Solving Applications 5.8 Applications The Pythagorean Theorem.

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Solving Applications 5.8 Applications The Pythagorean Theorem

The Mitchell’s are designing a garden The Mitchell’s are designing a garden. The garden will be in the shape of a rectangle and have an area of 270 square feet. The width of the garden is 3 feet less than the length. Find the length and width. Solution 1. Familiarize. We first make a drawing. Recall that the area of a rectangle is Length  Width. We let x = the length, in feet. The width is then x  3. x x  3

2. Translate. Rewording: The area of the rectangle is 270 ft2. Translating: x(x  3) = 270 3. Carry out. We solve the equation. x(x  3) = 270 x2  3x = 270 x2  3x  270 = 0 (x  18)(x + 15) = 0 x  18 = 0 or x + 15 = 0 x = 18 or x = 15

4. Check. The solutions of the equation are 18 and 15 4. Check. The solutions of the equation are 18 and 15. Since the length must be positive, 15 cannot be a solution. To check 18, we note that if the length is 18, then the width is x  3 or 15 and the area is 18 ft  15 ft = 270 ft2. Thus the solution checks. 5. State. The garden is 18 feet long and 15 feet wide.

The math club is designing a brochure The math club is designing a brochure. The design calls for a triangle to be placed on the front. The triangle has a base that is 6 centimeters less than the height. If the area of the triangle is 216 cm2. Find the height and base. Solution 1. Familiarize. We first make a drawing. The formula for the area of a triangle is A = ½ (base)(height). We let h = the height, in cm, and the base = h  6, in cm. h h  6

2. Translate. Rewording: The area of the triangle is 216 cm2. Translating: = 216 3. Carry out. We solve the equation.

3. Carry out. h  24 = 0 or h + 18 = 0 h = 24 or h = 18 4. Check 3. Carry out. h  24 = 0 or h + 18 = 0 h = 24 or h = 18 4. Check. The height must be positive, so 18 cannot be a solution. Suppose the height is 24 cm. The base would be 24  6, or 18 cm, and the area ½(24)(18), or 216 cm2. These numbers check in the original problem. 5. State. The height of the triangle would be 24 cm and the base would be 18 cm.

The Pythagorean Theorem In any right triangle, if a and b are the lengths of the legs and c is the length of the hypotenuse, then a2 + b2 = c2 or (Leg)2 + (Other leg)2 = (Hypotenuse)2. The equation a2 + b2 = c2 is called the Pythagorean equation.* *The converse of the Pythagorean theorem is also true. That is, if a2 + b2 = c2, then the triangle is a right triangle. c b a

A 13-ft ladder is leaning against a house A 13-ft ladder is leaning against a house. The distance from the bottom of the ladder to the house is 7 ft less than the distance from the top of the ladder to the ground. How far is the bottom of the ladder from the house? Solution 1. Familiarize. We first make a drawing. The ladder and the missing dimensions form a right triangle. x = distance from top of the ladder to the ground x  7 = distance from bottom ladder to house. The hypotenuse has length 13 ft. 13 ft x x  7

2. Translate. Since a right triangle is formed, we can use the Pythagorean theorem: 3. Carry out. We solve the equation.

3. Carry out. x  12 = 0 or x + 5 = 0 x = 12 or x = 5 4. Check 3. Carry out. x  12 = 0 or x + 5 = 0 x = 12 or x = 5 4. Check. The integer 5 cannot be a length of a side because it is negative. When x = 12, x  7 = 5, and 122 + 52 = 132. So 12 checks. 5. State. The distance from the bottom of the ladder to the house is 5 ft. The distance from the top of the ladder to the ground is 12 ft.