Core Connections Geometry, 2013
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Core Connections Geometry, 2013 View details
2. Section 8.2
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Exercise 76 Page 500

The formulas for the area of a square and the area of a rectangle are A = s^2 and A = l w, respectively.

x = 6 and x=2.5

Practice makes perfect

We want to find x knowing that the two figures below have equal areas.

Let's recall the formulas for the area of a square and the area of a rectangle. ccc Area of a Square:& & Area of a Rectangle: A = s^2 & & A = l wWe can use these formulas to write expressions for the areas of the figures from the given diagram. ccc Area of Given Square:& & Area of Given Rectangle: A = ( 2x)^2 && A = ( x+10)( 2x-3) Let's equate these expressions and solve the resulting equation for x.
(2x)^2 = (x+10)(2x-3)
Simplify
2^2x^2=(x+10)(2x-3)
4x^2=(x+10)(2x-3)
4x^2 = 2x(x+10)-3(x+10)
4x^2 = 2x^2+20x -3(x+10)
4x^2 = 2x^2+20x -3x - 30
4x^2 = 2x^2+17x-30
0 = -2x^2+17x-30
-2x^2+17x-30 = 0
The resulting equation is quadratic equation, so let's use the quadratic formula to solve it. ax^2+ bx+ c=0 ⇕ x=- b± sqrt(b^2-4 a c)/2 a Now let's the values of a, b, and c for our equation. -2x^2+17x-30 = 0 ⇕ -2x^2+ 17x+( - 30)=0 We see that a= -2, b= 17, and c= - 30. Let's substitute these values into the quadratic formula.
x=- b±sqrt(b^2-4ac)/2a
x=- 17±sqrt(17^2-4( -2)( -30))/2( -2)
Simplify right-hand side
x=-17±sqrt(289-4(-2)(- 30))/2(-2)
x=- 17±sqrt(289-240)/-4
x=- 17±sqrt(49)/-4
x = - 17 ± 7/-4
The solutions for this equation are x= -17 ± 7-4. Let's separate them into the positive and negative cases.
x=- 17± 7/-4
x_1=- 17+ 7/-4 x_2=- 17 - 7/-4
x_1=-10/-4 x_2=-24/-4
x_1= 2.5 x_2=6

We found that the solutions of the given equation are x_1=2.5 or x_2=6. The areas of the figures are the same when x equals 6 or when x equals 2.5.