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nika2105 [10]
3 years ago
11

HELP NEEDED. 37 POINTSI just need the answers

Mathematics
1 answer:
Juli2301 [7.4K]3 years ago
8 0

Answer:

Part 1) P=[2\sqrt{29}+\sqrt{18}]\ units or P=15.01\ units

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

Part 3) P=4[\sqrt{13}]\ units or P=14.42\ units

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

Part 5) P=2[\sqrt{17}+\sqrt{68}]\ units or P=24.74\ units

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

Part 7) A=20\ units^{2}

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

Part 9) A=10.5\ 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 1) we have the triangle ABC

A(0,3),B(5,1),C(2,-2)

step 1

Find the distance AB

A(0,3),B(5,1)

substitute in the formula

AB=\sqrt{(1-3)^{2}+(5-0)^{2}}

AB=\sqrt{(-2)^{2}+(5)^{2}}

AB=\sqrt{29}\ units

step 2

Find the distance BC

B(5,1),C(2,-2)

substitute in the formula

BC=\sqrt{(-2-1)^{2}+(2-5)^{2}}

BC=\sqrt{(-3)^{2}+(-3)^{2}}

BC=\sqrt{18}\ units

step 3

Find the distance AC

A(0,3),C(2,-2)

substitute in the formula

AC=\sqrt{(-2-3)^{2}+(2-0)^{2}}

AC=\sqrt{(-5)^{2}+(2)^{2}}

AC=\sqrt{29}\ units

step 4

Find the perimeter

The perimeter is equal to

P=AB+BC+AC

substitute

P=[\sqrt{29}+\sqrt{18}+\sqrt{29}]\ units

P=[2\sqrt{29}+\sqrt{18}]\ units

or

P=15.01\ units

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 3) we have the rhombus ABCD

A(-3,3),B(0,5),C(3,3),D(0,1)

Remember that  in a rhombus all sides are congruent

step 1

Find the distance AB

A(-3,3),B(0,5)

substitute in the formula

AB=\sqrt{(5-3)^{2}+(0+3)^{2}}

AB=\sqrt{(2)^{2}+(3)^{2}}

AB=\sqrt{13}\ units

step 2

Find the perimeter

The perimeter is equal to

P=4[AB]

substitute

P=4[\sqrt{13}]\ units

or

P=14.42\ 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 5) we have the quadrilateral ABCD

A(-1,5),B(3,6),C(5,-2),D(1,-3)

step 1

Find the distance AB

A(-1,5),B(3,6)

substitute in the formula

AB=\sqrt{(6-5)^{2}+(3+1)^{2}}

AB=\sqrt{(1)^{2}+(4)^{2}}

AB=\sqrt{17}\ units

step 2

Find the distance BC

B(3,6),C(5,-2)

substitute in the formula

BC=\sqrt{(-2-6)^{2}+(5-3)^{2}}

BC=\sqrt{(-8)^{2}+(2)^{2}}

BC=\sqrt{68}\ units

step 3

Find the distance CD

C(5,-2),D(1,-3)

substitute in the formula

CD=\sqrt{(-3+2)^{2}+(1-5)^{2}}

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

CD=\sqrt{17}\ units

step 4

Find the distance AD

A(-1,5),D(1,-3)

substitute in the formula

AD=\sqrt{(-3-5)^{2}+(1+1)^{2}}

AD=\sqrt{(-8)^{2}+(2)^{2}}

AD=\sqrt{68}\ units

step 5

Find the perimeter

The perimeter is equal to

P=\sqrt{17}+\sqrt{68}+\sqrt{17}+\sqrt{68}

substitute

P=2[\sqrt{17}+\sqrt{68}]\ units

or

P=24.74\ units

<h3>The complete answer in the attached file</h3>

Download docx
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