BOARD QUESTION PAPER : MARCH 2016 GEOMETRY


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1 BOARD QUESTION PAPER : MARCH 016 GEOMETRY Time : Hours Total Marks : 40 Note: (i) Solve All questions. Draw diagram wherever necessary. (ii) Use of calculator is not allowed. (iii) Diagram is essential for writing the proof of the theorem. (iv) Marks of constructions should be distinct. They should not be rubbed off. Q.1. Attempt any five of the following subquestions: [5] (1) DEF ~ MNK. If DE 5. MN 6, then find the value of A( DEF). A( MNK) () In the adjoining figure, in ABC, B 90 O, C60 O, A0 O, AC 18 cm. Find BC. () In the adjoining figure, m(arc PMQ) 10 o, find PQS. 0 (4) If the angle θ 60, find the value of cos θ. (5) Find the slope of the line with inclination 0 o. (6) Using Euler s formula, find V if E 0, F 1. Q.. Solve any four subquestions: [8] (1) In the following figure, in PQR, is the bisector of PRQ. If PS 6, SQ 8, PR 15, find QR. () In the following figure, a tangent segment PA touching a circle in A and a scant PBC shown. If AP 15, BP 10, find BC. () Draw an equilateral ABC with side 6. cm and construct is circumcircle. (4) For the angle in standard position if the initial arm rotates 5 o in anticlockwise direction, then state the quadrant in which terminal arm leis. (Draw the figure and write the answer.) (5) Find the area of sector whose arc length and radius are 10 cm and 5 cm respectively.
2 (6) Find the surface area of a sphere of radius 4. cm. π Q.. Attempt any three of the following subquestion: [9] (1) Adjacent sides of a parallelogram are 11 cm and 17 cm. of the length of one its diagonal is 6 cm, find the length of the other. () In the following figure, secants containing chords RS and PQ of a circle intersects each other in point A in the exterior of a circle. If m(arc PCR) 6 o, m(arc QDS) 48 o, then find. () Draw a circle of radius.5 cm. take any point K on it. Draw a tangent to the circle at K without using centre of the circle. (4) If sec α, then find the value of 1 cos 1+ cos 7 ec α, where α is in IV quadrant. ec α (5) Write the equation of the line passing through the pair of pints (, ) and (4, 7) in the form of y m x + c. Q.4. Attempt any two of the following subquestions: [8] (1) Prove that The length of the two tangent segments to a circle drawn from an external point are equal. () A Person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60 o. When he movies 40 m away from the bank, he finds the angle of elevation to be 0 o. Find the height of the tree and width of the river. ( 1.7) () A(5, 4), B(, ) and C(1, 8) are the vertices of a triangle ABC. Find the equations of median AD and line parallel to AC passing through the point B. Q.5. Attempt any two of the following subquestions: [10] (1) In the following figure, AE EF AF BE CF a, AT BC. Show that AB AC a. () SHR SVU. In SHR, SH 4.5 cm, HR HR 5. cm, SR 5.8 cm and SV 5. Construct SVU. () Water flows at the rate of 15 m per minute through a cylindrical pipe, having the diameter 0 mm. how much time will it take to fill a conical vessel of base diameter 40 cm and depth 45 cm?
3 BOARD ANSWER PAPER : MARCH 016 GEOMETRY Q.1. Attempt any five of the following subquestions: [5] (1) Solution: DEF ~ MNK () Solution: A( DEF) DE A( MNK) MN... (Given) ( 5 ) ( 6 ) A( DEF ) 5 A( MNK ) 6 In ABC, B 90 o, C 60 o, A 0 o... (Given) BC 1 AC... (Side opposite to 0 o )... (Ratio of areas of similar s ratio of squares of the corresponding sides.) BC 1 18 BC 9 cm. () Solution: m PQS 1 m( arc PMQ )... (Tangent  secant Th.) m PQS 1 10 (4) Solution: m PQS 60 o cos θ cos ( 60 o ) cos 60 o o cos 1 (5) Solution: Inclination (θ) 0 slope of line (m) tan θ 0 1 m tan 0 slope of line is 1
4 (6) Solution: Since solid figure is prism (polyhedron) By Eulers formula, F + V E 1 + V V V + 18 V 0 Q.. Attempt any four of the following subquestions: [8] (1) Solution: In PQR, seg RS bisects PRQ... (Given) PS PR... (Angle bisector property) QS QR QR QR QR 0 () Solution: Ray PA is a tangent and PBC is a secant PA PB PC [Tangent secant property] ( 15) 10 PC 10PC 5 PC 5 10 PC.5 PB + BC.5... (P  B  C) 10 + BC.5 BC.5 10 BC 1.5 () Solution: Steps of construction (a) To draw ABC. (b) To draw perpendicular bisectors of seg AC and seg BC intersecting at O. (c) To draw circumcircle.
5 (4) Solution: If the initial arm rotates 5 0 in anticlockwise direction, the terminal arm lies in I quadrant. 0 y 5 x (5) Solution: 1 Area of sector r length of arc cm The Area of sector is 5 cm (6) Solution: Surface area of a sphere 4πr cm The Total surface area of a sphere is 1.76 cm Q.. Attempt any three of the following subquestion: [9] (1) Solution : ABCD is a m AD 11 cm, AB 17 cm Diagonal AC 6 cm In a parallelogram diagonals bisect each other. O is the mid  point of AC and BD AO 1 AC In ADB, seg AO is the median By Apollonius principle, AB + AD AO + BO ( ) ( ) ( ) cm BO BO ( ) BO BO 7 BO BO 6 BO 6 BD BO 6 1 length of the other diagonal is 1 cm. 11 cm B A O 17 cm C D
6 () Solution : i) PQR 1 marcpcr ( )... (Inscribed angle Th.) 1 o o 6 1 ii) SPQ 1 marcsdq ( )... (Inscribed angle Th.) 1 o o 48 4 iii) SPQ SRQ... (Angles inscribed in the same arc) SRQ 4 o SRQ is an exterior angle of ARQ. () Solution : SRQ RAQ + RQA... (Remote interior angle Th.) 4 o RAQ + 1 o RAQ 11 o Steps of construction : (i) (ii) Draw a circle with.5 cm radius. Marks point K on the circle, draw chord KQ. (iii) Mark point P on the major segment. (iv) At K draw QKM congruent to QPK. (4) Solution : Line KM is tangent to the circle at K. As α is in IV quadrant is a negative acute angle x is positive & y is negative cos α and sec α are positive 1) 1 + tan α sec α [Identity] tan α sec α tanα 1 [Taking square root on both sides] 1 tan α [α is in IV quadrant]
7 ) cotα cotα 1 tanα 1 1 [Relation] cot α ) 1 + cot α cosec α [Identity] 1 + ( ) cosec α 1 + cosec α 4 cosec α cosecα cosecx The value of [Taking square root on both sides].[α is in IV quadrant] 1 cosecα 1 ( ) [Substituting the values] 1+ cos ecα 1 + ( ) 1 1 cosecα 1+ cos ecα (5) Solution: Let P (, ) (x1,y1), Q (4,7) (x, y) Since the line passes through points P and Q Equation of line in two point form is x x1 y y1 x x y y 1 1 x y 4 7 x y 4 (x ) y 4 4 (x ) y x 4 y x 4 + y x 1 y y x 1 Equation of line is y x 1.
8 Q.4. Attempt any two of the following subquestions: [8] (1) Solution : Given : A circle with centre O, an external point P of the circle. The two tangents through the point P touches the circle at the points A and B. O To prove : PA PB Construction : Draw seg OA, seg OB and seg OP. Proof : PAO PBO... (Radius is perpendicular to tangent) In the right angled PAO and the right angled PBO. Seg OA seg OB... (Radii of same circle) hypotenuse PO hypotenuse PO... (common side) PAO PBO... (Hypotenuse side theorem) seg PA seg PB... (c.s.c.t) PA PB Hence, the length of the two tangent segments to a circle drawn from an external point are equal. A B P () Solution: In the given figure, AB represents the height of the tree & BC represents the width of the river. The person at the beginning is at point C observing the top of the tree & making an angle of elevation. moving 40m away again makes an angle of elevation with the top of tree. ACB & ADB is an angle of elevation ACB 60 0 & BC x BD BC + CD (BCD) 1) BD (x + 40)m ) In ABC, ABC 90 0 & ACB 60 0 ) tan θ AB BC θ 60 0 D 0 60 C x A y B tan 60 y x y [ tan60 ] x x 4) In ABD y (i) ADB 0 0 & ABD 90 0 θ 0 0 5) tanθ AB BD tan0 y x + 40
9 6) 1 y x 40 + x + 40 y (ii) x + 40 x [from (i) & (ii)] x x + 40 x x 40 x 0 Substituting x 0 in eqn (i) [ tan0 ] x y 0 y y 17. y 4.6 y Height of tree 4.6 m & Width of river 0 m () Solution: A (5,4), B (, ), C (l, 8) seg AD is the median D is the midpoint of BC By midpoint formula D, 10 D, D ( 1, 5) For median AD, Let A (5, 4) (x1, y1), D ( 1, 5) (x, y) Equation of median AD in two point form is x x y y x x y y x 5 y 4 5 ( 1) 4 ( 5) x 5 y x 5 y (x 5) 6(y 4) 9x 45 6y 4
10 9x 6y x 6y 1 0 x y (divide by ) Equation of median AD is x y 7 0 Slope of line AC slope of line AC line BP line AC slope of line BP slope of line AC slope of line BP (m) For line BP,... [given] Equation of line BP with slope (m) and passing through point B(, ) (x1, y1) in slope point form is y y1 m(x x1) y ( ) [x ( )] y + (x + ) y + x + 9 x + y x + y 7 0 x y Equation of line parallel to AC is x y Q.5. Attempt any two of the following subquestions: [10] (1) Solution: Given: Proof: AT BC. AE EB AF RN FC a. To Prove : AB BC a. In ABF, EB EF a [Given] E is the midpoint of BF (i). AE is a median. [From (i)]. AB + AF AE + EB [By Appollonius Theorem] AB + a a + a AB + a 4a AB 4a a AB a AB a AB a. (ii)
11 Similarly, we can prove AC.a AB AC.a () Solution: (Analytical Figure) (iii) (From (ii) & (iii)) (Required triangle) () Solution : Cylindrical pipe, d 0 mm, r 10 mm 1 cm Rate of water flow 15 m / min h 1500 cm / min Conical vessel, D 40 cm R 0 cm and H 45 cm Let the conical vessel be filled in x minutes. Volume of water flowing through cylindrical pipe in x min. Volume of conical vessel 45 cm d 40 cm πr h x 1 πr H x x 1500 x 4 minutes. It will take 4 minutes to fill the conical vessel.
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