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Learn how to find the perimeter and area of a rectangular basketball court. Example problems and step-by-step solutions provided.
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Basketball Find the perimeter and area of the rectangular basketball court shown. Perimeter Area P A lw 2l + 2w = = 84(50) 2(84) + 2(50) = = 268 4200 = = The perimeter is 268feet and the area is 4200 square feet. ANSWER EXAMPLE 1 Find the perimeter and area of a rectangle 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. 1 2 First find the radius. The diameter is 9 centimeters, so the radius is (9) = 4.5centimeters. EXAMPLE 2 Find the circumference and area of a circle SOLUTION Then find the circumference and area. Use 3.14 to approximate the value of π.
C 2πr 2(3.14) (4.5) 28.26 = = A πr 3.14(4.5) 63.585 = = 2 2 The circumference is about 28.3cm2. The area is about 63.6cm . ANSWER EXAMPLE 2 Find the circumference and area of a circle
A lw = A 13 5.7 = A 74.1 = for Examples 1 and 2 GUIDED PRACTICE Find the area and perimeter (or circumference) of the figure. If necessary, round to the nearest tenth. Write area of a rectangle. Substitute values. Multiply.
p =213 + 2 5.7 The area is 74.1m2and the perimeter is 37.4m. ANSWER for Examples 1 and 2 GUIDED PRACTICE p = 2l + 2w Write perimeter of a rectangle. Substitute values. p = 26 + 11.4 Multiply. p = 37.4 Add.
A 2.6 A s2 = 1.62 = A 2.56 = for Examples 1 and 2 GUIDED PRACTICE Write area of a square. Substitute values. Simplify.
The area is 2.6cm2and the perimeter is 6.4cm. p = 4 1.6 ANSWER for Examples 1 and 2 GUIDED PRACTICE p =4s Write perimeter. Substitute value. p = 6.4 Simplify.
r2 A = A 3.14 (2)2 = A 12.56 = for Examples 1 and 2 GUIDED PRACTICE Write area of a circle. Subtract values. Simplify.
2 r c = The area is 12.6yd2and the perimeter is 12.6yd. c = 2 3.14 2 ANSWER for Examples 1 and 2 GUIDED PRACTICE Write perimeter of a circle. Substitute value. Simplify. c = 12.56