Rectangular Solids & Cylinders - GMAT Quantitative

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What is the length of the diagonal of a cube if its side length is ?

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The diagonal of a cube extends from one of its corners diagonally through the cube to the opposite corner, so it can be thought of as the hypotenuse of a right triangle formed by the height of the cube and the diagonal of its base. First we must find the diagonal of the base, which will be the same as the diagonal of any face of the cube, by applying the Pythagorean theorem:

Now that we know the length of the diagonal of any face on the cube, we can use the Pythagorean theorem again with this length and the height of the cube, whose hypotenuse is the length of the diagonal for the cube:

is a cube and face has an area of . What is the length of diagonal of the cube ?

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To find the diagonal of a cube we can apply the formula $d=e$\sqrt{3}$$ , where is the length of the diagonal and where is the length of an edge of the cube.

Since we are given an area of a face of the cube, we can find the length of an edge simply by taking its square root.

Here the length of an edge is 3.

Thefore the final andwer is $d=e\sqrt3=3$\sqrt{3}$$ .

What is the length of the diagonal of cube , knowing that face has diagonal equal to ?

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To find the length of the diagonal of the cube, we can apply the formula, however, we firstly need to find the length of an edge, by applying the formula for the diagonal of the square.

$d=s$\sqrt{2}$$ where is the diagonal of face ABCD, and , the length of one of the side of this square.

The length of must be $\sqrt{2}$ , which is the length of the edges of the square.

Therefore we can now use the formula for the length of the diagonal of the cube:

$e$\sqrt{3}$$ , where is the length of an edge.

Since $e=$\sqrt{2}$$ , we get the final answer $\sqrt{6}$ .

If a cube has a side length of , what is the length of its diagonal?

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The diagonal of a cube is the hypotenuse of a right triangle whose height is one side and whose base is the diagonal of one of the faces. First we must use the Pythagorean theorem to find the length of the diagonal of one of the faces, and then we use the theorem again with this value and length of one side of the cube to find the length of its diagonal:

So this is the length of the diagonal of one of the faces, which we plug into the Pythagorean theorem with the length of one side to find the length of the diagonal for the cube:

A given cube has an edge length of . What is the length of the diagonal of the cube?

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The diagonal of a cube with an edge length can be defined by the equation $d=s$\sqrt{3}$$ . Given in this instance, $d=5$\sqrt{3}$$ .

A given cube has an edge length of . What is the length of the diagonal of the cube?

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The diagonal of a cube with an edge length can be defined by the equation $d=s$\sqrt{3}$$ . Given in this instance, $d=9$\sqrt{3}$$ .

A given cube has an edge length of $\sqrt{3}$ . What is the length of the diagonal of the cube?

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The diagonal of a cube with an edge length can be defined by the equation $d=s$\sqrt{3}$$ . Given $s=$\sqrt{3}$$ in this instance, $d=$\sqrt{3}$\left ( $\sqrt{3}$ \right )=3$ .

The slant height of a pyramid is one and one-half times the perimeter of its square base. The base has sides of length 15 inches. What is the surface area of the pyramid?

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The square base of the pyramid has four sides with length 15 inches, making its perimeter four times that, or 60 inches. The slant height is

$1 $\frac{1}{2}$ \times 60 = 90$ inches.

Therefore, the area of the base is $B = 15 ^{2} = 225$ square inches.

Each of the four lateral triangles has area $$\frac{1}{2}$ \cdot 15 \cdot 90 = 675$ square inches.

The total surface area is $225 + 4 \cdot 675 = 2,925$ square inches.

In three-dimensional space, the four vertices of a tetrahedron - a solid with four faces - have Cartesian coordinates $\small (0,0,0), (60,0,0), (0,60,0), (60,0,0)$ .

Give the surface area of the tetrahedron.

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The tetrahedron looks like this:

is the origin and are the other three points, which are 60 units away from the origin on each of the three (perpendicular) axes.

The bottom, front, and left faces are each right triangles whose legs each measure 60. Each face has area

$\small $\frac{1}{2}$ \cdot 60 \cdot 60 = 1,800$ .

The remaining face has three edges each a hypotenuse of one of three congruent right triangles, so its sides are congruent, and it is an equilateral triangle. Its sidelength can be found via the 45-45-90 Theorem to be $\small 60 $\sqrt{ 2}$$ , so its area is

$\small $\frac{\left( 60 \sqrt{2}\right $)^{2}$ \cdot $\sqrt{3}$}{4} = $\frac{3,600\left ( \sqrt{2}\right $)^{2}$ \cdot $\sqrt{3}$}{4} = $\frac{3,600 \cdot 2 \cdot \sqrt{3}$}{4}=1,800 $\sqrt{3}$$

The total area is

$\small 3\cdot 1,800 + 1,800 $\sqrt{3}$ = 5,400+ 1,800 $\sqrt{3}$$

A regular tetrahedron comprises four faces, each of which is an equilateral triangle. If the sum of the lengths of its edges is 120, what is its surface area?

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As shown in the diagram below, a regular tetrahedron has six congruent edges, so each has length $\small 120 \div 6 = 20$ :

The area of one face is the area of an equilateral triangle with sidelength 20, which is

$\small $$\frac{20^{2}$$$\sqrt{3}$}{4} = $\frac{400\sqrt{3}$}{4} = 100 $\sqrt{3}$$

The total surface area is four times this, or $\small 400 $\sqrt{3}$$ .

Refer to the above diagram, which shows a tetrahedron.

, and . Give the surface area of the tetrahedron.

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Three of the surfaces of the tetrahedron - $\bigtriangleup ADB$ , $\bigtriangleup ADC$ , and $\bigtriangleup BDC$ - are isosceles right triangles with hypotenuse 30, so by the 45-45-90 Theorem, each leg measures this length divided by $\sqrt{2}$ , or

$\frac{30 }{\sqrt{2}$}= $\frac{30 \cdot \sqrt{2}$}{$\sqrt{2}$ \cdot $\sqrt{2}$}= $\frac{30 \cdot \sqrt{2}$}{2} = 15 $\sqrt{2}$ .

The area of each of these triangles is half the product of its legs, so each area is

$$\frac{1}{2}$\cdot 15 $\sqrt{2}$ \cdot 15 $\sqrt{2}$ = $\frac{1}{2}$\cdot 15 \cdot 15\cdot $\sqrt{2}$ \cdot $\sqrt{2}$= $\frac{1}{2}$\cdot 15 \cdot 15\cdot 2=225$

Also, the legs are of the same measure among the triangles, the hypotenuses are as well, so the fourth surface $\bigtriangleup ABC$ is an equilateral triangle. Its sidelength is , so we use the equilateral triangle area formula to calculate its area:

$\frac{30 ^{2}$\cdot $\sqrt{3}$}{4} = $\frac{900\cdot \sqrt{3}$}{4} = 225 $\sqrt{3}$

Add the areas of the faces:

$225 + 225 + 225+ 225 $\sqrt{3}$ = 675 + 225 $\sqrt{3}$$

A regular tetrahedron is a solid with four faces, each of which is an equilateral triangle.

If the lengths of all of the edges of a regular tetrahedron are added, the total length is 120. What is the surface area of the tetrahedron?

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A tetrahedron looks like this:

The tetrahedron has six edges, and in a regular tetrahedron, they are congruent, so each edge has length $120 \div 6 = 20$ .

The area of one face of the tetrahedron, it being an equilateral triangle, can be calulated using the formula

$A = $$\frac{s^{2}$$$\sqrt{3}$}{4}$

There are four congruent faces, so the total surface area is

$S= 4A = 4 \cdot $$\frac{s^{2}$$$\sqrt{3}$}{4} = $s^{2}$$\sqrt{3}$$

Since , the surface area of the tetrahedron is

$S= $20^{2}$\cdot $\sqrt{3}$ =400 $\sqrt{3}$$

A regular tetrahedron is a solid with four faces, each of which is an equilateral triangle.

Each edge of a regular tetrahedron has length $6 $\sqrt{3}$$ . What is the surface area of the tetrahedron?

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The area of one face of the tetrahedron, it being an equilateral triangle, can be calulated using the formula

$A = $$\frac{s^{2}$$$\sqrt{3}$}{4}$

There are four congruent faces, so the total surface area is

$S= 4A = 4 \cdot $$\frac{s^{2}$$$\sqrt{3}$}{4} = $s^{2}$$\sqrt{3}$$

Since $s = 6 $\sqrt{3}$$ , the surface area of the tetrahedron is

$S= \left( 6 $\sqrt{3}\right $)^{2}$\cdot $\sqrt{3}$ $=6^{2}$ \cdot \left ( $\sqrt{3}\right $)^{2}$\cdot $\sqrt{3}$ = 36 \cdot 3 \cdot $\sqrt{3}$ = 108 $\sqrt{3}$$

Evaluate the surface area of the above tetrahedron.

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Three of the faces of the tetrahedron are isosceles right triangles with legs of length 8, and, subsequently, by the 45-45-90 Theorem, hypotenuses of length $8 $\sqrt{2}$$ . The fourth face, consequently, is an equilateral triangle with three sides of length $8 $\sqrt{2}$$ .

Each of the three right triangle faces has area equal to half the product of its legs , which is

$$\frac{1}{2}$ \cdot 8 \cdot 8 = 32$ .

The equilateral face has as its area

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

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

$= $\frac{ 64\cdot 2\cdot \sqrt{3}$}{4}$

$= $\frac{ 128\cdot \sqrt{3}$}{4}$

$= 32 $\sqrt{3}$$

The sum of the areas of the faces is

$32 $\sqrt{3}$ + 32+32+32 = 96+ 32 $\sqrt{3}$$

The cube in the above figure has surface area 384. Give the surface area of the tetrahedron with vertices , shown in red.

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The surface area formula can be used to find the length of each edge of the cube:

Three faces of the tetrahedron - $\bigtriangleup DHE$ , $\bigtriangleup DHG$ , $\bigtriangleup EHG$ - are right triangles with legs of length 8, so the area of each is half the product of the lengths of their legs:

$$\frac{1}{2}$ \cdot 8 \cdot 8 =32$ .

Each triangle is isosceles, so, by the 45-45-90 Theorem, each of their hypotentuses measures $\sqrt{2}$ times a leg, or $8$\sqrt{2}$$ . is therefore an equilateral triangle with sidelenghth $8$\sqrt{2}$$ . Its area can be found as follows:

$A = $$\frac{s^{2}$$$\sqrt{3}$}{4}$

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

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

$= $\frac{64\cdot 2 \cdot \sqrt{3}$}{4}$

$= 32 $\sqrt{3}$$

The total surface area is

$32+32+32+ 32 $\sqrt{3}$ = 96 + 32 $\sqrt{3}$$

The above diagram shows a regular right triangular pyramid. Its base $\bigtriangleup ABC$ is an equilateral triangle; the other three faces are congruent isosceles triangles, with $\overline{VM}$ an altitude of $\bigtriangleup AVC$ . Give the surface area of the pyramid.

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The base is an equilateral triangle with sidelength 12, so its area can be calculated as follows:

.

Each of the three other faces is congruent, with base 12. The area of each is the product of its base and its height. To find the common height, we examine $\bigtriangleup VMC$ , which, since $\overline{VM}$ is an altitude of isosceles $\bigtriangleup AVC$ , is a right triangle with hypotenuse of length 18 and one leg of length $MC = $\frac{1}{2}$ AC = $\frac{1}{2}$ \cdot 12 = 6$ . We can find using the Pythagorean Theorem:

$VM = $\sqrt{\left(VC \right $)^{2}$$ - \left( CM\right $)^{2}$}$

$= $$\sqrt{18^{2}$$ - $6^{2}$}$

$= $\sqrt{288}$$

$= $\sqrt{144}$ \cdot $\sqrt{2}$$

$=12$\sqrt{2}$$

The area of $\bigtriangleup AVC$ is half the product of this height and the base:

$\frac{1}{2}$ \cdot 12 \cdot 12 $\sqrt{2}$ = 72 $\sqrt{2}$

All three lateral faces have this area.

Now add the areas of the four faces:

$36 $\sqrt{3}$+72 $\sqrt{2}$+72 $\sqrt{2}$+72 $\sqrt{2}$ = 36 $\sqrt{3}$+216 $\sqrt{2}$$

What is the volume of a cube whose diagonal measures 10 inches?

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By an extension of the Pythagorean Theorem, if is the length of an edge of the cube and is its diagonal length,

$s$\sqrt{3 }$= d$

$s= $\frac{d}{\sqrt{3 }$}$

The volume is therefore

$V = $s^{3}$ = \left ($\frac{d}{\sqrt{3}$} \right $)^{3}$ = \left ($\frac{10}{\sqrt{3}$} \right $)^{3}$\approx 192.5$

A cube of iron has a mass of 4 kg. What is the mass of a rectangular prism of iron that is the same height but has a width and length that are twice as long as the cube?

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Let the length of the sides of the cube equal 1. The volume of the cube is then 1times 1times 1=1. Therefore, the volume of the prism is 2times 2times 1=4. Therefore, the mass of the prism must be 4 times greater than the mass of the cube.

(4)(4)= 16

The sum of the length, the width, and the height of a rectangular prism is one yard. The length of the prism is eleven inches greater than its width, and the width is twice its height. What is the volume of the prism?

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Let be the height of the prism. Then the width is , and the length is . Since the sum of the three dimensions is one yard, or 36 inches, we solve for in this equation:

$5h \div 5 = 25 \div 5$

The height is 5 inches; the width is twice this, or 10 inches; the length is eleven inches greater than the width, or 21 inches. The volume is the product of the three dimensions:

$V = lwh = 5 \cdot10\cdot21 = 1,050$ cubic inches.

The length, width, and height of a rectangular prism, in inches, are three different prime numbers. All three dimensions are between one yard and four feet. What is the volume of the prism?

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One yard is equal to 36 inches; four feet are equal to 48 inches. There are four prime numbers between 36 and 48 - 37, 41, 43, and 47. Since we are only given that the dimensions are three different prime numbers between 36 and 48, we have no way of knowing which three they are.