Let's do this question, a semicircle is inscribed in a triangle, ABC The diameter of the same circle
is on ABC.
The sites, AC and Abbi, are tangents to the semicircle.
Amy is given us 13 centimetre acres equal to four centimetre and KB's equal to 15 centimetre.
Find the radius of the semicircle.
All right.
Was the video.
Give it a try and then let's do it together.
All right.
We're back.
Let's do it together.
Over here, we have Triangle ABC and this same circle is inscribed in Triangle ABC.
All right.
Now let's join this point to the center of the semicircle, as well as this point to the center of the
semicircle.
Let it be P and Q and let the center of the semicircle B point are.
Now, it's mentioned that the sides, AC and Ebbie are the engines to the semicircle right now.
Therefore, A, B and B R will make an angle of 90 degrees.
You can refer more on this when you started the engines under the section Sakalys.
So tangents to a circle will form a 90 degree angle when you join the point of intersection of the tangent
to the circle and the center of the circle.
That's why this angle over here is 90 degrees and this angle over here is 90 degrees.
All right.
Now, aspirin's formula.
We know that the area of Triangle ABC is equal to in the S minus E in the S minus B into S minus C right
now.
It's given that the sides of the triangle are 13.
Or and 15 centimeter, therefore, what's the same perimeter that would be 13 plus four plus 15 by two,
which gives you three by two, which is equal to 16.
All right.
Now, let's substitute this into this formula so you get area is equal to root of 16 into three in the
12 into one.
So that gives you the area as four in the two.
The three, which is equal to twenty four centimeter square.
All right.
Now, let me quickly mark the sides of this triangle so we know it's given 13, four and 15.
But I'm just marking them over here now.
Let's join a to our now you can see that the area of Triangle ABC can also be found by adding the area
of Triangle A, B, R and Triangle E, R, C, right.
So that's right that over here the area is equal to area of Triangle Ebarb plus area of Triangle ARCC,
and we have already found the areas.
Twenty four centimeter square over here.
Now we know that this line that is P R is perpendicular to Ebby, that is P R is the altitude through
point are similarly argue is perpendicular to AC, that is our Q is the altitude through our four triangle
area C therefore I can right the area of about as half in to A, B in the upper right.
Based on the formula, that area is equal to half in the base in the height.
Therefore over here the area is half in the 13 into our similarly over here the area is half Endou or
into R and over here I'm digging out an hour because it's the same, same circle, right.
Radius remains the same.
So we use the same variable.
Therefore, the area of Triangle ABC can also be mentioned as half into 13 into our plus avenue, four
into our.
Now that is equal to off into our into 13 plus four, which gives you are by two in to seventeen and
that is equal to twenty four, which is what we have found over here.
Therefore from this equation you can find that R is equal to forty eight by 17 centimeter.
And that is your answer.