Answer:
Hey
So my brother posted this on Yahoo
Draw a line from the center of the circle to one of the ends of the chord (water surface) and another to the point at greatest depth. A right-angled triangle is formed. Length of side to the water-surface is 5 cm, the hypot is 7 cm.
What you do now is the following:
Calculate the angle θ in the corner of the right-angled triangle by: cos θ = 5/7 ⇒ θ = cos ˉ¹ (5/7)
So θ is approx 44.4°, so the angle subtended at the center of the circle by the water surface is roughly 88.8°
The area shaded will then be the area of the sector minus the area of the triangle above the water in your diagram.
Shaded area ≃ 88.8/360*area of circle - ½*7*7*sin88.8°
= 88.8/360*π*7² - 24.5*sin 88.8°
≃ 13.5 cm²
(using area of ∆ = ½.a.b.sin C for the triangle)
b)
volume of water = cross-sectional area * length
≃ 13.5 * 30 cm³
≃ 404 cm³
Hoped it Helped
Explanation:
Hey
So my brother posted this on Yahoo
Draw a line from the center of the circle to one of the ends of the chord (water surface) and another to the point at greatest depth. A right-angled triangle is formed. Length of side to the water-surface is 5 cm, the hypot is 7 cm.
What you do now is the following:
Calculate the angle θ in the corner of the right-angled triangle by: cos θ = 5/7 ⇒ θ = cos ˉ¹ (5/7)
So θ is approx 44.4°, so the angle subtended at the center of the circle by the water surface is roughly 88.8°
The area shaded will then be the area of the sector minus the area of the triangle above the water in your diagram.
Shaded area ≃ 88.8/360*area of circle - ½*7*7*sin88.8°
= 88.8/360*π*7² - 24.5*sin 88.8°
≃ 13.5 cm²
(using area of ∆ = ½.a.b.sin C for the triangle)
b)
volume of water = cross-sectional area * length
≃ 13.5 * 30 cm³
≃ 404 cm³
Hoped it Helped