CCE RF CCE RR RF & RR-00 [ Turn over O⁄´¤%lO⁄ ÆË√v⁄ ÃO⁄–y Æ⁄¬fiO¤– »⁄flMs⁄ÿ, »⁄fl≈Ê«fiÀ⁄ ¡⁄M, ∑ÊMV⁄◊⁄‡¡⁄fl — 560 003 KARNATAKA SECONDARY EDUCATION EXAMINATION BOARD, MALLESWARAM, BANGALORE – 560 003 G—È.G—È.G≈È.“. Æ⁄¬fiOÊ⁄–, »⁄·¤^È% / HØ√≈È — 2018 S. S. L. C. EXAMINATION, MARCH/APRIL, 2018 »⁄·¤•⁄¬ D}⁄ °¡⁄V⁄◊⁄fl MODEL ANSWERS ¶´¤MO⁄ : 26. 03. 2018 ] —⁄MOÊfi}⁄ —⁄MSÊ¿ : 81-E Date : 26. 03. 2018 ] CODE NO. : 81-E …Œ⁄æ⁄fl : V⁄{}⁄ Subject : MATHEMATICS ( ‘ʇ—⁄ Æ⁄p⁄¿O⁄√»⁄fl / New Syllabus ) ( À¤≈¤ @∫⁄¥¿£% & Æ⁄‚¥´⁄¡¤»⁄~%}⁄ À¤≈¤ @∫⁄¥¿£% / Regular Fresh & Regular Repeater ) (BMW«ŒÈ ∫¤Œ¤M}⁄¡⁄ / English Version ) [ V⁄¬Œ⁄r @MO⁄V⁄◊⁄fl : 80 [ Max. Marks : 80 Qn. Nos. Ans. Key Value Points Marks allotted I. 1. In the given Venn diagram n ( A ) is Ans. : A 3 1 2. Sum of all the first ‘n’ terms of even natural number is Ans. : A n ( n + 1 ) 1
31
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36. A solid is in the form of a cone mounted on a right circular cylinder, both having same radii as shown in the figure. The radius of the base and height of the cone are 7 cm and 9 cm respectively. If the total height of the solid is 30 cm, find the volume of the solid.
OR
The slant height of the frustum of a cone is 4 cm and the perimeters of its circular bases are 18 cm and 6 cm respectively. Find the curved surface area of the frustum. Ans. :
r = 7 cm Let 1h = 21 cm for cylinder
r = 7 cm 2h = 9 cm for cone
Volume of solid = Volume of cylinder + Volume of cone ½
= 22
12
31 hrhr π+π ½
= )31( 21
2 hhr +π ½
= )93121(7
722 32 ×+× ½
= 722
× 7 × 7 ( 24 ) ½
= 3696 c.c. ½ 3
Direct substitution of 1h and 2h value can also be considered.
OR
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12 rπ = 18 cm 22 rπ = 6 cm l = 4 cm ½
1r =
π=
π9
218 cm
π=
π=
326
2r cm
Curved Surface Area = lrr )( 21 +π 1
= ⎟⎟⎠
⎞⎜⎜⎝
⎛π
+π
π39 4 1½
= 48 cm2. 3 OR
CSA = l ][ 21 rr π+π
= 4 [ 9 + 3 ] = 4 [ 12 ] = 48 cm2
V. 37. Solve the equation 022 =−− xx graphically.
Ans. :
Let y = 0
2x – x – 2 = 0 given
∴ y = 2x – x – 2
x 0 1 – 1 2 3 – 2
y – 2 – 2 0 0 4 4
1) x = 0
y = 20 – 0 – 2
y = – 2
4) x = 2
y = 22 – 2 – 2
y = 0
2) x = 1
y = 21 – 1 – 2 y = – 2
5) x = 3
y = 23 – 3 – 2
y = 9 – 5 y = 4
3) x = – 1
y = ( – 1 )2 – ( – 1 ) – 2 = 1 + 1 – 2 y = 0
6) x = – 3
y = ( – 3 )2 – ( – 3 ) – 2 y = 9 + 3 – 2 = 10
7) x = – 2
y = ( – 2 )2 – ( – 2 ) – 2 y = 4 + 2 – 2 y = 4
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Graph roots
Table — 2
Parabola — 1
Roots — ½ + ½ 4
Alternate Method :
Given 2x – x – 2 = 0
2x = x + 2
Consider y = 2x and y = x + 2
(i) y = 2x
x 0 1 – 1 2 – 2 3 − 3
y 0 1 1 4 4 9 9
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(ii) y = x + 2
x 0 1 2 – 1 2
y 2 3 4 1 0
Tables — 2 ( 1 + 1 ) Line — ½ Parabola — ½ Roots — ( ½ + ½ )
4
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38. Construct a direct common tangent to two circles of radii 4 cm and 2
cm whose centres are 9 cm apart. Measure and write the length of the
tangent.
Ans. :
R = 4 cm r = 2 cm d = 9 cm
R – r = 2 cm
Length of the tangent = 8·7 cm
Drawing four circles 2
Drawing tangent 1½
Finding the length ½ 4
39. State and prove Basic Proportionality ( Thale’s ) Theorem.
Ans. :
If a straight line is drawn parallel to a side of a triangle, then it divides
the other two sides proportionally. 1
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Data : In Δ ABC, DE || BC
To prove : CEAE
BDAD
= ½
Construction : Join DC and EB
Draw EL ⊥ AB and DN ⊥ AC. ½
Proof :
BDEADE of Area of Area
ΔΔ =
ELBD
ELAD
××
××
2121
⎥⎦⎤
⎢⎣⎡ = bhA
21
Q ½
∴
BDAD
BDEADE
=ΔΔ
... (i)
CDEADE of Area of AreaΔΔ =
DNEC
DNAE
××
××
2121
½
ECAE
CDEADE
=ΔΔ
⇒
CEAE
BDAD
= Q ½
4
Area Δ BDE = area
of Δ CDE and Axiom-1
½
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40. A vertical tree is broken by the wind at a height of 6 metre from its foot
and its top touches the ground at a distance of 8 metre from the foot of
the tree. Calculate the distance between the top of the tree before
breaking and the point at which tip of the tree touches the ground,
after it breaks.
OR
In Δ ABC, AD is drawn perpendicular to BC. If BD : CD = 3 : 1, then prove that )(2 222 ACABBC −= .
Ans. :
In the figure,
Let AC represents the tree h.
B is the point of break BC = 6 m
E is the top of the tree touches the ground CE = 8 m
AE is the distance between the top of the tree before break and after