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MAVERICK [17]
3 years ago
6

Three spherical balls with radius r are contained in a rectangular box. two of the balls are each touching 5 sides of the rectan

gular box and the middle ball. the middle ball also touches four sides of the rectangular box. What is the volume of the space between the balls and the rectangular box? This is a SAT question and is no calculator. Show all the work Answer is 4r^3(6-pi)

Mathematics
1 answer:
tangare [24]3 years ago
3 0

Answer:

\bold{4r^3 (6-\pi)}

Step-by-step explanation:

Let us try to visualize the given situation in the form of 2 Dimensional image as shown in the attached diagram.

Let the Diameter of spherical balls be D.

As the spherical balls are completely fit in the rectangular box, the sides of box become:

D \times D\times (3 \times D) OR D\times D\times 3D

We know that Diameter is twice of radius.

Therefore D = 2r

So, the dimensions of rectangular box becomes:

2r \times 2r \times 6r

Volume of a rectangular box is given as:

V = Length \times Width \times Height\\\Rightarrow V = 2r\times 2r\times 6r = \bold{24r^3}

Now, let us find out the volume of each spherical ball.

Volume of a sphere with radius 'r' is given as:

V_{sphere} = \dfrac{4}{3}\pi r^3

Volume of 3 spheres = 3\times \frac{4}{3}\pi r^3 =4\pi r^3

Now, the volume of space between the balls and rectangular box is = Volume of rectangular box - Volume of 3 balls

24r^3 - 4\pi r^3\\\Rightarrow \bold{4r^3 (6-\pi)}

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2 years ago
Solve the following problems. Remember that all reasoning must be explained, and all steps of math work must be shown!
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<h2>Question 1:</h2>

<h3>How to solve Part A</h3>

Since the diagram shows a right triangle and gives out some measurements of the sides, you can use the Pythagorean Theorem (a^{2}+b^{2}=c^{2}) to find the length of MP.

<u>Given:</u>

hypotenuse = 7  – because the radius is 7 units and it can be used as the hypotenuse

One side of the triangle = 4  – because the length of PO is 4 units and it can be used as one sides of the triangle

<u>To Solve:</u>

Since we know two values of the triangle, you can use the Pythagorean Theorem (a^{2}+b^{2}=c^{2}) to find what the other value is. The variables <em>c</em> and <em>b</em> in the formula are already given, and we need to find what <em>a</em> is to find the length of MP. So, plug the given values of <em>c</em> and <em>b</em> into the formula:

a^{2}+4^{2}=7^{2}

Simplify:

a^{2}+16=49

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a^{2}=33

Do the square root of both sides to isolate the variable <em>a</em>:

\sqrt{a^{2}}=\sqrt{33}

Evaluate:

a=\sqrt{33}  Which simplifies to a= 5.744562646.... and round it to the nearest tenth to finally get a=5.7

Answer: The length of MP is 5.7 units

<h3>How to solve Part A</h3>

If you look at the diagram in the question, you can see that the length of MN is twice the length of MP. So multiply the length of MP, which is 5.7, by 2:

5.7 · 2 = 11.4

Answer: The length of MN is 11.4 units

___________________________________________________________

<h2>Question 2:</h2>

<h3>How to solve Part A</h3>

Look at the <u>first image</u> below ↓

<u>Description of the first image:</u>

The first image shows a drawn image of the circle with labeled parts and measurements from the infomation given from the question.

The diameter is 26cm, which means that the radius is 13cm because the radius half of the diameter in a circle. The very top line is the surface of the water where it’s filled and it has a length of 20cm. Side <em>b</em> of the triangle is 10cm becuase it’s half the length of the the surface of the water bowl. The hypotenuse (side <em>c</em>) is 13cm because it’s the length of the radius. And side <em>a </em>is what what we need to find because it’s the distance between the surface of the water and the center of the bowl. But after when you find the length of side <em>a</em>, you need to add the length of the radius (which is 13cm) to the length of side <em>a</em> because that’s the rest of the length/depth of the water that’s filled in the bowl. The length of the radius is included with the depth of the water.

<u>Total information given:</u>

⇒ radius = 13cm  – because the radius is half of the diameter, and the diameter has a length of 26cm

⇒ length of the surface where the water’s filled = 20cm

⇒ hypotenuse (side <em>c</em>) = 13cm  – because it’s the length of the radius

⇒ side <em>b</em> = 10cm  – because it’s half the length of the surface where the water’s filled

<u>To Solve</u>

Since there is a right triangle shown in the image, you can use the Pythagorean Theorem  (a^{2}+b^{2}=c^{2}) to find the missing length of side <em>a</em>. Plug the given values of <em>c</em> and <em>b</em> into the formula:

a^{2}+10^{2}=13^{2}

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Subtract 100 from both sides of the equation:

a^{2}=69

Do the square root of both sides to isolate the variable <em>a</em>:

\sqrt{a^{2}}=\sqrt{69}

Evaluate:

a=\sqrt{69}  Which simplifies to a= 8.30662386....  and round it to the nearest tenth to finally get a ≈ 8.3

This means that 8.3 is just the length of side <em>a</em>, which is the length between the surface of where the water’s filled and the center of the circle. But, now you need to add 13 to 8.3 to find the total depth of the water because the length of the radius is included with the depth of the water.

So,

13 + 8.3 = 21.3

Answer: The depth of the water is approximately 21.3 cm

<h3>How to solve Part B (is in the image below)</h3>

<em />

<em>Sorry for the very, VERY long explanation and the long solving process, but I really, really hope you understand and that this helps with your question! </em>:)

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Answer:

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