Liza is designing a calendar as a fund-raising project for Math Class. The cost of printing
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Answer:
The lower supports are <u>Congruent Rectangles</u> and the area of the two supports is<u> 24 square meters.</u>
The upper arch can be decomposed as one semicircle with radius<u> 6 </u>meters minus a semicircle with radius 3 meters.
The area of the archway is <u>(13.5 π</u> + 24) square meters.
Step-by-step explanation:
See attachment for the firgure,
In order to determine the area of archway = The area of upper support + area of lower support.
As given, the lower support are two congruent rectangles consists of dimension 3 m × 4 m
Therefore, the area of lower support can be written as,
The area of lower support = 3 × 4 + 3 × 4 = 12 + 12 = 24 square m.
Now, the upper support arch can be decomposed as the two concentric semi circles having radius 3 m and 6 m,
Hence, the area of the upper support = Area of semi circle having radius i.e 6 m - Area of semi circle having radius i.e 3 m
=> π6²/2 - π3²/2= 27π/2= 13.5π square m
Therefore, the area of the archway = (13.5 π + 24) square meters
