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vampirchik [111]
3 years ago
14

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.
Mathematics
1 answer:
shepuryov [24]3 years ago
5 0

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

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