Mathematics - Properties of Triangle Question with Solution | TestHub

Problem: Triangle $ABC$ is right-angled at point $A$, radii of described circles are $r_1,r_2,r_3$, given that $r_2=2,r_3=3$ then which of the following hold?

Options:

A. $a=5$

B. $r_1=6$

C. $r=1$

D. None of these

Correct Answer: A, B, C

Explanation:

Let the sides opposite to vertices $A,B,C$ be $a,b,c$ respectively.

Given that triangle $ABC$ is right-angled at $A$.

The radii of the excircles opposite to vertices $A,B,C$ are $r_1,r_2,r_3$ respectively.

The inradius is $r$.

We are given $r_2=2$ and $r_3=3$.

For a right-angled triangle with the right angle at $A$, we have the following relations:

1. $r_1=s$ (where $s$ is the semi-perimeter)

2. $r_2=s-c$

3. $r_3=s-b$

4. $r=s-a$

5. $r_1=r+r_2+r_3$ (Euler's theorem for a right-angled triangle)

6. $a^2=b^2+c^2$ (Pythagorean theorem)

Let's use the given values and the relations.

From $r_2=s-c$, we have $2=s-c$.

From $r_3=s-b$, we have $3=s-b$.

Adding these two equations:

$2+3=(s-c)+(s-b)$

$5=2s-(b+c)$

We know that the semi-perimeter $s=\frac{a+b+c}{2}$, so $2s=a+b+c$.

Substitute $2s$ into the equation:

$5=(a+b+c)-(b+c)$

$5=a$

So, the length of the hypotenuse $a=5$. This confirms option A.

Now, let's find $r_1$ and $r$.

Using the relation $r_1=r+r_2+r_3$:

$r_1=r+2+3$

$r_1=r+5$

We also know that for a right-angled triangle, the inradius $r=\frac{b+c-a}{2}$.

And the exradius $r_1=\frac{b+c+a}{2}=s$.

Let's use the relations for $r_2$ and $r_3$ again:

$s-c=2⟹c=s-2$

$s-b=3⟹b=s-3$

Substitute $b$ and $c$ into the Pythagorean theorem $a^2=b^2+c^2$:

$5^2=(s-3{)}^2+(s-2{)}^2$

$25=(s^2-6s+9)+(s^2-4s+4)$

$25=2s^2-10s+13$

$2s^2-10s+13-25=0$

$2s^2-10s-12=0$

Divide by 2:

$s^2-5s-6=0$

Factor the quadratic equation:

$(s-6)(s+1)=0$

Since $s$ (semi-perimeter) must be positive, $s=6$.

Therefore, $r_1=s=6$. This confirms option B.

Now we can find $r$:

$r=s-a$

$r=6-5$

$r=1$

This confirms option C.

Let's verify the side lengths $b$ and $c$:

$b=s-3=6-3=3$

$c=s-2=6-2=4$

Check Pythagorean theorem: $a^2=b^2+c^2$

$5^2=3^2+4^2$

$25=9+16$

$25=25$

The side lengths are consistent.

All options A, B, and C are correct.

The final answer is $\boxed{A,B,C}$.