All related to triangles Part 2
Find the area of a triangle with sides 1 3 , 6 1 , 8 0 .
The answer is 14.
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5 solutions
There you go!
Moderator note:
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.