Question

someone please explain to me how to find the area PLEASE! Find the area and perimeter of quadrilateral ABCD below. Explain your process for finding both the area and the perimeter, and show your mathematical steps clearly.

Answer 1

Answer:

Part A) The area of the figure is $24\ units^{2}$

Part B) The perimeter of the figure is $20\ units$

Step-by-step explanation:

step 1

Find the area of the figure

we know that

The area of the figure is equal to the area of triangle ABD plus the area of triangle BCD

The area of triangle is equal to

$A=\frac{1}{2}bh$

Area of triangle ABD

Observing the graph

$b=BD=(-2+8)=6\ units$

$h=(9-5)=4\ units$

substitute

$A=\frac{1}{2}(6)(4)=12\ units^{2}$

Area of triangle BCD

Observing the graph

$b=BD=(-2+8)=6\ units$

$h=(5-1)=4\ units$

substitute

$A=\frac{1}{2}(6)(4)=12\ units^{2}$

The area of the figure is

$12\ units^{2}+12\ units^{2}=24\ units^{2}$

step 2

Find the perimeter of the figure

we know that

The perimeter of the figure is equal to

$P=AB+BC+CD+AD$

we have

$A(-5,1),B(-8,5),C(-5,9),D(-2,5)$

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

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

Find the distance AB

$d=\sqrt{(5-1)^{2}+(-8+5)^{2}}=5\ units$

Find the distance BC

$d=\sqrt{(9-5)^{2}+(-5+8)^{2}}=5\ units$

Find the distance CD

$d=\sqrt{(5-9)^{2}+(-2+5)^{2}}=5\ units$

Find the distance AD

$d=\sqrt{(5-1)^{2}+(-2+5)^{2}}=5\ units$

substitute the values

$P=5+5+5+5=20\ units$