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2 years ago

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Nequalities in Non Standard Form

1 answer:

Dominik [7]2 years ago

7 0

Answer:

x < 31 I think

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The vertices of ∆ABC are A(-2, 2), B(6, 2), and C(0, 8). The perimeter of ∆ABC is units is? What is the area?

11111nata11111 [884]

we have

$A(-2, 2),B(6, 2),C(0, 8)$

see the attached figure to better understand the problem

we know that

The perimeter of the triangle is equal to

$P=AB+BC+AC$

and

the area of the triangle is equal to

$A=\frac{1}{2}*base *heigth$

in this problem

$base=AB\\heigth=DC$

we know that

The distance between two points is equal to

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

Step $1$

<u>Find the distance AB</u>

$A(-2, 2),B(6, 2)$

Substitute the values in the formula

$d=\sqrt{(2-2)^{2}+(6+2)^{2}}$

$d=\sqrt{(0)^{2}+(8)^{2}}$

$dAB=8\ units$

Step $2$

<u>Find the distance BC</u>

$B(6, 2),C(0, 8)$

Substitute the values in the formula

$d=\sqrt{(8-2)^{2}+(0-6)^{2}}$

$d=\sqrt{(6)^{2}+(-6)^{2}}$

$dBC=6\sqrt{2}\ units$

Step $3$

<u>Find the distance AC</u>

$A(-2, 2),C(0, 8)$

Substitute the values in the formula

$d=\sqrt{(8-2)^{2}+(0+2)^{2}}$

$d=\sqrt{(6)^{2}+(2)^{2}}$

$dAC=2\sqrt{10}\ units$

Step $4$

<u>Find the distance DC</u>

$D(0, 2),C(0, 8)$

Substitute the values in the formula

$d=\sqrt{(8-2)^{2}+(0-0)^{2}}$

$d=\sqrt{(6)^{2}+(0)^{2}}$

$dDC=6\ units$

Step $5$

<u>Find the perimeter of the triangle</u>

$P=AB+BC+AC$

substitute the values

$P=8\ units+6\sqrt{2}\ units+2\sqrt{10}\ units$

$P=22.81\ units$

therefore

The perimeter of the triangle is equal to $22.81\ units$

Step $6$

<u>Find the area of the triangle</u>

$A=\frac{1}{2}*base *heigth$

in this problem

$base=AB=8\ units\\heigth=DC=6\ units$

substitute the values

$A=\frac{1}{2}*8*6$

$A=24\ units^{2}$

therefore

the area of the triangle is $24\ units^{2}$

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