Math question?

Question

Architects and others use drafting tiangles which a pictured example shows two isosceles right triangles. The larger isosceles triangle has legs of length 8 inches. The distance between any pair of parallel sides is 1 inch. Find the lengths of the three sides of the smaller triangle.

Answer 1

The outside triangle has legs of 8 and a hypotenuse of 8 * sqrt(2).

The hardest part is figuring out the "1 inch" from each side. For the right angle sides, that is easy. For the hypotenuse, we have to think.

Easiest way is to split the triangle into two smaller 45-45-90 triangles. In other words bisect the 90 degree angle.

These two triangles will have a *hypotenuse* of 8 (same as the original legs) and legs of 4 sqrt(2) which is half of 8 sqrt(2).

Thus the diagonal bisector is 4 sqrt(2). From this you need to subtract 1" (distance from the hypotenuse) and sqrt(2) which is the distance from the right angle:

4 sqrt(2) - 1 - sqrt(2) = 3 sqrt(2) - 1

Working backwards from this result, you can see that the hypotenuse of the smaller triangle is double this:
6 sqrt(2) - 2

And using the ratio of 1 : 1 : sqrt(2), the legs will be:
6 - 2/sqrt(2)
= 6 - sqrt(2)

Turning these into numbers:
The legs of the small triangle:
6 - sqrt(2) ≈ 4.6"
The hypotenuse of the small triangle:
6 sqrt(2) - 2 ≈ 6.5"

Answer 2

This is purely a guess, based on your statement (smaller triangle inscribed within larger triangle, each side being 1" away from its parallel side).

By drawing smaller triangles in the 45° corner of the larger triangle, you can find that the legs of smaller triangle = 6 - √2

Therefore, the third side = √(2(6 - √2)²) = √(2(36 - 12√2 + 2)) = √(76 - 24√2)

Or, (6 - √2) * √2

Take your pick.

Answer 3

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Answer 4

you forgot to put the "pictured example" here