Question

Calculate the area of triangle ABC with altitude BD, given A (−6, 0), B (0, 0), C (0, 6), and D (−3, 3).

Answer 1

Calculate distances

$\boxed{\sf \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$

AC:-

$\\ \sf\longmapsto \sqrt{(0+6)^2+(6-0)^2}$

$\\ \sf\longmapsto \sqrt{6^2+6^2}$

$\\ \sf\longmapsto \sqrt{36+36}$

$\\ \sf\longmapsto \sqrt{72}$

$\\ \sf\longmapsto 6\sqrt{2}$

BD:-

$\\ \sf\longmapsto \sqrt{(-3-0)^2+(3-0)^2}$

$\\ \sf\longmapsto \sqrt{(-3)^2+(3)^2}$

$\\ \sf\longmapsto \sqrt{9+9}$

$\\ \sf\longmapsto \sqrt{18}$

$\\ \sf\longmapsto 3\sqrt{2}$

BD is height and AC is base.

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

$\\ \sf\longmapsto Area=\dfrac{1}{2}(6\sqrt{2})(3\sqrt{2})$

$\\ \sf\longmapsto Area=18units^2$

Answer 2

Answer:

• 18 unit²

Step-by-step explanation:

If BD is altitude then AC is the base.

The length of AC is:

• AC = $\sqrt{6^2+6^2} = 6\sqrt{2}$

The length of BD is:

• BD = $\sqrt{3^2 + 3^2} = 3\sqrt{2}$

The area is:

• A = 1/2bh
• A = 1/2 * $6\sqrt{2} *3\sqrt{2}$ = 18 unit²