math

SDN Members don't see this ad. (About Ads)

what is the max number of isosceles right triangles whose legs are 3 inches long that can be cut from a rectangular sheet of paper measuring 14 x 10 inches.

a. 12
b. 15
c. 24
d. 30
e. 31

so 1. how do you do this problem?
2. why can't i compare areas? like the way i did it below....
my way: i compared areas: area of rectangular sheet/area of triangle
140/(.5) (3)(3)=140 (2/9) = approximately 15 x 2=30 this is under approximation so I choose E as the answer

answer to problem: C: 24
instead, kaplan compared sides 14/3 =4 and 10/3= 3 4x 3 =12 so if you double this you get 24 if you think about 2 isosceles =1 square.

Originally Posted by BrownieDDD

what is the max number of isosceles right triangles whose legs are 3 inches long that can be cut from a rectangular sheet of paper measuring 14 x 10 inches.

a. 12
b. 15
c. 24
d. 30
e. 31

so 1. how do you do this problem?
2. why can't i compare areas? like the way i did it below....
my way: i compared areas: area of rectangular sheet/area of triangle
140/(.5) (3)(3)=140 (2/9) = approximately 15 x 2=30 this is under approximation so I choose E as the answer

answer to problem: C: 24
instead, kaplan compared sides 14/3 =4 and 10/3= 3 4x 3 =12 so if you double this you get 24 if you think about 2 isosceles =1 square.

I'd do it just like how kaplan did. if you put two isosceles triangles together against their hypotenuse, it'll give you 3x3 square. you can fit 4 squares on a row and 3 squares on a column so you'll make 12 squares which is equal to 24 triangles.

You can't compare areas because after you cut out the triangles there will be paper left that is 2x10 and 14x1 rectangle (they have intersection where they meet so it's technically not 34 but number still works, it would be 2x9 and 14x1 or 2x10 and 12x1 = 32), which adds up to 34 and theoretically capable of making 3 more squares (9x3) = 6 more triangles. But that can't happen because triangle must have 3 inches by 3 inches but they only have 2 inches on one side and 1 inch on one side of the rectangle.

I'm sorry my explanation is very confusing it'll be easier if you visualize it


Edit: Sorry for my poor drawing but it'll help you probably. The side of the rectangle isn't big enough to construct triangles

Oh i get it i think...

so kaplan's way was just simplifying it...bc thinking about this in terms of area can be confusing..

bc if you think about the areas, the area of the triangle overestimates how much can fit in bc of the amount of each side left over.

So the area of the rectangle divided by area of triangle = 34 (this is overestimate)
Like you said 14/3=4 but u have about 2 left over
the other side 10/3=3 so 1 left over.

take the 2 left oer and multiply that by the 10= 20 ft^2
1 left over multiply by 14=14 ft^2

so total left over is 34 ft^2

see how many triangles canfit into 34 ft^2= about 7

and since the 34 ft^2 represents UNUSABLE parts that a full triangle can't fit (essentially it's "trash"), we must SUBTRACT 7 from 34! thus, # triangles that can fit fully =24

ah thanks so much.