Area of a Triangle enclosed by Two Cir-cum-radius and a side.

Hello maths bugs(🐞) & hivers(🐝)

Well come back to another intersting geometric problem with it's beautiful solution. There is large ∆ABC whose whose sides are AB= 13 cm, AC = 14 cm and BC = 15 cm. You have to find area of smaller ∆BOC enclosed by BO, CO and sideBC. Before heading to the solution, give it a try first.

We know that area of a triangle can be given by:

(∆) = 1/2 × Base × Height

In the problem figure the height of ∆BOC is not visible but base 15 cm is given. So we need to find the height. If we draw a perpendicular from centre O to the side BC, it will be the height of the ∆BOC. Thus we will get two similiar right angle triangle but in each case we just know about one side.
We also need to find out Cir-cum-radius. So if we get the value of Cir-cum-radius(R), problem will be solve. To find out R we gonna use the following formula.

R = (abc)/(4∆)
Here, R = Cir-cum-radius.
a,b & c are the three sides of ∆ABC
∆ prepresents the area of ∆ABC.

To find out area we gonna use Heron's Formula

Actually the area of ∆ABC is crammed. It is 84 cm²

Let's compute it in the following figure:

If you wanna know derivation of above formula , hot a comment below , I'll bring it for you in next post.

Now let's draw the perpendicular OD on BC. Here OD will divide BC into two equal half as ∆BOC is a isosceles Triangle where BO = CO = R. Let's check the following figure:

In the above figure from ∆BOD:

        BO² = OD² + BD²

    Or, (65/8)² = OD²+ (15/2)²

    Or, OD² = (65/8)² - (15/2)²
    
    Or, OD² = (65/8+15/2 )(65/8-15/2)
 
    [  As,a²-b² = (a+b)(a-b)]
   
    Or, OD² = 125/8 × 5/8

    Or, OD = 25/8 [cm]

Now we got the height of ∆BOC and the base is already given. So, area of the ∆BOC can be found using basic formula as follows: