Spherical water tank of radius R = 5m is emptied through a small circular hole of radius r = 0.03 m at the bottom. The top of th
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Ok so this question is a bit complicated, but it's easier to understand if you break it down into smaller parts!
1) First, you know that ABGF is half the perimeter of ACDE. This means that the length of one side of ABGF must be 1/2 the length of one side of ACDE.
>> You can think of this by putting in random numbers. Say the perimeter of the larger square is 24 and the perimeter of the smaller square is 12. That means one side of the larger square of 24/4 (b/c four sides) = 6 and one side of the smaller square is 12/4 = 3!
2. Ok know you know the lengths of the sides relative to each other, but you're only given one value: 4in. Since the smaller square has sides that are 1/2 the larger squares, you know that it makes up 1/4 of the larger square! So imagine 4 of those smaller squares filling up that larger square to make a 2 by 2. It just so happens that 4in is the diagonal going through one of our imaginary squares, which is equal in size to ABGF!
3. Now use the 45-45-90 rule to figure out the length of one side of that imaginary square because the 4in diagonal splits that imaginary square into two of those 45-45-90 triangles. You know the hypotenuse of that triangle is 4in. That means one of the legs is 4/✓2 (since the rule says that the hypotenuse and the leg are in a ✓2:1 ratio). And like we said before the length of that leg is the length of the side of our imaginary square. And our imaginary square must be the same size as ABGF! So now we know the side of the smaller square to be 4/✓2!
4. Multiply the side of the smaller square by 2 to get the side of our larger square. (4/✓2)*2=8/✓2
5. Now to find the area of the shaded region, just find the area of the smaller square ABGF and subtract from the larger square ACDE. Use equation for the area of a square!
where s=the length of one side.
The length of one side of the smaller square is 4/✓2. So it's area is:
The length of one side of the larger square is 8/✓2. So it's area is:
Now subtract. 32-8=24! :)
Hope this helps! Let me know if you have any questions.
1) First, you know that ABGF is half the perimeter of ACDE. This means that the length of one side of ABGF must be 1/2 the length of one side of ACDE.
>> You can think of this by putting in random numbers. Say the perimeter of the larger square is 24 and the perimeter of the smaller square is 12. That means one side of the larger square of 24/4 (b/c four sides) = 6 and one side of the smaller square is 12/4 = 3!
2. Ok know you know the lengths of the sides relative to each other, but you're only given one value: 4in. Since the smaller square has sides that are 1/2 the larger squares, you know that it makes up 1/4 of the larger square! So imagine 4 of those smaller squares filling up that larger square to make a 2 by 2. It just so happens that 4in is the diagonal going through one of our imaginary squares, which is equal in size to ABGF!
3. Now use the 45-45-90 rule to figure out the length of one side of that imaginary square because the 4in diagonal splits that imaginary square into two of those 45-45-90 triangles. You know the hypotenuse of that triangle is 4in. That means one of the legs is 4/✓2 (since the rule says that the hypotenuse and the leg are in a ✓2:1 ratio). And like we said before the length of that leg is the length of the side of our imaginary square. And our imaginary square must be the same size as ABGF! So now we know the side of the smaller square to be 4/✓2!
4. Multiply the side of the smaller square by 2 to get the side of our larger square. (4/✓2)*2=8/✓2
5. Now to find the area of the shaded region, just find the area of the smaller square ABGF and subtract from the larger square ACDE. Use equation for the area of a square!
where s=the length of one side.
The length of one side of the smaller square is 4/✓2. So it's area is:
The length of one side of the larger square is 8/✓2. So it's area is:
Now subtract. 32-8=24! :)
Hope this helps! Let me know if you have any questions.