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Fofino [41]
3 years ago
9

What is the area of a parallelogram whose vertices are A(−4, 9) , B(11, 9) , C(5, −1) , and D(−10, −1) ?

Mathematics
2 answers:
o-na [289]3 years ago
8 0
AB = 15
CD = 15
Distance from AB to CD = 10
the formula of area K= b*h
K= 15*10 = 150

anzhelika [568]3 years ago
6 0

Let

A(-4,9)\\B(11,9)\\C(5,-1) \\D(-10,-1)\\E(-4.-1)          

using a graphing tool

see the attached figure to better understand the problem

we know that

Parallelogram is a quadrilateral with opposite sides parallel and equal in length

so

AB=CD \\AD=BC

The area of a parallelogram is equal to

A=B*h  

where

B is the base

h is the height

the base B is equal to the distance AB

the height h is equal to the distance AE

Step 1

<u>Find the distance AB</u>

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

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

A(-4,9)\\B(11,9)      

substitute the values

d=\sqrt{(9-9)^{2}+(11+4)^{2}}

d=\sqrt{(0)^{2}+(15)^{2}}

dAB=15\ units

Step 2

<u>Find the distance AE</u>

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

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

A(-4,9)\\E(-4.-1)      

substitute the values

d=\sqrt{(-1-9)^{2}+(-4+4)^{2}}

d=\sqrt{(-10)^{2}+(0)^{2}}

dAE=10\ units

Step 3

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

The area of a parallelogram is equal to

A=B*h

A=AB*AE

substitute the values

A=15*10=150\ units^{2}

therefore

<u>the answer is</u>

the area of the parallelogram is 150\ units^{2}

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