Question

For questions 2 and 4 calculate the perimeter and 4 questions 6/8 and 10 calculate the area

Answer 1

Answer:

Part 2) $P=2[\sqrt{20}+\sqrt{45}]\ units$ or $P=22.36\ units$

Part 4) $P=[19+\sqrt{17}]\ units$ or $P=23.12\ units$

Part 6) $A=36\ units^{2}$

Part 8) $A=16\ units^{2}$

Part 10) $A=6.05\ units^{2}$  

Step-by-step explanation:

we know that

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

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$  

Part 2) we have the rectangle ABCD

$A(-4,-4),B(-2,0),C(4,-3),D(2,-7)$

Remember that in a rectangle opposite sides are congruent

step 1

Find the distance AB

$A(-4,-4),B(-2,0)$

substitute in the formula

$AB=\sqrt{(0+4)^{2}+(-2+4)^{2}}$  

$AB=\sqrt{(4)^{2}+(2)^{2}}$  

$AB=\sqrt{20}\ units$  

step 2

Find the distance BC

$B(-2,0),C(4,-3)$

substitute in the formula

$BC=\sqrt{(-3-0)^{2}+(4+2)^{2}}$  

$BC=\sqrt{(-3)^{2}+(6)^{2}}$  

$BC=\sqrt{45}\ units$  

step 3

Find the perimeter

The perimeter is equal to

$P=2[AB+BC]$

substitute

$P=2[\sqrt{20}+\sqrt{45}]\ units$

or

$P=22.36\ units$

Part 4) we have the quadrilateral ABCD

$A(-2,-3),B(1,1),C(7,1),D(6,-3)$

step 1

Find the distance AB

$A(-2,-3),B(1,1)$

substitute in the formula

$AB=\sqrt{(1+3)^{2}+(1+2)^{2}}$  

$AB=\sqrt{(4)^{2}+(3)^{2}}$  

$AB=5\ units$  

step 2

Find the distance BC

$B(1,1),C(7,1)$

substitute in the formula

$BC=\sqrt{(1-1)^{2}+(7-1)^{2}}$  

$BC=\sqrt{(0)^{2}+(6)^{2}}$  

$BC=6\ units$  

step 3

Find the distance CD

$C(7,1),D(6,-3)$

substitute in the formula

$CD=\sqrt{(-3-1)^{2}+(6-7)^{2}}$  

$CD=\sqrt{(-4)^{2}+(-1)^{2}}$  

$CD=\sqrt{17}\ units$  

step 4

Find the distance AD

$A(-2,-3),D(6,-3)$

substitute in the formula

$AD=\sqrt{(-3+3)^{2}+(6+2)^{2}}$  

$AD=\sqrt{(0)^{2}+(8)^{2}}$  

$AD=8\ units$  

step 5

Find the perimeter

The perimeter is equal to

$P=AB+BC+CD+AD$

substitute

$P=[5+6+\sqrt{17}+8]\ units$

$P=[19+\sqrt{17}]\ units$

or

$P=23.12\ units$

Part 6) Calculate the area of rectangle ABCD

$A(-1,5),B(3,5),C(3,-4),D(-1,-4)$

Remember that in a rectangle opposite sides are congruent

step 1

Find the distance AB

$A(-1,5),B(3,5)$

substitute in the formula

$AB=\sqrt{(5-5)^{2}+(3+1)^{2}}$  

$AB=\sqrt{(0)^{2}+(4)^{2}}$  

$AB=4\ units$  

step 2

Find the distance BC

$B(3,5),C(3,-4)$

substitute in the formula

$BC=\sqrt{(-4-5)^{2}+(3-3)^{2}}$  

$BC=\sqrt{(-9)^{2}+(0)^{2}}$  

$BC=9\ units$  

step 3

Find the area

The area is equal to

$A=[AB*BC]$

substitute

$A=[4*9]=36\ units^{2}$

Part 8) Calculate the area of right triangle ABC

$A(-3,3),B(-3,-1),C(5,-1)$

step 1

Find the distance AB

$A(-3,3),B(-3,-1)$

substitute in the formula

$AB=\sqrt{(-1-3)^{2}+(-3+3)^{2}}$  

$AB=\sqrt{(-4)^{2}+(0)^{2}}$  

$AB=4\ units$  

step 2

Find the distance BC

$B(-3,-1),C(5,-1)$

substitute in the formula

$BC=\sqrt{(-1+1)^{2}+(5+3)^{2}}$  

$BC=\sqrt{(0)^{2}+(8)^{2}}$  

$BC=8\ units$  

step 3

Find the distance AC

$A(-3,3),C(5,-1)$

substitute in the formula

$AC=\sqrt{(-1-3)^{2}+(5+3)^{2}}$  

$AC=\sqrt{(-4)^{2}+(8)^{2}}$  

$AC=\sqrt{80}\ units$ -----> is the hypotenuse

step 4

Find the area

The area is equal to

$A=(1/2)AB*BC$

substitute

$A=(1/2)(4*8)=16\ units^{2}$

Part 10) Calculate the area of triangle ABC

$A(3,0),B(1,8),C(2,10)$

step 1

Find the distance AB

$A(3,0),B(1,8)$

substitute in the formula

$AB=\sqrt{(8-0)^{2}+(1-3)^{2}}$  

$AB=\sqrt{(8)^{2}+(-2)^{2}}$  

$AB=\sqrt{68}\ units$  

step 2

Find the distance BC

$B(1,8),C(2,10)$

substitute in the formula

$BC=\sqrt{(10-8)^{2}+(2-1)^{2}}$  

$BC=\sqrt{(2)^{2}+(1)^{2}}$  

$BC=\sqrt{5}\ units$  

step 3

Find the distance AC

$A(3,0),C(2,10)$

substitute in the formula

$AC=\sqrt{(10-0)^{2}+(2-3)^{2}}$  

$AC=\sqrt{(10)^{2}+(-1)^{2}}$  

$AC=\sqrt{101}\ units$  

step 4

we know that  

Heron's Formula is a method for calculating the area of a triangle when you know the lengths of all three sides.  

Let  

a,b,c be the lengths of the sides of a triangle.  

The area is given by:  

$A=\sqrt{p(p-a)(p-b)(p-c)}$  

where  

p is half the perimeter  

p=$\frac{a+b+c}{2}$  

we have  

$a=AB=\sqrt{68}=8.25\ units$  

$b=BC=\sqrt{5}=2.24\ units$  

$c=AC=\sqrt{101}=10.05\ units$  

p=$\frac{8.25+2.24+10.05}{2}=10.27\ units$  

Find the area

$A=\sqrt{10.27*(10.27-8.25)(10.27-2.24)(10.27-10.05)}$  

$A=\sqrt{10.27*(2.02)(8.03)(0.22)}$  

$A=6.05\ units^{2}$