Pyramids - ISEE Upper Level (grades 9-12) Quantitative Reasoning

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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

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

What is the length of one side of a cube that has a volume of ?

Answer

We must begin by using the equation for the volume of a cube:

(It is like doing the area of a square, then adding another dimension!)

We know that the volume is . Therefore, we can rewrite our equation:

Using your calculator, we can find the cube root of . It is . (If you get just round up to . This is a calculator issue!).

This is the side length you need!

Another way you could do this is by cubing each of the possible answers to see which gives you a volume of .

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Question

What is the length of one side of a cube that has a surface area of ?

Answer

Recall that the formula for the surface area of a cube is:

, where is the length of a side of the cube. This equation is easy to memorize because it is merely a multiplication of a single side () by because a cube has equal sides.

Now, we know that is ; therefore, we can write:

Solve for :

Take the square root of both sides:

This is the length of one of your sides.

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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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