Question

In △ABC,∠B is a right angle. The coordinates for each point are A(10, 7), B(5, 9), and C(3, 4). ​ Rounded to the nearest tenth, what is the area, in square units, of △ABC ? ​Enter the area in the box.

Answer 1

Answer:

$A=14.5\ units^2$

Step-by-step explanation:

we know that

The area of the right triangle ABC is equal to

$A=\frac{1}{2}(AB)(BC)$

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have

A(10, 7), B(5, 9), and C(3, 4)

step 1

Find the distance AB

A(10, 7), B(5, 9)

substitute in the formula

$d=\sqrt{(9-7)^{2}+(5-10)^{2}}$

$d=\sqrt{(2)^{2}+(-5)^{2}}$

$d_A_B=\sqrt{29}\ unjts$

step 2

Find the distance BC

B(5, 9),C(3, 4)

substitute in the formula

$d=\sqrt{(4-9)^{2}+(3-5)^{2}}$

$d=\sqrt{(-5)^{2}+(-2)^{2}}$

$d_B_C=\sqrt{29}\ unjts$

step 3

Find the area

substitute the values

$A=\frac{1}{2}(\sqrt{29})(\sqrt{29})$

$A=14.5\ units^2$