143 1.5 Applications and Modeling with Quadratic Equations PROBLEM-SOLVING HINT As seen in Example 1, discard any solution that does not satisfy the physical constraints of a problem. The Pythagorean Theorem Example 2 requires the use of the Pythagorean theorem for right triangles. Recall that the legs of a right triangle form the right angle, and the hypotenuse is the side opposite the right angle. Pythagorean Theorem In a right triangle, the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse. a2 +b2 =c2 Leg b Hypotenuse c Leg a 2x + 30 x 2x + 20 x is in meters. Figure 7 EXAMPLE 2 Applying the Pythagorean Theorem A piece of property has the shape of a right triangle. The longer leg is 20 m longer than twice the length of the shorter leg. The hypotenuse is 10 m longer than the length of the longer leg. Find the lengths of the sides of the triangular lot. SOLUTION Step 1 Read the problem. We must find the lengths of the three sides. Step 2 Assign a variable. Let x = the length of the shorter leg (in meters). Then 2x + 20 = the length of the longer leg, and 12x + 202 + 10, or 2x + 30 = the length of the hypotenuse. See Figure 7. Step 3 Write an equation. a2 + b2 = c2 x2 + 12x + 2022 = 12x + 3022 Substitute into the Pythagorean theorem. Step 4 Solve the equation. x2 + 14x2 + 80x + 4002 = 4x2 + 120x + 900 Square the binomials. Remember the middle terms. x2 - 40x - 500 = 0 Standard form 1x - 5021x + 102 = 0 Factor. x - 50 = 0 or x + 10 = 0 Zero-factor property x = 50 or x = -10 Solve each equation. Step 5 State the answer. Because x represents a length, -10 is not reasonable. The lengths of the sides of the triangular lot are 50 m, 21502 + 20 = 120 m, and 21502 + 30 = 130 m. Step 6 Check. The lengths 50, 120, and 130 satisfy the words of the problem and also satisfy the Pythagorean theorem. S Now Try Exercise 35. The hypotenuse is c.
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