Maharashtra State Board Class X Mathematics - Geometry Board Paper 016 Solution 1. i. ΔDEF ΔMNK (given) A( DEF) DE A( MNK) MN A( DEF) 5 5 A( MNK) 6 6...(Areas of similar triangles) ii. ΔABC is 0-60 -90 triangle. BC is the side opposite to 0. 1 1 1 BC hypotenuse AC 18 9 cm iii. m(arc PMQ) = 10.(given) 1 1 PQS m(arcpmq) 10 65 iv. cos( θ) = cosθ cos( 60 ) = cos 60 cos 60 o 1 v. Inclination of the line = θ = 0 1 slope tan0 1 Thus, the slope of the line is. vi. E = 0, F = 1 Euler s formula: F + V = E + 1 + V = 0 + 1 + V = V = 1 = 0
. i. Seg RS bisects PRQ.(given) PR PS QR SQ...(angle bisector property) 15 6 QR 8 15 8 QR 0 6 ii. PA is a tangent segment and PBC is the secant. PB PC = PA 10 PC 15 5 PC.5 10 Now, PB BC PC...(P B C) 10 BC.5 BC.5 10 1.5 iii. Steps of construction: 1. Construct the equilateral XYZ of side equal to 6. cm.. Draw the perpendicular bisectors PR and QS of sides XY and YZ respectively.. Mark the point of intersection as O. 4. Draw a circle with centre O and radius OX or OY or OZ. This is the required circumcircle.
iv. The initial arm rotates 5 in the anticlockwise direction. The angle is positive and the measure of the angle is between 0 and 90. Thus, the terminal arm lies in quadrant I. v. Length of an arc = 10 cm Radius (r) = 5 cm r Area of the sec tor length of an arc 5 10 5 cm Thus,the area of the sector is 5 cm. vi. Radius (r) of a sphere = 4. cm Surface area of a sphere = 4πr 4 (4.) 7 1.76 cm Thus, the surface area of sphere is 1.76 cm.
. i. Let PQRS be a parallelogram. Then, PQ = 17 cm, QR = 11 cm and diagonal PR = 6 cm The diagonals of a parallelogram bisect each other. Point M is the point of intersection of diagonals PR and QS. 1 1 PM MR PR 6 PM MR 1 cm...(1) 1 QM MS = QS QS QM...() In PQR, QM is the median. PQ QR PM QM... By Apollonius theorem (17) (11) (1) QM 89 11 (169) QM 410 (169) QM Diving by, we get 05 169 + QM QM 05 169 6 QM 6 QS QM 6 1 cm Thus, the length of the other diagonal is 1 cm. ii. Given: m(arc PCR) = 6, m(arc QDS) = 48 By Inscribed Angle Theorem,we get 1 1 (i) PQR marc PCR 6 1...(1) 1 1 (ii) SPQ = marc QDS 48 4...() (iii) In AQR, by the Remote Interior Angle theorem, RAQ AQR SRQ SRQ SPQ...(Angles subtended by the same arc) i.e. RAQ AQR SPQ mraq mspq maqr 4 1 11...[From (1) and ()]
iii. Steps of construction: 1. Draw a circle of radius.5 cm. Take any point K on it.. Draw a chord KL through K. Take any point M on the major arc KL.. Join KM and ML. 4. Draw an arc of the same radius taking M and L as the centres. Taking H as the centre and radius equal to IJ, draw an arc intersecting the previous arc at G. KML = KLR Join LG and extend it on both the sides to draw PR which is the required tangent to the circle at K. iv. Given that is in quadrant IV, where x is positive and y is negative. sec r x Let r k, then x k r x y k k y y 4k k k y k Now, y is negative. y k r k cosec =. y k Substituting the value of cosec, we get 1 cosec 1 ( ) 1 1+cosec 1 ( ) 1 1 1 cosec 1+cosec
v. Let (, ) (x 1, y 1 ) and (4, 7) (x, y ). The equation of a line passing through a pair of points is y y1 x x1 y y x x 1 1 y x 7 4 y x 4 y (x ) [Multiplying both the sides by (-4)] y = x 4 y = x 4 + y = x 1 The equation of the line is y = x 1 4. i. Given: A circle with centre O and an external point P are given. AP and BP are the two tangents drawn from an external point P. To prove: AP = BP Construction: Draw seg OA, seg OB and seg OP. Proof: In ΔOBP and ΔOAP, OA = OB (Radii of the same circle) OP = OP (Side common to both the triangles) OAP = OBP = 90 (tangent is perpendicular to the radius at the point of contact) ΔOBP ΔOAP (By R.H.S) AP = BP (corresponding sides of congruent triangles) Thus, the lengths of two tangent segments to a circle drawn from an external point are equal.
ii. Let AB = height of the tower = h metres In the right angled ABC and right angled ABD, h h tan60 BC BC h 1 tan0 BD h BD Now, BD BC = 40 h h 40 h h 40 h = 40 h = 0 metres In right angled ABC, 1 tan0 BC 1 h BC 1 0 BC b 0 m Width of the river = BC = 0 m Thus, the height of the tree is 0 metres and width of the river is 0 metres. iii. AD is the median CD BD... D is the midpoint of BC Coordinates of D can be found by using section formula. Let x,y be coordinates of the centre of the circle. 1 8 1 1 1 1 x,y, 1, 5 Coordinates of point D are 1, 5. Let m be the slope of AD. Coordinates of A(5, 4) and D( 1, 5) y 4 5 y 1 9 m x x 5 1 6 1 Equation of line is y mx c, where c is the y int ercept. 5 4 c... Substituting the coordinates of A 7 c
x 7 Equation of line of AD is y y x 7 is equation of the median AD. 1 The coordinates of A 5,4 and C 1, 8 y y1 8 4 1 Slope of AC x x 1 5 4 Substituting the coordinates of A 5, 4 in the equation y mx c 4 5 c c 11 Line parallel to AC and pas sing through B, has slope Substituting the coordinates of B, in the equation y mx c c c 7 Equation of line parallel to AC and pas sing through B, is y x 7. 5. i. Given: AE = EF = AF = BE = CF, AT EF ΔAEF is equilateral triangle. a ET = TF a BT CT a...(1) In right triangles, ΔATB and ΔATC, AT = AT (Side common to both triangles) ATB = ATC (Right angles) BT = CT. (from 1) ΔATB ΔATC..(by SAS) AB = AC In ΔAEF, AE = AF = EF (Given) ΔAEF is an equilateral triangle. AT = a...(altitude of equilateral triangle) In ΔATB, AB AT BT a a 9a 1a AB a a a 4 4 4 AB a i.e., AB AC a
ii. Steps of construction: 1. Construct the Δ SHR with the given measurements. For this draw SH of length 4.5 cm.. Taking S as the centre and radius equal to 5.8 cm draw an arc above SH.. Taking H as the centre and radius equal to 5. cm draw an arc to intersect the previous arc. Name the point of intersection as R. 4. Join SR and HR. Δ SHR with the given measurements is constructed. Extend SH and SR further on the right side. 5. Draw any ray SX making an acute angle with SH on the side opposite to the vertex R. 6. Locate 5 points. (the ratio of old triangle to new triangle is 5 and 5 > ) Locate A 1, A, A, A 4 and A 5 on AX so that SA 1 = A 1 A = A A = A A 4 = A 4 A 5. 7. Join A H and draw a line through A 5 parallel to A H, intersecting the extended part of SH at V. 8. Draw a line VU through V parallel to HR. Δ SVU is the required triangle.
iii. Diameter of a pipe = 0 mm.(given) 0 Radius of the pipe mm 10 mm 1 cm Speed of water 15 m/min 1500 cm/min Volume of cylinder r h 1 1500 Volume of water that flows in pipe in 1 minute 7 000 cm 7 Radius of conical vessel Depth 45 cm... Given 40 0 cm, 1 Capacity of the conical vessel r h 1 0 0 45 7 96000 cm 1 Time required to fill the vessel Capacity of the vessel Volume of water flowing per minute 96000 1 000 7 96 99 4 minutes Thus, the time required to fill the conical vessel is 4 minutes.