Question

Segment BD is an altitude of triangle ABC. Find the area of the triangle.

Answer 1

We know that side AC=5 and BD=3. The area of a triangle is base*height/2, so the area of this triangle is 3*5=2=7.5

Answer 2

Answer:

The area of the triangle is 7.5 sq units.

Step-by-step explanation:

The coordinates of the vertices of triangle ABC is,

A = (2, 1)

B = (-1, 4)

C = (2, 6)

D = (2, 4)

So,

$AC=\sqrt{(2-2)^2+(6-1)^2}=\sqrt{0^2+5^2}=\sqrt{5^2}=5$

$BD=\sqrt{(-1-2)^2+(4-4)^2}=\sqrt{(-3)^2+0^2}=\sqrt{9}=3$

we know that,

$\text{Area}=\dfrac{1}{2}\times \text{Base}\times \text{Height}$

$=\dfrac{1}{2}\times \text{AC}\times \text{BD}$

$=\dfrac{1}{2}\times 5\times 3$

$=7.5$