SOLUTION: The hypotenuse of a right triangle is 12 inches and the area is 24 square inches. Find the dimension of the triangle, correct to one decimal place.

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Question 1040514: The hypotenuse of a right triangle is 12 inches and the area is 24 square inches. Find the dimension of the triangle, correct to one decimal place.
Found 2 solutions by Alan3354, josmiceli:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
The hypotenuse of a right triangle is 12 inches and the area is 24 square inches. Find the dimension of the triangle, correct to one decimal place.
==========
Area = b*h/2 = 24
Using 12 for the base, h = 4.
The altitude from the right angle = 4.
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Label the 2 parts of the base c & d, and the 2 sides e & f
c + d = 12
e^2 - c^2 = 16
f^2 - d^2 = 16
e^2 + f^2 = 12^2
------
Sub for c: c = 12 - d
e^2 - d^2 + 24d - 144 = 16
e^2 = d^2 - 24d + 160
---
Sub for f: f^2 = d^2 + 16
e^2 + d^2 + 16 = 144
--> e^2 = 128 - d^2
----
128 - d^2 = d^2 - 24d + 160
2d^2 - 24d + 32 = 0
d^2 - 12d + 16 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-12x%2B16+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-12%29%5E2-4%2A1%2A16=80.

Discriminant d=80 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--12%2B-sqrt%28+80+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-12%29%2Bsqrt%28+80+%29%29%2F2%5C1+=+10.4721359549996
x%5B2%5D+=+%28-%28-12%29-sqrt%28+80+%29%29%2F2%5C1+=+1.52786404500042

Quadratic expression 1x%5E2%2B-12x%2B16 can be factored:
1x%5E2%2B-12x%2B16+=+%28x-10.4721359549996%29%2A%28x-1.52786404500042%29
Again, the answer is: 10.4721359549996, 1.52786404500042. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-12%2Ax%2B16+%29

=================
The 2 solutions are c & d. Their sum is 12, the hypotenuse.
6 + sqrt(20) and 6 - sqrt(20)
---------
e^2 = c^2 + 16
= 36 + 20 + 12sqrt(20)
e+=+sqrt%2856+%2B+12sqrt%2820%29%29
f+=+sqrt%2856+-+12sqrt%2820%29%29

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +A+ = the area of the right triangle in square inches
Let +b+ = the base of the right triangle in inches
Let +c+ = the height of the right triangle in inches
----------------------------------------------
+A+=+%281%2F2%29%2Ab%2Ac+
+24+=+%281%2F2%29%2Ab%2Ac+
+b%2Ac+=+48+
+c+=+48%2Fb+
-------------------
Use pythagorean theorem

+b%5E4+%2B+2304+=+144b%5E2+
+144b%5E2+-+b%5E4+=+2304+
Let +z+=+b%5E2+
+-z%5E2+%2B+144z+-+2304+=+0+
Use quadratic formula
+z+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
+a+=+-1+
+b+=+144+
+c+=+-2304+
+z+=+%28+-144+%2B-+sqrt%28+144%5E2-4%2A%28-1%29%2A%28-2304%29+%29%29%2F%282%2A%28-1%29%29+
+z+=+%28+-144+%2B-+sqrt%28+20736+-+9216+%29%29%2F%28%28-2%29+%29+
+z+=+%28+-144+%2B-+sqrt%28+11520+%29+%2F+%28-2%29+%29+
+z+=+%28+-144+%2B+107.3313+%29+%2F+%28-2%29+
+z+=+36.6687%2F2+
+z+=+18.3344+
and, also:
+z+=+%28+-144+-+107.3313+%29+%2F+%28-2%29+
+z+=+251.3313%2F2+
+z+=+125.6657+
Since +z+=+b%5E2+
+b+=+sqrt%28z%29+
+b+=+sqrt%28+18.3344+%29+
+b+=+4.2819%0D%0Aand%0D%0A%7B%7B%7B+c+=+48%2Fb+
+c+=+48%2F4.2819+
+c+=+11.21+
--------------------
check:
+24+=+%281%2F2%29%2Ab%2Ac+
+48+=+b%2Ac+
+48+=+4.2819%2A11.21+
+48+=+48.0001+
close enough
----------------------

+144+=+143.9988+
close enough
-----------------------
The triangle is:
12 x 4.3 x 11.2
Hope I got it!