How to get exact surface of triangle if i only have hypotenuse and legs.
c=13
a+b=17
I tryed to do it by the system but it is not correct and it can't get correct...
+118687 Hi guest,
You have shown that the area will be 30 if the legs are 12 and 5 but the legs do not have to be 12 and 5.
You have not shown that the area will always be 30.
+37153 Hey Melody..... kinda just figured that one in my head (I do not think you can find two other values for a and b that add to 17 with a hyp of 13 ....try) ....here is the method:
a+b =17 so b = 17-a (1)
Pythagorean theorem: a^2 + b^2 = c^2 = 13^2= 169 Substitute (1) into this
a^2 + (17-a)^2 = 169
a^2 +a^2 - 34a + 289 = 169
2 a^2 -34a + 120 = 0
a^2 -17a +60 = 0 Use Quadratic formula to get (17+- sqrt49)/2 results in 12 and 5
12-5-13 triangle area = 1/2 a x b = 1/2 x 12 x 5 = 30 sq units
+118687 How to get exact surface of triangle if i only have hypotenuse and legs.
c=13
a+b=17
I tryed to do it by the system but it is not correct and it can't get correct...
This is a REALLY neat question. I do not remember seeing one like it before.
13 is the length of the hypotenuse.
a and b are the legs.
so using pythagoras's theorem we can say that
\(a^2+b^2=13^2\\ a^2+b^2=169 \qquad (1)\\~\\ \mbox{We also know that }\\ a+b=17\\ \mbox{Squaring both sides we get}\\ a^2+b^2+2ab=289\qquad (2)\\~\\ \mbox{Solve equation 1 and 2 simultaneously}\\ (2)-(1)\\ 2ab=289-169\\ 2ab=120\\ ab=60\\ \frac{1}{2}ab=30\\ \mbox{This is the formula for the area of the triangle so the area is }30units^2 \)
