A Problem on Rhombus and Circle

Hello math bugs(🐞) and hivers(🐝)
I hope evrything is alright with you.
Well come to another interesting geometric problem.

You must have seen the question figure. Originally it was a verbal question without any drawing. I have just draw the figure for your easier understanding.

The Original question:

There is a Rhombus ABCD. The cir-cum-radius of ∆BDC and ∆ABC are 12.5 unit and 25 unit respectively. Find the area of the Rhombus ABCD.

To solve it you need to have deep understanding of rombus and circle. Before heading to the solution , you may give it a try first. The more you try yours, the more mature you will be at solving problems.

This really hard for me to make you understand in written format because I have to let you go through a couple of facts and I am not aware of how many of them are allready known to you. So, I will touch everything that we need here to solve it.

We know that when a Square is pressurised from side, it turn into a Rhombus with all equal sides. If it is Rhombus, we can have the following consideration:

1️⃣ All sides are equal 2️⃣ All smaller triangle inside are equal and each of them is a right triangle 3️⃣ Diagonal divide a Rhombus into two equal area of triangles and in that case all of the triangle are also equal 4️⃣ Diagonals bisect each other at 90° 5️⃣ But diagonals are not equal. Check the following figure also:

The most important formula we need to know is the relation between sides, Cir-cum-radius and the area of a triangle. It is given by R = (abc)/(4∆).
Where,
R= Ciur-cum-radius.
a, b and c are sides of the triangle.
∆ = Area of the triangle.

I am not gonna put details here. It was discussed and proved here in previous post.

Obcourse we also need to know the Pythagoras' formula H² = B² + p²

Where,

H = Hypotenuse of a right triangle.
B = Base of the right triangle.
P = perpendicualar of the triangle.
Details of pythagoras' formula or triplet here.

Finally, We must know the baic formula for area of a triangle which is given by ∆=1/2 × Base × height unit².

I have already mentioned all the related staffs and now it's time for the solution.

SOLUTION:

We have proper knowledge of a Rhombus and three relations (formulae) mentioned above. We will start from R = abc/4∆. In the problem, the only values given are two Cir-cum-radius. The smaller one (R1) is 12.5 unit and the larger one (R2) is 25 units. The ratio of these two Radius is 1:2. Now check the problem figure; two triangle angles and two cir-cum-circles are mentioned. In case of both of the triangle two sides equals. So in both cases we can repalce side b by a.

So, the two Cir-cum-radius will be:

R1 = a²c1/4∆
R2 =a²c2/4∆

You can say why should I not use ∆1 and ∆2 also if R split into R1 and R2 & for c it is c1 and c2 ? It is because in both of the cases, only Cir-cum-radius(R) and the third side(c) of the mentioned triangles are not equal. The area will be the same. If you can't relate, re-visit the Rhombus talks mentioned above. So , what we can conclude here is R1 : R2 = c1 : c2. For clarity check the fillowing figure:

Check previos figure, we already have a relation between a and x uisng Pythagoras' formula in smaller triangle. Now check the following figure.

We are almost done. Now need to put value of x in formula of area of Rhombus. Check it below👇

Bravo!! It was hard work to make you understand without verbal expression.

NOTE: You can use other formula to find the area of the Rhombus but you must know one most important consideration that if diagonals of any quadrilateral meet at 90° , the formula of area of the quadrilateral can be given by 1/2× D1 × D2 unit². Here D means diagonal.

Thank you so much frineds staying till end. I hope, my explanation is clear to you. For any querry , feel free to hit a comment below.

All the shapes and figures used in this article are made by me using android cellphone. If you find any silly mistake, please try ignoring it. Re-editing hurts.😭

Have a good day.

All is well

Regards: @meta007