Question

Solve the formula for the indicated variable. Then
evaluate the rewritten formula for the given values. Use 3.14 for Π.

1)  Area: A = (1/2)bh  

Solve for b. Find the value of b.  When A=160 and h=20

2)  Volume:V = (1/3)Πr²h  

Solve for h. Find the value of h.  When V=84 and r=4.

3)  Volume: V = Πr²h  

Solve for h. Find the value of h.  When V=790 and r=5.

4)  Area: A = (1/2)(b₁+b₂)

Solve for b₁. Find the value of b₁.  When A=30,b₂=8 and h=5.

5)  Surface Area: S =Πr²+Πrl

Solve for l. Find the value of l.  When S=100 and r=5

Answer 1

1)A = (1/2)bh

Substitute the given values in A = (1/2)bh

160 = (1/2)×b×20

160 = b×20/2

160 = b×10

Apply division property divide each sides by 10

160/10 = b×10/10

16 = b

⇒b = 16

2)  Volume:V = (1/3)Πr²h

Substitute the given values

84 = 1/3×3.14×4×4×h

By divisible property divide both sides 3/3.14×4×4

84×3/3.14×4×4 = h

84×3/3.14×16 = h

21×3/3.14×4 = h

5.25×3/3.14 = h

15.75/3.14 = h

5 = h

⇒h = 5

3)  Volume: V = Πr²h

Substitute the given values

790 = 3.14×5×5×h

Apply division property divide each sides by 3.14×5×5

790/3.14×5×5 = h×3.14×5×5/3.14×5×5

158/3.14×5 = h

30.16/3.14 =h

9.5 = h

⇒h = 9.5

4)  Area: A = (1/2)(b₁+b₂)

Substitute the given values

30 = 1/2(b₁+8)5

Apply multiplication property multiply each sides 2/5

30×2/5=2×5/5×2(b₁+8)

60/5 = b₁+ 8

Subtract each sides by 8

12 - 8 = b₁

_{4 =} b₁

_⇒ b_{1 = 4}

5)  Surface Area: S =Πr²+Πrl

Substitute the given values

100 = 3.14×5×5+3.14×5×L

100 = 3.14×5(5+L)

Divide each sides by 3.14×5

100/3.14×5 = 5+ L

100/15.70 = 5+L

6.3 = 5+L

Subtract each sides by 5

6.3-5 = 5-5+L

1.3 = L

⇒ L = 1.3

 

Comments

2).

Volume : V = (1/3)Πr²h.

Solve for h :

h = (3V)/Πr² .

Substitute the values V = 84, Π = 3.14, and r = 4.

h = (3 * 84)/(3.14 * 4 * 4)

= (3 * 21)/(3.14 * 4)

= 63/12.56

= 5.0159.

Therefore, h = 5.015.

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3).

h = V/Πr².

= 790 / (3.14 * 5 * 5)

= 158 / (3.14 * 5)

= 158 / (15.7)

= 10.06.

Therefore, h = 10.06.