YOU TRY 3 Find the length of the missing side. 4 5 EXAMPLE 4 Sketching a picture and labeling it is very useful in visualizing the application problem. This may make it easier to set up the equation. Solve. An animal holding pen situated between two buildings at a right angle with each other will have walls as two of its sides and a fence on the longest side. The side with the fence is 20 ft longer than the shortest side, while the third side is 10 ft longer than the shortest side. Find the length of the fence. Solution Step 1: Read the problem carefully, and identify what we are being asked to fi nd. Draw a picture. We must fi nd the length of the fence. Step 2: Choose a variable to represent an unknown, and defi ne the other unknowns in terms of this variable. Draw and label the picture. x length of the shortest side (a leg) x 10 length of the side along other building (a leg) x 20 length of the fence (hypotenuse) x 20 x 10 x Step 3: Translate the information that appears in English into an algebraic equation. We will use the Pythagorean theorem. a2 b2 c2 Pythagorean theorem x2 (x 10)2 (x 20)2 Substitute. Step 4: Solve the equation. x2 (x 10)2 (x 20)2 x2 x2 20x 100 x2 40x 400 Multiply using FOIL. 2x2 20x 100 x2 40x 400 x2 20x 300 0 Write in standard form. (x 30)(x 10) 0 Factor. b R x 30 0 or x 10 0 Set each factor equal to 0. x 30 or x 10 Solve. Step 5: Check the answer, and interpret the solution as it relates to the problem. The length of the shortest side, x, cannot be a negative number, so x cannot equal 10. Therefore, the length of the shortest side must be 30 ft. The length of the side along the other building is x 10, so 30 10 40 ft. The length of the fence is x 20, so 30 20 50 ft. Do these lengths satisfy the Pythagorean theorem? Yes. a2 b2 c2 (30)2 (40)2 (50)2 900 1600 2500 ✓ Therefore, the length of the fence is 50 ft. 406 CHAPTER 7 Factoring Polynomials www.mhhe.com/messersmith
messersmith_power_intermediate_algebra_1e_ch4_7_10
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