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Question 1 of 3
Find x x if the perimeter of the triangle below is 44 44 cm.
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Inverse Operations
When moving a term to the other side of an equation, the operation is inversed.
First, form an equation knowing that the Perimeter of a shape is the sum of all sides.
P = 44 P = 44 cm
Side 1 = x 1 = x cm
Side 2 = ( x + 6 ) 2 = ( x + 6 ) cm
Side 3 = ( 3 x − 2 ) 3 = ( 3 x - 2 ) cm
P P
= =
Side 1 1 + + Side 2 2 + + Side 3 3
44 44
= =
x x + + ( x + 6 ) ( x + 6 ) + + ( 3 x − 2 ) ( 3 x - 2 )
Substitute the values
44 44
= =
5 x + 4 5 x + 4
Simplify
5 x + 4 5 x + 4
= =
44 44
To solve for x x , it needs to be alone on one side.
Start by moving 4 4 to the other side by subtracting 4 4 from both sides of the equation.
5 5 x x + 4 + 4
= =
44 44
5 5 x x + 4 + 4 − 4 - 4
= =
44 44 − 4 - 4
5 5 x x
= =
40 40
4 − 4 4 - 4 cancels out
Finally, remove 5 5 by dividing both sides of the equation by 5 5 .
5 5 x x
= =
40 40
5 5 x x ÷ 5 ÷ 5
= =
40 40 ÷ 5 ÷ 5
x x
= =
8 8
5 ÷ 5 5 ÷ 5 cancels out
Question 2 of 3
An isosceles triangle has a perimeter of 55 55 cm and a base that is 5 5 cm less than either of the other 2 2 equal sides. What are the lengths of all three sides?
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Inverse Operations
When moving a term to the other side of an equation, the operation is inversed.
First, label the values and form an equation knowing that the Perimeter of a shape is the sum of all sides.
Equal Side 1 = x 1 = x
Equal Side 2 = x 2 = x
Base = ( x − 5 ) = ( x - 5 )
P = 55 P = 55 cm
Equal Side 1 1 + + Equal Side 2 2 + + Base
= =
P P
x x + + x x + + x − 5 x - 5
= =
55 55
Substitute the values
3 x − 5 3 x - 5
= =
55 55
Simplify
To solve for x x , it needs to be alone on one side.
Start by moving 5 5 to the other side by adding 5 5 to both sides of the equation.
3 3 x x − 5 - 5
= =
55 55
3 3 x x − 5 - 5 + 5 + 5
= =
55 55 + 5 + 5
3 3 x x
= =
60 60
− 5 + 5 - 5 + 5 cancels out
Finally, remove 3 3 by dividing both sides of the equation by 3 3 .
3 3 x x
= =
60 60
3 3 x x ÷ 3 ÷ 3
= =
60 60 ÷ 3 ÷ 3
x x
= =
20 20
3 ÷ 3 3 ÷ 3 cancels out
Substitute x = 20 x = 20 to find each side length.
Equal Side 1 1
= =
x x
= =
20 20 cm
Equal Side 2 2
= =
x x
= =
20 20 cm
Base
= =
x x − 5 - 5
= =
20 20 − 5 - 5
= =
15 15 cm
Check our work
To confirm our answer, add the 3 3 numbers and check if the sum is 55 55 .
20 + 20 + 15 20 + 20 + 15
= =
55 55
55 55
= =
55 55
Since the equation is true, the answer is correct.
Equal Side 1 = 20 1 = 20 cm
Equal Side 2 = 20 2 = 20 cm
Base = 15 = 15 cm
Question 3 of 3
A rectangle is twice as long as it is wide. The total perimeter is 126 126 cm. What are the dimensions of this rectangle?
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Inverse Operations
When moving a term to the other side of an equation, the operation is inversed.
First, draw a rectangle and label it using the given statement: “A rectangle is twice as long as it is wide.”
This means if the width is w w , the length should be 2 w 2 w
Form an equation knowing that the Perimeter of a shape is the sum of all sides.
P = 126 P = 126 cm
Side 1 1 and 2 = w 2 = w cm
Side 3 3 and 4 = 2 w 4 = 2 w cm
P P
= =
Side 1 1 + + Side 2 2 + + Side 3 3 + + Side 4 4
126 126
= =
w w + + w w + + 2 w 2 w + + 2 w 2 w
Substitute the values
126 126
= =
6 w 6 w
Simplify
6 w 6 w
= =
126 126
Now, to solve for w w , it needs to be alone on one side.
Remove 6 6 by dividing both sides of the equation by 6 6 .
6 6 w w
= =
126 126
6 6 w w ÷ 6 ÷ 6
= =
126 126 ÷ 6 ÷ 6
w w
= =
21 21
6 ÷ 6 6 ÷ 6 cancels out
Finally, substitute w = 21 w = 21 to the dimensions of the rectangle.
Length
= =
2 2 w w
= =
2 ( 2 ( 21 21 ) )
= =
42 42 cm
Length= 42 = 42 cm
Width= 21 = 21 cm