How to find the area of an acute / obtuse triangle

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SAT Math › How to find the area of an acute / obtuse triangle

Questions 1 - 2

If triangle ABC has vertices (0, 0), (6, 0), and (2, 3) in the xy-plane, what is the area of ABC?

Explanation

Sketching ABC in the xy-plane, as pictured here, we see that it has base 6 and height 3. Since the formula for the area of a triangle is 1/2 * base * height, the area of ABC is 1/2 * 6 * 3 = 9.

The height, , of triangle in the figure is one-fourth the length of . In terms of h, what is the area of triangle ?

$2h^{2}$

$\frac{1}{2}h^{2}$

$h^{2}$

$3h^{2}$

Explanation

If $\dpi{100} \small h=\frac{1}{4}$ * $\dpi{100} \small \overline{PQ}$ , then the length of $\dpi{100} \small \overline{PQ}$ must be $\dpi{100} \small 4h$ .

Using the formula for the area of a triangle ( $\frac{1}{2}bh$ ), with $\dpi{100} \small b=4h$ , the area of the triangle must be $2h^{2}$ .

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