Question

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

Answer 1

Part 2)
`P=2[√(20)+√(45)]\ units` or
`P=22.36\ units`

Part 4)
`P=[19+√(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=√(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=√(45)\ units`

step 3

Find the perimeter

The perimeter is equal to

`P=2[AB+BC]`

substitute

`P=2[√(20)+√(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=√(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+√(17)+8]\ units`

`P=[19+√(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=√(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=√(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=√(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=√(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=√(p(p-a)(p-b)(p-c))`

where

p is half the perimeter

p=
`(a+b+c)/(2)`

we have

`a=AB=√(68)=8.25\ units`

`b=BC=√(5)=2.24\ units`

`c=AC=√(101)=10.05\ units`

p=
`(8.25+2.24+10.05)/(2)=10.27\ units`

Find the area

`A=√(10.27*(10.27-8.25)(10.27-2.24)(10.27-10.05))`

`A=√(10.27*(2.02)(8.03)(0.22))`

`A=6.05\ units^(2)`