A triangle with an area of 228 square inches has a height that is 10 less than 4 times its base. Find the base and height, in inches, of the triangle.

Answer 1

Area of a triangle formula: 1/2(b)(h)

Area is given: 228

Let b = base

write h in terms of the base: height is 10 less than 4 times its base means h= 4b-10

Plug each in the formula to solve for b

228 = 1/2 (b)(4b-10)

228 = 2b^2-5b

0=2b^2-5b-228

factor by grouping: 2b^2 -24b+19b-228=0

2b(b-12)+19(b-12)=0

(b-12)(2b+19)=0

Solve: b=12. Substitute it in for h: h=4(12)-10 = 48-10=38. h=38

Check: 1/2(b)(h)

1/2(12)(38) = 228

Answer 2

We need 2 equations in order to solve for the 2 unknowns, b = base of triangle and h = height of triangle.

We know that the following 2 equations have to be used:

A = (1/2) * b * h = 228 in²

h = 4 * b - 10

Let's plug h into b and solve:

228 = (1/2) * b * (4b - 10)

456 = b * (4b - 10)

456 = 4b² - 10b

4b² - 10b - 456 = 0

We can use our quadratic formula to solve now where:

a = 4

b = -10

c = -456

-(-10) ± √(-10)² - 4(4)(-456))/(2*4)

10 ± √(100 + 7296)/8

(10 ± 86)/8 = 96/8, -76/8

We reject the negative because we cannot have a negative base, therefore:

b = 96/8 = 12 in

Now we can solve for h by plugging it into one of our original formulas:

h = 4 * 12 - 10

h = 48 - 10

h = 38 in

We can double check our work by seeing if the area formula works now:

A = (1/2) * b * h

A = (1/2) * 12 * 38

A = 228 in² which means our solution is correct

Answer 3

Area of a triangle, A = base•height/2 = bh/2.

From the description, h = 4b-10

,

So, A = b(4b-10)/2 = 228 ---> 2b² - 5 - 228 = 0

by the quadratic equation: b = (5 ± √(25 - 4(2)(-228))/(2•2) = (5 ± 43)/4 = 12, -9.5

Since this is length we can disregard the negative value, therefore b = 12

h = 4b-10 = 48 - 10 = 38

Thus, base = 12 inches and height = 38 inches