Set 67
- 22 feet.The area of trapezoid RMNO plus the area of trapezoid
RQPO equals the area of figure RMNOPQ. Since trapezoids
RMNO and trapezoid RQPO are congruent, their areas are equal:
^12 (320 sq. ft.) = 160 sq. ft. The congruent height of each trapezoid
is known, and one congruent base length is known (MAΜΝ).
Using the equation to find the area of a trapezoid, create the
equation: 160 sq. ft. = ^12 (10 ft.)(10 ft. + x). 160 sq. ft. = 50 sq. ft. +
5 x ft. 110 sq. ft. = 5x ft. 22 feet = x. - 10 2 feet.Work backwards using the given area of ΔRMA: 50 sq. ft.
= ^12 b(10 ft.). 50 sq. ft. = 5 ft. ×b. 10 ft. = b. Once the base and
height of ΔRMA are established, use the Pythagorean theorem to
find RM: 10^2 + 10^2 = c^2. 100 + 100 = c^2. 200 = c^2. 10 2 = c.
RM= 10 2 feet. - 2 26 feet.Imagine a perpendicular line from vertex N to the
base of trapezoid RMNO. This imaginary line divides ROinto
another 10-foot segment. The remaining portion of line ROis 2
feet long. Use the Pythagorean theorem to find the length of NO:
(10 ft.)^2 + (2 ft.)^2 = z^2. 100 sq. ft. + 4 sq. ft. = z^2. 104 sq. ft. = z^2.
2 26 feet = z.
501 Geometry Questions