Solution: Steps to construct: Step 1: Draw a line segment BC = 4cm. Step 2: At point B, draw a perpendicular and cut off BE = 2.5cm. Step 3: From, E, draw a line EF parallel to BC. Step 4: From...
Solution : (a) Given: O is the center of the circle. To Prove : ∠AOC = 2 (∠ACB + ∠BAC). Proof: In ∆ABC, ∠ACB + ∠BAC + ∠ABC = 180° (Angles of a triangle) ∠ABC = 180o – (∠ACB + ∠BAC)….(i) In the...
(B) inthe figure given below, AB is the diameter of the semi-circle ABCDE with centre O. If AE = ED and ∠BCD = 140°, find ∠AED and ∠EBD. Also Prove that OE is parallel to BD. Solution: (a) Join DB,...
Solution: (a) triangle ABC is an equilateral triangle Each angle = 60o ∠A = 60o But ∠A = ∠D (Angles in the same segment) ∠D = 600 Now ABEC is a cyclic quadrilateral, ∠A = ∠E = 180o 60o + ∠E = 180o...
Solution: Three circles with centers A, B and C touch each other externally at P, Q and R respectively and the radii of these circles are 2 cm, 3 cm and 4 cm. By joining the centers of triangle ABC...
Solution: (i) Join OB ∠OBA = 90° (Radius through the point of contact is perpendicular to the tangent) OB2 = OA2 – AB2 r2 = (r + 7.5)2 – 152 r2 = r2 + 56.25 + 15r – 225 15r = 168.75 r = 11.25 Hence,...
Solution: (a) From A, AP and AS are the tangents to the circle ∴AS = AP = 6 From B, BP and BQ are the tangents ∴BQ = BP = 5 From C, CQ and CR are the tangents CR = CQ From D, DS and DR are the...
(a) In figure (i) given below, triangle ABC is circumscribed, find x. (b) In figure (ii) given below, quadrilateral ABCD is circumscribed, find x. Solution: (a) From A, AP and AQ are the tangents...
Solution: Radii of the circles are 5 cm and 2.8 cm. i.e. OP = 5 cm and CP = 2.8 cm. (i) When the circles touch externally, then the distance between their centers = OC = 5 + 2.8 = 7.8 cm. (ii) When...
Solution: Two concentric circles with center O OP and OB are the radii of the circles respectively, then OP = 5 cm, OB = 13 cm. Ab is the chord of outer circle which touches the inner circle at P....
Solution: In a circle with center O and radius 3cm and p is at a distance of 5cm. That is OT = 3 cm, OP = 5 cm OT is the radius of the circle OT ⊥ PT Now in right ∆ OTP, by Pythagoras axiom, OP2 =...
Solution: (a) Construction: Join BC, and AC then ABCD is a cyclic quadrilateral. Now in ∆DCF Ext. ∠2 = x + z and in ∆CBE Ext. ∠1 = x + y Adding (i) and (ii) x + y + x + z = ∠1 + ∠2 2 x + y + z =...
(b) In the figure given below, if ∠ACE = 43° and ∠CAF = 62°, find the values of a, b and c (2007) Solution: (a) In ∆PQR, ∠PRQ = 90° (Angle in a semi-circle) and ∠PQR = 58° ∠RPQ = 90° – ∠PQR = 90° –...
Solution: (a) ADFE is a cyclic quadrilateral Ext. ∠FEB = ∠ADF ⇒ ∠ADF = 80° ABCD is a parallelogram ∠B = ∠D = ∠ADF = 80° or ∠ABC = 80° (b)In trapezium ABCD, AD || BC (i) ∠B + ∠A = 180° ⇒ 70° + ∠A =...
(b) In the figure given below, O is the center of the circle. ∠AOE =150°, ∠DAO = 51°. Calculate the sizes of ∠BEC and ∠EBC. Solution: (a) In the given figure, ABCD is a cyclic quadrilateral ∠ADC =...
(b) In the figure (if) given below, AB is parallel to DC, ∠BCE = 80° and ∠BAC = 25°. Find: (i) ∠CAD (ii) ∠CBD (iii) ∠ADC (2008) Solution: (a) ∠DBC = 58° BD is diameter ∠DCB = 90° (Angle in...
Solution: From the figure (i) ABCD is a cyclic quadrilateral Ext. ∠DCE = ∠BAD ∠BAD = xo Now arc BD subtends ∠BOD at the center And ∠BAD at the remaining part of the circle. ∠BOD = 2 ∠BAD = 2 x 2 x =...
(b) In the figure given below, O is the circumcenter of triangle ABC in which AC = BC. Given that ∠ACB = 56°, calculate (i)∠CAB (ii)∠OAC Solution: Given that (a) Arc AB subtends ∠APB at the center...
Solution: In ∆APB, ∠APB = 90° (Angle in a semi-circle) But ∠A + ∠APB + ∠ABP = 180° (Angles of a triangle) ∠A + 90° + 42°= 180° ∠A + 132° = 180° ⇒ ∠A = 180° – 132° = 48° But ∠A = ∠PQB (Angles in the...
Solution: (a) Construction: Join AB ∠A = ∠C = 350 (Alt Angles) ∠ABC = 35o (b) ∠AOC + reflex ∠AOC = 360o 130o + Reflex ∠AOC = 360o Reflex ∠AOC = 360o – 130o = 230o Now arc BC Subtends reflex ∠AOC at...
Solution: (i) ∠ACB = ∠ADB (Angles in the same segment of a circle) But ∠ADB = x° ∠ABC = xo Now in ∆ABC ∠CAB + ∠ABC + ∠ACB = 180o 40o + 900 + xo = 180o (AC is the diameter) 130o + xo = 180o xo =...
Solution: (i) ∠ADB and ∠ACB are in the same segment. ∠ADB = ∠ACB = 50° Now in ∆ADB, ∠DAB + X + ∠ADB = 180° = 42o + x + 50o = 180o = 92o + x = 180o x = 180o – 92o x = 88o (ii) In the given figure we...