GEOMETRY ALGEBRA REVIEW TEST FOR GRADE 8

Problem 1 :

The area of a trapezoid is found using the formula A = 1/2 h(b1 + b2), where A is the area, h is the height, and b1 and b2 are the lengths of the bases.

the-area-of-trapezoid

What is the area of the above trapezoid ?

A)  A = 4x + 2               B) A = 4x + 8

C) A = 2x2 + 4x - 21      D) A = 2x2 + 8x  - 42

Solution :

A = 1/2 h(b1 + b2)

height = 4, b1 = x - 3, b2 = x + 7

A = ?

A = 1/2 × 4(x - 3 + x + 7)

= 1/2 × 4(2x + 4)

= 2(2x + 4)

A = 4x + 8

Area of the trapezoid is 4x + 8.

So, option B is correct.

Problem 2 :

What is the sum of -3x2 - 7x + 9 and -5x2 + 6x - 4 ?

(1) -8x2 - x + 5

(2) -8x4 - x + 5

(3) -8x2 - 13x + 13

(4) -8x4 - 13x2 + 13

Solution :

= -3x2 - 7x + 9 + (-5x2 + 6x - 4)

= -3x2 - 7x + 9 - 5x2 + 6x - 4

= -8x2 - x + 5

So, option (1) is correct.

Problem 3 :

Which expression is equivalent to (3x5 + 17x3 - 1) + (-2x5 - 6) ?

A) x5 + 17x3 - 7         B) x5  - 11x3 - 1

C) 5x3 + 17x3 + 7      D) 6x5 + 17x3 + 6

Solution :

= 3x5 + 17x3 - 1 + (-2x5 - 6)

= 3x5 + 17x3 - 1 - 2x5 - 6 

= 3x5 + 17x3 - 7

So, option A is correct.

Problem 4 :

Find the length of the missing side of the triangle below.

Perimeter = x2 + 5x + 11 cm

length-of-the-missing-side-of-the-triangle-q4

Solution :

P = h + b + c

x2 + 5x + 11 = 2x - 4 + 3x - 5 + c

x2 + 5x + 11 = 5x - 9 + c

x2 + 5x + 11 - 5x + 9 = c

x2 + 20 = c

So, the length of the missing side is x2 + 20 cm.

Problem 5 :

Approximate the length of the perimeter of the quadrilateral below.

length-of-the-perimeter-of-the-quadrilateral-q5

Solution :

length-of-the-perimeter-of-the-quadrilateral-S5-

MQ = 2

NM = √[(x2 - x1)2 + (y2 - y1)2]

N = (0, 3) = x1, y1

M = (2, 1) = x2, y2

NM = √[(2 - 0)2 + (1 - 3)2]

= √[(2)2 + (-2)2]

= √(4 + 4)

NM = √8

Q = (4, 1) = x1, y1

P = (5, 2) = x2, y2

QP = √[(5 - 4)2 + (2 - 1)2]

= √[(1)2 + (1)2]

= √(1 + 1)

QP = √2

N = (0, 3) = x1, y1

P = (5, 2) = x2, y2

NP = √[(5 - 0)2 + (2 - 3)2]

= √[(5)2 + (-1)2]

= √(25 + 1)

NP = √26

Perimeter of the quadrilateral = MQ + NM + QP + NP

= 2 + √8 + √2 + √26 

= 2 + 2.8 + 1.4 + 5.1

= 11.3

Problem 6 :

Silvia worked in a store that sold cylinder - shaped children's pools. She made a sign relating the volumes of these two pools.

length-of-the-missing-side-of-the-triangle-q6
6

The volume of the Paddler Pool is 108π cubic feet.

The Splasher Pool holds which percent of the water the Paddler Pool holds ?

A. 33%         B 75%          C 133%              D 300%

Solution :

volume of the cylinder = πr2h

Given Paddler Pool = 108 π

Splasher Pool =  π × (12)2 × 1

= 144 π

Volume of the Splasher Pool is 144π.

Percent of the water = Splasher Pool/Paddler Pool

= 144 π/108 π

= 1.33

Percent of the water = 1.33 × 100

= 133%

So, option C is correct.

Problem 7 :

A line segment has endpoints J(2, 4) and L(6, 8). The point K is the midpoint of JL What is an equation of a line perpendicular to JL and passing through K ?

A) y = -x + 10           B) y = -x - 10

C) y = x + 2              D) y = x - 2

Solution :

J(2, 4) = (x1, y1

L(6, 8) = (x2, y2)

(m1,m2) = x1 + x22, y1 + y22(xk,yk) = 2 + 62, 4 + 82= 82, 122k = (4, 6)m =y2 - y1x2 - x1 = 8 - 46 - 2 = 44m = 1

y = mx + b

6 = -1(4) + b

6 = -4 + b

b = 6 + 4

b = 10

y = -1x + 10

y = -x +10

So, option A is correct.

Problem 8 :

A triangle has vertices at (1, 3), (2, -3), and (-1, -1). What is the approximate perimeter of the triangle ?

A) 10     B 14       C 15           D 16

Solution :

perimeter-of-the-triangle

A(1, 3) = (x1, y1)

B(2, -3) = (x2, y2)

AB = √[(x2 - x1)2 + (y2 - y1)2]

= √[(2 - 1)2 + (-3 - 3)2]

= √[(1)2 + (-6)2]

= √(1 + 36)

= √37

AB = 6.1

B(2, -3) = (x1, y1)

C(-1, -1) = (x2, y2)

BC = √[(x2 - x1)2 + (y2 - y1)2]

= √[(-1 - 2)2 + (-1 + 3)2]

= √[(-3)2 + (2)2]

= √(9 + 4)

= √13

BC = 3.6

A(1, 3) = (x1, y1)

C(-1, -1) = (x2, y2)

CA = √[(x2 - x1)2 + (y2 - y1)2]

= √[(-1 - 1)2 + (-1 - 3)2]

= √[(-2)2 + (-4)2]

= √(4 + 16)

= √20

AC = 4.5

Perimeter = AB + BC + CA

= 6.1 + 3.6 + 4.5

= 14.2

So, perimeter of the triangle is 14.2.

So, option B is correct.

Problem 9 :

The vertices of quadrilateral EFGH are E(-7, 3), F(-4, 6), G(5, -3), and H(2, -6), What kind of quadrilateral is EFGH ? 

A) trapezoid       B) square   C) rectangle that is not a square     D) rhombus that is not a square

Solution :

rectangle

So, option C is correct.

Problem 10 :

R is the midpoint of segment PS, Q is the midpoint of segment RS.

P is located at (8, 10),and S is located at (12, -6). What are the coordinates of Q ?

A) (4, 2)         B (2, -8)        C (11, -2)            D (10, 2)

coordinates-of-q

Solution :

P is located at (8, 10) = x1, y1

S is located at (12, -6) = x2, y2

(m1,m2) = x1 + x22, y1 + y22(xQ,yQ) = 10 + 122, 2 - 62= 222, -42Q = (11, -2)

So, the coordinates of Q is (11, -2).

So, option C is correct.

Problem 11 :

County X has a population density of 250 people per square mile. The total population of the county is 150,000. Which geometric model could be the shape of county X?

A)  a parallelogram with a base of 25 miles and a height of 25 miles.

B) a rectangle that is 15 miles long and 45 miles wide

C) a right triangle with a leg that is 30 miles long and a hypotenuse that is 50 miles long 

D) a trapezoid with base lengths of 10 miles and 30 miles and a height of 25 miles.

Solution :

Population density = 250

Total population of the county = 150,000

Total area which represents the population = Total population/Population density

= 150000/250

= 600 square miles

A) area of parallelogram = Base × Height

= 25 × 25

= 625 square miles.

Which is not equal to 600 square miles.

So it's not the correct option.

B) area of a rectangle = Length × Width

= 15 × 45

= 675 square miles

675 square miles not equal to 600 square miles.

So it's not the correct option.

C) area of a right triangle = Base × Height

Since length of hypotenuse is given.

So we will find the length of height.

Height = √[Hypotenuse)2 - (Base)2]

= √(50)2 - (30)2

= √1600

Height = 40 miles

So, area of the  right triangle= 1/2 (Base × Height)

= 1/2 (40 × 30)

= 1/2 (1200)

= 600 square miles

Which matches the area of country x.

So, it's correct option.

Problem 12 :

A company is designing a cylinder to hold marbles for a new game it is inventing. The cylinder has a height of 18 inches and a diameter of 6 inches. Find the volume of the cylinder to the nearest of a cubic inch.

[1] 108.0      [2] 508.9      [3] 678.6     [4] 1065.92

Solution :

height = 18 inches

diameter = 6 inches

v = πr2h

d = 2r

6 = 2r

6/2 = r

3 = r

v = 3.14 × (3)2 × 18

= 3.14 × 9 × 18

v = 508.68

So, option [2] is correct.

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