Remark
This problem is one of those little gems you come across every so often. It has a couple of things worth knowing even if it is a little late.
Step One
Find a + b [which are the two legs of the triangle.]
r = the inradius
r = 4 in this question.
r = (a + b - c)/2 where c is the hypotenuse. This is easily proved. If you can't, google inradius and perimeter.
4 = (a+b - 20)/2 Multiply by 2
4*2 = a+b - 20
8 = a+b - 20 Add 20 to both sides.
8 + 20 = a + b
28 = a + b
Step two
Set up the Pythagorean Theorem
a^2 + b^2 = c^2
a^2 + b^2 = 20^2
a^2 + b^2 = 400
Step three
Expand (a + b)^2 to find ab
(a + b)^2 = a^2 + 2ab + b^2
(28)^2 = a^2 + b^2 + 2ab
Remember that a^2 + b^2 = 400
28^2 = 400 + 2ab
784 = 400 + 2ab Subtract 400 from both sides.
784 - 400 = 2ab
384 = 2ab divide by 2
384/2 = ab
ab = 192
Step 4
Find a and b
ab = 192
a + b = 28
b = 28 - a
ab = 192 Substitute 28 - a for b
a(28 - a) = 192
-a^2 + 28a = 192
-a^2 + 28a - 192 = 0
You have to use the quadratic formula.
a = -1
b = 28
c = - 192
When you solve this you get 2 roots.
a = 12 and b = 16 These are the two sides of the triangle or
a = 16 and b = 12
Check
a^2 + b^2 =? 20^2
12^2 + 16^2 = ?400
144 + 256 =? 400
400 = 400
The problem does check
Neat little problem Thanks for posting.
I would solve 2*r = (a + b - c) for you but I can't get the many diagram programs to work properly for labeling these diagrams.
There are great discussions on how to make the incircle radius an integer value by choosing the triangle sides with an integer value.
This problem is one of those little gems you come across every so often. It has a couple of things worth knowing even if it is a little late.
Step One
Find a + b [which are the two legs of the triangle.]
r = the inradius
r = 4 in this question.
r = (a + b - c)/2 where c is the hypotenuse. This is easily proved. If you can't, google inradius and perimeter.
4 = (a+b - 20)/2 Multiply by 2
4*2 = a+b - 20
8 = a+b - 20 Add 20 to both sides.
8 + 20 = a + b
28 = a + b
Step two
Set up the Pythagorean Theorem
a^2 + b^2 = c^2
a^2 + b^2 = 20^2
a^2 + b^2 = 400
Step three
Expand (a + b)^2 to find ab
(a + b)^2 = a^2 + 2ab + b^2
(28)^2 = a^2 + b^2 + 2ab
Remember that a^2 + b^2 = 400
28^2 = 400 + 2ab
784 = 400 + 2ab Subtract 400 from both sides.
784 - 400 = 2ab
384 = 2ab divide by 2
384/2 = ab
ab = 192
Step 4
Find a and b
ab = 192
a + b = 28
b = 28 - a
ab = 192 Substitute 28 - a for b
a(28 - a) = 192
-a^2 + 28a = 192
-a^2 + 28a - 192 = 0
You have to use the quadratic formula.
a = -1
b = 28
c = - 192
When you solve this you get 2 roots.
a = 12 and b = 16 These are the two sides of the triangle or
a = 16 and b = 12
Check
a^2 + b^2 =? 20^2
12^2 + 16^2 = ?400
144 + 256 =? 400
400 = 400
The problem does check
Neat little problem Thanks for posting.
I would solve 2*r = (a + b - c) for you but I can't get the many diagram programs to work properly for labeling these diagrams.
There are great discussions on how to make the incircle radius an integer value by choosing the triangle sides with an integer value.