 # EXAMPLE 3 Use a ratio of areas Cooking SOLUTION First draw a diagram to represent the problem. Label dimensions and areas. Then use Theorem 11.7. If the.

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EXAMPLE 3 Use a ratio of areas Cooking SOLUTION First draw a diagram to represent the problem. Label dimensions and areas. Then use Theorem 11.7. If the area ratio is a 2 : b 2, then the length ratio is a:b. A large rectangular baking pan is 15 inches long and 10 inches wide. A smaller pan is similar to the large pan. The area of the smaller pan is 96 square inches. Find the width of the smaller pan.

EXAMPLE 3 Use a ratio of areas Area of smaller pan Area of large pan = 150 96 = 25 16 Write ratio of known areas. Then simplify. = 4 5 Find square root of area ratio. Length in smaller pan Length in large pan ANSWER Any length in the smaller pan is, or 0.8, of the corresponding length in the large pan. So, the width of the smaller pan is 0.8(10 inches ) 8 inches. 4 5 =

EXAMPLE 4 Solve a multi-step problem The floor of the gazebo shown is a regular octagon. Each side of the floor is 8 feet, and the area is about 309 square feet. You build a small model gazebo in the shape of a regular octagon. The perimeter of the floor of the model gazebo is 24 inches. Find the area of the floor of the model gazebo to the nearest tenth of a square inch. Gazebo

EXAMPLE 4 SOLUTION All regular octagons are similar, so the floor of the model is similar to the floor of the full-sized gazebo. STEP 1 Find the ratio of the lengths of the two floors by finding the ratio of the perimeters. Use the same units for both lengths in the ratio. Perimeter of full-sized Perimeter of model = 8(8 ft ) 24 in. = 64 ft 2 ft = 32 1 So, the ratio of corresponding lengths (full-sized to model) is 32:1. Solve a multi-step problem

EXAMPLE 4 Solve a multi-step problem STEP 2 (length of full-sized) 2 (length of model) 2 Calculate the area of the model gazebo’s floor. Let x be this area. Area of full-sized Area of model = 309 ft 2 x ft 2 32 2 1212 = Theorem 11.7 Substitute. 1024x 309= Cross Products Property x ≈ 0.302 ft 2 Solve for x.

EXAMPLE 4 Solve a multi-step problem STEP 3 Convert the area to square inches. 0.302 ft 2 144 in. 2 1 ft. 2 ≈ 43. 5 in. 2 The area of the floor of the model gazebo is about 43.5 square inches. ANSWER

GUIDED PRACTICE for Examples 3 and 4 2. The ratio of the areas of two regular decagons is 20:36. What is the ratio of their corresponding side lengths in simplest radical form? SOLUTION Then use Theorem 11.7. If the area ratio is a 2 : b 2, then the length ratio is a:b. Area of smaller decagon Area of large decagon = 9 5 Write ratio of known areas. Then simplify. Find square root of area as ratios. Length in smaller decagon Length in large decagon = 36 20 = 5 3 √

GUIDED PRACTICE for Examples 3 and 4 3. Rectangles I and II are similar. The perimeter of Rectangle I is 66 inches. Rectangle II is 35 feet long and 20 feet wide. Show the steps you would use to find the ratio of the areas and then find the area of Rectangle I. Perimeter of Rectangle I = 66 inches SOLUTION Perimeter of Rectangle II = 2(l + b) = 2(35 + 20) = 2 35 + 2 20 = 110 feet.

GUIDED PRACTICE for Examples 3 and 4 STEP 1 Find the ratio of the lengths of the two rectangles by finding the ratio of the perimeters. = 66 1320 = 1 20 Perimeter of Rectangle I Perimeter of Rectangle II So, the ratio of the corresponding lengths to is 1:20. Convert 110 feet to inches. 110 ft= 110 12 = 1320 inches.

GUIDED PRACTICE for Examples 3 and 4 STEP 2 STEP 3 Find the area of Rectangle II. Area of Rectangle II = lb = 35(20) = 700 feet 2 Convert 700 feet 2 to inches 2. 700 ft 2 = 700 144 = 100,800 inches 2. Find the ratio of their Areas (sides of Rectangle I ) 2 (sides of Rectangle II ) 2 = Area of Rectangle I Area of Rectangle II 1 20 2 = 1 400

GUIDED PRACTICE for Examples 3 and 4 STEP 4 Find the area of Rectangle I. Area of Rectangle I = Area of Rectangle II · (ratio of areas) = 252 in. 2 Area of Rectangle I = 100,800 · 1 400

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