Question

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

Answer 1

Answer:

 The lower supports are Congruent Rectangles and the area of the two supports is 24 square meters.

The upper arch can be decomposed as one semicircle with radius 6 meters minus a semicircle with radius 3 meters.

The area of the archway is (13.5 π + 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