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Mathematics 19 Online
OpenStudy (anonymous):

What are the dimensions of a rectangle of 12*10 inches, if resized to 1/8th the area (15 square inches) and scaled proportionally?

OpenStudy (amistre64):

the area is 120. if we scale that to 1/8 we get: 120/8 = 60/4 = 30/2 = 15 1,15; 3,5 are the factors of 15. Id take a gander and say 3 and 5; but that s just a guess really....

OpenStudy (anonymous):

A 3 x 5 would proportionally be a 6 x 10

OpenStudy (amistre64):

blex, area is proportional, not side length. for example: 2*4 = 8 lets scale the side by 3 6*12 = 72 but the area is not 8(3) = 24...not 72

OpenStudy (anonymous):

What about when you have two similar triangles? You take the corresponding sides and make them proportional to the other sides... don't you?

OpenStudy (amistre64):

if you want to scale "sides" then yes; but if you want to scale "area" then no.... because area does not scale in the same manner as sides. but I think I see a point you are making... if we want the sides to stay proportional as well, then then we should not restrict our answers to "intergers" alone....

OpenStudy (anonymous):

That's what I am thinking...

OpenStudy (amistre64):

then we might want to setup a few equation to help keep things "in order ":)

OpenStudy (anonymous):

My real question is what, let what be x, times itself multiplied by 1.2, will yield 15. So 3*5 does equal 15, but 5 is not 1.2(3). Solve for x if x(1.2x)=15. That's my question.

OpenStudy (amistre64):

12(x) * 10(x) = 15 would that be a good equation?

OpenStudy (amistre64):

this would factor each side proportionally and get us an area of 15... i think

OpenStudy (amistre64):

120x^2 = 15 x^2 = 15/120 = 5/40 = 1/8 x^2 = 1/8 x = sqrt(1/8)

OpenStudy (anonymous):

The area of the original rectangle is given as 12*10, i.e 12 is the length and 10 is the width. So, the smaller rectangle should also have the length and width proportional to 12:10 and area 15 sq.inches

OpenStudy (amistre64):

12*sqrt(8)/8 = side "a" 10*sqrt(8)/8 = side "b"

OpenStudy (anonymous):

x*1.2x = 15or x^2 = 15/1.2 or x = sqrt(15/1.2)

OpenStudy (anonymous):

i.e length = 4.24 and width = 3.53

OpenStudy (amistre64):

4.2426406871192851464050661726291 and 3.5355339059327376220042218105242 so its the same either way ;)

OpenStudy (anonymous):

I think the dimensions of the rectangle would be [sqrt{18}] * 5/[sqrt{2}]. Just divide both - length and breadth by [sqrt{8}]. Here sqrt means a squareroot

OpenStudy (amistre64):

wpg; that is correct.

OpenStudy (anonymous):

I know. It was not any complicated. I've joined openstudy.com just now.. Looks great place.

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