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What is the area of a triangle whose ver...

What is the area of a triangle whose vertices are `(0, 6 sqrt(3)), (sqrt(35, 7))`, and (0, 3) ?

A

`15.37`

B

`17.75`

C

`21.87`

D

`25.61`

Text Solution

Verified by Experts

The correct Answer is:
C
Draw a picture showing the triangle.

The area of a triangle is `(1)/(2)xx` base `xx` height. The base of this triangle can be considered to be the side along the y - axis. The length of this side is `6sqrt(3)-3`. The height of this triangle drawn to the side along the y - axis is the distance from the point `(sqrt(35),7)` to the y-axis. This distance is the x - coordinate of the point `(sqrt(35),7)`, which is `sqrt(35)`. The area of the triangle is `(1)/(2)(6sqrt(3)-3)(sqrt(35))~~21.87`.
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