Question

Frank's painting is on a large triangular canvas with side lengths 4 meters,

5 meters and 7 meters, as shown. He wants to create a frame such that its inner
border is flush with each side of the canvas and every point on its outer border
is a distance of exactly 2 cm from the closest point on the canvas. In square
centimeters, what is the area of the frame? Express your answer to the nearest
whole number
4 m
5
m

Answer 1

3213 cm

Explanation:

First draw the frame around the triangle. The circular edges added together give you 360 degrees. Use pi*r^2 to get 4pi which is around 12.56. 100cm=m. 2*500+2*400*2*700=3200.

3200+ 13 = 3213

Answer 2

`A_(frame)=958 cm^(2)`

Explanation:

The triangle is attached.

According to the problem, the inner frame is 2 cm wide.

To find the area of that inner frame, we just have to subtract the outer triangle area with the inner triangle area.

Now, we know that the outer part of the triangular canvas has dimensions of 4m, 5m and 7, which are equivalent to 400cm, 500cm, and 700cm, because 1 meter is 100 centimeters.

Now, the area of the other triangle is

`A_(outer)=√(p((p-a)(p-b)(p-c))`

Where
`p` is half the perimeter. So, if the perimeter is

`P=400+500+700=1600 cm`

`p=(1600cm)/(2)=800cm`

And
`a, b, c` are the sides of the triangle.

Replacing all these value, we have

`A_(outer)=√(800((800-400)(800-500)(800-700))\\A_(outer)\approx 97979.59 cm^(2)`

Then, we do the same process, but in this case, the inner canvas has dimensions of 398cm, 498cm and 698 cm.

So, the perimeter is

`P=398+498+698=1594cm`

`p=797cm`

And the area would be

`A_(inner)=√(797(797-398)(797-498)(797-698)) \\A_(inner)\approx 97022cm^(2)`

So, the area of the frame is

`A_(frame)=A_(outer)-A_(inner)=97980-97022\\A_(frame)=958 cm^(2)`