81279 + Interview Questions in Mathematics QA in Mathematics

1.	What per cent is612 people of 720 people
Answer»

Discussion

2.	ABCD is a quadrilateral. The diagonal AC bisects the diagonal BD at E. Prove that AABC andAADC are equal in area.
Answer» thanks

Discussion

3.	,In ΔΑΒΟ, if AD is the median, show that AB + AC2-2 (AD2 + BD2)
Answer» Like if you find it useful

Discussion

4.	In the adjoining figure, the point D divides theside BC of AABC in the ratio m:n. Prove that4.ar(AABD): ar(AADC)m:n.
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Discussion

5.	In the given figure, BC is a common and AB= AC, ABC=BCD then prove thatAC-=AD
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6.	25' 10dd the following78710 5 2
Answer»

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7.	Find the value of k for the following system of equation have infinite solution6x + (2x-1)y2k + 5
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Discussion

8.	In ΔΑΒ. if AD is the median, then show that AB-AC2-2(AD2-BD2)
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Discussion

9.	In ÎABC, if AD is the median, then show that AB2+AC2-2(AD2-Bd')
Answer» thank you

Discussion

10.	I. Determinee whether the solution-set is finite or infinite or empty:(i) r < 1000. r e N
Answer» Given : x < 1000 and x belongs to N So, solution set S is S = { 1, 2, 3, 4, 5 ...... 999} This is finite set

10.

I. Determinee whether the solution-set is finite or infinite or empty:(i) r < 1000. r e N

Answer»

Given : x < 1000

and x belongs to N

So, solution set S is

S = { 1, 2, 3, 4, 5 ...... 999}

This is finite set

Discussion

11.	In AABC, if AD is the median, then show that AB+AC2 2(AD2+BD2)
Answer» construction :- draw a line AM perpendicular to BC. we have to prove : AB² + AC² = 2(AD² + BD²) proof :- case 1 :- when , it means AD is perpendicular on BC and both angles are right angle e.g., 90° then, from ∆ADB, according to Pythagoras theorem, AB² = AD² + BD² ..... (1) from ∆ADC ,according to Pythagoras theorem, AC² = AD² + DC² ...... (2) AD is median. so, BD = DC .......(3) from equations (1) , (2) and (3), AB² + AC² = AD² + AD² + BD² + BD² AB² + AC² = 2(AD² + BD²) [hence proved ] case 2 :- when Let us consider that, ADB is an obtuse angle. from ∆ABM, from Pythagoras theorem, AB² = AM² + BM² AB² = AM² + (BD + DM)² AB² = AM² + BD² + DM² + 2BD.DM ......(1) from ∆ACM, according to Pythagoras theorem, AC² = AM² + CM² AC² = AM² + (DC - DM)² AC² = AM² + DC² + DM² - 2DC.DM ......(2) from equations (1) and (2), AB² + AC² = 2AM² + BD² + DC² + 2DM² + 2BD.DM - 2DC.DM AB² + AC² = 2(AM² + DM²) + BD² + DC² + 2(BD.DM - DC.DM) ...........(3) a/c to question, AD is median on BC.so, BD = DC .....(4)and from ADM, according to Pythagoras theorem, AD² = AM² + DM² ........(5) putting equation (4) and equation (5) in equation (3), AB² + AC² = 2AD² + 2BD² + 2(BD.DM - BD.DM) AB² + AC² = 2(AD² + BD²) [hence proved].

11.

In AABC, if AD is the median, then show that AB+AC2 2(AD2+BD2)

Answer»

construction :- draw a line AM perpendicular to BC.

we have to prove : AB² + AC² = 2(AD² + BD²)

proof :- case 1 :- when , it means AD is perpendicular on BC and both angles are right angle e.g., 90°

then, from ∆ADB, according to Pythagoras theorem, AB² = AD² + BD² ..... (1)

from ∆ADC ,according to Pythagoras theorem, AC² = AD² + DC² ...... (2)

AD is median. so, BD = DC .......(3)

from equations (1) , (2) and (3), AB² + AC² = AD² + AD² + BD² + BD² AB² + AC² = 2(AD² + BD²) [hence proved ]

case 2 :- when Let us consider that, ADB is an obtuse angle. from ∆ABM, from Pythagoras theorem, AB² = AM² + BM² AB² = AM² + (BD + DM)² AB² = AM² + BD² + DM² + 2BD.DM ......(1)

from ∆ACM, according to Pythagoras theorem, AC² = AM² + CM² AC² = AM² + (DC - DM)² AC² = AM² + DC² + DM² - 2DC.DM ......(2)

from equations (1) and (2), AB² + AC² = 2AM² + BD² + DC² + 2DM² + 2BD.DM - 2DC.DM

AB² + AC² = 2(AM² + DM²) + BD² + DC² + 2(BD.DM - DC.DM) ...........(3)

a/c to question, AD is median on BC.so, BD = DC .....(4)and from ADM, according to Pythagoras theorem, AD² = AM² + DM² ........(5)

putting equation (4) and equation (5) in equation (3),

AB² + AC² = 2AD² + 2BD² + 2(BD.DM - BD.DM)

AB² + AC² = 2(AD² + BD²) [hence proved].

Discussion

12.	In AABC, if AD is the median, then show that AB2+AC2 2(AD2+BD2)
Answer» construction :- draw a line AM perpendicular to BC. we have to prove : AB² + AC² = 2(AD² + BD²) proof :- case 1 :- when , it means AD is perpendicular on BC and both angles are right angle e.g., 90° then, from ∆ADB, according to Pythagoras theorem, AB² = AD² + BD² ..... (1) from ∆ADC ,according to Pythagoras theorem, AC² = AD² + DC² ...... (2) AD is median. so, BD = DC .......(3) from equations (1) , (2) and (3), AB² + AC² = AD² + AD² + BD² + BD² AB² + AC² = 2(AD² + BD²) [hence proved ] case 2 :- when Let us consider that, ADB is an obtuse angle. from ∆ABM, from Pythagoras theorem, AB² = AM² + BM² AB² = AM² + (BD + DM)² AB² = AM² + BD² + DM² + 2BD.DM ......(1) from ∆ACM, according to Pythagoras theorem, AC² = AM² + CM² AC² = AM² + (DC - DM)² AC² = AM² + DC² + DM² - 2DC.DM ......(2) from equations (1) and (2), AB² + AC² = 2AM² + BD² + DC² + 2DM² + 2BD.DM - 2DC.DM AB² + AC² = 2(AM² + DM²) + BD² + DC² + 2(BD.DM - DC.DM) ...........(3) a/c to question, AD is median on BC.so, BD = DC .....(4)and from ADM, according to Pythagoras theorem, AD² = AM² + DM² ........(5) putting equation (4) and equation (5) in equation (3), AB² + AC² = 2AD² + 2BD² + 2(BD.DM - BD.DM) AB² + AC² = 2(AD² + BD²) [hence proved].

12.

In AABC, if AD is the median, then show that AB2+AC2 2(AD2+BD2)

Answer»

construction :- draw a line AM perpendicular to BC.

we have to prove : AB² + AC² = 2(AD² + BD²)

proof :- case 1 :- when , it means AD is perpendicular on BC and both angles are right angle e.g., 90°

then, from ∆ADB, according to Pythagoras theorem, AB² = AD² + BD² ..... (1)

from ∆ADC ,according to Pythagoras theorem, AC² = AD² + DC² ...... (2)

AD is median. so, BD = DC .......(3)

from equations (1) , (2) and (3), AB² + AC² = AD² + AD² + BD² + BD² AB² + AC² = 2(AD² + BD²) [hence proved ]

case 2 :- when Let us consider that, ADB is an obtuse angle. from ∆ABM, from Pythagoras theorem, AB² = AM² + BM² AB² = AM² + (BD + DM)² AB² = AM² + BD² + DM² + 2BD.DM ......(1)

from ∆ACM, according to Pythagoras theorem, AC² = AM² + CM² AC² = AM² + (DC - DM)² AC² = AM² + DC² + DM² - 2DC.DM ......(2)

from equations (1) and (2), AB² + AC² = 2AM² + BD² + DC² + 2DM² + 2BD.DM - 2DC.DM

AB² + AC² = 2(AM² + DM²) + BD² + DC² + 2(BD.DM - DC.DM) ...........(3)

a/c to question, AD is median on BC.so, BD = DC .....(4)and from ADM, according to Pythagoras theorem, AD² = AM² + DM² ........(5)

putting equation (4) and equation (5) in equation (3),

AB² + AC² = 2AD² + 2BD² + 2(BD.DM - BD.DM)

AB² + AC² = 2(AD² + BD²) [hence proved].

Discussion

13.	1. In the figure 10.9, lll m ll n and p is a transversal. If A = 110°, find angles x, y, z,and wW.pn110°ラ> mrnFig. 10.9
Answer» x = 180° -110° = 70°y = 70° ( vertically opposite angle)z = 70° (corresponding angle)w = 180-70 = 110° ( linear pair)

Discussion

14.	thePolyno rYLaln 3-6x-1rngivenven, polynomlal 612Find PC)
Answer» To find P(1) put value of x as 1. (1)^3 -6(1)^2 -11(1)-6=1-6-11-6=1-23 =-22.

Discussion

15.	22 / ( 11 x 10 ^ - 19)
Answer» 2000000000= 2×10^19 is the answer

Discussion

16.	For which value of K, (2, 2) is a solutionof the equation x +2y-K-0?
Answer» Put in (2,2) in the equation as it is a solution2+2(2)-k=02+4-k=0k=6

Discussion

17.	admiring sums in maths
Answer» Ans :- When these numbers are added together (represented by the "+" sign), the resulting answer is called the "sum." When a number is subtracted from another number (represented by the "-" sign), the result is known as the "difference."

Discussion

18.	7. Do these sums(a) 23+ 36+ 74
Answer» 59 And 79 is the ans of this sums

Discussion

19.	Q3- Find the value of k show that the following system of question has infinitesolution?3x-y-5-0,6x-2y+k o
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Discussion

20.	Solve the following : A circle touches all the four sides of quadABCD. ProveIf the cloth is AB+CD =AD+BC
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21.	Show that exactly one of the numbers n, n + 2 or n + 4 is divisible by 3.
Answer» let n be any positive integer and b=3n =3q+rwhere q is the quotient and r is the remainder0_ <r<3so the remainders may be 0,1 and 2so n may be in the form of 3q, 3q=1,3q+2 CASE-1 IF N=3qn+4=3q+4n+2=3q+2here n is only divisible by 3 CASE 2if n = 3q+1n+4=3q+5n+2=3q=3here only n+2 is divisible by 3 CASE 3IF n=3q+2n+2=3q+4n+4=3q+2+4=3q+6here only n+4 is divisible by 3 HENCE IT IS JUSTIFIED THAT ONE AND ONLY ONE AMONG n,n+2,n+4 IS DIVISIBLE BY 3 IN EACH CASE

21.

Show that exactly one of the numbers n, n + 2 or n + 4 is divisible by 3.

Answer»

let n be any positive integer and b=3n =3q+rwhere q is the quotient and r is the remainder0_ <r<3so the remainders may be 0,1 and 2so n may be in the form of 3q, 3q=1,3q+2

CASE-1

IF N=3qn+4=3q+4n+2=3q+2here n is only divisible by 3

CASE 2if n = 3q+1n+4=3q+5n+2=3q=3here only n+2 is divisible by 3

CASE 3IF n=3q+2n+2=3q+4n+4=3q+2+4=3q+6here only n+4 is divisible by 3

HENCE IT IS JUSTIFIED THAT ONE AND ONLY ONE AMONG n,n+2,n+4 IS DIVISIBLE BY 3 IN EACH CASE

Discussion

22.	2x-128
Answer» 2x-128= 02x = 128x = 128/2x = 64

Discussion

23.	If the mid-point of the line joining (3, 4 ) and (13, 7 ) is (x, y) and2x+ 2y+k11.0, find the value of k.
Answer»

Discussion

24.	Show that any one of the numbers (n+2), n and (n + 4) is divisible by 3
Answer» Solution: let n be any positive integer and b=3n =3q+rwhere q is the quotient and r is the remainder0_ <r<3so the remainders may be 0,1 and 2so n may be in the form of 3q, 3q=1,3q+2 CASE-1 IF N=3qn+4=3q+4n+2=3q+2here n is only divisible by 3 CASE 2if n = 3q+1n+4=3q+5n+2=3q=3here only n+2 is divisible by 3 CASE 3IF n=3q+2n+2=3q+4n+4=3q+2+4=3q+6here only n+4 is divisible by 3 HENCE IT IS JUSTIFIED THAT ONE AND ONLY ONE AMONG n,n+2,n+4 IS DIVISIBLE BY 3 IN EACH CASE

24.

Show that any one of the numbers (n+2), n and (n + 4) is divisible by 3

Answer»

Solution:

let n be any positive integer and b=3n =3q+rwhere q is the quotient and r is the remainder0_ <r<3so the remainders may be 0,1 and 2so n may be in the form of 3q, 3q=1,3q+2

CASE-1

IF N=3qn+4=3q+4n+2=3q+2here n is only divisible by 3

CASE 2if n = 3q+1n+4=3q+5n+2=3q=3here only n+2 is divisible by 3

CASE 3IF n=3q+2n+2=3q+4n+4=3q+2+4=3q+6here only n+4 is divisible by 3

HENCE IT IS JUSTIFIED THAT ONE AND ONLY ONE AMONG n,n+2,n+4 IS DIVISIBLE BY 3 IN EACH CASE

Discussion

25.	18.A quadrilateral ABCD is drawn to circumscribe a circle. Prove that: AB + CD = AD + BC
Answer»

Discussion

26.	9. A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.17). Show thatAB + CD AD BC
Answer»

Discussion

27.	07 2.031-V) = 501) 2 (0+32) = 50v.salvez
Answer» 4259+634258#6/786344 2(3u-v)=5uv; 2(u+3v)=5uv. 6u-2v=5 uv; 2u+6uv=5uv. 3(2u+6uv=5uv)=6u+18uv=15uv; 6u+18uv=15uv; 6u-2v=15vu/18uv+2v=0; 2v(9u+1)

27.

07 2.031-V) = 501) 2 (0+32) = 50v.salvez

Answer»

4259+634*258*#6/786344

2(3u-v)=5uv; 2(u+3v)=5uv. 6u-2v=5 uv; 2u+6uv=5uv. 3(2u+6uv=5uv)=6u+18uv=15uv; 6u+18uv=15uv; 6u-2v=15vu/18uv+2v=0; 2v(9u+1)

Discussion

28.	हि 5600 =v2 tang न? ] .. 207 )3 OF1 . cosec’ ©-sec’ 8 e
Answer» Please hit the like button if this helped you

Discussion

29.	2.ली ती CG-\|रे " G३ टूर 541ल) (1) ले २01) 1दिवा' » १ , है { 07, 2
Answer»

Discussion

30.	0, 2. Find the sums givenb(07+ 107 + 14+84
Answer» a= 7d= 21/2-7= 7/2last term = 84

Discussion

31.	07 ला 2894 ००१ एक § o 2" वL v oo cOw 14
Answer» let x be the common ratiothenx+3x=1804x=180x=453x=3x45=135angles are ,45,45,135,135

Discussion

32.	A quadrilateral ABCD is drawn to circumscribe a circle (as shown in figure). Prove thatAB CD AD BC.18.
Answer»

Discussion

33.	In an examination,70% of the candidates passed in english,65% in mathematics,26% failed in both subjects and 248 passed in both subjects.Find the total number of candidates.
Answer» If there are 100 candidates, thencandidates failed in english =30candidates failed in Maths = 35candidates failed in both = 27candidates failed in atleast one subject = 30+35-27 = 38candidates passed in both subjects = 62If 62 candidates passed in both subjects, then total candidates = 100If 248 candidates passed in both subjects, then total candidates = 100*248/62=400

Discussion

34.	Find the equation of the circle havingCentre at (1, 3) & radius- 3
Answer»

Discussion

35.	ŕ¤ŕ¤ż 21. 2a+ 3 =K: ; 3a - 4 = - 5
Answer» Multiplying eq 1 by 3 and eq 2 by 2, we get 6a+9b = 246a-8b = -10 solving we get, 17b = 34 b = 2 Substituting b= 2, in eq 1 2a + 3(2) = 8 2a = 8-6 a = 2/2 a= 1

35.

ŕ¤ŕ¤ż 21. 2a+ 3 =K: ; 3a - 4 = - 5

Answer»

Multiplying eq 1 by 3 and eq 2 by 2, we get

6a+9b = 246a-8b = -10

solving we get,

17b = 34

b = 2

Substituting b= 2, in eq 1

2a + 3(2) = 8

2a = 8-6

a = 2/2

a= 1

Discussion

36.	Show that in a quadrilateral ABCD, AB+ BC+ CD + AD> AC+BD
Answer» that's another question

Discussion

37.	hink Discuss and write:Is every rectangle a parallelogram? Is every parallelogram a rectangle?
Answer» Every rectangle has 2 sets of parallel lines so it is parallelogram. But a parallelogram does not necessarily have sides at right angles so every parallelogram is not a rectangle

Discussion

38.	Using opposite angles test for parallelogram , prove that every rectangle is a parallelogram.
Answer» we can easily prove that ABM ≡ CDM using the SAS congruence test. (alternate angles are equal). Hence, ABCD is a parallelogram, because one pair of opposite sides are equal and parallel. Thereafter, ABCD is also a rectangle, because it is a parallelogram with one right angle it's not right because one pair is also parallel in trapezium

38.

Using opposite angles test for parallelogram , prove that every rectangle is a parallelogram.

Answer»

we can easily prove that ABM ≡ CDM using the SAS congruence test. (alternate angles are equal).

Hence, ABCD is a parallelogram, because one pair of opposite sides are equal and parallel.

Thereafter, ABCD is also a rectangle, because it is a parallelogram with one right angle

it's not right because one pair is also parallel in trapezium

Discussion

39.	6. Show that the following points form a equilateraltriangle A(a,0), B(-a, 0), C(0, a 13)
Answer»

Discussion

40.	18. Find the values of a and b for which the following system of linear equations has infinitely manysolutions(0) 2x + 3y 7 (a+b+1) + (a 2b+ 2)y 4la+ b)+1(a) 2x+3y-7, (a+ b)x (2a - b)y 3a+ b+ 1)Soluti
Answer»

Discussion

41.	The diameter of Venus planet is 12278 km. Find its circumference. (Ď = 22
Answer» Radius = Diameter/ 2 = 12278/2 = 6139 km Circumference of the circle = 2πr= 222/76139= 38552.92 km Hence, circumference of the circle is 38552.92 km

41.

The diameter of Venus planet is 12278 km. Find its circumference. (Ď = 22

Answer»

Radius = Diameter/ 2 = 12278/2 = 6139 km

Circumference of the circle = 2πr= 2*22/7*6139= 38552.92 km

Hence, circumference of the circle is 38552.92 km

Discussion

42.	1285 i< ol डाWidh side i cCm = Teyn séuapd — 88 °t, 07 — s ChlGiads Jhe O eqotane shade dEnEgiah comen
Answer»

Discussion

43.	5+7+2+-07-8
Answer» -1 is the right answer

Discussion

44.	Cu0 + 50,$07 5 cuo*Cus8'+ #
Answer» 3 Cu2S + 8 SO4{-}2 + 8 H{+} = 6 CuO + 11 SO2 + 4 H2O

Discussion

45.	Draw a circle of radius 3.6 cm. In the circle, draw a chord AB = 5 cm. Now shade theminor segment of the circle.
Answer» 👍 👍

45.

Draw a circle of radius 3.6 cm. In the circle, draw a chord AB = 5 cm. Now shade theminor segment of the circle.

Answer»

👍

Discussion

46.	on,84%ofthecandidatespassedand 780 failed. Find the number of candidatamination.
Answer» 84% passed so 16% failed16% = 780so total students=(780×100)/16=4875

Discussion

47.	16. Find the values of(1) sin 150°(2) cos 120°(3) tan 135°(4) cos 780°(5) sec 210°(6) cot 330°
Answer»

Discussion

48.	(a)C.P=70, S.P=780
Answer»

Discussion

49.	\operatorname { sin } 780 ^ { \circ } \operatorname { sin } 480 ^ { \circ } + \operatorname { cos } 240 ^ { \circ } \operatorname { cos } 300 ^ { \circ } = \frac { 1 } { 2 }
Answer» sin780. sin480 + cos120.sin150=sin(8 x 90+60).sin(5 x90+30) + cos(2 x90 -60).sin(2 x90 +30)=sin60.cos30 + cos60.(-sin30)=sin60.cos30 - cos60.sin30=sin(60 -30)=sin30=1/2

Discussion

50.	\| Two: Anges of a quadrilateral are 780 5and 12200he other tree angles areTequal. Find the messuie de these twoAnglese
Answer» find zero of following question polynomial and verify the relationship between the zeros and their coefficient

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

What per cent is612 people of 720 people

ABCD is a quadrilateral. The diagonal AC bisects the diagonal BD at E. Prove that AABC andAADC are equal in area.

,In ΔΑΒΟ, if AD is the median, show that AB + AC2-2 (AD2 + BD2)

In the adjoining figure, the point D divides theside BC of AABC in the ratio m:n. Prove that4.ar(AABD): ar(AADC)m:n.

In the given figure, BC is a common and AB= AC, ABC=BCD then prove thatAC-=AD

25' 10dd the following78710 5 2

Find the value of k for the following system of equation have infinite solution6x + (2x-1)y2k + 5

In ΔΑΒ. if AD is the median, then show that AB-AC2-2(AD2-BD2)

In ÎABC, if AD is the median, then show that AB2+AC2-2(AD2-Bd')

I. Determinee whether the solution-set is finite or infinite or empty:(i) r &lt; 1000. r e N

In AABC, if AD is the median, then show that AB+AC2 2(AD2+BD2)

In AABC, if AD is the median, then show that AB2+AC2 2(AD2+BD2)

1. In the figure 10.9, lll m ll n and p is a transversal. If A = 110°, find angles x, y, z,and wW.pn110°ラ&gt; mrnFig. 10.9

thePolyno rYLaln 3-6x-1rngivenven, polynomlal 612Find PC)

22 / ( 11 x 10 ^ - 19)

For which value of K, (2, 2) is a solutionof the equation x +2y-K-0?

admiring sums in maths

7. Do these sums(a) 23+ 36+ 74

Q3- Find the value of k show that the following system of question has infinitesolution?3x-y-5-0,6x-2y+k o

Solve the following : A circle touches all the four sides of quadABCD. ProveIf the cloth is AB+CD =AD+BC

Show that exactly one of the numbers n, n + 2 or n + 4 is divisible by 3.

2x-128

If the mid-point of the line joining (3, 4 ) and (13, 7 ) is (x, y) and2x+ 2y+k11.0, find the value of k.

Show that any one of the numbers (n+2), n and (n + 4) is divisible by 3

18.A quadrilateral ABCD is drawn to circumscribe a circle. Prove that: AB + CD = AD + BC

9. A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.17). Show thatAB + CD AD BC

07 2.031-V) = 501) 2 (0+32) = 50v.salvez

हि 5600 =v2 tang न? ] .. 207 )3 OF1 . cosec’ ©-sec’ 8 e

2.ली ती CG-|रे " G३ टूर 541ल) (1) ले २01) 1दिवा' » १ , है { 07, 2

0, 2. Find the sums givenb(07+ 107 + 14+84

07 ला 2894 ००१ एक § o 2" वL v oo cOw 14

A quadrilateral ABCD is drawn to circumscribe a circle (as shown in figure). Prove thatAB CD AD BC.18.

In an examination,70% of the candidates passed in english,65% in mathematics,26% failed in both subjects and 248 passed in both subjects.Find the total number of candidates.

Find the equation of the circle havingCentre at (1, 3) &amp; radius- 3

ŕ¤ŕ¤ż 21. 2a+ 3 =K: ; 3a - 4 = - 5

Show that in a quadrilateral ABCD, AB+ BC+ CD + AD&gt; AC+BD

hink Discuss and write:Is every rectangle a parallelogram? Is every parallelogram a rectangle?

Using opposite angles test for parallelogram , prove that every rectangle is a parallelogram.

6. Show that the following points form a equilateraltriangle A(a,0), B(-a, 0), C(0, a 13)

18. Find the values of a and b for which the following system of linear equations has infinitely manysolutions(0) 2x + 3y 7 (a+b+1) + (a 2b+ 2)y 4la+ b)+1(a) 2x+3y-7, (a+ b)x (2a - b)y 3a+ b+ 1)Soluti

The diameter of Venus planet is 12278 km. Find its circumference. (Ď = 22

1285 i&lt; ol डाWidh side i cCm = Teyn séuapd — 88 °t, 07 — s ChlGiads Jhe O eqotane shade dEnEgiah comen

5+7+2+-07-8

Cu0 + 50,$07 5 cuo*Cus8'+ #

Draw a circle of radius 3.6 cm. In the circle, draw a chord AB = 5 cm. Now shade theminor segment of the circle.

on,84%ofthecandidatespassedand 780 failed. Find the number of candidatamination.

16. Find the values of(1) sin 150°(2) cos 120°(3) tan 135°(4) cos 780°(5) sec 210°(6) cot 330°

(a)C.P=70, S.P=780

\operatorname { sin } 780 ^ { \circ } \operatorname { sin } 480 ^ { \circ } + \operatorname { cos } 240 ^ { \circ } \operatorname { cos } 300 ^ { \circ } = \frac { 1 } { 2 }

| Two: Anges of a quadrilateral are 780 5and 12200he other tree angles areTequal. Find the messuie de these twoAnglese

In ÎABC, if AD is the median, then show that AB2+AC2-2(AD2-Bd')

I. Determinee whether the solution-set is finite or infinite or empty:(i) r < 1000. r e N

1. In the figure 10.9, lll m ll n and p is a transversal. If A = 110°, find angles x, y, z,and wW.pn110°ラ> mrnFig. 10.9

Find the equation of the circle havingCentre at (1, 3) & radius- 3

ŕ¤ŕ¤ż 21. 2a+ 3 =K: ; 3a - 4 = - 5

Show that in a quadrilateral ABCD, AB+ BC+ CD + AD> AC+BD

The diameter of Venus planet is 12278 km. Find its circumference. (Ď = 22

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