In the figure below, PQ and RQ are tangents to the circle with centre O and radius 6cm i) Prove that △ PQR is an equilateral triangle. ii) Find the length of RP.

Given that PQ and RQ are tangents drawn from an external point Q and ∠ PQR = 60 ° We know that the lengths of the tangents drawn from an external point to a circle are equal. ∴ PQ = RQ ⇒ △ PQR is an isosceles triangle. Let ∠ QPR = ∠ QRP = x ° ∴ ∠ PQR + ∠ QPR + ∠ QRP = 180 ° ⇒ 60 ° + x + x = 180 ° ⇒ 2 x = 120 ° ⇒ x = 60 ° All the angles are equal to 60 ° . ∴ △ PQR is an equilateral triangle. ii) To find the length of RP we join OP and OQ We know that radius is perpendicular to tangent. So, ∠ OPQ = 90 ° ⇒ ∠ OPS = ∠ OPQ - ∠ QPR ⇒ ∠ OPS = 90 - 60 = 30 ° Given that radius is 6 cm In △ OPS , cos 30 ° = PS OP 3 2 = PS 6 ⇒ PS = 3 3 cm We know that the perpendicular from the centre of a circle to a chord bisects the chord. So, RP = 2 PS = 6 3 cm .

In the figure given below, PA is a tangent to the circle with centre O and PCB is a straight line. Find the measure of ∠ OBC . Show your steps and give valid reasons.

Given that PA is a tangent to the circle with centre O and PCB is a straight line. ∠ BCA + ∠ PCA = 180 ° ∵ PCB is a straight line 70 ° + ∠ PCA = 180 ° ∠ PCA = 180 ° - 70 ° = 110 ° In ∆ ACP , 30 ° + 110 ° + ∠ PAC = 180 ° ∠ PAC = 180 ° - 140 ° ∠ PAC = 40 ° We know that radius is perpendicular to the tangent. So, ∠ OAP = 90 ° ∴ ∠ OAC = 90 ° - 40 ° ⇒ ∠ OAC = 50 ° Join OC . In △ OAC , OA = OC ∠ OAC = ∠ OCA = 50 ° ∴ ∠ OCB = 70 ° - 50 ° = 20 ° In ∆ OBC , OC = OB ∠ OBC = ∠ OCB = 20 ° ∴ x = 20 °

ABCD is a square. CD is a tangent to the circle with centre O as shown in the figure below. If OD = CE, what is the ratio of the area of the circle and the area of the square? Show your steps and give valid reasons.

Given, A circle with centre O is drawn such that CD is the tangent to the circle such that OD=CE We know that Area of the square = s 2 Area of circle = πr 2 Let the side of square is 's’ and 'r' be the radius of circle. In triangle DOC: OD 2 + DC 2 = OC 2 ⇒ r 2 + s 2 = 2 r 2 ⇒ r 2 + s 2 = 4 r 2 ⇒ s 2 = 3 r 2 ⇒ s = 3 r ∴ Area of circle Area of square = πr 2 s 2 Substitute the value of s in the formula. ⇒ πr 2 3 r 2 = π 3 Hence, the required ratio is π : 3 .