Unit 3 Exam Review

Question

The height of the triangle is 11 inches longer than its base. The area of the triangle is 40 inches squared. How long is the base of the triangle?

Answer 1

H = 11 + B

40 = 1/2BH

40=1/2B(11+B). Substitution

80=B(11+B)

80=11B+B^2

0=B^2+11B-80

0=(B+16)(B-5) Solve by factoring

B=-16 and B=5

Base must be 5, since the base cannot be a negative number

Substitute 5, back into the original two equations to make sure it works!

Answer 2

Hi Sanyu,

This is a pretty straight-forward problem as long as you know the formula for the area of a triangle, A = bh/2 where base is the length of the base and h is the height of the triangle.

Since the height is 11 inches longer than the base and the area is 40 square inches,

A = bh/2 becomes

40 = b(b + 11)/2 I replaced the height with b + 11 since it is 11 inches longer than the base.

Multiply both sides by 2 and get

80 = b(b + 11)

Distribute the b, and then solve the equation.

80 = b² + 11b

0 = b² +11b - 80

Now factor

0 = (b + 16)(b - 5)

So b + 16 = 0, which makes b = -16. The base can't be negative, so this answer doesn't make sense.

Therefore b - 5 = 0, which makes b = 5. So, the base is 5 inches.

Hope this helps!

:)

Answer 3

A = 40 = b(b + 11)/2

A = 80 = b² + 11b

b² + 11b - 80 = 0

Factor (b+16)(b-5) = 0

b ≠ -16, so b = 5