MAHESH TUTORIALS. GEOMETRY Chapter : 1, 2, 6. Time : 1 hr. 15 min. Q.1. Solve the following : 3
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1 S.S.C. Test - III atch : S Marks : 30 Date : MHESH TUTORILS GEOMETRY Chapter : 1,, 6 Time : 1 hr. 15 min. Q.1. Solve the following : 3 (i) The radius of the base of a cone is 7 cm and its height is 4 cm. What is its slant height? In QR, m Q = 90º, m = 30º, m R = 60º. If R = 8 cm, find QR. In the adjoining figure, if m = 30º, then m O? O Q.. Solve the following : 6 (i) Find the measure of the central angle subtended by an arc of length 6.05 m and radius 5.5 m. ( = 7 ) In C, is a median. If = 7, + C = 60 then find C. C The volume of a cube is 1000 cm 3. Find its total surface area. Q.3. Solve the following : 9 (i) The curved surface area of the frustum of a cone is 180 sq. cm and the perimeters of its circular bases are 18 cm and 6 cm respectively. Find the slant height of the frustum of a cone. CD is a rectangle. Taking D as a diameter, a semicircle XD is drawn which intersects the diagonal D at X. If = 1 cm, D = 9 cm then find the values of D and X.
2 Determine the values of x, y and z with the help of information given in the adjoing figure. 4 6 y S X Q z R Q.4. Solve the following : 1 (i) Suppose and C are equal chords of a circle and a line parallel to the tangent at intersects the chords at D and E. rove that D = E. test tube has diameter 0 mm and height is 15 cm. The lower portion is a hemisphere in the adjoining figure. Find the capacity of the test tube. ( = 3.14) In the adjoining figure, QR = 60º and ray QT bisects QR. seg ray Q and seg C ray QR. If C = 8 find the perimeter of CQ. Q C T R est Of Luck
3 S.S.C. MHESH TUTORILS Test - III atch : S GEOMETRY Marks : 30 Date : Chapter : 1,, 6 Time : 1 hr. 15 min. MODEL NSWER ER.1. Solve the following : (i) Radius of base of cone (r) = 7 cm its height (h) = 4 cm l = r + h l = l = l = 65 l = 5 [Taking square roots] Slant height of cone is 5 cm 1 In QR, m = 30º m R = 60º m Q = 90º QR is a 30º - 60º - 90º triangle y 30º - 60º - 90º triangle theorem, QR = 1 R [Side opposite to 30º] m = 1 m (arc ) [Inscribed angle theorem] 30º = 1 O m (arc ) m (arc ) = 60º m O = m (arc ) [Definition of measure of minor arc] m O = 60º.. Solve the following : (i) l = = = 360 r
4 = = 63º Measure of the arc is 63º. 1 In C, seg is the median + C = + [y ppollonius theorem] 60 = (7) + 60 = (49) + 60 = = = 16 = 16 C = 81 = 9 units [Taking square roots] = 1 C [ is the midpoint of seg C] 9 = 1 C C = 18 units Volume of a cube = 1000 cm 3 Volume of a cube = l 3 l 3 = 1000 l = 10 cm [Taking cube roots] 1 Total surface area of a cube = 6l = 6 10 = = 600 cm Total surface area of a cube is 600 cm Solve the following : (i) Curved surface area of the frustum of a cone = 180 cm erimeters of circular bases are 18 cm and 6 cm r 1 = 18...(i) r = 6... dding (i) and, we get r 1 + r = (r 1 + r ) = 4 (r 1 + r ) = 4 (r 1 + r ) = 1...
5 Curved surface area of the frustum of a cone = (r 1 + r ) l 180 = (r 1 + r ) l 180 = 1 l [From ] l = 15 cm Slant height of the frustum of a cone is 15 cm. In D m D = 90º [ngle of a rectangle] 9 D D = + D [y pythagoras theorem] D = D = D = 5 X D = 15 cm [Taking square roots] m D = 90º [ngle of a rectangle] C line is a tangent to the circle at point [ line perpendicular to the radius at its 1 outer end is a tangent to the circle] Line is a tangent and line XD is a secant intersecting at points X and D = X. D [Tangent secant property] 1 = X = X. 15 X = X = 9.6 cm In SQ, m SQ = 90º Q = S + QS [y ythagoras theorem] 6 = 4 + y 6 36 = 16 + y y = Q y = 0 y = 4 5 [Taking square roots] 1 4 y z S X R y = 5 In QR, m QR = 90º seg QS hypotenuse R QS = S SR [y property of geometric mean] y = 4 x ( 5 ) = 4 x
6 = 4 x x = 0 4 x = 5 1 In QSR, m QSR = 90º QR = QS + SR [y ythagoras theorem] z = y + x z = ( 5 ) + (5) z = z = 45 z = 9 5 [Taking square roots] z = Solve the following : (i) Construction : Draw seg C roof : Take points R and S on the tangent C at as shown in the figure line DE line RS On transversal D, ED DR [Converse of D E ED R alternate angles test]...(i) [ - D - ] R S R C... [ngles in alternate segment] ED C... [From (i) and ] Similarly, we can prove that DE C...(iv) In C, seg seg C C C...(v) [Isosceles triangle theorem] In DE, ED DE [From, (iv) and (v)] seg D seg E [Converse of isosceles triangle theorem] D = E 1 Diameter of a test tube = 0 mm its radius (r) = 0 15 cm = 10 mm = 1 cm Its height (h) = 15 cm
7 Height of hemispherical part (h 1 ) = radius of hemisphere = 1 cm Height of cylindrical part (h ) = h h = 15 1 = 14 cm Volume of test tube = Volume of cylindrical part + Volume of hemispherical part = r h + 3 r3 = r h + r 3 = 3.14 (1) = = Volume of test tube = cm 3 Capacity of a test tube is cm 3 Ray QT is the angle bisector of QR lies on ray QT seg ray Q T seg C ray QR = C [ngle bisector theorem] ut, C = 8 units...(i) = 8 units... Q C R m QR = 60º m QT = m RQT = 1 60º [ Ray QT bisects QR] m QT = m RQT = 30º... In Q, m Q = 90º m Q = 30º [From, - - Q and Q - - T] m Q = 60º [Remaining angle] 1 Q is a 30º - 60º - 90º triangle y 30º - 60º - 90º triangle theorem = 1 Q [Side opposite to 30º] 8 = 1 Q [From ] Q = 16 units
8 Q = 3 Q [Side opposite to 60º] 3 Q = 16 Q = 8 3 units...(iv) Similarly, QC = 8 3 cm...(v) erimeter of CQ = + C + QC + Q = [From (i),, (iv) and (v)] = erimeter of CQ = units
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