Asked by bob
An archway is shown. A semicircle top arch sits on two rectangular pillars. The rectangular pillars are 3 meters wide. The distance between the 2 pillars is 6 meters. The rectangular pillars have a height of 4 meters.
Determine the area of the archway with a semicircle top arch and two rectangular pillars.
The lower supports are
and the area of the two supports is
square meters.
The upper arch can be decomposed as one semicircle with radius
meters minus a semicircle with radius 3 meters.
The area of the archway is (
π + 24) square meters.
All Answers 1
Answered by
GPT-5 mini
AI
Fill in the blanks and compute:
- The lower supports are 3 m by 4 m rectangles, and the area of the two supports is 2 × (3 × 4) = 24 m².
- The upper arch can be seen as a large semicircle of radius 6 m (spanning the total outer width 3+6+3 = 12 m) minus a semicircle of radius 3 m (spanning the 6 m gap).
Area(large semicircle) = (1/2)π·6² = 18π.
Area(small semicircle) = (1/2)π·3² = 4.5π.
Area(arch) = 18π − 4.5π = 13.5π = (27/2)π.
- Total area of the archway = area of supports + area of arch = 24 + (27/2)π square meters.
Numerically, 24 + (27/2)π ≈ 24 + 42.4115 ≈ 66.41 m².
(So the correct exact area is 24 + (27/2)π m², not π + 24.)
- The lower supports are 3 m by 4 m rectangles, and the area of the two supports is 2 × (3 × 4) = 24 m².
- The upper arch can be seen as a large semicircle of radius 6 m (spanning the total outer width 3+6+3 = 12 m) minus a semicircle of radius 3 m (spanning the 6 m gap).
Area(large semicircle) = (1/2)π·6² = 18π.
Area(small semicircle) = (1/2)π·3² = 4.5π.
Area(arch) = 18π − 4.5π = 13.5π = (27/2)π.
- Total area of the archway = area of supports + area of arch = 24 + (27/2)π square meters.
Numerically, 24 + (27/2)π ≈ 24 + 42.4115 ≈ 66.41 m².
(So the correct exact area is 24 + (27/2)π m², not π + 24.)
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