Question

The length of each equal side of an isosceles triangle is 20 cm and the angle included between them is 30° , find the area of this triangle.

Answer 1

Answer:

The constraint here is - NOT to use Trigonometry.

I have tried not to use any trigonometric property or angle formulas.

The method goes as follows -

Triangle is Isosceles, means it 2 sides are equal.

Now, if its 2 sides are equal, we have to think of 2 line segments of equal lengths.

Question arises, where we can find it…!!

A simple solution to question is, Radii of a Circle .

2 radius of the same circle, are always equal.

So, lets assume a circle with 2 radii.

These radii, touch each other at O, at 30°. ——————-(1)

Let these 2 radii be 2 sides of the given triangle. (Length = 20 Cm)

(Given that these 2 sides make 30° angle between them). ——— (2)Now we have a circle. Inside it we have a triangle which 2 sides are the 2 different radii of the circle making an angle 30°.

Suppose we have a triangle which has 90° angle making between its 2 radii. Here also side = 20 Cms.

Area of this triangle (Making 90°) will be = (1/2)*20*20 = 200 cm*cm. ——(3)

Now we are talking about a circle where the area is always a function of angle. Area of Circle = π*r*r

And Area of Segment of Circle = π*r*r*Θ/360 ——————-(4)

Here except Θ, everything is constant, for a given circle.

Area α Θ (Proportional). ———————— (5)

But here we need Area of triangle, which is not Area of Segment.

Since, the triangle is a part of the segment itself.

We can write, Area of Triangle α Θ (Proportional). ————(6)

Using the above direct proportionality, from Eqn(6).

From Eqn(3), we know,

For 90°, we have 200.

For 30°, we will get 200/3 cm*cm

So, Answer 200/3 cm*cm

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