Trigonometry! #24

Geometry Level 3

If in a triangle $ABC$ , right angled at $B$ , $s-a=3$ , $s-c=2$ , then the value of $a$ is

Note: $s$ denotes the semiperimeter of the triangle.

This problem is part of the set Trigonometry .

The answer is 3.

1 solution

Janardhanan Sivaramakrishnan
Jan 16, 2015

Because of the right-angle, $b^2=c^2+a^2$

and we also have $s=\frac{a+b+c}{2}$

Hence, $s-a=\frac{b+c-a}{2}=3$ and $s-c=\frac{b+a-c}{2}=2$

Adding both equations $2s-(a+c)=b=5$

Substituting for $b=5$ in the first equation $5+(c-a)=6$ , or $c-a=1 \rightarrow c=(a+1)$

Since, $b^2=a^2+c^2$ , $25=a^2+(a+1)^2$

Which gives a quadratic equation for $a$ as $a^2+a-12=(a+4)(a-3)=0$

Since, lengths need to be positive, the only valid solution for $a$ would be $\boxed{a=3}$

THE SOLUTION IS INCORRECTLY STATED IN THE QUESTION.

Point. Sorry! I must have made a mistake somewhere.

Omkar Kulkarni - 6 years, 4 months ago

I am getting a = 12, b = 5, c = 13

s = (a + b + c)/2 = 30/2 = 15

s-a = 15 - 12 = 3 (Check)

s-c = 15 - 13 = 3 (Check)

Your solution is WRONG in this problem !

Vijay Simha - 1 year, 8 months ago

The right triangle should be right-angled at $B$ as per the problem. Hence, the side $b$ should be the longest side (as the other two angles would be lesser than 90 degrees). In your solution, $b$ is the smallest side. Thus, your approach is incorrect.

Janardhanan Sivaramakrishnan - 1 year, 8 months ago

Trigonometry! #24

The answer is 3.

1 solution

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