Chapter 4 Geometric Constructions Practice Set 4.2 Solutions

Analysis: seg PM ⊥ line l ….[Tangent is perpendicular to radius] The perpendicular to seg PM at point M will give the required tangent at M.

Analysis: seg OM ⊥ line l …[Tangent is perpendicular to radius] The perpendicular to seg OM at point M will give the required tangent at M.

Analysis: As shown in the figure, line lis a tangent to the circle at point K. seg BK is a chord of the circle and LBAK is an inscribed angle. By tangent secant angle theorem, ∠BAK = ∠BKR By converse of tangent secant angle theorem, If we draw ∠BKR such that ∠BKR = ∠BAK, then ray KR i.e. (line l) is a tangent at point K.

Analysis: seg OP ⊥ line l …[Tangent is perpendicular to radius] seg OQ ⊥ line m The perpendicular to seg OP and seg OQ at points P and Q respectively will give the required tangents at P and Q. Radius = 3.3 cm ∴ Diameter = 2 × 3.3 = 6.6 cm ∴ Chord PQ is the diameter of the circle. ∴ The tangents through points P and Q (endpoints of diameter) are parallel to each other.

Analysis: seg ON ⊥ linel l seg OM ⊥ Iine m …….[Tangent is perpendicular to radius] The perpendicular to seg ON and seg 0M at points N and M respectively will give the required tangents at N and M.

Analysis: As shown in the figure, let Q be a point in the exterior of circle at a distance of 5.5 cm. Let QR and QS be the tangents to the circle at points R and S respectively. ∴ seg PR ⊥ tangent QR …[Tangent is perpendicular to radius] ∴ ∠PRQ = 90° ∴ point R is on the circle having PQ as diameter. …[Angle inscribed in a semicircle is a right angle] Similarly, point S also lies on the circle having PQ as diameter. ∴ Points R and S lie on the circle with PQ as diameter. On drawing a circle with PQ as diameter, the points where it intersects the circle with centre P, will be the positions of points R and S respectively. Ray QR and QS are the required tangents to the circle from point Q.

Analysis: As shown in the figure, let Q be a point in the exterior of circle at a distance of 5.5 cm. Let QR and QS be the tangents to the circle at points R and S respectively. ∴ seg PR ⊥ tangent QR …[Tangent is perpendicular to radius] ∴ ∠PRQ = 90° ∴ point R is on the circle having PQ as diameter. …[Angle inscribed in a semicircle is a right angle] Similarly, point S also lies on the circle having PQ as diameter. ∴ Points R and S lie on the circle with PQ as diameter. On drawing a circle with PQ as diameter, the points where it intersects the circle with centre P, will be the positions of points R and S respectively. Ray QR and QS are the required tangents to the circle from point Q.