CH 10

Ch 10 Study Guide A

Solve.

1. Identify the base and height of the triangle.

base = __________

height = __________

2. Find the area of the triangle.

3. Find the area of the trapezoid.

4. Find the area of the parallelogram.

5. The area of the regular hexagon is 240 square meters. Find the area of the shaded region.

6. The area of the trapezoid is 72 square centimeters. Find the height of the trapezoid.

7. Figure ABCD is a parallelogram. Find the length of AB.

8. Figure PQRS is a trapezoid. The area of triangle RST is 150 square centimeters. Find the total area of the shaded regions.

9. Mike drew a regular octagon with side lengths of 10 inches. He divided the octagon into two identical trapezoids and a rectangle. He measured the height of one of the trapezoids to be 6 inches, and the width of the rectangle to be 24 inches. Find the area of the octagon.

ANSWERS:

1. Base = CB

height = AS

2. A = 1/2ab =1/2 x 5 x 8 = 1/2 x 40 = 20 sq cm

3. A = h x 1/2(b₁ + b₂)

= 6 x 1/2 (4 +9)

= 6 x 1/2 (13)

= 6 x 6.5

= 39 sq ft.

4. A = bxh = 15 x 8 = 120 sq. in.

5. Each triangle is 240/6 = 40 sq.m. The shaded part is 1/2 a triangle so it's area is 40/2 = 20 sq.m.

6. 72 = h x 1/2( 9 + 15)

72 = h x 1/2(24)

72 = h x 12

12 12

6 = h or h = 6

7. Triangle BDC area = 1/2 x 14 x 12 = 84 sq.in.

Area of Parallelogram = 2 triangles = 84 + 84 = 168 sq.in.

168 = h x b

168 = 10 x b

b = 168/10 = 16.8 inches

8. Step 1: Triangle RST = 150

150 = 1/2 x20 x h

150 = 10 x h

h = 15

Step 2: Area of trapezoid = h x 1/2(b1 + b2)

= 15 x 1/2(32 + 20)

= 15 x 1/2(52)

= 15 x 26

= 390 sq. in.

Step 3: Area of shaded triangles

= all - small

= 390 - 150

= 240 sq.in.

9. Area Rectangle = 10 x 24 = 240 sq.in.

Area top trapezoid = h x 1/2(b1+b2)= 6 x 1/2(10+24)

= 6 x 17 = 102 sq. in

Area bottom trapezoid (same) = 102 sq. in.

Total area = 102 + 102 = 240 = 444 sq.in.