Joshua W. answered • 04/09/15
Tutor
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Joshua, math tutor
Say that the diagonals of the rhombus are m and n. The area of the rhombus is A. Now the formula is A = (m*n)/2. We want to multiply both m and n by 8, so we have A = (8m*8n)/2. Do the arithmetic, and you have A=64(m*n)/2, but the formula that is being multiplied by 64 is the original area, so what happens is, the area of the new rhombus is the the area of the old rhombus, times 64. We could say that the new area is bold, and the old area is regular. A=64*A.
This is a little different (the circle problem). A=(pi)(r)^2. We will do C=2(pi)r. Thus, C/(2pi)=r, so A=(pi)(C/2pi)^2. Now we can multiply C by 2.4 and see how that affects A. A=((2.4)^2)(pi)(C/2pi)^2. That is, A=((2.4)^2)*A
Remember, bold A means new area. Non-bold A means old area.
A=b*h is the formula for the area of the rectangle. b is base, and h is height. We multiply b by 4 and h by 7. So A=(4b*7h)=28(b*h)=28*A
A=nsa/2 where n is the number of a polygon's sides (in this case, 8), s is the side length, a is the apothem. It's done the same way the rectangle problem was done. A=3*A
With the last one, maybe draw a picture. A square has equal sides, so its diagonal is 1.4142*a, where a is the side length. Now draw another square. If you divide the diagonal's length from 1.4142*a, to (1.4142*a)/4, then the side of the square is now a/4 instead of just a. You can prove this to yourself by using a right triangle, with two equal sides. The hypotenuse will be like the diagonal. By the way, 1.4142 is the square root of 2. The area of the old square was a^2, and what we've changed is the side length, from a to a/4. S0, A=(a/4)^2=A/16.
I hope you enjoyed this. Maybe we can meet in person sometime.
