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SAT II Math I : Surface Area

Study concepts, example questions & explanations for SAT II Math I

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Example Questions

Example Question #771 : Geometry

The radius of the base of a cone is ; its slant height is two-thirds of the diameter of that base. Give its surface area in terms of .

Possible Answers:

$\frac{\left (6- \sqrt{7} \right )\pi r ^{2}}{3}$

$\frac{5\pi r ^{2}}{3}$

$\frac{7\pi r ^{2}}{3}$

$\frac{\left (3- \sqrt{5} \right )\pi r ^{2}}{3}$

$\frac{4\pi r ^{2}}{3}$

Correct answer:

$\frac{7\pi r ^{2}}{3}$

Explanation:

The formula for the surface area of a cone with base of radius and slant height is

$A = \pi r^{2} + \pi r l$ .

The diameter of the base is twice radius , or , and its slant height is two-thirds of this diameter, which is $\frac{2}{3}\cdot 2r = \frac{4}{3}r$ . Substitute this for in the formula:

$A = \pi r^{2} + \pi r \cdot \frac{4}{3}r$

$A = \pi r^{2} + \frac{4}{3} \pi r ^{2}$

$A = \frac{7}{3} \pi r ^{2} = \frac{7\pi r ^{2}}{3}$

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Example Question #2 : How To Find The Surface Area Of A Cone

The radius of the base of a cone is ; its height is twice of the diameter of that base. Give its surface area in terms of .

Possible Answers:

$3 \pi r^{2}$

$5 \pi r^{2}$

$\left ( 1+ \sqrt{5} \right ) \pi r^{2}$

$\left ( 1+2 \sqrt{5} \right ) \pi r^{2}$

$\left ( 1+ \sqrt{17} \right ) \pi r^{2}$

Correct answer:

$\left ( 1+ \sqrt{17} \right ) \pi r^{2}$

Explanation:

The formula for the surface area of a cone with base of radius and slant height is

$A = \pi r^{2} + \pi r l$ .

The base has radius and diameter . The height is twice the diamter, which is . Its slant height can be calculated using the Pythagorean Theorem:

$l = \sqrt{r^{2}+ h^{2}}$

$l = \sqrt{r^{2}+ (4r)^{2}}$

$l = \sqrt{r^{2}+ 16r^{2}}$

$l = \sqrt{17r^{2}}$

$l = r \sqrt{17 }$

Substitute $r \sqrt{17 }$ for in the surface area formula:

$A = \pi r^{2} + \pi r \cdot r \sqrt{17}$

$A = \pi r^{2} + \pi r^{2} \sqrt{17}$

$A = \left ( 1+ \sqrt{17} \right ) \pi r^{2}$

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Example Question #16 : Solid Geometry

The height of a cone is ; the diameter of its base is twice the height. Give its surface area in terms of .

Possible Answers:

$10 \pi h^{2}$

$4 \pi h^{2}$

$5 \pi h^{2}$

$\left (2 +2 \sqrt{2} \right ) \pi h^{2}$

$\left (1 + \sqrt{2} \right ) \pi h^{2}$

Correct answer:

$\left (1 + \sqrt{2} \right ) \pi h^{2}$

Explanation:

The formula for the surface area of a cone with base of radius and slant height is

$A = \pi r^{2} + \pi r l$ .

The diameter of the base is twice the height, which is ; the radius is half this, which is .

The slant height can be calculated using the Pythagorean Theorem:

$l = \sqrt{r^{2}+ h^{2}}$

$l = \sqrt{h^{2}+ h^{2}}$

$l = \sqrt{2h^{2}}$

$l =h \sqrt{2}$

Substitute $h \sqrt{2}$ for and for in the surface area formula:

$A = \pi h^{2} + \pi h \cdot h \sqrt{2}$

$A = \pi h^{2} + \pi h ^{2} \sqrt{2}$

$A = \left (1 + \sqrt{2} \right ) \pi h^{2}$

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Example Question #123 : Geometry

The circumference of the base of a cone is 80; the slant height of the cone is equal to twice the diameter of the base. Give the surface area of the cone (nearest whole number).

Possible Answers:

2,039

2,485

2,546

4,077

1,975

Correct answer:

2,546

Explanation:

The formula for the surface area of a cone with base of radius and slant height is

$A = \pi r^{2} + \pi r l$ .

The slant height is twice the diameter, or, equivalently, four times the radius, so

l = 4r

and

$A = \pi r^{2} + \pi r \cdot 4r$

$A = \pi r^{2} +4 \pi r^{2}$

$A = 5 \pi r^{2}$

The radius of the base is the circumference divided by $2\pi$ , which is

$80 \div 2 \pi \approx 80 \div (2 \times 3.14 )\approx 12.73$

Substitute:

$A = 5 \pi r^{2}$

$A \approx 5 \cdot 3.14 \cdot (12.73)^{2}$

$A \approx 2,546$

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Example Question #2 : How To Find The Surface Area Of A Cone

The circumference of the base of a cone is 100; the height of the cone is equal to the diameter of the base. Give the surface area of the cone (nearest whole number).

Possible Answers:

2,575

2,385

1,779

8,443

1,589

Correct answer:

2,575

Explanation:

The formula for the surface area of a cone with base of radius and slant height is

$A = \pi r^{2} + \pi r l$ .

The diameter of the base is the circumference divided by $\pi$ , which is

$d = C \div (2\pi ) \approx 100 \div 3.14 \approx 31.83$

This is also the height .

The radius is half this, or

$r = d \div 2 \approx 31.83 \div 2 \approx 15.92$

The slant height can be found by way of the Pythagorean Theorem:

$l = \sqrt{r^{2}+ h ^{2}}$

$l \approx \sqrt{15.92^{2}+ 31.83 ^{2}}$

$l \approx \sqrt{253.30+ 1,013.21}$

$l \approx \sqrt{1266.51}$

$l \approx 35.60$

Substitute in the surface area formula:

$A = \pi r^{2} + \pi r l$

$A \approx 3.14 \cdot 15.92 ^{2} + 3.14 \cdot 15.92 \cdot 35.60$

$A \approx 795.8+ 1,780.1$

$A \approx 2,575$

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Example Question #21 : Surface Area

If a cone were unfurled into a 2-dimensional figure. The lateral area of the cone would look most like which figure?

Possible Answers:

Rectangle

Sector of a Circle

Triangle

Circle

Correct answer:

Sector of a Circle

Explanation:

When creating a net image of a 3D figure - one imagines it is made of paper and is unfurled into its' 2D form. The lateral portion of the cone cone would be unfurled into the image of a Sector of a Circle. To include the full surface area of the cone a circle is included to form the base of the cone as in the figure below. The lateral area portion is the top part of the figure below.

Cone net

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Example Question #22 : Surface Area

Find the surface area of a cube with side length of 6in.

Possible Answers:

250cm^2

18cm^2

36cm^2

24cm^2

216cm^2

Correct answer:

216cm^2

Explanation:

A cube is made up of 6 identical sides. Find the area of one side and multiplying it by 6 will result in the surface area of a cube.

S=6a^2

a is the side length, in this case 6in.

S=6(6cm)^2=6*36cm^2=216cm^2

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Example Question #27 : Surface Area

What is the surface area of a cylinder with height 10 and radius 7?

Possible Answers:

$238\pi$

$163\pi$

$140\pi$

$300\pi$

$98\pi$

Correct answer:

$238\pi$

Explanation:

The formula of the surface area of a cylinder is:

$SA=2\pi rh+2\pi r^2$

In this problem h is 10 and r is 7. Substituting in those values gives us:

$SA=2\pi 7*10+2\pi 7^2$

Simplifying and combining the like terms results in:

$SA=140\pi+98\pi=238\pi$

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Example Question #28 : Surface Area

Figure not drawn to scale.

Box example

Find the surface area of the box above.

Possible Answers:

Surface area: 48 yd²

Surface area: 84 yd²

Surface area: 82 yd³

Surface area: 84 yd³

Surface area: 82 yd²

Correct answer:

Surface area: 82 yd²

Explanation:

Box example

You can find the surface area of a box by following the equation below:

$\small \small \small Surface Area = 2(length\times height) + 2(length\times depth) + 2(height\times depth)$

$\small \small \small 2(7\times 3) + 2(7\times 2) + 2(3\times 2)=$

$\small 2(21) + 2(14) + 2(6)=$

$\small \small 42 + 28 + 12= 82$

The surface area of the box is 82yd²(remember that area measurements are square units NOT cubic units)

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Example Question #29 : Surface Area

Box 3

The shaded face of the provided rectangular prism is a square. If the surface area of the prism is , give in terms of .

Possible Answers:

$x = \frac{-25+\sqrt{4A+625}}{4}$

$x = \frac{A}{625}$

$x=\frac{ A - 1,250}{100}$

$x = \frac{-50+\sqrt{2A+2,500}}{2}$

$x=\frac{ A -2,500}{50}$

Correct answer:

$x=\frac{ A - 1,250}{100}$

Explanation:

Since the top face of the prism is a square, the common sidelength - and the missing dimension - is 25.

The surface area of a rectangular prism with length , width , and height can be found using the formula

2LW + 2 WH + 2LH = A .

Setting H= x, L=W= 25 , and solving for :

$2 \cdot 25 \cdot 25 + 2 \cdot 25 \cdot x + 2 \cdot 25 \cdot x = A$

1,250 + 50x + 50 x = A

100 x + 1,250 = A

100 x + 1,250 - 1,250 = A - 1,250

100 x = A - 1,250

$\frac{100 x }{100}=\frac{ A - 1,250}{100}$

$x=\frac{ A - 1,250}{100}$

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SAT II Math I : Surface Area

Study concepts, example questions & explanations for SAT II Math I

All SAT II Math I Resources

Example Questions

Example Question #771 : Geometry

Example Question #2 : How To Find The Surface Area Of A Cone

Example Question #16 : Solid Geometry

Example Question #123 : Geometry

Example Question #2 : How To Find The Surface Area Of A Cone

Example Question #21 : Surface Area

Example Question #22 : Surface Area

Example Question #27 : Surface Area

Example Question #28 : Surface Area

The surface area of the box is 82yd²(remember that area measurements are square units NOT cubic units)

Example Question #29 : Surface Area

All SAT II Math I Resources

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DMCA Complaint

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HIGH SCHOOL

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K-8

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SAT II Math I : Surface Area

Study concepts, example questions & explanations for SAT II Math I

All SAT II Math I Resources

Example Questions

Example Question #771 : Geometry

Example Question #2 : How To Find The Surface Area Of A Cone

Example Question #16 : Solid Geometry

Example Question #123 : Geometry

Example Question #2 : How To Find The Surface Area Of A Cone

Example Question #21 : Surface Area

Example Question #22 : Surface Area

Example Question #27 : Surface Area

Example Question #28 : Surface Area

The surface area of the box is 82yd2 (remember that area measurements are square units NOT cubic units)

Example Question #29 : Surface Area

All SAT II Math I Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

The surface area of the box is 82yd²(remember that area measurements are square units NOT cubic units)