Answer: Option (B)$\frac{\varepsilon_{1} r_{2}+\varepsilon_{2} r_{1}}{r_{1}+r_{2}}$
Using ruler and compasses only, draw an equilateral triangle of side 5 cm and draw its inscribed circle. Measure the radius of the circle.
Solution: Steps to construct: Step 1: Draw a line segment BC = 5cm. Step 2: With Center as B and radius 5cm, with center as C and radius 5cm draw two arcs which intersect each other at point A. Step...
(a) In the figure (i) given below, AB is a diameter of the circle. If ∠ADC = 120°, find ∠CAB. (b) In the figure (ii) given below, sides AB and DC of a cyclic quadrilateral ABCD are produced to meet at E, the sides AD and BC are produced to meet at F. If x : y : z = 3 : 4 : 5, find the values of x, y and z.
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 =...
(a) In the figure (i) given below, M, A, B, N are points on a circle having centre O. AN and MB cut at Y. If ∠NYB = 50° and ∠YNB = 20°, find ∠MAN and the reflex angle MON. (b) In the figure (ii) given below, O is the centre of the circle. If ∠AOB = 140° and ∠OAC = 50°, find (i) ∠ACB (ii) ∠OBC (iii) ∠OAB (iv) ∠CBA
Solution (a) ∠NYB = 50°, ∠YNB = 20°. In ∆YNB, ∠NYB + ∠YNB + ∠YBN = 180o 50o + 20o + ∠YBN = 180o ∠YBN + 70o = 180o ∠YBN = 180o – 70o = 110o But ∠MAN = ∠YBN (Angles in the same segment) ∠MAN = 110o...
(a) In the figure (i) given below, AD || BC. If ∠ACB = 35°. Find the measurement of ∠DBC. (b) In the figure (ii) given below, it is given that O is the centre of the circle and ∠AOC = 130°. Find ∠ ABC
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...
Using the given information, find the value of x in each of the following figures:
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...