Posted by **sheila** on Tuesday, January 29, 2008 at 9:33pm.

i need help in solving these 2 problems...please help..thank you...

Your home is built on a square lot. To add more space to your yard, you purchase an additional 4 feet along the side of the property. The area of the lot is now 9600 square feet. What are the dimentions of the new lot??

The Garys have a triangular pennant of area 420in.^2 flying from the flagpole in their yard. The height of the triangle is 10 in. than 5 times the base of the triangle. What are the dimensions of the pennant??

- math -
**Quidditch**, Tuesday, January 29, 2008 at 10:42pm
For the first problem:

x=length of side before the addition

x+4=the length of the longer side after the the additional 4 feet is purchased.

The new area of the property is:

x(x+4)=9600

x^2 + 4x =9600

x^2 + 4x - 9600=0

which is:

(x-96)(x+100)=0

Ignore the negative solution for x since a negative number would not make much sense for a lot length.

x=96 and x+4=100

- math -
**Damon**, Tuesday, January 29, 2008 at 10:46pm
x * ( x + 4 ) = 9600

x^2 + 4 x - 9600 = 0

solve quadratic

area = (1/2) b h

420 = (1/2) b (5b+10)

840 = 5 b^2 + 10 b

b^2 + 2 b - 168 = 0

solve quadratic

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