Question

adjacent figure is triangular in shape and is semi circular at the bottom .find the total area of this figure

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Answer 1

Given :

• OB = 10m
• AB = AC = 6m
• CB = 12m

To find :

Area of given figure

Formula used :

• Pythagoras theorem ➝ (Perpendicular)² + (Base)² = (Hypoteneus)²
• Area of triangle = (1/2)×Base × Height
• Area of semi-circle = (1/2)πr²

Solution :

Total area of given figure = Area of triangle + Area of semi-circle

First of all we need to find, h that is OA

Using Pythagoras theorem in ∆OAB

➝ (OA)² + (AB)² = (OB)²

➝ (OA)² + (6)² = (10)²

➝ (OA)² + 36 = 100

➝ (OA)² = 100 - 36

➝ (OA)² = 64

➝ (OA) = √64

➝ (OA) = ±8

{ As OA is side , it's value can't be negative }

➝ (OA) = +8 m

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Area of triangle ∆OAB = (1/2)×(OA)×(CB)

Area of triangle ∆OAB = (1/2) × (8m) × (12m)

Area of triangle ∆OAB = 48 m²

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Area of semi-circle = (1/2)×π×(6m)²

Area of semi-circle = (1/2) × 3.14 × 36 m²

Area of semi-circle = 56.52 m²

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Total area of figure = 48m² + 56.52m²

Total area of figure = 104.52 m²

ANSWER : 104.52 m²

Answer 2

Answer:

Total area=area of triangle+area of semicircle

Area of triangle=(b x h)/2

h2=L2-r2

h2=100-36

h2=64

h=8

Area of triangle=(8x6)/2

Area=24m2

Area of circle=(πr2)/2

Area. =(3.14x36)/2

Area. =56.52

Total area=24+56.52

=80.52m2