Question

Question

Determine what shape is formed for the given coordinates for ABCD, and then find the perimeter and

area as an exact value and rounded to the nearest tenth.

A (-29,0), B (-22, -24), C (26,-10), D (-5,7)

The shape is a square

The perimeter is

or approximately

The area is

, or approximately




Please helpppp

Answer 1

Answer:

Part 1) The shape is a trapezoid

Part 2) The perimeter is

$P=(100+25\sqrt{2})\ units$  

or

$P=135.4\ units$

Part 3) The area is $A=937.5\ units^2$

Step-by-step explanation:

Part 1) Determine what shape is formed for the given coordinates for ABCD

Plot the given coordinates to better understand the problem

we have

A (-29,0), B (-22, -24), C (26,-10), D (-5,7)

using a graphing tool

The shape is a trapezoid

see the attached figure

Part 2) Find the perimeter

The perimeter of the trapezoid is the sum of its four length sides

$P=AB+BC+CD+AD$

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

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

step 1

Find the distance AB

A (-29,0), B (-22, -24)

$d=\sqrt{(-24-0)^{2}+(-22+29)^{2}}$

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

$d=\sqrt{625}\ units$

$d_A_B=25\ units$

step 2

Find the distance BC

B (-22, -24), C (26,-10)

$d=\sqrt{(-10+24)^{2}+(26+22)^{2}}$

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

$d=\sqrt{2,500}\ units$

$d_B_C=50\ units$

step 3

Find the distance CD

C (26,-10), D (-5,7)

$d=\sqrt{(7+10)^{2}+(-5-26)^{2}}$

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

$d=\sqrt{1,250}\ units$

$d_C_D=25\sqrt{2}\ units$

step 4

Find the distance AD

A (-29,0), D (-5,7)

$d=\sqrt{(7-0)^{2}+(-5+29)^{2}}$

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

$d=\sqrt{625}\ units$

$d_A_D=25\ units$

step 5

Find the perimeter

$P=AB+BC+CD+AD$

substitute the given values

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

$P=(100+25\sqrt{2})\ units$ ---> exact value

$P=135.4\ units$

Part 3) Find the area

The area of trapezoid is equal to

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

substitute the given values

$A=\frac{1}{2}[25+50]25$

$A=937.5\ units^2$