In the following figure, shows the cross-section of railway tunnel. The radius OA of the circular part is 2 m. If ∠AOB = 90°, calculate:
(i)the height of the tunnel
(ii)the perimeter of the cross-section
(iii)the area of the cross-section.
figure
Answers
Answer:
The height of the tunnel is (2 + √2)m , Perimeter of cross-section is (3π + 2√2) m and Area of cross section = (3π + 2)m.
Step-by-step explanation:
Given :
Radius OA of the circular part = 2 m
∠AOB = 90°
Let OM ⊥ AB.
(i)
In ∆OAB ,
AB² = OA² + OB²
AB² = 2² + 2²
AB² = 4 + 4
AB² = 8
AB = √8
AB = √4×2
AB = 2√2 cm
Let the height of the tunnel to be h.
Area of ∆ OAB = ½ × Base × height
Area of ∆ OAB = ½ × OA × OB
½ × 2 × 2
Area of ∆ OAB = 2
½ × AB × OM = 2
½ × 2√2 × OM = 2
√2 OM = 2
OM = 2/√2
OM = 2 × √2 / √2×√2
[By rationalising the denominator]
OM = 2√2/2
OM = √2 m
Height of the tunnel ,h = OM + Radius
h = √2 + 2
h = (2 + √2)m
Height of the tunnel = (2 + √2)m
(ii) Central angle of major arc ,θ = 360° - 90° = 270°
Perimeter of cross-section , P = length of the major Arc AB + AB
P = θ/360° × 2πr + 2√2
P = 270°/360° × 2π × 2 + 2√2
P = ¾ × 4π + 2√2
P = (3π + 2√2) m
Perimeter of cross-section = (3π + 2√2) m
(iii) Area of cross section,A = θ/360° × Area of circle + area of ∆AOB
A = θ/360° × πr² + ½ × base × height
A = 270°/360° × π× 2² + ½ × 2 × 2
A = ¾ × π × 4 + 2
A = (3π + 2)m
Area of cross section = (3π + 2)m
Hence, the height of the tunnel is (2 + √2)m , Perimeter of cross-section is (3π + 2√2) m and Area of cross section = (3π + 2)m.
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