Question

Compute the perimeter of the circles shown below, correct to a millimetre.

(i)

(ii)

(iii)

(iv)

Answer 1

(i)

Construction: Join the diagonal AC of the given rectangle.

The diagonal AC will pass through the centre of the circle. So, AC is the diametre of the circle.

In ΔADC, ∠ADC = 90°.

Using Pythagoras theorem:

AC² = AD² + DC²

= (3² + 4²) cm²

= (9 + 16) cm²

= 25 cm²

Perimeter of the circle = π × Diametre

= π × 5 cm

= (3.141 × 5) cm

= 15.705 cm

(ii)

Construction: Join the diagonals of the regular hexagon.

The diagonals of the regular hexagon pass through the centre of the circle.

It can be observed that the diagonals of the regular hexagon divide it into six equilateral triangles.

∴ ΔOAB is an equilateral triangle with OA = OB = AB = 2 cm

∴ Radius (r) of the circle = 2 cm

Perimeter of the circle = 2πr

= (2 × 3.141 × 2) cm

= 12.564 cm

(iii)

Diagonal of the given circle = Side of the square = 4 cm

Perimeter of the circle = π × diametre

= (3.141 × 4) cm

= 12.564 cm

(iv)

The given circle is the circumcircle for ΔABC. Therefore, AD is the perpendicular bisector of side BC.

∴ DC = BD =

Using Pythagoras theorem in ΔADC:

AC² = AD² + DC²

= (4² + 2²) cm²

= (16 + 4) cm²

= 20 cm²

Similarly, AB =

∴ ΔABC is an isosceles triangle.

We know that perpendicular bisector and median of an isosceles triangle coincide with each other.

∴ AD is the median and O is the centroid.

Also, centroid divides the median in the ratio 2:1.

∴ AO =

∴ Radius of the circle is

Perimeter of the circle = 2πr

= 2 × 3.141 ×

= 16.752 cm