Chapter 4 Geometric Constructions Problem Set 4 Solutions

Analysis: As shown in the figure, line l is a tangent to the circle at point A. seg BA is a chord of the circle and ∠BCA is an inscribed angle. By tangent secant angle theorem, ∠BCA = ∠BAR By converse of tangent secant angle theorem, If we draw ∠BAR such that ∠BAR = ∠BCA, then ray AR (i.e. line l) is a tangent at point A.

Diameter of circle = 6.4 cm 6.4 Radius of circle = = 3.2 cm Analysis: As shown in the figure, let R be a point in the exterior of circle at a distance of 6.4 cm. Let RQ and RS be the tangents to the circle at points Q and S respectively. ∴ seg PQ ⊥tangent RQ …[Tangent is perpendicular to radius] ∴ ∠PQR = 90° ∴ point Q is on the circle having PR as diameter. …[Angle inscribed in a semicircle is a right angle] Similarly, Point S also lies on the circle having PR as diameter. ∴ Points Q and S lie on the circle with PR as diameter. On drawing a circle with PR as diameter, the points where it intersects the circle with centre P, will be the positions of points Q and S respectively.

m(arc AB) = ∠APB = 100° Analysis: seg PA ⊥ line l seg PB ⊥line m … [Tangent is perpendicular to radius] The perpendicular to seg PA and seg PB at points A and B respectively will give the required tangents at A and B. Steps of construction: i. With centre P, draw a circle of any radius and take any point A on it. ii. Draw ray PA. iii. Draw ray PB such that ∠APB = 100°. iv. Draw line l ⊥ray PA at point A. v. Draw line m ⊥ ray PB at point B. Lines l and m are tangents at points A and B to the circle.

Analysis: Draw a circle of radius 3.4 cm As shown in the figure, let A be a point in the exterior of circle at a distance of (3.4 + 4.1) = 7.5 cm. Let AP and AQ be the tangents to the circle at points P and Q respectively. ∴ seg EP ⊥ tangent PA … [Tangent is perpendicular to radius] ∴ ∠EPA = 90° ∴ point P is on the circle having EA as diameter. …[Angle inscribed in a semicircle is a right angle] Similarly, point Q also lies on the circle having EA as diameter. ∴ Points P and Q lie on the circle with EA as diameter. On drawing a circle with EA as diameter, the points where it intersects the circle with centre E, will be the positions of points P and Q respectively.

Analysis: As shown in the figure, Let B – C – N and B – A – L. ∆ABC ~ ∆LBN …[Given] ∴ ∠ABC ≅ ∠LBN …[Corresponding angles of similar triangles] = = …(i)[Corresponding sides of similar triangles] But. = …(ii)[Given] ∴ = = = …[From(i)and(ii)] ∴ sides of ∆LBN are longer than corresponding sides of ∆ABC. ∴ If seg BC is divided into 4 equal parts, then seg BN will be 7 times each part of seg BC. So, if we construct ∆ABC, point N will be on side BC, at a distance equal to 7 parts from B. Now, point L is the point of intersection of ray BA and a line through N, parallel to AC. ∆LBN is the required triangle similar to ∆ABC. Steps of construction: i. Draw ∆ABC of given measure. Draw ray BD making an acute angle with side BC. ii. Taking convenient distance on compass, mark 7 points B1, B2, B3, B4, B5, B6 and B7 such that BB1 = B1B2 = B2B3 B3= B44 = B4B5 = B5B6 = B6B7. iii. Join B4C. Draw line parallel to B4C through B7 to intersects ray BC at N. iv. Draw a line parallel to side AC through N. Name the point of intersection of this line and ray BA as L. ∆LBN is the required triangle similar to ∆ABC.

Analysis: As shown in the figure, Let Y – Q – Z and Y – P – X. ∆XYZ ~ ∆PYQ …[Given] ∴ ∠XYZ ≅ ∠PYQ …[Corresponding angles of similar triangles] = = …(i)[Corresponding sides of similar triangles] But, = ,..(ii)[Given] ∴ = = = …[From (i) and (ii)] ∴ sides of ∆XYZ are longer than corresponding sides of ∆PYQ. ∴ If seg YQ is divided into 5 equal parts, then seg YZ will be 6 times each part of seg YQ. So, if we construct ∆PYQ, point Z will be on side YQ, at a distance equal to 6 parts from Y. Now, point X is the point of intersection of ray YP and a line through Z, parallel to PQ. ∆XYZ is the required triangle similar to ∆PYQ. Steps of construction: i. Draw ∆ PYQ of given measure. Draw ray YT making an acute angle with side YQ. ii. Taking convenient distance on compass, mark 6 points Y1, Y2, Y3, Y4, Y5 and Y6 such that YY1 = Y1Y2 = Y2Y3 = Y3Y4 = Y4Y5 = Y5Y6. iii. Join Y5Q. Draw line parallel to Y5Q through Y6 to intersects ray YQ at Z. iv. Draw a line parallel to side PQ through Z. Name the point of intersection of this line and ray YP as X. ∆XYZ is the required triangle similar to ∆PYQ.