Area of Triangle & Quadratic Function ?

**Monocerotis** · October 29th 2009, 10:48 AM

Question
The base of a triangle is 2cm more than the height. The area of the triangle is $5cm^2$ . Find the base to the nearest tenth of a centimeter.

Attempt @ Solution
Area of Triangle = $1/2((b)(h))$

Let x represent h (height), and let x represent b (base)
h = x
b = x + 2

$2x+x^2 / 2=5$
$2x+x^2=10$
$x^2+2x-10=0$

Standard Form: a = 1, b= 2, c= 10

Sub in values into quadratic equation and solve for x which would give you the height, then add 2 to that value which would give you the base.

Is this correct ?

**masters** · October 29th 2009, 11:17 AM

Originally Posted by Monocerotis

Question
The base of a triangle is 2cm more than the height. The area of the triangle is $5cm^2$ . Find the base to the nearest tenth of a centimeter.

Attempt @ Solution
Area of Triangle = $1/2((b)(h))$

Let x represent h (height), and let x represent b (base)
h = x
b = x + 2

$2x+x^2 / 2=5$
$2x+x^2=10$
$x^2+2x-10=0$

Standard Form: a = 1, b= 2, c= 10

Sub in values into quadratic equation and solve for x which would give you the height, then add 2 to that value which would give you the base.

Is this correct ?

Hi Monocerotis,

Almost. c = -10

**Monocerotis** · October 29th 2009, 11:28 AM

Originally Posted by masters

Hi Monocerotis,

Almost. c = -10

right right, simple mistake lol

thanks bones.

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