Question

In triangle abc, measure of angle a = 90 degrees and line an is an altitude. if ab = 20 in. and ac = 15 in., find bc, bn, nc, an

Answer 1

Answer:

BC = 25 in

AN = 12 in

BN = 16 in

NC = 9 in

Formulas:

Pythagorean Theorem

$c^2 = a^2 + b^2$

c ... hypotenuse

a ... one leg

b ... another leg

Pythagorean theorem is used in right triangles (triangles in which one angle is 90°).

Area of triangle

$A = \frac{1}{2}bh$

A ... area of triangle

b ... base (one of the sides of the triangle)

h ... height perpendicular to base

Step-by-step explanation:

Given:

A = 90°

AB = 20 in

AC = 15 in

See attached picture.

1) BC

To find BC we can use Pythagorean theorem, since ABC is a right triangle. We have the lengths of two legs (AB, AC) and we are finding the hypotenuse.

$c^2 = a^2 + b^2\\\text{BC}^2 = \text{AB}^2 + \text{AC}^2\\\text{BC}^2 = 20^2 + 15^2\\\text{BC}^2 = 625\\\sqrt{\text{BC}^2} = \sqrt{625}\\\text{BC} = 25 \text{ in}$

2) AN

AN is the height corresponding to base BC. To find it we can use formula for the area of the triangle. But first let's find the area of the triangle.

Notice how AB is perpendicular to AC. We can use AB as base and AC as height.

$A = \frac{1}{2}bh\\\\A = \frac{1}{2} \times \text{AB} \times \text{AC}\\\\A = \frac{1}{2} \times 20 \times 15\\\\A = 150 \text{ in}^2$

Now that we have area, we can use BC as base and AN as height and solve for AN.

$A = \frac{1}{2}bh\\\\A = \frac{1}{2} \times \text{BC} \times \text{AN}\\\\150 = \frac{1}{2} \times 25 \times \text{AN}\\\\\ 300 = 25 \times \text{AN}\\\\12 = \text{AN}\\\\\text{AN} = 12 \text{ in}$

3) BN

Let's use Pythagorean theorem in right triangle ABN. Hypotenuse is AB.

$c^2 = a^2 + b^2\\\text{AB}^2 = \text{AN}^2 + \text{BN}^2\\20^2 = 12^2 + \text{BN}^2\\400 = 144 + \text{BN}^2\\256 = \text{BN}^2\\\sqrt{256} = \sqrt{\text{BN}^2}\\16 \text{ in} = \text{BN}$

4) NC

We know the lengths of BC an BN, so we can just subtract.

$\text{NC} = \text{BC} - \text{BN}\\\text{NC} = 25 - 16\\\text{NC} = 9 \text{ in}$