BRD NSWER PPER :CTBER 04 GEETRY. Solve any five sub-questions: BE i. BE ( BD) D BE 6 ( BD) 9 ΔBE (ΔBD) ----[Ratio of areas of two triangles having equal base is equal to the ratio of their corresponding heights.] ii. If two circles touch externally, then the distance between their centres is sum of their radii. the distance between their centres 5 + 8 cm iii. The terminal arm is in II quadrant. P Y X X Y If the initial arm rotates in anticlockwise direction, angle lies between 90 and 80 If the initial arm rotates in clockwise direction, angle lies between 70 and 80 the possible angles lies between 90 and 80 or 70and 80 iv. slope tan tan 60 v. Radius (r) 6 cm length of arc (l) 5 cm rea of sector radius length of arc 6 5 45 rea of the sector is 45 cm. vi. The longest side is 5 cm. (5) 65 ----(i) Now, sum of the squares of the other two sides will be (7) + (4) 49 + 576 65 ----(ii) (5) (7) + (4) ----[From (i) and (ii)] Yes, the given sides form a right angled triangle. ----[By converse of Pythagoras theorem]
Board nswer Paper: ctober 04. Solve any four sub-questions: i. In LN, Ray LS is the angle bisector of LN. --- [Given] S SN L LN --- (i) [By property of angle bisector of a triangle] seg L seg LN ----[Given] L LN L LN ----(ii) ii. S SN ----[From (i) and (ii)] S SN seg S seg SN Construction: Join seg N. L N Proof: Line L is a tangent to the smaller circle at N. ----[Given] seg N seg L ----[Tangent is perpendicular to radius.] LN N ----[Perpendicular drawn from the centre of the circle to the chord bisects the chord.] N is the midpoint of seg L. iii. Let P (, ) (x, y ), Q (4, ) (x, y ) y y Slope of line PQ x x 4 iv. + tan sec + (4) sec sec + 6 7 sec 7 cos sec cos 7
v..5 cm Draw the circle of radius.5 cm Extend the line through Draw the perpendicular at point [] vi. The equation of line is y 5x 4 i.e., y 5x + 4 Comparing this equation with y mx + c, we get m 5, c 4 slope 5 and y-intercept 4. Solve any three sub-questions: i. In BC, seg P is the median. ----[Given] B + C P + PC ----[By pollonius theorem] 60 (7) + PC ----[Substituting the given values] 60 98 + PC 60 98 PC 6 PC 6 PC 8 PC PC 9 units ----(i)[taking square root on both sides] BC PC ----[P is the midpoint of seg BC.] BC 9 ----[From(i)] BC 8 units
Board nswer Paper: ctober 04 ii. Construction: Draw seg GF. E F G H Proof: Chord EF chord GH, GF is the transversal EFG HGF ----(i) [lternate angles] EFG m(arc GE) ----(ii) HGF [Inscribed angle theorem] m(arc HF) ----(iii) m(arc GE) m(arc HF) ----(iv) [From (i), (ii) and (iii)] lso, EG m(arc GE) ----(v) [Definition of measure of arc] FH m(arc HF) ----(vi) EG FH ----[From (iv), (v) and (vi)] iii. T P 60 5.6 cm 70 Draw PT of given measure [] Draw the perpendicular bisectors of side TP and side T [] Draw circumcircle by taking as centre [] iv. L.H.S. cos cos cos cos cos cos ----[n rationalising the denominator] ( cos ) cos 4
( cos ) ----[ sin cos ] sin cos sin sin + cos sin cosec + cot R.H.S. +cos cosec + cot -cos v. Let x-intercept of the line be a and y-intercept of the line be b. b a Equation of the line in double intercept form is x y a b x y a a x y a The equation of line passes through (, 4). a 4 a Equation of the line is x y x y + 0 The required equation of line is x y + 0. 4. Solve any two sub-questions: i. Given: circle with centre, an external point P of the circle. The two tangents through the point P touch the circle at the points and B. To prove: P PB Construction: Draw seg, seg B and seg P. B P Proof: P PB 90 ----[Tangent is perpendicular to radius.] In the right angled P and the right angled PB, seg seg B ----[Radii of the same circle] hypotenuse P hypotenuse P ----[Common side] P PB ----[Hypotenuse side test] seg P seg PB ----[c.s.c.t.] P PB Hence, the lengths of the two tangent segments drawn to a circle from an external point are equal. 5
Board nswer Paper: ctober 04 ii. Let B represent the height of the tree. Let the tree breaks at point C. C 0 B 0m D C is the broken part of the tree which takes position CD such that CDB 0 C CD ----(i) In right angled CBD, tan CDB tan 0 tan 0 BC BD BC ----[substituting the given value] 0 BC 0 m lso, cos 0 BD CD 0 CD CD 0 CD 60 ----(ii) ----(iii) B C + BC ----[CB] B CD + BC ----[From (i)] B 60 + 0 ----[From (ii) and (iii)] B 90 90 ----[n rationalizing the denominator] B 90 0 0.7 5.9 m height of the tree is 5.9 m. iii. The volume of cuboid l b h (44 4 ) cm 8808 cm Since, the metallic cuboid is molten and recasted into a sphere, volume of sphere volume of cuboid volume of sphere 8808 cm 6
But, volume of sphere 4 r 4 r 8808 r 8808 4 7 r r cm surface area of sphere 4r 4 7 () 5544 cm The surface area of the sphere is 5544 cm. 5. Solve any two sub-questions: i. Given: In CB, m 90, mb 60, mc 0 To prove : i. B BC ii. C BC Construction: Take a point D on ray B such that B D. Join point C to point D. C 0 D Proof: In CBD, D B B ----[By construction] is midpoint of seg BD ----(i) lso,m CB 90 ----[Given] seg C seg BD ----(ii) seg C is perpendicular bisector of seg BD ----[From (i) and (ii)] CD CB ----[By perpendicular bisector theorem] CDB is an isosceles triangle CDB CBD ----(iii) [By isosceles triangle theorem] But, CBD 60 ----(iv) [Given] CDB 60 ----[From (iii) and (iv)] BCD 60 ----[Remaining angle of a triangle] CDB is an equilateral triangle ----[ll angles are 60] BD BC CD ----(v) [Sides of equilateral triangle] B BD 60 ----(vi) [By construction] B BC ----(vii) [From (v) and (vi)] In CB, CB 90 ----[Given] BC C + B ----[By Pythagoras theorem] BC C + BC ----[From (vii)] 7
Board nswer Paper: ctober 04 BC C + 4 BC C BC 4 BC C 4BC BC 4 C BC 4 C ii. BC ----[Taking square root on both sides] T T E E 0 H 6. cm [] nalytical figure 0 6. cm H 4 5 6 7 X Draw T of given measure [] Draw a ray making an acute angle at with side, mark points,,., 7 such that 4 4 5 5 6 6 7 Join 7 and and draw seg 5 H parallel to 7, where H is the point on [] Draw HE side T [] HE is the required triangle similar to T. 8
iii. For the cylindrical pipe: diameter 0 mm radius diameter 0 0 mm cm Rate of flow of water through the pipe 0m/minute 0 00 cm/minute 000 cm/minute Water flow s through a distance (h) of 000 cm in a minute volume of water flowing through the pipe in minute r h 000 000 cm For the conical vessel: diameter 40 cm radius diameter 40 0 cm depth (h) 4 cm Volume of conical vessel r h 004 cm volume of conical vessel Time taken to fill the conical vessel volume of water flowing through pipe inmin 004 000 004 000 00. minutes + 0. 60 min sec 000 Time taken to fill the conical vessel is min sec. 9