Find the area of a triangle with sides 1 3 , 6 1 , 8 0 .
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Good. It would better to clarify that 1 3 is the hypotenuse of right triangle with sides 2 and 3. This goes for the other sides.
Bonus question: Using the same approach, find the area of a triangle with sides 2 9 , 8 5 , 1 4 6 .
Reply to Challenge Master:
We can imagine the following figure where the triangle's sides are the hypotenuse of other 3 triangles:
Then, the triangle area is:
7 7 − 5 − 2 1 − 2 7 . 5 = 2 3 . 5
Reply to Challenge Master: 77 square units if I'm not mistaken. I used pythagorean theorem btw :) and looked for what will fit with everything :3
Reply to Challenge Master:
I get the answer as:
frac{7\sqrt{{{179}}{4}}
which is around 23.413 (3dp)
We can use the alternative form of Heron's formula:
A = 4 1 2 ( a 2 b 2 + a 2 c 2 + b 2 c 2 ) − ( a 4 + b 4 + c 4 )
By squaring the side lengths, the square roots will be gone and the answer 14 can be easily obtained through some calculator spamming.
I solved it by using the formula 2 1 a b s i n C for the area of triangle and calculated s i n C by using cosine law.
assume the Angle between two sides √13&√80 is P so cos(P)=(√80^2+√13^2-√61^2)(2×√80×√13)=4÷√65 so sin(P)=7÷√65 .area of∆=0.5×√80×√13×(7÷√65)=14#####
This one was the exact same Same as the last one. When trying to solve the last one , instead of starting of by finding the areas of those smaller outside triangles, I used Pythagoras to find each f those outside triangles's hypotenuses. These were route 13 route 61 and route 80 so as soon as I saw this I knew the answer would be the same.
Problem Loading...
Note Loading...
Set Loading...
There you go!