Answer #1: they could make the frame with an area of 65 square inches the same area as the triangle the cost of the frame would be low and there would be no wasted space the fame would be 3 sided 20*14*10

Explanation: i thought of both the different ways the problem showed but the area would be much smaller if you just framed the painting in a triangle frame it they said it could be framed 3 different ways and it only showed 2 ways of framing the pictured i am right unless the frame had to be a rectangle

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Answer #2: They need to frame the picture with 10 inches as the base

Explanation: I know that the area of the triangle is twelve square inches. I also know that to get the area of a triangle you need base times height divided by two equals the area. (b*h)/2
Then I plugged all the known bases (10, 14, and 20) into the formula, 10*h/2 = 65, 20*h/2 = 65 and 14*h/2 = 65. Simplified that's
10*h = 130
14*h = 130 and
20*h = 130.
That comes out to the heights being 13, 9 and two sevenths, and 6 and a half. The frames will be the perimeter of the picture. Perimeter is 2*Base+2*Height.
2*13+2*10 = 46
2*(9+2/7)+2*14 = (46+4/7)
2*(6+1/2)+2*20 = 53. The first one is the lowest and therefore the smallest perimeter.

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Answer#3: The 10 inch side would be the cheapest frame for the triangle because the perimeter is 46 inches which is the smallest of all the combinations.

Explanation: I used the equation bh/2 (Base times height divided by 2). For the triangle with the 20 inch side used as the base, I found that the height is 6.5 inches and the perimeter equals 53 inches. For the triangle with the 14 inch side used as base, I found that the height is around 9.3 inches and the perimeter is 46.6 inches. For the triangle with the base of 10 inches, I found that the height equal 13 inches and the perimeter equals 46 inches. This is the smallest perimeter out of the three possible triangle bases that I found. How I found the height of each triangle and its frame is that I used the equation bh/2. I plugged in the area for triangle given in the problem, and then for each triangle, I plugged in the base side used for each of the three possible solutions/frames. Then I just added up the sides of each triangle using the base and the height. BONUS: If they took account the glass used in the frame, all you would have to do is add the area of the rectangle plus the perimeter of the frame. Therefore, if you add something extra to each of the frames, the cost would obviously go up. The 10 inch one would still be the cheapest of them all.

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Answer #4: The frame that would use the least amount of material is the frame with the 14 inch side flush to a side of the triangle.

Explanation: The area of the triangle is given as 65 and to find the area of a triangle it is half the product of the base and the height (A=1/2*b*h). So I replaced A with 65 inches and b with 14 in., 20 in., or with 10 in. After finding the height of each triangle which is 6.5 in. for the frame with a side of 20 in., 7 2/7 in. for the frame with a side 14 in., and for the frame with a side of 10 in. the height is 13 in. All the frames are rectangles so I multiplied the height of the triangle, which becomes the height of the frame, by the side given. The three areas were 130 in. squared for the frame with 20 in. as a side, 102 in. squared for the frame with 14 in. as a side, and 130 in. squared for the side with 10 in. as the side given.

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Answer #5: the frame with the dimensions of 10 inches by 13 inches will take the least amount of material

Explanation: Let the length (the base of the triangle) of the frame be b inches, and the width (the height of the triangle) be h inches.
Case 1: b = 20
A = (1/2)bh
65 = (1/2)(20)h
h = 6.5
P = 2b + 2h
P = 2(20) + 2(6.5) = 53

Case 2:
b = 10
A = (1/2)bh
65 = (1/2)(10)h
h = 13
P = 2b + 2h = 2(10) + 2(13) = 46

Case 3:
b = 14
A = (1/2)bh
65 = (1/2)(14)h
h = 65/7
P = 2b + 2h = 2(14) + 2(65/7) = approximately 47

the frame with the dimensions of 10 inches by 13 inches will take the least amount of material

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Answer#6: All 3 framing posibilities would use equal amounts of framing material

Explanation: First, you need to find the height of each each scenario using the area formula for a triange. With a base of 20,the height is 6.5, with a base of 10, the height is 13, and with a base of 14, the height is 65/7. After you do this step, you need to find the area of all three frames and they all equal 130 sq. in., therefore all three frames are equal

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Answer#: TO HAVE THE SMALLEST FRAME, YOU PUT THE 20 INCH SIDE DOWN, THE FRAME WILL HAVE A PERIMITER OF 53.64 INCHES, HOWEVER, IF YOU WERE TO BREAK WITH CONVENTION AND HAVE A TRIANGULAR FRAME, THEN THE PERIMETER WOULD BE 48 INCHES.

Explanation: TO FIND THE PERIMETER OF A RECTANGLE YOU ADD ALL THE SIDES TOGETHER, WE KNOW THAT TWO OF THE SIDES WERE 20 INCHES THE TRICK IS TO FIND THE LENGTH OF THE OTHER SIDE. WE USED THE LAW OF COSINES TO FIND THE MEASUERE'S OF ALL THE ANGLES. WE THEN USED THE LAW OF SINES AND RIGHT ANGLE TRIG TO FIND TEH LENGTH OF THE ALTITUDE PERPENDICULAR TO THE SIDE WITH A MEASURE OF 20 INCHES. WE FOUND THAT THIS ALTITUDE WAS 6.82. ADDING 20+20+6.82+6.82 GIVES A MEASURE OF 53.64 INCHES. THIS IS A SMALLEST PERIMETER POSSIBLE WITH A TRIANGLE OF THIS SIZE.
IF YOU HAVE A TRIANGULAR FRAME THOUGH, THEN YOU CAN BUILD THE BUILD THE FRAME TO FIT THE EXACT OUTER DIMENTIONS OF THE ART, IN THIS CASE THE PERIMETER OF SUCH A FRAME WOULD BE 48 INCHES.
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