Question

Two circles with centres a and b and radius 2 units touch each other externally at c. A third circle with centre c and radius 2 units meets other two at d and e. Then the area of the quadrilateral abde is

Answer 1

Answer:

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Answer 2

Given:

2 circles touch each other at C, they have a radius of 2cm.

One more circle is drawn by taking C as center with radius as 2 cm.

To find:

The area of quadrilateral ABDE.

Formula to be used:

The area of an equilateral triangle is $\frac{\sqrt{3} }{4} (side)^{2}$

Step-by-step explanation:

Step 1 of 1

Draw 2 circles with radius 2 cm so that they touch each other at point C. Let the centers of circle 1 and 2 be A and B, respectively.

Now take C as center and draw another circle with 2cm radius, it will cross through the center of circle 1 and 2. And touch them at D and E.

Now join AD, DE, EB, DC, EC, AC, and CB.

THE AREA OF ABDE = AREA OF ΔADC + AREA OF ΔDCE +AREA OF ΔCEB

The three triangles formed are equilateral triangles.

THE AREA OF ABDE = $3*\frac{\sqrt{3} }{4} (side)^{2}$

$=3*\frac{\sqrt{3} }{4} (2)^{2}\\\\=3*\frac{\sqrt{3} }{4} (4)\\\\=3\sqrt{3}cm^{2}$

THEREFORE, AREA OF ABDE IS $3\sqrt{3}cm^{2}$.