- 60.0 feet^2 .The simplest way to find the area of polygon
BCDEFGHI is to find the area of rectangle BGHI: 10 ft. ×7 ft. =
70 sq. ft. Subtract the area of rectangular void CFED: 5 ft. ×2 ft. =
10 sq. ft. Then subtract the two areas. 70 sq. ft. – 10 sq. ft. = 60
square feet. - 70 feet^2 .Again, the simplest way to the find the area of polygon
MNOPQR is to find the area of trapezoid MPQR. ^12 ×8 feet
(4 ft. + 15 ft.) = ^12 ×8(19) = 76 sq. ft. Subtract the area of ΔNPO:
^12 ×3 ft. ×4 ft. = 6 sq. ft. 76 sq. ft. – 6 sq ft. = 70 square feet.
Set 66
- DE= 5 feet.To find DE, use the given area of hexagon
HCDEFG and work backwards. The area of a regular polygon
equals half the product of its perimeter by its apothem: 45 sq. ft. =
^12 p×3 ft.; p= 30 ft. The perimeter of a regular polygon equals the
length of each side multiplied by the number of sides: 30 ft. =
sft. ×6.; s= 5 ft. - 6 feet^2 .ΔACH is an isosceles triangle. A line drawn from its vertex
to ACbisects the line segment, which means mAI= mCI, or ^12 of 8
feet long. Since question 316 found the measurement of HC, only
the measurement of HIremains unknown. Plug the given
measurements for ΔCHI into the Pythagorean theorem. 4^2 + b^2 = - 16 + b^2 = 25. b^2 = 9. b= 3. Once the height is established, find
the area of ΔCHI: ^12 ×4 ft. ×3 ft. = 6 feet^2. - 24 feet^2. Using the Pythagorean theorem with ΔABC shows that
the height of the triangle is 6: 10^2 = 8^2 + 6^2. Therefore, A= ^12 bh=
^12 (8)(6) =
24 feet^2. - 81 square feet.The areas within the entire figure are the sum of
its parts: 24 sq. ft. + 6 sq. ft. + 6 sq. ft. + 45 sq. ft. = 81 square feet.
501 Geometry Questions