McGraw Hill Glencoe Geometry, 2012
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McGraw Hill Glencoe Geometry, 2012 View details
1. Areas of Parallelograms and Triangles
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Exercise 24 Page 784

The area of a parallelogram is the product of its base and its height.

Base: 13mm
Height: 17mm

Practice makes perfect

We are told that the height of a parallelogram is 4mm more than its base. Therefore, if we let b be the base, the height can be expressed as h=b+4. We are also told that the area is 221 square millimeters. Let's draw a diagram to illustrate the situation.

The area of a parallelogram is the product of its base and its height. A=bhWe can substitute A= 221 and h= b+4 into this formula and solve for b, the base of the parallelogram. Let's do it!
A=bh
221=b( b+4)
â–Ľ
Solve for b
221=b^2+4b
0=b^2+4b-221
b^2+4b-221=0
Note it's a quadratic equation. To solve this equation we will use the Quadratic Formula. ax^2+ bx+ c=0 ⇕ x=- b± sqrt(b^2-4 a c)/2 a The variable in our equation is b. Let's replace it with x not to mistake it with the b in the Quadratic Formula. b^2+4b-221=0 ⇕ x^2+4x-221=0 Next, we need to identify the values of a, b, and c. x^2+4x-221=0 ⇕ 1x^2+ 4x+( - 221)=0 We see that a= 1, b= 4, and c= - 221. Let's substitute these values into the Quadratic Formula.
x=- b±sqrt(b^2-4ac)/2a
x=- 4±sqrt(4^2-4( 1)( - 221))/2( 1)
â–Ľ
Solve for x
x=- 4±sqrt(16-4(1)(- 221))/2(1)
x=- 4±sqrt(16+884)/2
x=- 4±sqrt(900)/2
x=- 4 ± 30/2
x=2(- 2 ± 15)/2
x=- 2 ± 15
Now, we can go back to the previous notation and replace x with b. The solutions of our equation are b=- 2 ± 15. Let's separate them into the positive and negative cases.
b=- 2 ± 15
b_1=- 2+15 b_2=- 2-15
b_1=13 b_2=- 17

Since b is the length base of a parallelogram it must be positive. Therefore, the base of the parallelogram is 13mm. The height is 4mm more than its base, so we can add 4 to 13 to obtain the height. Height: 13+4=17 mm