Karnataka SSLC Question Paper 2026 Mathematics

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Joár ÊÜáá©ÅñÜ ±ÜâoWÜÙÜ ÓÜíTæ : 16 ]
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A
Joár ±ÜÅÍæ°WÜÙÜ ÓÜíTæ : 38 ] CCE RF/RR/
PF/PR
Total No. of Questions : 38 ]

—⁄MOÊfi}⁄ —⁄MSÊ¿ : 81-E 81-E

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Subject : MATHEMATICS
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1. This question paper consists of 38 questions.
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10/402 1 of 16

CCE RF/RR/PF/PR(A) 81-E

I. Four alternatives are given for each of the following

questions / incomplete statements. Choose the correct

alternative and write the complete answer along with its

letter of alphabet. 8×1=8

1. The probability of an impossible event is

(A) 1 (B) 0

(C) 1·5 (D) – 1

5
2. If cosec θ = , then the value of sin θ is
3

5 5
(A) (B)
3 4

3 3
(C) (D)
4 5

3. The quadratic equation among the following is

(A) x − 2 – 3x + 2 = 0 (B) 2x + 3 = 0

(C) x 2 – 5x + 6 = 0 (D) 2 x 3 + 7x + 1 = 0

10/402 2 of 16

CCE RF/RR/PF/PR(A) 81-E

4. In the pair of linear equations a1x + b1y + c1 = 0 and
a1 b1
a 2 x + b 2y + c 2 = 0 if ≠ , then the equations
a2 b2

(A) have unique solution

(B) do not have any solution

(C) have three solutions

(D) have infinitely many solutions

5. The volume of a sphere of radius ‘r’ units is
2
(A) π r 2 cubic units
3
1
(B) π r 3 cubic units
3
4
(C) π r 3 cubic units
3
4
(D) π r 2 cubic units
3

6. In an arithmetic progression if a n = 2n + 1, then the

common difference of the arithmetic progression is

(A) 1 (B) 2

(C) 3 (D) 4

10/402 3 of 16

CCE RF/RR/PF/PR(A) 81-E

7. The coordinates of the mid-point of the line segment joining

the points ( 2, 3 ) and ( 4, 7 ) are

(A) ( 3, 5 ) (B) ( 5, 3 )

(C) ( 1, 2 ) (D) ( 6, 10 )

8. In trapezium ABCD, AB || DC and P, Q are points on the

sides AD and BC respectively. If PQ || DC, PD = 18 cm,

BQ = 35 cm and QC = 15 cm, then length of AD is

(A) 5 cm (B) 50 cm

(C) 57 cm (D) 60 cm

10/402 4 of 16

CCE RF/RR/PF/PR(A) 81-E

II. Answer the following questions : 8×1=8

9. State the ‘fundamental theorem of arithmetic’.

10. If the first term of an arithmetic progression is ‘a’ and the
last term is ‘ an ’, then write the formula to find the sum of

the first ‘n’ terms of the arithmetic progression.

11. Write the degree of a cubic polynomial.

12. In the figure, D and E are the mid-points of sides AB and AC

respectively. If DE = 4 cm, then write the measure of BC.

13. Write the formula to find the total surface area of a cube of

edge of length ‘a’ units.

14. If the lines represented by the equations 4x + py + 8 = 0 and

4x + 4y + 2 = 0 are parallel, then find the value of ‘p’.

10/402 5 of 16

CCE RF/RR/PF/PR(A) 81-E

15. In the given cumulative frequency distribution table, write

the median class.

Marks Number of Cumulative
students frequency

0 – 10 3 3

10 – 20 4 7

20 – 30 7 14

30 – 40 6 20

n = 20

16. Write the coordinates of the point where the graph of linear
polynomial p ( x ) = x + 5 intersects the x-axis.

III. Answer the following questions : 8 × 2 = 16

17. Prove that 5 + 3 is an irrational number.

18. Solve the given pair of linear equations by elimination
method :

2x + y = 8

3x – y = 7

19. Find the sum of the first 20 terms of the arithmetic
progression 5, 8, 11, .......... using formula.

OR

Find the number of terms of the arithmetic progression
100, 96, 92, .... 12 using formula.

10/402 6 of 16

CCE RF/RR/PF/PR(A) 81-E

20. In the figure, ABC = 90°. Write the values of the

following :

i) sin α

ii) tan θ

21. Find the roots of the quadratic equation x 2 + 4x – 60 = 0 by

factorisation method.

OR

Find the discriminant of the quadratic equation

2x 2 + 3x − 7 = 0 and write the nature of the roots.

22. Find the HCF of 135 and 75 by prime factorisation method

and then find the LCM of HCF ( 135, 75 ) and 20.

23. A ( x, 5 ) and B ( 2, y ) are the coordinates of the end points

of the diameter AB of a circle with centre P ( 4, 3 ). Find the

values of ‘x’ and ‘y’. Also find the length of the diameter of

the circle.

10/402 7 of 16

CCE RF/RR/PF/PR(A) 81-E

24. AB is a tangent to the circle with centre ‘O’ and ‘A’ is the

point of contact. If OBA = 45°, then prove that ∆ AOB is

an isosceles triangle. If OB = 5 2 cm, then find the radius

of the circle.

IV. Answer the following questions : 9 × 3 = 27

25. Find the zeroes of the quadratic polynomial

p ( x ) = x 2 – 2x – 8 and verify the relationship between the

zeroes and the coefficients.

26. Prove that “The tangent at any point of a circle is

perpendicular to the radius through the point of contact”.

27. ∆ ABC ~ ∆ PQR. In ∆ ABC, A = 55°, C = 35° and the

sides of ∆ ABC are in the ratio 3 : 4 : 5. If the perimeter of

∆ PQR is 60 cm, then find the measure of sides and angles

of ∆ PQR.

10/402 8 of 16

CCE RF/RR/PF/PR(A) 81-E

28. Find the mean for the following data :

Class-interval Frequency

0 – 10 4

10 – 20 9

20 – 30 15

30 – 40 14

40 – 50 8

OR

Find the mode for the following data :

Class-interval Frequency

5 – 15 3

15 – 25 4

25 – 35 8

35 – 45 7

45 – 55 3

29. Prove that,
1
( cosec A – sin A ) ( sec A – cos A ) =
tan A + cot A

OR

Prove that,
sec 60 o 2 sin 90 o tan 45 o
− + = sin 2 45° ( cosec 2 45° – sec 2 0° )
o o o
cot 45 cos 0 cosec 30

10/402 9 of 16

CCE RF/RR/PF/PR(A) 81-E

30. A hand fan is made up of cloth fixed in between the metallic

wires. It is in the shape of a sector of a circle of radius

21 cm and of an angle 120° as shown in the figure.

Calculate the area of the cloth used and also find the total

length of the metallic wire required to make such a fan.

OR

In the given figure, ‘O’ is the centre of the circle of radius
21 cm. If AOB = 60°, then find the area of the segment

APB.
[ Take 3 = 1·73 ]

10/402 10 of 16

CCE RF/RR/PF/PR(A) 81-E

31. The perimeter of square ABCD is 12 m less than the

perimeter of the square PQRS. The sum of the areas of these

two squares is 117 m 2 . Find the ratio of the area of the

square ABCD to the area of the square PQRS.

32. In the figure, find the coordinates of the point which divides

the line segment AB internally in the ratio 1 : 2 using

section formula.

OR

10/402 11 of 16

CCE RF/RR/PF/PR(A) 81-E

Verify whether the points A ( 1, 5 ), B ( 2, 4 ) and C ( 9, – 3 )

are collinear with the help of ‘distance formula’.

33. Two fair dice whose faces are numbered from 1 to 6 are

rolled once. Find the probability

a) that the sum of the numbers on the top faces is 6

b) of getting a square number on one of the faces only.

V. Answer the following questions : 4 × 4 = 16

34. Find the solution of the given pair of linear equations by

graphical method :

x + 2y = 6

x+y = 5

35. The 7th term of an arithmetic progression is four times

its 2 nd term and the 12 th term is 2 more than thrice its

4th term. Find the progression.

36. Prove that, “If in two triangles, corresponding angles are

equal, then their corresponding sides are in the same ratio

( or proportion ) and hence the two triangles are similar”.

OR

10/402 12 of 16

CCE RF/RR/PF/PR(A) 81-E

Prove that, “If one angle of a triangle is equal to one angle of

the other triangle and the sides including these angles are

proportional, then the two triangles are similar”.

37. In the given figure, a rope is tightly stretched and tied from

the top of a vertical pole on a level ground to a peg on the

same level ground such that the length of the rope is 20 m.

The angle made by the rope with the ground is 30°. A circus

artist climbs the rope, reaches the top of the pole and from

there he observes that the angle of elevation of the top of

another pole on the same ground to be 60°. If the distance

of the foot of the longer pole from the peg is 30 m, then find

the height of this pole. ( Take 3 = 1·73 )

OR

10/402 13 of 16

CCE RF/RR/PF/PR(A) 81-E

A tower AB and a building DE are standing vertically on a

level ground. The angles of elevation of the top of the tower

from a point on the same ground and from the top of the

building are found to be 30° and 60° respectively as shown

in the figure. If the distance of the point from the foot of the

tower is 30 3 m and height of the building is 10 m, then

find the distance ( BE ) between the foot of the tower and

the building and also the distance between their tops ( AD ).

VI. Answer the following question : 1×5=5

38. From a solid cuboidal metallic block of dimensions

15 cm × 10 cm × 8 cm, three cylindrical depressions each

having diameter 4·2 cm and height 6 cm are drilled out as

10/402 14 of 16

CCE RF/RR/PF/PR(A) 81-E

shown in the figure. Find the total surface area and volume

of the remaining metallic block.

OR

In a cylinder of height 24 cm and radius 10 cm, water is
filled up to the height of 14·4 cm. A solid is made by placing
a cone of slant height 10 cm and curved surface area
60 π cm 2 on the hemisphere having the same radius as
that of the cone. If this solid is completely immersed in the
cylinder as shown in the figure, then find the rise in the
water level in the cylinder. Also find the amount of water
still to be filled in the cylinder so that the water level rises
up to the brim of the cylinder. ( Take π = 3·14 )

10/402 15 of 16

CCE RF/RR/PF/PR(A) 81-E

10/402 16 of 16

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Total No. of Questions : 38 ] PF/PR

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ÓÜíPæàñÜ ÓÜíTæ : 81-K Code No. : 81-K

×ÊÜáá¾S hÝPæp… Öæãí©ÃÜáÊÜ ±ÜÅÍ氱ܣÅPæ¿á®Üá° ñæÃæ¿áÆá CÈÉ PÜñܤÄÔ
ËÐÜ¿á : WÜ~ñÜ
Subject : MATHEMATICS
PܮܰvÜ ÊÜÞ«ÜÂÊÜá / Kannada Medium
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©®ÝíPÜ 28. 03. 2026 ] [ Date : 28. 03. 2026
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WÜÄÐÜu AíPÜWÜÙÜá 80 ] [ Max. Marks : 80

±ÜÄàûݦìWÝX ÓÝÊÜޮܠÓÜãaÜ®æWÜÙÜá Cut here/CÈÉ PÜñܤÄÔ
1. D ±ÜÅÍ氱ܣÅPæ¿áá Joár 38 ±ÜÅÍæ°WÜÙÜ®Üá° Öæãí©¨æ.
2. D ±ÜÅÍ氱ܣÅPæ¿á®Üá° ×ÊÜáá¾S hÝPæp… ÊÜáãÆPÜ ÊæãÖÜÃÜá ÔàÇ… ÊÜÞvÜÇÝX¨æ. ±ÜÄàûæ
±ÝÅÃÜí»ÜÊÝWÜáÊÜ ÓÜÊÜá¿áPæR ¯ÊÜá¾ ±ÜÅÍ氱ܣÅPæ¿á ŸÆŸ© ±ÝÍÜÌìÊÜ®Üá° PÜñܤÄÔ ¸Ý|¨Ü
WÜáÃÜáñÜ®Üá° A®ÜáÓÜÄÔ . GvÜŸ© ±ÝÍÜÌìÊÜ®Üá° PÜñܤÄÓܸæàw. ±ÜÅÍ氱ܣÅPæ¿áÈÉ GÇÝÉ
±ÜâoWÜÙÜá CÊæÁáà Gí¨Üá ±ÜÄàüÔPæãÚÛ.
3. ±ÜÅÍæ°WÜÚWæ PæãqrÃÜáÊÜ ÓÜãaÜ®æWÜÙÜ®Üá° ±ÝÈÔ.
Tear here/CÈÉí¨Ü PÜñܤÄÔ

4. ŸÆ»ÝWܨÜÈÉ PæãqrÃÜáÊÜ AíQWÜÙÜá ±ÜÅÍæ°WÜÚXÃÜáÊÜ ±Üä|ì AíPÜWÜÙÜ®Üá° ÓÜãbÓÜáñܤÊæ.
5. ±ÜÅÍ氱ܣÅPæ¿á®Üá° K©PæãÙÜÛÆá 15 ¯ËáÐÜWÜÙÜ PÝÇÝÊÜPÝÍÜÊÜâ ÓæàĨÜíñæ, EñܤÄÓÜÆá
¯WÜ©±ÜwÓÜÇÝ¨Ü ÓÜÊÜá¿áÊÜ®Üá° ±ÜÅÍ氱ܣÅPæ¿á ÊæáàÇݽWܨÜÈÉ ¯àvÜÇÝX¨æ.

10/401 1 of 16

CCE RF/RR/PF/PR(A) 81-K
I. PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ A¥ÜÊÝ A±Üä|ì ÖæàÚPæWÜÚWæ ®ÝÆáR ±Ü¿Þì¿á EñܤÃÜWÜÙÜ®Üá°

¯àvÜÇÝX¨æ. AÊÜâWÜÙÜÈÉ ÓÜãPܤÊÝ¨Ü EñܤÃÜÊÜ®Üá° BÄÔ, A¨ÜÃÜ PÜÅÊÜÞûÜÃܨæãvÜ®æ ±Üä|ì

EñܤÃÜÊÜ®Üá° ŸÃæÀáÄ 8×1=8

1. Jí¨Üá AÓÜí»ÜÊÜ Zo®æ¿á ÓÜí»ÜÊܯà¿áñæ¿áá

(A) 1 (B) 0

(C) 1∙5 (D) – 1

5
2. cosec  =
3
B¨ÝWÜ, sin  ¨Ü ¸æÇæ¿áá

5 5
(A) (B)
3 4

3 3
(C) (D)
4 5

3. D PæÙÜX®ÜÊÜâWÜÙÜÈÉ ÊÜWÜìÓÜËáàPÜÃÜ|ÊÜâ

(A) x  2 – 3x + 2 = 0 (B) 2x + 3 = 0

(C) x 2 – 5x + 6 = 0 (D) 2 x 3 + 7x + 1 = 0

10/401 2 of 16

CCE RF/RR/PF/PR(A) 81-K
4. a1x  b1y  c1  0 ÊÜáñÜᤠa 2x  b2y  c 2  0 D hæãàw ÃæàTÝñܾPÜ
a1 b1
ÓÜËáàPÜÃÜ|WÜÙÜÈÉ  B¨ÜÃæ, ÓÜËáàPÜÃÜ|WÜÙÜá
a2 b2

(A) A®Ü®Ü ±ÜÄÖÝÃÜÊÜ®Üá° Öæãí©ÃÜáñܤÊæ

(B) ¿ÞÊÜâ¨æà ±ÜÄÖÝÃÜÊÜ®Üá° Öæãí©ÃÜáÊÜâ©ÆÉ

(C) ÊÜáãÃÜá ±ÜÄÖÝÃÜWÜÙÜ®Üá° Öæãí©ÃÜáñܤÊæ

(D) A±ÜÄËáñÜ ÓÜíTæÂ¿á ±ÜÄÖÝÃÜWÜÙÜ®Üá° Öæãí©ÃÜáñܤÊæ

5. ‘r’ ÊÜÞ®Ü £ÅgÂÊÜ®Üá° Öæãí©ÃÜáÊÜ WæãàÙÜ¨Ü Z®Ü¶ÜÆÊÜâ

2 1
(A)  r 2 Z®ÜÊÜÞ®Ü (B)  r 3 Z®ÜÊÜÞ®Ü
3 3

4 4
(C)  r 3 Z®ÜÊÜÞ®Ü (D)  r 2 Z®ÜÊÜÞ®Ü
3 3

6. Jí¨Üá ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿áÈÉ a n = 2n + 1 B¨ÝWÜ, ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿á

ÓÝÊÜޮܠÊÜÂñÝÂÓÜÊÜâ

(A) 1 (B) 2

(C) 3 (D) 4

10/401 3 of 16

CCE RF/RR/PF/PR(A) 81-K
7. ( 2, 3 ) ÊÜáñÜᤠ( 4, 7 ) D ¹í¨ÜáWÜÙÜ®Üá° ÓæàÄÓÜáÊÜ ÃæàTÝSívܨÜ

ÊÜá«Ü¹í¨Üá訆 ¯¨æàìÍÝíPÜWÜÙÜá

(A) ( 3, 5 ) (B) ( 5, 3 )

(C) ( 1, 2 ) (D) ( 6, 10 )

8. ñÝŲg ABCD ¿áÈÉ AB || DC ÊÜáñÜᤠP, Q WÜÙÜá PÜÅÊÜáÊÝX AD ÊÜáñÜá¤

BC ¸ÝÖÜáWÜÙÜ Êæáà騆 ¹í¨ÜáWÜÙÜá. PQ || DC, PD = 18 cm,

BQ = 35 cm ÊÜáñÜᤠQC = 15 cm B¨ÜÃæ, AD ¿á E¨ÜªÊÜâ,

(A) 5 cm (B) 50 cm

(C) 57 cm (D) 60 cm

10/401 4 of 16

CCE RF/RR/PF/PR(A) 81-K
II. PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ EñܤÄÔ 8×1=8

9. AíPÜWÜ~ñÜ¨Ü ÊÜáãÆ ±ÜÅÊæáà¿á ÊÜ®Üá° ¯ÃÜã²Ô.

10. Jí¨Üá ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿á Êæã¨ÜÆ ±Ü¨Ü ‘a’ ÊÜáñÜᤠPæã®æ¿á ±Ü¨Ü ‘ an ’ B¨ÜÃæ,
ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿á Êæã¨ÜÆ ‘n’ ±Ü¨ÜWÜÙÜ ÊæãñܤÊÜ®Üá° PÜívÜá×w¿ááÊÜ ÓÜãñÜÅ
ŸÃæÀáÄ.

11. Jí¨Üá Z®Ü ŸÖÜá±Ü¨æãàQ¤¿á ÊÜáÖÜñܤÊÜá [ÝñÜÊÜ®Üá° wXÅ ŸÃæÀáÄ.

12. bñÜŨÜÈÉ, D ÊÜáñÜᤠE WÜÙÜá PÜÅÊÜáÊÝX AB ÊÜáñÜᤠAC ¸ÝÖÜáWÜÙÜ
ÊÜá«Ü¹í¨ÜáWÜÙÜá. DE = 4 cm B¨ÜÃæ, BC ¿á AÙÜñæ¿á®Üá° ŸÃæÀáÄ.

13. Aíb®Ü E¨Üª ‘a’ ÊÜÞ®Ü CÃÜáÊÜ Jí¨Üá aèPÜZ®Ü¨Ü ±Üä|ì ÊæáàÇ澄 ËÔ¤à|ìÊÜ®Üá°
PÜívÜá×w¿ááÊÜ ÓÜãñÜÅ ŸÃæÀáÄ.

14. 4x + py + 8 = 0 ÊÜáñÜᤠ4x + 4y + 2 = 0 D ÓÜËáàPÜÃÜ|WÜÙÜá ÓÜÊÜÞíñÜÃÜ

ÃæàTæWÜÙÜ®Üá° ±ÜÅ£¯˜Ô¨ÜÃæ, ‘p’ ¸æÇæ¿á®Üá° PÜívÜá×wÀáÄ.

10/401 5 of 16

CCE RF/RR/PF/PR(A) 81-K
15. PæÙÜWæ ¯àwÃÜáÊÜ ÓÜíbñÜ BÊÜ꣤ ËñÜÃÜOÝ PæãàÐÜrPܨÜÈÉ ÊÜá«ÝÂíPÜËÃÜáÊÜ
ÊÜWÝìíñÜÃÜÊÜ®Üá° ŸÃæÀáÄ

AíPÜWÜÙÜá ˨ݦìWÜÙÜ ÓÜíTæ ÓÜíbñÜ BÊÜ꣤

0 ― 10 3 3

10 ― 20 4 7

20 ― 30 7 14

30 ― 40 6 20

n = 20
16. p ( x ) = x + 5 GíŸ ÃæàTÝñܾPÜ ŸÖÜá±Ü¨æãàQ¤¿á ®Üûæ¿áá, x-AûÜÊÜ®Üá°
dæà©ÓÜáÊÜ ¹í¨Üá訆 ¯¨æàìÍÝíPÜWÜÙÜ®Üá° ŸÃæÀáÄ.

III. PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ EñܤÄÔ 8 × 2 = 16

17. 5 + 3 Jí¨Üá A»ÝWÜÆŸœ ÓÜíTæ Gí¨Üá ÓݘÔ.

18. PæãqrÃÜáÊÜ ÃæàTÝñܾPÜ ÓÜËáàPÜÃÜ|WÜÙÜ hæãàw¿á®Üá° ÊÜiìÓÜáÊÜ Ë«Ý®Ü©í¨Ü
¹wÔ

2x + y = 8

3x – y = 7
19. 5, 8, 11, ..... D ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿á Êæã¨ÜÆ 20 ±Ü¨ÜWÜÙÜ ÊæãñܤÊÜ®Üá° ÓÜãñÜÅ
E±ÜÁãàXÔ PÜívÜá×wÀáÄ.
A¥ÜÊÝ
100, 96, 92, .... 12 D ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿á ±Ü¨ÜWÜÙÜ ÓÜíTæ¿á®Üá° ÓÜãñÜÅ
E±ÜÁãàXÔ PÜívÜá×wÀáÄ.

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20. bñÜŨÜÈÉ, ABC = 90° BX¨æ. PæÙÜX®ÜÊÜâWÜÙÜ ¸æÇæWÜÙÜ®Üá° ŸÃæÀáÄ

i) sin 

ii) tan 

21. A±ÜÊÜñÜì®Ü ˫ݮܩí¨Ü x 2 + 4x – 60 = 0 ÊÜWÜìÓÜËáàPÜÃÜ|¨Ü ÊÜáãÆWÜÙÜ®Üá°
PÜívÜá×wÀáÄ.

A¥ÜÊÝ

2x 2  3x  7  0 D ÊÜWÜìÓÜËáàPÜÃÜ|¨Ü Íæãà«ÜPÜÊÜ®Üá° PÜívÜá×wÀáÄ ÊÜáñÜá¤

ÊÜáãÆWÜÙÜ ÓÜÌ»ÝÊÜÊÜ®Üá° ŸÃæÀáÄ.

22. 135 ÊÜáñÜᤠ75 ÃÜ ÊÜá.ÓÝ.A.ÊÜ®Üá° AË»Ýg A±ÜÊÜñÜì®Ü ˫ݮܩí¨Ü
PÜívÜá×w¨Üá, ®ÜíñÜÃÜ ÊÜá.ÓÝ.A. ( 135, 75 ) ÊÜáñÜᤠ20 ÃÜ Æ.ÓÝ.A.ÊÜ®Üá°
PÜívÜá×wÀáÄ.

23. P ( 4, 3 ) Pæàí¨ÜÅÊÝXÃÜáÊÜ ÊÜêñܤ¨Ü ÊÝÂÓÜ AB ¿á AíñÜ ¹í¨ÜáWÜÙÜ
¯¨æàìÍÝíPÜWÜÙÜá A ( x, 5 ) ÊÜáñÜᤠB ( 2, y ) BXÊæ. ‘x’ ÊÜáñÜᤠ‘y’ ¸æÇæWÜÙÜ®Üá°
PÜívÜá×wÀáÄ ÖÝWÜã ÊÜêñܤ¨Ü ÊÝÂÓÜ¨Ü E¨ÜªÊÜ®Üá° PÜívÜá×wÀáÄ.

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24. ‘O’ Pæàí¨ÜÅÊÜâÙÜÛ ÊÜêñܤPæR AB ¿áá ÓܳÍÜìPÜÊÝX¨æ. ‘A’ ¿áá ÓܳÍÜì¹í¨ÜáÊÝX¨æ.

OBA = 45° B¨ÜÃæ,  AOB Jí¨Üá ÓÜÊÜá©Ì¸ÝÖÜá £Å»Üág Gí¨Üá ÓݘÔ.

OB = 5 2 cm B¨ÜÃæ, ÊÜêñܤ¨Ü £ÅgÂÊÜ®Üá° PÜívÜá×wÀáÄ.

IV. PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ EñܤÄÔ 9 × 3 = 27

25. p ( x ) = x 2 – 2x – 8 D ÊÜWÜìŸÖÜá±Ü¨æãàQ¤¿á ÍÜã®ÜÂñæWÜÙÜ®Üá°

PÜívÜá×wÀáÄ ÖÝWÜã ÍÜã®ÜÂñæWÜÙÜá ÊÜáñÜᤠÓÜÖÜWÜá|PÜWÜÙÜ ®ÜvÜá訆 ÓÜíŸí«ÜÊÜ®Üá°

ñÝÙæ ®æãàw.

26. ÊÜêñܤ¨Ü Êæáà騆 ¿ÞÊÜâ¨æà ¹í¨ÜáË®ÜÈÉ GÙæ¨Ü ÓܳÍÜìPÜÊÜâ, ÓܳÍÜì ¹í¨ÜáË®ÜÈÉ

GÙæ¨Ü £ÅgÂPæR ÆíŸÊÝXÃÜáñܤ¨æ Gí¨Üá ÓݘÔ.

27.  ABC ~  PQR.  ABC¿áÈÉ A = 55°, C = 35° ÊÜáñÜá¤

 ABC ¿á ¸ÝÖÜáWÜÙÜá 3 : 4 : 5 A®Üá±ÝñܨÜÈÉÊæ.  PQR ®Ü ÓÜáñܤÙÜñæ¿áá

60 cm B¨ÜÃæ,  PQR ®Ü ¸ÝÖÜáWÜÙÜá ÊÜáñÜᤠPæãà®ÜWÜÙÜ AÙÜñæ¿á®Üá°

PÜívÜá×wÀáÄ.

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28. D PæÙÜX®Ü ¨ÜñݤíÍÜWÜÚWæ ÓÜÃÝÓÜÄ¿á®Üá° PÜívÜá×wÀáÄ

ÊÜWÝìíñÜÃÜ BÊÜ꣤

0 ― 10 4

10 ― 20 9

20 ― 30 15

30 ― 40 14

40 ― 50 8

A¥ÜÊÝ
D PæÙÜX®Ü ¨ÜñݤíÍÜWÜÚWæ ŸÖÜáÆPÜÊÜ®Üá° PÜívÜá×wÀáÄ
ÊÜWÝìíñÜÃÜ BÊÜ꣤

5 ― 15 3

15 ― 25 4

25 ― 35 8

35 ― 45 7

45 ― 55 3
1
29. ( cosec A – sin A ) ( sec A – cos A ) = Gí¨Üá
tan A  cot A
ÓݘÔ.
A¥ÜÊÝ
sec 60 2 sin 90 tan 45

  = sin 2 45° ( cosec 2 45° – sec 2 0° )
cot 45 cos 0 cosec 30
Gí¨Üá ÓݘÔ.

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30. ÇæãàÖÜ¨Ü ñÜí£¿á AíaÜáWÜÙÜ ®ÜvÜáÊæ Ÿpær¿á®Üá° Coár Jí¨Üá Pæç ¹àÓÜ~Wæ¿á®Üá°
ñÜ¿ÞÄÓÜÇÝX¨æ. A¨Üá bñÜŨÜÈÉ ñæãàÄÔ¨Üíñæ ÊÜêñܤ¨Ü £ÅhÝÂíñÜÃÜ SívܨÜ
BPÝÃܨÜÈɨÜáª, A¨ÜÃÜ £ÅgÂÊÜâ 21 cm ÊÜáñÜᤠPæãà®Ü 120° BX¨æ. ¹àÓÜ~WæWæ
E±ÜÁãàXÔÃÜáÊÜ Ÿpær¿á ËÔ¤à|ì ÊÜáñÜᤠAíñÜÖÜ ¹àÓÜ~Wæ¿á®Üá° ñÜ¿ÞÄÓÜÆá
¸æàPÝWÜáÊÜ ÇæãàÖÜ¨Ü ñÜí£¿á Joár E¨ÜªÊÜ®Üá° PÜívÜá×wÀáÄ.

A¥ÜÊÝ

PæãqrÃÜáÊÜ bñÜŨÜÈÉ, ‘O’ Pæàí¨ÜÅËÃÜáÊÜ ÊÜêñܤ¨Ü £Åg 21 cm BX¨æ.
AOB = 60° B¨ÜÃæ, APB ÊÜêñܤSívÜ¨Ü ËÔ¤à|ìÊÜ®Üá° PÜívÜá×wÀáÄ.

[ 3 = 1·73 Gí¨Üá ñæWæ¨ÜáPæãÚÛ ]

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31. ABCD aèPÜ¨Ü ÓÜáñܤÙÜñæ¿áá PQRS aèPÜ¨Ü ÓÜáñܤÙÜñæXíñÜ 12 m PÜwÊæá C¨æ. D

GÃÜvÜã aèPÜWÜÙÜ ËÔ¤à|ìWÜÙÜ Êæãñܤ 117 m 2 BX¨æ. ABCD aèPÜ¨Ü ËÔ¤à|ì
ÊÜáñÜᤠPQRS aèPÜ¨Ü ËÔ¤à|ìWÜÙÜ A®Üá±ÝñÜÊÜ®Üá° PÜívÜá×wÀáÄ.

32. bñÜŨÜÈÉ, AB ÃæàTÝSívÜÊÜ®Üá° 1 : 2 A®Üá±ÝñܨÜÈÉ BíñÜÄPÜÊÝX Ë»ÝXÓÜáÊÜ
¹í¨Üá訆 ¯¨æàìÍÝíPÜWÜÙÜ®Üá° »ÝWÜ ±ÜÅÊÜÞ| ÓÜãñÜÅÊÜ®Üá° E±ÜÁãàXÔ
PÜívÜá×wÀáÄ.

A¥ÜÊÝ

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A ( 1, 5 ), B ( 2, 4 ) ÊÜáñÜᤠC ( 9, – 3 ) D ¹í¨ÜáWÜÙÜá
ÓÜÃÜÙÜÃæàTÝWÜñÜÊÝXÊæÁáà Gí¨Üá ¨ÜãÃÜÓÜãñÜÅ ¨Ü ÓÜÖÝ¿á©í¨Ü ±ÜÄàüÔ.

33. ÊÜááSWÜÙÜ ÊæáàÇæ 1 Äí¨Ü 6 ÃÜÊÜÃæWæ ÓÜíTæÂWÜÚÃÜáÊÜ GÃÜvÜá PÜáí©ÆÉ¨Ü ¨ÝÙÜWÜÙÜ®Üá°
Jí¨Üá ¸ÝÄ EÃÜáÚÓÜÇÝX¨æ.

a) Êæáà騆 ÊÜááSWÜÙÜÈÉ ŸÃÜáÊÜ ÓÜíTæÂWÜÙÜ Êæãñܤ 6 BXÃÜáÊÜ

b) Jí¨Üá ÊÜááS¨Ü ÊæáàÇæ ÊÜÞñÜÅ ÊÜWÜìÓÜíTæ ŸÃÜáÊÜ

ÓÜí»ÜÊܯà¿áñæ¿á®Üá° PÜívÜá×wÀáÄ.

V. PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ EñܤÄÔ 4 × 4 = 16

34. PæãqrÃÜáÊÜ ÃæàTÝñܾPÜ ÓÜËáàPÜÃÜ|WÜÙÜ hæãàw¿á ±ÜÄÖÝÃÜÊÜ®Üá° ®Üûæ¿á
˫ݮܩí¨Ü PÜívÜá×wÀáÄ

x + 2y = 6

x+y = 5

35. Jí¨Üá ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿á 7®æà ±Ü¨ÜÊÜâ A¨ÜÃÜ 2 ®æà ±Ü¨Ü¨Ü ®ÝÆRÃÜÑr¨æ. ÖÝWÜã
12®æà ±Ü¨ÜÊÜâ 4 ®æà ±Ü¨Ü¨Ü ÊÜáãÃÜÃÜÐÜrQRíñÜ 2 ÖæaÝcX¨æ. ÖÝWݨÜÃæ, B
ÍæÅà{¿á®Üá° PÜívÜá×wÀáÄ.

36. GÃÜvÜá £Å»ÜágWÜÙÜÈÉ A®ÜáÃÜã±Ü Pæãà®ÜWÜÙÜá ÓÜÊÜáÊݨÜÃæ, AÊÜâWÜÙÜ A®ÜáÃÜã±Ü
¸ÝÖÜáWÜÙÜ A®Üá±ÝñÜWÜÙÜá ÓÜÊÜá A¥ÜÊÝ ÓÜÊÜÞ®Üá±ÝñܨÜÈÉÃÜáñܤÊæ B¨ÜªÄí¨Ü B
£Å»ÜágWÜÙÜá ÓÜÊÜáÃÜã±ÜÊÝXÃÜáñܤÊæ Gí¨Üá ÓݘÔ.

A¥ÜÊÝ

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£Å»Üág¨Ü Jí¨Üá Pæãà®ÜÊÜâ ÊÜáñæã¤í¨Üá £Å»Üág¨Ü Jí¨Üá Pæãà®ÜPæR ÓÜÊÜá®ÝX¨Üáª,
B Pæãà®ÜWÜÙÜ®Üá° Eíoá ÊÜÞwÃÜáÊÜ ¸ÝÖÜáWÜÙÜá ÓÜÊÜÞ®Üá±ÝñܨÜÈɨܪÃæ, B GÃÜvÜá
£Å»ÜágWÜÙÜá ÓÜÊÜáÃÜã²WÜÙÝXÃÜáñܤÊæ. Gí¨Üá ÓݘÔ.

37. bñÜŨÜÈÉ, ÓÜÊÜáñÜpÝr¨Ü ®æÆ¨Ü ÊæáàÈÃÜáÊÜ ®æàÃÜÊÝ¨Ü PÜíŸ¨Ü ñÜá©Àáí¨Ü A¨æà
®æÆ¨Ü Êæáà騆 Jí¨Üá WÜãoPæR 20 m E¨ÜªËÃÜáÊÜíñæ Jí¨Üá ÖÜWÜYÊÜ®Üá° GÙæ¨Üá
¹X¿ÞX PÜorÇÝX¨æ. ÖÜWÜYÊÜâ ®æƨæãí©Wæ 30° Pæãà®ÜÊÜ®Üá° EíoáÊÜÞw¨æ.
JŸº ÓÜPÜìÓ… PÜÇÝ˨ܮÜá D ÖÜWÜYÊÜ®Üá° ÖÜ£¤ PÜíŸ¨Ü ñÜá©¿á®Üá° ñÜÆá² AÈÉí¨Ü
A¨æà ®æÆ¨Ü ÊæáàÇæ ®æàÃÜÊÝX ¯í£ÃÜáÊÜ ÊÜáñæã¤í¨Üá PÜíŸ¨Ü ñÜ á©¿á®Üá°
ËàüÔ¨ÝWÜ EípÝWÜáÊÜ E®Ü°ñÜ Pæãà®ÜÊÜâ 60° BXÃÜáñܤ¨æ. ®æÆ¨Ü ÊæáàÈ®Ü
WÜão©í¨Ü ¨æãvÜx PÜíŸ¨Ü ±Ý¨ÜPæR CÃÜáÊÜ ¨ÜãÃÜÊÜâ 30 m B¨ÜÃæ, D PÜퟨÜ
GñܤÃÜÊÜ®Üá° PÜívÜá×wÀáÄ.
[ 3 = 1∙73 Gí¨Üá ñæWæ¨ÜáPæãÚÛ ]

A¥ÜÊÝ

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Jí¨Üá Wæãà±ÜâÃÜ AB ÊÜáñÜᤠPÜorvÜ DE WÜÙÜá ÓÜÊÜáñÜpÝr¨Ü ®æÆ¨Ü ÊæáàÇæ ®æàÃÜÊÝX
¯í£Êæ. bñÜŨÜÈÉ ñæãàÄÔÃÜáÊÜíñæ A¨æà ®æÆ¨Ü Êæáà騆 Jí¨Üá ¹í¨Üá˯í¨Ü
ÖÝWÜã PÜorvÜ¨Ü ñÜá©Àáí¨Ü Wæãà±ÜâÃÜ¨Ü ñÜá©¿á®Üá° ®æãàw¨ÝWÜ, E®Ü°ñÜ
Pæãà®ÜWÜÙÜá PÜÅÊÜáÊÝX 30° ÊÜáñÜᤠ60° BXÊæ. Wæãà±ÜâÃÜ¨Ü ±Ý¨Ü©í¨Ü
¹í¨ÜáËXÃÜáÊÜ ¨ÜãÃÜÊÜâ 30 3 m ÖÝWÜã PÜorvÜ¨Ü GñܤÃÜÊÜâ 10 m B¨ÜÃæ,
Wæãà±ÜâÃÜ ÊÜáñÜᤠPÜorvÜ¨Ü ±Ý¨ÜWÜÙÜ ®ÜvÜá訆 ¨ÜãÃÜ ( BE ) ÖÝWÜã AÊÜâWÜÙÜ ñÜá©WÜÙÜ
®ÜvÜá訆 ¨ÜãÃÜÊÜ®Üá° ( AD ) PÜívÜá×wÀáÄ.

VI. PæÙÜX®Ü ±ÜÅÍæ°Wæ EñܤÄÔ 1×5=5

38. ±ÜÅ£Áãí¨ÜÃÜ ÊÝÂÓÜ 4∙2 cm ÊÜáñÜᤠGñܤÃÜ 6 cm CÃÜáÊÜíñæ ÔÈívÜÃ… BPÜꣿá
ÊÜáãÃÜá ñÜWÜáYWÜÙÜ®Üá°, 15 cm  10 cm  8 cm AÙÜñæWÜÙÜáÙÜÛ Jí¨Üá Z®Ü
ÇæãàÖÜ¨Ü B¿áñÜ Z®ÝPÜê£Àáí¨Ü bñÜŨÜÈÉ ñæãàÄÔÃÜáÊÜíñæ PæãÃæ¨Üá
ÖæãÃÜñæWæ¿áÇÝX¨æ. EÚ¨Ü ÇæãàÖÜ¨Ü Z®ÝPÜꣿá Joár ÊæáàÇ澄 ËÔ¤à|ì
ÊÜáñÜᤠZ®Ü¶ÜÆÊÜ®Üá° PÜívÜá×wÀáÄ.

A¥ÜÊÝ

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24 cm GñܤÃÜ ÊÜáñÜᤠ10 cm £ÅgÂËÃÜáÊÜ Jí¨Üá ÔÈívÜÃ…®ÜÈÉ, 14∙4 cm

GñܤÃܨÜÊÜÃæWæ ¯àÃÜ®Üá° ñÜáíŸÇÝX¨æ. KÃæ GñܤÃÜ 10 cm ÊÜáñÜᤠÊÜPÜÅ ÊæáàÇ澄

ËÔ¤à|ì 60  cm 2 CÃÜáÊÜ Jí¨Üá ÍÜíPÜáÊÜ®Üá°, ÍÜíPÜáË®ÜÐærà £ÅgÂËÃÜáÊÜ

Jí¨Üá A«ÜìWæãàÙÜ¨Ü ÊæáàÇæ CÄÔ, Jí¨Üá Z®ÝPÜꣿá®Üá° ñÜ¿ÞÄÓÜÇÝX¨æ. D

Z®ÝPÜꣿá®Üá° bñÜŨÜÈÉ ñæãàÄÔÃÜáÊÜíñæ, ÔÈívÜÃ…®ÜÈÉ ÓÜí±Üä|ìÊÝX

ÊÜááÙÜáXÔ¨ÝWÜ, ÔÈívÜÃ…®ÜÈÉ ¯àÄ®Ü ÊÜáorÊÜâ GÐÜár ÖæaÝcWÜáñܤ¨æ GíŸá¨Ü®Üá°

PÜívÜá×wÀáÄ. ÖÝWÜã ÔÈívÜÃ…®Ü Aíb®ÜÊÜÃæWæ ¯àÃÜá HÃܸæàPݨÜÃæ,

ÔÈívÜÃ…®ÜÈÉ C®Üã° ñÜáퟸæàPÝ¨Ü ¯àÄ®Ü ±ÜÅÊÜÞ|ÊÜ®Üá° PÜívÜá×wÀáÄ.

[  = 3∙14 Gí¨Üá ñæWæ¨ÜáPæãÚÛ ]

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