Exercise 62 Page 885

We are told that the perimeter of a right triangle is 36 inches. Twice the length of the longer leg minus twice the length of the shorter leg is 6 inches. Let's plot a right triangle. We will call the sides x, y, and z.

Now, let's write the equations in terms of x, y, and z. Recall that the perimeter is the sum of all side lengths. x+y+z=36 & (I) 2x-2y=6 & (II) Since we have 3 variables, we will need 3 equations to find the lengths of all sides. Recall the Pythagorean Theorem. It states that the sum of the squares of the leg lengths equals the square of the length of the hypotenuse. Let's use it to write the third equation. x+y+z=36 & (I) 2x-2y=6 & (II) x^2+y^2=z^2 & (III) Now, we will solve the system of equations. Since in equation II there are only 2 variables, we will isolate one of them in this equation. Then we will substitute the obtained expression into equations I and III.

x+y+z=36 & (I) 2x-2y=6 & (II) x^2+y^2=z^2 & (III)

AddEqn

(II): LHS+2y=RHS+2y

x+y+z=36 2x=2y+6 x^2+y^2=z^2

DivEqn

.LHS /2.=.RHS /2.

x+y+z=36 x=y+3 x^2+y^2=z^2

(I), (III): x= y+3

y+3+y+z=36 x=y+3 ( y+3)^2+y^2=z^2

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(I), (III): Simplify

AddTerms

(I):Add terms

2y+3+z=36 x=y+3 (y+3)^2+y^2=z^2

SubEqn

(I): LHS-3=RHS-3

2y+z=33 x=y+3 (y+3)^2+y^2=z^2

ExpandPosPerfectSquare

(III): (a+b)^2=a^2+2ab+b^2

2y+z=33 x=y+3 y^2+6y+9+y^2=z^2

AddTerms

(III): Add terms

2y+z=33 x=y+3 2y^2+6y+9=z^2

Now we will do the same process as before. We will isolate z in the first equation and then substitute the obtained expression into the third equation.

2y+z=33 x=y+3 2y^2+6y+9=z^2

SubEqn

(I): LHS-2y=RHS-2y

z=33-2y x=y+3 2y^2+6y+9=z^2

Substitute

(III): z= 33-2y

z=33-2y x=y+3 2y^2+6y+9=( 33-2y)^2

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(III): Simplify

ExpandNegPerfectSquare

(III): (a-b)^2=a^2-2ab+b^2

z=33-2y x=y+3 2y^2+6y+9=1089-132y+4y^2

SubEqn

(III): LHS-4y^2=RHS-4y^2

z=33-2y x=y+3 -2y^2+6y+9=1089-132y

AddEqn

(III): LHS+132y=RHS+132y

z=33-2y x=y+3 -2y^2+138y+9=1089

SubEqn

(III): LHS-1089=RHS-1089

z=33-2y x=y+3 -2y^2+138y-1080=0

DivEqn

(III): .LHS /(-2).=.RHS /(-2).

z=33-2y x=y+3 y^2-69y+540=0

We have obtained an equation with only one variable, so we can solve it. Note that it is a quadratic equation. We will use the Quadratic Formula to solve this equation. Let's start by identifying the values of a, b, and c. y^2-69y+540=0 ⇕ 1y^2+( -69)y+ 540=0 We have that a= 1, b= -69, and c= 540. Let's substitute these values into the Quadratic Formula and simplify.

y=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

y=- ( -69)±sqrt(( -69)^2-4( 1)( 540))/2( 1)

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Simplify

NegNeg

- (- a)=a

y=69±sqrt((-69)^2-4(1)(540))/2(1)

NegBaseToPosBase

(- a)^2=a^2

y=69±sqrt(4761-4(1)(540))/2(1)

Multiply

Multiply

y=69±sqrt(4761-2160)/2

SubTerm

Subtract term

y=69±sqrt(2601)/2

CalcRoot

Calculate root

y=69± 51/2

Notice that if we simplify this we will obtain 2 different solutions: one considering the plus sign, and one considering the minus sign.

y_1=69+51/2	y_2=69-51/2
y_1=120/2	y_2=18/2
y_1=60	y_2=9

We have obtained two different solutions and both are possible. Start by considering y_1= 60. Let's substitute this value into equations I and II to find x and z.

z=33-2y x=y+3 y=60

(I), (II): y= 60

z=33-2( 60) x= 60+3 y=60

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Simplify

Multiply

(I): Multiply

z=33-120 x=60+3 y=60

(I), (II): Add and subtract terms

z=-87 x=63 y=60

Note that since a negative side length does not make sense, this option does not work in our case. Let's now try substituting y_2= 9.

z=33-2y x=y+3 y=9

(I), (II): y= 9

z=33-2( 9) x= 9+3 y=9

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Simplify

Multiply

(I): Multiply

z=33-18 x=9+3 y=9

(I), (II): Add and subtract terms

z=15 x=12 y=9

Therefore, 9 inches, 12 inches, and 15 inches are the lengths of our triangle's sides. This corresponds to option C.