Question

pLeAse AnSwEr ThiS QuEsTiOn.........❣️

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Answer 1

Question :–

ABC is right angle triangle with sides AB = 8 cm, BC = 8 cm and angle B = 90°. Two semi circles are drawn with BC and AC as diameters. A structure of circle of radius 4 cm is drawn with B as centre as shown in the figure. Find the area of shaded region.

Answer –

Let us Find the shaded area one by one.

Let's firstly fund the hypotenuse of the right angle triangle with base and altitude as 8 cm.

We know that, Base² + Altitude ² = Hypotenuse ².

➸ 8² + 8² = Hypotenuse²

➸ 64 + 64 = Hypotenuse²

➸ Hypotenuse² = 128

➸ Hypotenuse = √128

➸ H = 11.38 cm

Hence, radius of the semicircle formed on hypotenuse of the right angled triangle is : 11.38/2 = 5.69 cm.

Hence, the area of the shaded semicircle formed on hypotenuse of the right angled triangle = πr²/4.

➸ π (5.69)²/2

➸ 3.14 * 32.3761/2

➸ 101.66/2

➸ 50.83 cm²

Now, we will find the area of the semicircle formed on the basis of the right angle triangle.

The radius of the semi circle is 4 cm.

➸ π r²/2

➸ 3.14 * (4)²/2

➸ 3.14 * 16/2

➸ 50.24/2

➸ 25.12 cm²

Now, we will find the area of the sector formed inside the right angle triangle.

The radius of the sector is 4 cm.

Also, the angle of the sector given in the question is 90°, hence, it is quarter of the circle.

Hence, the area = πr²/4.

➸ 3.14 * (4)²/4

➸ 3.14 * 4

➸ 12.56 cm²

Hence, Total shaded area = 50.83 cm² + 25.12 cm² + 12.56 cm².

➸ 75.95 + 12.56

➸ 88.51 cm²

Hence, 88.51 cm² is the total shaded area.

Answer 2

GIVEN:

• ABC is a right ∆
• AB = 6 cm
• BP = PC
• BC = 8 cm
• Angle B = 90°
• Two semicircle's are drawn with BC & AC

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First finding the sector:

• Radius ( r ) = 4cm
• Angle B = 90°

Area of sector = πr²θ/360°

Where

θ = angle B = 90°

r = 4cm

π = 3.14

Substituting we get

→ (3.14)(4)²(90)/360

→ 3.14(4)

→ 12.56

Hence area of sector ≈ 12.56 cm²

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Finding area of semicircle which has diameter AC

Finding AC

Using Pythagoras theorem

(Hypotenuse)² = (Base)² + (height)²

→ AC² = AB² + BC²

→ AC² = 8² + 8²

→ AC² = 64 + 64

→ AC² = 128

→ AC = 11.38

Now ,

Diameter = AC = 11.38

Radius (r) = diameter/2

r = 11.38/2

r = 5.69 cm

Area of semicircle = πr²/2

→ Area of semicircle = (3.14)(5.69)²/2

→ Area of semicircle = 50.8 cm²

Hence area of semicircle which has diameter AC is = 50.8 cm²

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Finding area of semicircle which has Diameter BC

→ Area of semicircle = πr²/2

→ Area of semicircle = (3.14)(4)²/2

→ Area of semicircle = 50.24/2

→ Area of semicircle = 25.12 cm²

Hence area of semicircle= 25.12cm²

_______________________________

Finding the area of shaded region:

Area of shaded region = Ar(sector) + Ar(semicircle) + Ar(semicircle)

→ Area of shaded region = 12.56 + 50.8 + 25.12

→ Area of shaded region = 88.48 cm²

Hence area of shaded region = 88.48 cm²