Construction: Construct an equilateral triangle (1) in and (2) about a cricle.
Steps of Construction: Since each side of the equilateral triangle subtend an angle (360°)/3 = 120° at the centre O therefore, we construct as under:
1. (i) Draw the given circle and in it draw three radii OA, OB and OC making angle of 1200 with one another.
(ii) Join AB, BC and CA. Then ABC is the inscribed triangle.
2. (i) Draw tangents at A, B and C.
(ii)The tangents meet each other at P, Q and R. DPQR is the circumscribed traingle.
Construction: Construct a sqaure (1) in and (2) about a circle.
Steps of Construction: Since the diagonals of a square are at right angles and each diagonal is equal to the diameter of the circumscribed circle, therefore, we construct as under:
1. (i) Draw two diameters, perpendicular to each other to meet circle at A, B, C and D.
(ii) Join AB, BC, CD and DA Then ABCD is the inscribed square.
2. Draw tangents at points A, B, C, D and then intersect at points P, Q, R and S respectively. Then PQRS is the circumscribed sqaure.
Construction: Construct a required pentagon about a given circle.
Steps of Construction:
1. Draw radii OA and OB such that 360°
∠AOB = (360°)/5=72°
2. Cut arcs BC, CD, DE and EA equal to arc AB.
3. Draw tangents to the circle at points A, B, C, D and E.
Let these tangents intersect each other at points P, Q, R, Sand T. Then pa RST is, the required pentagon about the given circle.
Construction: Construct a regular hexagon about a given circle,
Or
Steps of Construction:
1. Draw radii OA and OB such that ∠AOB= (360°)/6=60°
2. Cut arcs BC, CD, DE, EF and FA equal to arc AB.
3. Draw tangents to the circle at points A, B, C, D, E and F. Let these tangents intersect each other at points P, Q, R, S, T and U.
Then PQRSTU is the required regular hexagon about the given circle.
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