S.S.C. Test - III Batch : SB Marks : 0 Date : MHESH TUTORILS GEOMETRY Chapter : 1,, 6 Time : 1 hr. 15 min..1. Solve the following : (i) The dimensions of a cuboid are 5 cm, 4 cm and cm. Find its volume. If DEF ~ MNK, DE = 5, MN = 6, find the value of (ΔDEF) (ΔMNK). Line B is a tangent and line BCD is a secant. If B = 6 units, BC = 4 units, find BD. D B C.. Solve the following : 6 (i) The dimensions of a cuboid in cm are 16 14 0. Find its total surface area. Find the perimeter of an isosceles right triangle with each of its congruent sides is 7 cm. Find the length of the arc of a circle with radius 0.7 m and area of the sector is 0.49 m... Solve the following : 9 (i) The surface area of a sphere is 616cm. What is its volume? In the adjoining figure, in the isosceles triangle R, the vertical = 50º. The circle passing through and R cuts. in S and R in T. ST is joined. Find ST. S 50º T R = 7
...... rove that three times the square of any side of an equilateral triangle is equal to four times the square of an altitude..4. Solve the following : 1 (i) Let the circles with centre and touch each other at point. Let the extended chord B intersect the circle with centre at point E E and the chord BC touches the circle with centre at the point D.Then prove that ray D is an angle bisector of the CE. D M C B In the adjoining figure, m O = 0º and radius O = 1 cm. Find the following (Given =.14) (a) rea of sector O-R (b) rea of O (c) rea of segment R O 1 cm 0º M R In the adjoining figure, if LK = 6 find MK, ML, KN, MN and the perimeter of MNKL. M 45º N L 6 0º K Best Of Luck
S.S.C. MHESH TUTORILS Test - III Batch : SB GEOMETRY Marks : 0 Date : Chapter : 1,, 6 Time : 1 hr. 15 min. MODEL NSWER ER.1. Solve the following : (i) Length of a cuboid (l) = 5 cm Its breadth (b) = 4 cm Its height (h) = cm Volume of a cuboid = l b h = 5 4 = 60 cm Volume of cuboid is 60 cm. DEF ~ MNK ( DEF) ( MNK) = DE MN [rea of similar triangles] ( DEF) ( MNK) = 5 6 ( DEF) ( MNK) = 5 6 Line BCD is a secant intersecting the circle at points C and D and line B is a tangent at B = BC BD 6 = 4 BD 6 = 4 BD BD = 6 4 B C D BD = 9 units.. Solve the following : (i) Length of a cuboid (l) = 16 cm its breadth (b) = 14 cm its height (h) = 0 cm Total surface area of a cuboid = (lb + bh + lh) = (16 14 + 14 0 + 16 0)
...... = (4 + 80 + 0) = 84 = 1648 cm Total surface area of a cuboid is 1648 cm. Given : In BC, m BC = 90º B = BC = 7 cm To find : erimeter of BC 7 cm Sol. In BC, m BC = 90º B C C = B + BC [By ythagoras theorem] 7 cm C = (7) + (7) C = 49 + 49 C = 98 C = 49 [Taking square roots] C = 7 cm erimeter of BC = B + BC + C = 7 7 7 = 14 7 erimeter of BC = 7 cm Radius of a circle = 0.7 cm rea of the sector = 0.49 m rea of the sector = r l 0.49 = 0.7 l 49 7 = 100 0 l 49 0 100 7 = l l = 1.4 The length of the arc is 1.4 m... Solve the following : (i) Surface area of sphere = 616 cm Surface area of a sphere = 4r 616 = 4 7 r 616 7 4 = r
...... r = 49 r = 7 cm [Taking square roots] 4 Volume of a sphere = r 4 = 7 7 7 7 41 = = 147. cm The volume is 147. cm. S 50º T In R seg seg R R R R...(i) [Isosceles triangle theorem] m R + m R + m R = 180º [Sum of the measures of angles of a triangle is 180º] m R + m R + 50 = 180º [From (i) and Given] m R = 180º 50º m R = 10º m R = 65º... SRT is cyclic ST TR [n exterior angle of cyclic quadrilateral is congruent to the angle opposite to adjacent interior angle] ST R [ - T - R] ST = 65º [From ] To prove : B = 4D roof : BC is an equilateral triangle In DB, m DB = 90º m BD = 60º [ngle of an equilateral triangle] m BD = 0º [Remaining angle] B D C 1
... 4... DB is a 0º - 60º - 90º triangle By 0º - 60º - 90º triangle theorem, D = B [Side opposite to 60º] D = B 4D = B [Squaring both sides] B = 4D.4. Solve the following : (i) E B In MD, M C M = MD D [The lengths of two tangent segments from an external point to a circle are equal] m MD = m MD [Isosceles triangle theorem] Let, m MD = m MD = xº...(i) CM BC [ngles in alternate segments] Let, m CM = m BC = yº... m CD = m CM + m MD [ngles ddition property] m CD = (x + y)º... [From (i) and ] DE is an exterior angle of DB m DE = m DB + mbd [Remote Interior angles theorem] m DE = m DM + m BC [D - M - C - B] mde = (x + y)º...(iv) [From (i) and ] m CD = mde [From and (iv)] ray D is an angle bisector of CE. Radius of the circle (r)= 1 cm O Measure of arc () = 0º rea of sector O - R = r 1 cm 0º M 60 = 0.14 1 1 R 60 = 7.68 cm
... 5... In OM, m OM = 90º m OM = 0º [Given and O - M - ] m OM = 60º [Remaining angle] 1 OM is 0º - 60º - 90º triangle By 0º - 60º - 90º triangle theorem M = 1 O = 1 1 M = 6 cm. O = O = 1 cm [Radii of same circle] rea of O = 1 base height = 1 O M = 1 1 6 = 6 cm rea of segment R = rea of sector O-R rea of O = 7.68 6 = 1.68 cm (a) rea of sector O-R is 7.68 cm (b) rea of O is 6 cm (c) rea of segment R is 1.68 cm In MLK, m MLK = 90º m MKL = 0º M m LMK = 60º [Remaining angle] MLK is a 0º - 60º - 90º triangle. By 0º - 60º - 90º triangle theorem, L 0º K 6 LK = MK [Side opposite to 60º] 45º N 6 = MK MK = 6 MK = 1 units...(i)
... 6... ML = 1 MK [Side opposite to 0º] ML = 1 1 ML = 6 units... In MKN, m MKN = 90º m MNK = 45º m NMK = 45º [Remaining angle] MKN is a 45º - 45º - 90º triangle By 45º - 45º - 90º triangle theorem, MK = KN = 1 MN... 1 MK = MN [From ] 1 1 = MN MN = 1 units...(iv) KN = 1 units...(v) [From (i) and ] erimeter of MNKL = MN + KN + KL +ML = 1 1 6 6 [From, (iv) and (v) and given] = 18 1 6 1 erimeter of MNKL = 6 units