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McGraw Hill Glencoe Algebra 1, 2012

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McGraw Hill Glencoe Algebra 1, 2012 View details

Preparing for Standardized Tests

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Exercise 2 Page 613

To find the area of a triangle, we need one half of the base multiplied by the height.

J

Practice makes perfect

To find the value of x and the dimensions of the given triangle, recall that the area of a triangle is found by taking one half of the base multiplied by the height. We know that the area is A=80cm^2, the height is h=xcm, and the base is b =2x+4cm.

Solving for x

We can create an equation to solve for x by substituting the given expressions into the formula for the area of a triangle. A=1/2bh ⇒ 80=1/2(2x+4)xLet's simplify the equation. We will start by applying Distributive Property on the right-hand side.

80=1/2(2x+4)x

Distribute x

80=1/2(2x^2+4x)

Distribute 1/2

80=x^2+2x

LHS-80=RHS-80

0=x^2+2x-80

Rearrange equation

x^2+2x-80=0

We will use the Quadratic Formula to solve the given quadratic equation. We first need to identify the values of a, b, and c. x^2+2x-80=0 ⇕ 1x^2+ 2x+( - 80)=0 We see that a= 1, b= 2, and c= - 80. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- 2±sqrt(2^2-4( 1)( - 80))/2( 1)

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Solve for x and Simplify

Calculate power

x=- 2±sqrt(4-4(1)(- 80))/2(1)

Identity Property of Multiplication

x=- 2±sqrt(4-4(- 80))/2

MultNegNegOnePar

- a(- b)=a* b

x=- 2±sqrt(4+320)/2

Add terms

x=- 2±sqrt(324)/2

Calculate root

x=- 2 ± 18/2

Factor out 2

x=2(- 1 ± 9)/2

CancelCommonFac

Cancel out common factors

x=- 1 ± 9

Using the Quadratic Formula, we found that the solutions of this equation are x_1=- 1+9 and x_2=-1-9. Therefore, the solutions are x_1=8 and x_2=- 10.

Finding the Dimensions

We need to determine which of the solutions that we found will satisfy the given conditions of our triangle. To do this, let's substitute these values into the expressions for the base and the height of the triangle. Then, we can evaluate the rationality of each measurement.

	Base (b)	Height (h)
x= 8	2( 8)+4= 20	8
x= - 10	2( - 10)+4= - 16	- 10

If x=- 10, the base and the height are both negative. This does not make sense as a triangle cannot have negative dimensions. Therefore, x=8, the length of the base is b=20cm, and the correct option is J.

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Exercises p.613