Solid Geometry - ISEE Upper Level Quantitative Reasoning

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Question

What is the surface area of a cube on which one face has a diagonal of ?

Answer

One of the faces of the cube could be drawn like this:

Notice that this makes a triangle.

This means that we can create a proportion for the sides. On the standard triangle, the non-hypotenuse sides are both , and the hypotenuse is $\sqrt{2}$ . This will allow us to make the proportion:

$\frac{1}{\sqrt{2}} = \frac{x}{5}$

Multiplying both sides by , you get:

$x=\frac{5}{\sqrt{2}}$

To find the area of the square, you need to square this value:

$(\frac{5}{\sqrt{2}})^2 = \frac{25}{2}$

Now, since there are sides to the cube, multiply this by to get the total surface area:

$\frac{25}{2} * \frac{6}{1} = 75$

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Question

A cube has sidelength one and one-half feet; a rectangular prism of equal volume has length 27 inches and height 9 inches. Give the width of the prism in inches.

Answer

One and one half feet is equal to eighteen inches, so the volume of the cube, in cubic inches, is the cube of this, or

$V = 18^{3} = 5,832$ cubic inches.

The volume of a rectangular prism is

Since its volume is the same as that of the cube, and its length and height are 27 and 9 inches, respectively, we can rewrite this as

$5,832= 27 \cdot 9 \cdot W$

$5,832= 243 \cdot W$

$W = 5,832 \div 243 = 24$

The width is 24 inches.

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Question

A cube has sidelength one and one-half feet; a rectangular prism of equal surface area has length 27 inches and height 9 inches. Give the width of the prism in inches.

Answer

One and one half feet is equal to eighteen inches, so the surface area of the cube, in square inches, is six times the square of this, or

$6 \cdot 18^{2} = 6 \cdot 324 = 1,944$ square inches.

The surface area of a rectangular prism is determined by the formula

.

So, with substitutiton, we can find the width:

$1,944= 2 \cdot 27 \cdot W + 2\cdot 27 \cdot 9 + 2 \cdot W \cdot 9$

$72W\div 72 = 1,458 \div 72$

$W = 20 \frac{1}{4}$ inches

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Question

A rectangular prism has volume one cubic foot; its length and width are, respectively, 9 inches and inches. Which of the following represents the height of the prism in inches?

Answer

The volume of a rectangular prism is the product of its length, its width, and its height. The prism's volume of one cubic foot is equal to $12^{3} = 1,728$ cubic inches.

Therefore, can be rewritten as $9 \cdot t \cdot H = 1,728$ .

We can solve for as follows:

$\frac{9 t \cdot H }{9t}= \frac{1,728 }{9t}$

$H = \frac{1,728 \div 9 }{9t \div 9 }$

$H = \frac{192}{t}$

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Question

Which is the greater quantity?

(a) The volume of a cube with surface area inches

(b) The volume of a cube with diagonal inches

Answer

The cube with the greater sidelength has the greater volume, so we need only calculate and compare sidelengths.

(a) $A=6s^{2}$ , so the sidelength of the first cube can be found as follows:

$A=6s^{2}$

$6s^{2}= 864$

$6s^{2} \div 6= 864 \div 6$

$s^{2} = 144$

$s = \sqrt{144 }= 12$ inches

(b) $d^{2} = s^{2}+ s^{2}+ s^{2} = 3 s^{2}$ by an extension of the Pythagorean Theorem, so the sidelength of the second cube can be found as follows:

$3 s^{2}= d^{2}$

$3 s^{2}= 21^{2}= 441$

$3 s^{2}\div 3= 441 \div 3$

$s^{2}= 147$

$s=\sqrt{ 147}$

Since , $\sqrt{147 }> \sqrt{144}$ . The second cube has the greater sidelength and, subsequently, the greater volume. This makes (b) greater.

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Question

is a positive number. Which is the greater quantity?

(A) The surface area of a rectangular prism with length , width , and height

(B) The surface area of a rectangular prism with length , width , and height .

Answer

The surface area of a rectangular prism can be determined using the formula:

Using substitutions, the surface areas of the prisms can be found as follows:

The prism in (A):

$S = 2( 16t \cdot 9t +9t \cdot t +16t \cdot t)$

$S = 2( 144t ^{2}+9t^{2} +16t^{2})$

$S = 2( 169t ^{2})$

$S = 338t ^{2}$

$S = 2( 8t \cdot 6t +6t \cdot 3t +8t \cdot 3 t)$

$S = 2( 48t^{2} +18t ^{2}+24 t^{2})$

$S = 2( 90 t^{2})$

$S = 180 t^{2}$

Regardless of the value of , $338t ^{2} > 180t ^{2}$ - that is, the first prism has the greater surface area. (A) is greater.

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Question

A large crate in the shape of a rectangular prism has dimensions 5 feet by 4 feet by 12 feet. Give its volume in cubic yards.

Answer

Divide each dimension by 3 to convert feet to yards, then multiply the three dimensions together:

$V = \frac{5}{3} \cdot \frac{4}{3} \cdot \frac{12}{3}$

$V = \frac{5}{3} \cdot \frac{4}{3} \cdot \frac{4}{1}$

$V = \frac{80}{9 } = 8 \frac{8}{9 } \textrm{ yd}^{3}$

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Question

Which is the greater quantity?

(A) The volume of a rectangular solid ten inches by twenty inches by fifteen inches

(B) The volume of a cube with sidelength sixteen inches

Answer

The volume of a rectangular solid ten inches by twenty inches by fifteen inches is

$V = 10 \cdot 20 \cdot15 = 3,000$ cubic inches.

The volume of a cube with sidelength 13 inches is

$V = 16^{3} = 16 \cdot 16 \cdot 16 = 4,096$ cubic inches.

This makes (B) greater

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Question

A pyramid with a square base and a cone have the same height and the same volume. Which is the greater quantity?

(A) The perimeter of the base of the pyramid

(B) The circumference of the base of the cone

Answer

The volume of a pyramid or a cone with height and base of area is

$V = \frac{1}{3} Bh$ ,

so in both cases, the area of the base is

$\frac{1}{3} Bh = V$

$\frac{3}{h} \cdot \frac{1}{3} Bh =\frac{3}{h} \cdot V$

$B = \frac{3V}{h}$

Since the pyramid and the cone have the same volume and height, their bases has the same area .

The length of one side of the square base of the pyramid is the square root of this, or $\sqrt{B}$ , and the perimeter is four times this, or $P = 4 \sqrt{B}$ .

The radius and the area of the base of the cone are related as follows:

$\pi r ^{2} = B$

$\sqrt{\pi r ^{2} }= \sqrt{B}$

$\sqrt{\pi } \cdot r = \sqrt{B}$

Multiply both sides by $2 \sqrt{\pi}$ to get:

$2 \sqrt{\pi} \cdot \sqrt{\pi } \cdot r =2 \sqrt{\pi} \cdot \sqrt{B}$

$2\pi r =2 \sqrt{\pi} \cdot \sqrt{B}$

$C =2 \sqrt{\pi} \cdot \sqrt{B}$

$2 \sqrt{\pi} < 2 \sqrt{4} = 2 \times 2 = 4$ , so

$2 \sqrt{\pi} \cdot \sqrt{B} < 4 \sqrt{B}$ , and

The perimeter of the base of the pyramid, which is (A), is greater than the circumference of the base of the cone.

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Question

A pyramid with a square base has height equal to the perimeter of its base. Which is the greater quantity?

(A) Twice the area of its base

(B) The area of one of its triangular faces

Answer

Since the answer is not dependent on the actual dimensions, for the sake of simplicity, we assume that the base has sidelength 2. Then the area of the base is the square of this, or 4.

The height of the pyramid is equal to the perimeter of the base, or $4 \times 2 = 8$ . A right triangle can be formed with the lengths of its legs equal to the height of the pyramid, or 8, and one half the length of a side, or 1; the length of its hypotenuse, which is the slant height, is

$\sqrt{8^{2} + 1 ^{2}} = \sqrt{64+ 1 } = \sqrt{65}$

This is the height of one triangular face; its base is a side of the square, so the length of the base is 2. The area of a face is half the product of these dimensions, or

$\frac{1}{2} \times 2 \times \sqrt{65} = \sqrt{65}$

Since twice the area of the base is $8 = \sqrt{64}$ , the problem comes down to comparing $\sqrt{64}$ and $\sqrt{65}$ ; the latter, which is (B), is greater.

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Question

Pyramid 1 has a square base with sidelength ; its height is .

Pyramid 2 has a square base with sidelength ; its height is .

Which is the greater quantity?

(a) The volume of Pyramid 1

(b) The volume of Pyramid 2

Answer

Use the formula $V = \frac {1}{3} Bh$ on each pyramid.

(a) $B = x^{2}; h = 2x$

$V = \frac {1}{3} Bh = \frac {1}{3} \cdot x^{2} \cdot 2x = \frac {2}{3} x^{3}$

(b) $B = (2 x)^{2} = 4x^{2}; h = x$

$V = \frac {1}{3} Bh = \frac {1}{3} \cdot 4x^{2} \cdot x = \frac {4}{3} x^{3}$

Regardless of , (b) is the greater quantity.

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Question

Which is the greater quantity?

(a) The volume of a pyramid with height 4, the base of which has sidelength 1

(b) The volume of a pyramid with height 1, the base of which has sidelength 2

Answer

The volume of a pyramid with height and a square base with sidelength is

$V = \frac{1}{3} Bh = \frac{1}{3} s^{2}h$ .

(a) Substitute : $V = \frac{1}{3} s^{2}h = \frac{1}{3} \cdot 1^{2} \cdot 4 = \frac{4}{3}$

(b) Substitute : $V = \frac{1}{3} s^{2}h = \frac{1}{3} \cdot 2^{2} \cdot 1 = \frac{4}{3}$

The two pyramids have equal volume.

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Question

Which is the greater quantity?

(a) The volume of a pyramid whose base is a square with sidelength 8 inches

(b) The volume of a pyramid whose base is an equilateral triangle with sidelength one foot

Answer

The volume of a pyramid is one-third of the product of the height and the area of the base. The areas of the bases can be calculated, but no information is given about the heights of the pyramids. There is not enough information to determine which one has the greater volume.

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Question

A pyramid with a square base has height equal to the perimeter of its base. Its volume is . In terms of , what is the length of each side of its base?

Answer

The volume of a pyramid is given by the formula

$V = \frac{1}{3} A h$

where is the area of its base and is its height.

Let be the length of one side of the square base. Then the height is equal to the perimeter of that square, so

and the area of the base is

$A = s^{2}$

So the volume formula becomes

$\frac{1}{3} \cdot s^{2} \cdot 4s = V$

Solve for :

$\frac{4}{3} \cdot s^{3} = V$

$\frac{3}{4} \cdot \frac{4}{3} \cdot s^{3} =\frac{3}{4} \cdot V$

$s^{3} =\frac{3V}{4}$

$s =\sqrt[3]{\frac{3V}{4}}$

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Question

Cube 2 has twice the sidelength of Cube 1; Cube 3 has twice the sidelength of Cube 2; Cube 4 has twice the sidelength of Cube 3.

Which is the greater quantity?

(a) The mean of the volumes of Cube 1 and Cube 4

(b) The mean of the volumes of Cube 2 and Cube 3

Answer

The sidelengths of Cubes 1, 2, 3, and 4 can be given values , respectively.

Then the volumes of the cubes are as follows:

Cube 1: $V= s^{3}$

Cube 2: $V= (2s)^{3} = 8s^{3}$

Cube 3: $V= (4s)^{3} = 64s^{3}$

Cube 4: $V= (8s)^{3} = 512s^{3}$

In both answer choices ask for a mean, so we can determine which answer (mean) is greater simply by comparing the sums of volumes.

(a) The sum of the volumes of Cubes 1 and 4 is $s^{3}+ 512^{3} = 513s^{3}$ .

(b) The sum of the volumes of Cubes 2 and 3 is $8s^{3}+ 64^{3} = 72s^{3}$ .

Regardless of , the sum of the volumes of Cubes 1 and 4 is greater, and therefore, so is their mean.

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Question

What is the length of the diagonal of a cube with a side length of ? Round to the nearest hundreth.

Answer

It is easiest to think about diagonals like this by considering two points on a cube (if it were drawn in three dimensions). We could draw it like this:

(Note that not all points are drawn in on the cube.)

The two points we are looking at are:

and

Solve this by using the distance formula. This is very easy since one point is all s. It is merely:

$\sqrt{12.5^2+12.5^2+12.5^2}=\sqrt{468.75}$

This is approximately .

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Question

What is the length of the diagonal of a cube with a side length of ? Round to the nearest hundreth.

Answer

It is easiest to think about diagonals like this by considering two points on a cube (if it were drawn in three dimensions). We could draw it like this:

(Note that not all points are drawn in on the cube.)

The two points we are looking at are:

and

Solve this by using the distance formula. This is very easy since one point is all s. It is merely:

$\sqrt{2.75^2+2.75^2+2.75^2}=\sqrt{22.6875}$

This is approximately .

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Question

What is the length of the diagonal of a cube with a volume of ? Round to the nearest hundredth.

Answer

First, you need to find the side length of this cube. We know that the volume is:

, where is the side length.

Therefore, based on our data, we can say:

Solving for by taking the cube-root of both sides, we get:

Now, it is easiest to think about diagonals like this by considering two points on a cube (if it were drawn in three dimensions). We could draw it like this:

(Note that not all points are drawn in on the cube.)

The two points we are looking at are:

and

Solve this by using the distance formula. This is very easy since one point is all s. It is merely:

$\sqrt{7^2+7^2+7^2}=\sqrt{147}$

This is approximately .

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Question

What is the length of the diagonal of a cube with a surface area of ? Round your answer to the nearest hundredth.

Answer

First, you need to find the side length of this cube. We know that the surface area is defined by:

, where is the side length. (This is because the cube is sides of equal area).

Therefore, based on our data, we can say:

Take the square root of both sides and get:

Now, it is easiest to think about diagonals like this by considering two points on a cube (if it were drawn in three dimensions). We could draw it like this:

(Note that not all points are drawn in on the cube.)

The two points we are looking at are:

and

Solve this by using the distance formula. This is very easy since one point is all s. It is merely:

$\sqrt{5^2+5^2+5^2}=\sqrt{75}$

This is approximately .

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Question

What is the volume of a cube on which one face has a diagonal of ?

Answer

One of the faces of the cube could be drawn like this:

Notice that this makes a triangle.

This means that we can create a proportion for the sides. On the standard triangle, the non-hypotenuse sides are both , and the hypotenuse is $\sqrt{2}$ . This will allow us to make the proportion:

$\frac{1}{\sqrt{2}} = \frac{x}{2}$

Multiplying both sides by , you get:

$x=\frac{2}{\sqrt{2}}$

Recall that the formula for the volume of a cube is:

Therefore, we can compute the volume using the side found above:

$V = (\frac{2}{\sqrt{2}})^3=\frac{8}{(\sqrt{2})^3}=\frac{8}{2\sqrt{2}}=\frac{4}{\sqrt2}$

Now, rationalize the denominator:

$\frac{4}{\sqrt2} * \frac{\sqrt{2}}{\sqrt{2}} = \frac{4\sqrt{2}}{2}= 2\sqrt{2}$

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