EXAMPLE 1 Find the perimeter and area of a rectangle SOLUTION Basketball Find the perimeter and area of the rectangular basketball court shown. PerimeterArea 2l + 2w = PA = lw 2(84) + 2(50) = = 84(50) 268 = = 4200 The perimeter is 268 feet and the area is 4200 square feet. ANSWER
EXAMPLE 2 Find the circumference and area of a circle SOLUTION Team Patch You are ordering circular cloth patches for your soccer team’s uniforms. Find the approximate circumference and area of the patch shown. Use 3.14 to approximate the value of π. First find the radius. The diameter is 9 centimeters, so the radius is (9) = 4.5 centimeters. 1 2 Then find the circumference and area.
EXAMPLE 2 Find the circumference and area of a circle = = Aπr (4.5) 2 = C = 2πr2(3.14) (4.5)28.26 The circumference is about 28.3 cm 2. The area is about 63.6 cm. ANSWER
GUIDED PRACTICE for Examples 1 and 2 Find the area and perimeter (or circumference) of the figure. If necessary, round to the nearest tenth. lw = A = A 74.1 = A Write area of a rectangle. Substitute values. Multiply.
GUIDED PRACTICE for Examples 1 and 2 p = 2l + 2w p = p = 37.4 p = Write perimeter of a rectangle. Substitute values. Multiply. Add. The area is 74.1m 2 and the perimeter is 37.4m. ANSWER
GUIDED PRACTICE for Examples 1 and 2 s2s2 = A = A 2.56 = A Write area of a square. Substitute values. Simplify. 2.6
GUIDED PRACTICE for Examples 1 and 2 p = 4s p = 6.4 Write perimeter. Substitute value. Simplify. p = The area is 2.6cm 2 and the perimeter is 6.4cm. ANSWER
GUIDED PRACTICE for Examples 1 and (2) 2 = A = A Write area of a circle. Subtract values. Simplify. r 2 = A
GUIDED PRACTICE for Examples 1 and 2 c = Write perimeter of a circle. Substitute value. Simplify. c = r 2 c = The area is 12.6yd 2 and the perimeter is 12.6yd. ANSWER