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inysia [295]
3 years ago
7

Can someone help . 8th grade math.

Mathematics
1 answer:
SVETLANKA909090 [29]3 years ago
8 0

the answer should be if i did my math right is 0.25


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Step-by-step explanation:

$0.75 because 7/10 is 0.7

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Determine what shape is formed for the given coordinates for ABCD, and then find the perimeter and area as an exact value and ro
Helga [31]

Answer:

Part 1) The shape is a trapezoid

Part 2) The perimeter is 25(4+\sqrt{2})\ units   or approximately  135.4\ units

Part 3) The area is 937.5\ units^2

Step-by-step explanation:

step 1

Plot the figure to better understand the problem

we have

A(-28,2),B(-21,-22),C(27,-8),D(-4,9)

using a graphing tool

The shape is a trapezoid

see the attached figure

step 2

Find the perimeter

we know that

The perimeter of the trapezoid is equal to

P=AB+BC+CD+AD

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

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

Find the distance AB

we have

A(-28,2),B(-21,-22)

substitute in the formula

d=\sqrt{(-22-2)^{2}+(-21+28)^{2}}

d=\sqrt{(-24)^{2}+(7)^{2}}

d=\sqrt{625}

d_A_B=25\ units

Find the distance BC

we have

B(-21,-22),C(27,-8)

substitute in the formula

d=\sqrt{(-8+22)^{2}+(27+21)^{2}}

d=\sqrt{(14)^{2}+(48)^{2}}

d=\sqrt{2,500}

d_B_C=50\ units

Find the distance CD

we have

C(27,-8),D(-4,9)

substitute in the formula

d=\sqrt{(9+8)^{2}+(-4-27)^{2}}

d=\sqrt{(17)^{2}+(-31)^{2}}

d=\sqrt{1,250}

d_C_D=25\sqrt{2}\ units

Find the distance AD

we have

A(-28,2),D(-4,9)

substitute in the formula

d=\sqrt{(9-2)^{2}+(-4+28)^{2}}

d=\sqrt{(7)^{2}+(24)^{2}}

d=\sqrt{625}

d_A_D=25\ units

Find the perimeter

P=25+50+25\sqrt{2}+25

P=(100+25\sqrt{2})\ units

simplify

P=25(4+\sqrt{2})\ units ----> exact value

P=135.4\ units

therefore

The perimeter is 25(4+\sqrt{2})\ units   or approximately  135.4\ units

step 3

Find the area

The area of trapezoid is equal to

A=\frac{1}{2}[BC+AD]AB

substitute the given values

A=\frac{1}{2}[50+25]25=937.5\ units^2

4 0
3 years ago
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