Segment bd is an altitude of triangle abc. find the area of the triangle.

Asked by maham237 @ 25/12/2020 in Mathematics viewed by 332 People

Segment BD is an altitude of triangle ABC. Find the area of the triangle.
a. 7.5

b. 9.5

c. 10

d. 15

Answered by maham237 @ 25/12/2020

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,

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

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

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

$=7.5$

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