If the length and width of a rectangular solid are each decreased by 20%, by what percent must the height be increased for the volume to remain unchanged? Give answer the nearest whole percent..
please and thanks! try and show the work if you can, im desperate!!!
2007-05-30 12:37:39 · 5 answers · asked by Joanna 2 in Science & Mathematics ➔ Mathematics
the volume of the rectangular solid is length X width X height.
If you've reduced the length and width by 20%, then each are being multiplied by .8. So that will, by itself reduce the volume to .64 times the original. To maintain the original volume you'll want to increase the height by the reciprocal of that, or 1.5625, which when multiplied by .64 will cancel to 1.
So the height must be increased to 1.5625 times the original height, or increased by 56.25 %
2007-05-30 12:51:20 · answer #1 · answered by Phaedrus 3 · 0⤊ 0⤋
Volume = length x width x height, so if you decrease two of the variables by 20%, the volume would be .8 x .8 = .64 times smaller if the height remained the same. To keep the volume constant, you would need to increase the volume by 1/.64 = 1.5625, which is the same as saying 56.25% more.
2007-05-30 12:48:41 · answer #2 · answered by brubeck_take5 4 · 0⤊ 0⤋
volume = length * width * height
new volume is (length * 0.8) * (width * 0.8) * (height * x)
We need to figure out what x is, such that:
length * width * height = (length * 0.8) * (width * 0.8) * (height * x)
We can cancel out length, width & height on both sides of the equation, and get:
1 = 0.8 * 0.8 * x
x = 1/(0.8 * 0.8) = 1/0.64 = 100/64 = 25/16
The height must increase by 9/16.
In percentage terms, that's 900/16 = 56.25 percent.
Rounding to the nearest whole percent: Height must be increased by 56 percent.
Hope that helps!
2007-05-30 12:46:43 · answer #3 · answered by Bramblyspam 7 · 1⤊ 0⤋
The radius of the inscribed sphere ("r") is sqrt(6)/12 circumstances the area length (which we are going to call "a"). the quantity of the tetrahedron is sqrt(2)/12 circumstances the cube of the area length. So... r = sqrt(6)/12 * a V = sqrt(2)/12 * a^3 resolve the 1st equation for a: r = sqrt(6)/12 * a a = 12/sqrt(6) * r a = 2 * sqrt(6) * r Now plug that into the quantity equation: V = sqrt(2)/12 * a^3 V = sqrt(2)/12 * (2*sqrt(6)*r)^3 V = sqrt(2)/12 * 2^3 * sqrt(6)^3 * r^3 V = sqrt(2)/12 * 8 * 6 * sqrt(6) * r^3 V = sqrt(2) * 4 * sqrt(6) * r^3 V = 8 * sqrt(3) * r^3
2016-11-23 19:40:03 · answer #4 · answered by ? 3 · 0⤊ 0⤋
V= wlh
V = (.8w)(.8l)h = .64wlh
So we must multiply h by 100/64 = 1.5625 to get the same V
So .64wl(1.5625h) = wlh , which is the original volume.
The height must be multiplied by 1.5625, or if you prefer, the height must be increased by 56.25% = 56% to the nearest %..
2007-05-30 12:53:19 · answer #5 · answered by ironduke8159 7 · 0⤊ 0⤋