) = 24 sq. units

Q. Find the area of the shaded portion.

∴ Area of the shaded region is 24 sq. units

Area of the shaded region

=

Sol.

ar(semicircle with diameter PR)

+ ar(semicircle with diameter PQ)

– ar(semicircle with diameter RQ)

=

=

(r₁²

+ r₂²

– r₃²)

=

(3²

+ 4²

– 5²)

=

(9

+ 16

– 25)

= 24

(0)

= 24 sq. units

= 24

(25

– 25)

r₁ = 3

r₂ = 4

r₃ = 5

What is formula for

finding area of semicircle ?

+ ar(ΔPRQ)

+ 24

+ 24

+ 24

+ 24