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Given:AD, BE and CF are the medians of ΔABC.

To prove:(1) AD + BE > CF

(2) BE + CF > AD

(3) AD + CF > BE

Construction:Produce AD to H, such that AG = GH.

Join BH and CH

Proof:In ΔABH, F is the mid-point of AB and G is the mid point of AH.

∴FG||BH (Mid-point theorem)

∴ GC||BH (1)

Similarly, BG||HC (2)

From (1) and (2), we get

BGCH is a parallelogram (Both pair of opposite sides are parallel)

∴ BH = CG (3) (Opposite sides of parallelogram are equal)

In ∆BGH,

BG + GH > BH (Sum of any to sides of a triangle is always greater than the third side)

⇒BG + AG > CG (GH = AG and BH = CG)

Similarly, BE + CF > AD and AD + CF > BE.

thanks

she scored 90% in her test

in triangle CPA AP= AC= 7

(i) = 180°×3 = 540°so, the larger angle is 540°.(ii) = 63°÷3 = 21°so, the smaller angle is 21°. please like my answer ☺☺

21° is the right answer..

i)180×3=540°larger angle id 540°ii)63÷3=21the smaller angle is 21

correct ans is 1 is the

4x² + 3x+ 5= 0First, divide throughout with 4 as the coefficient for x² is 4.→ x2+ 3/4x + 5/4=0now, take 5/4 on the other side....→ x2+ 3/4x = -5/4by using completing square method add the square of the half of the middle term...→x^2+ 3/4x+ (3/8)^2= -5/4 + (3/8)^2we know that, (a + b)2= a^2+ 2ab +b^2. so, by using that........→(x+3/8)^2= -5/4 + 9/64→​ (x+3/8)^2 = (-80+9)/64→ (x+3/8)^2= -71/64now, take square root on both sides...→x+3/8 =√-71/8therefore,x=+√-71/8 - 3/8HENCE,x=(+√-71 - 3)/8 OR x= (-√-71-3)/8

Given sinxlim ---------- = 1x->0 x

Consider

tanxlim ---------- x->0 x

sinxlim ---------- = 1x-->0 xcosx

Use algebra of limits cosx --> 1 as x --> 2

So,

tanxlim ---------- = 1x-->0 x

I didn't understand, sorry

in triangle ABC, AD is median

then , BD = BC

in triangles ABD and ADC

AB^2= AD^2+ BD^2 .........(1)

AC^2= AD^2+ BC^2 ............(2)

now add both the equations,

AB^2+ AC^2= AD^2+ AD^2+ BD^2+ BC^2

AB^2+ AC^2= 2AD^2+ 2BD^2 [ since BD = BC]

AB^2+ AC^2= 2[AD^2+ BD^2]

hence proved.

let score of xi ping is xthen saina score 2x2x+x=453x=45 x=45/3 =15 which is score of xi pingthen score of saina =45-15=30

Geeta scored 90 percent of total marks

She scored 90 percent Marks

54/60*100%=9*1090%Geete scored

Geeta scored 90% marks

Geeta scored 90% in her test

Geeta score 90% marks

the total percant is 90

90% of markssssss.....

Geeta scored

90% total marks

90 percent is right answer

90% is correct answer

marks scored = 54total marks = 60percentage = out of 100so 54/60 = 100 = 90therefore she scored 90%

1.45 2.104 3.72 4.60 5.420 6.891

9×5=4513×8=1049×8=8212×5=6060×7=42099×9=891

2)454)1046)728)6010)42012)881

As AD is the median of ΔABC,Area of ΔABD=Area of ΔACDTherefore, Area of ΔACD=25cm²

Area of ΔABC= Area of ΔABD+Area of ΔACD=25cm²+25cm²=50cm²

See figure, here it is clearly shown that ABC is a triangle, in which AD is a median on BC.

construction :- draw a line AM perpendicular to BC.

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

proof :- case 1 :- when angle ADB = angle ADC, 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 angle ADB is not equal to angle ADCLet 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].

like my answer if you find it useful!

See figure, here it is clearly shown that ABC is a triangle, in which AD is a median on BC.

construction :- draw a line AM perpendicular to BC.

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

proof :- case 1 :- when angle ADB is equals to angle ADC, 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 angle ADC is not equal to angle ADBLet 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].

construction :- draw a line AM perpendicular to BC.

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

proof :- case 1 :- when angle adb is equal to angle adc, 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 angle adb is not equal to angle adcLet 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].

Like my answer if you find it useful!

AD is median, So BD=DC.

AB^2= AE^2+BE^2

AC^2 = AE^2+EC^2

Adding both,

AB^2+AC^2 = 2AE^2+BE^2+CE^2

= 2(AD^2-ED^2)+(BD-ED)^2+(DC+ED)^2

= 2AD^2-2ED^2+BD^2+ED^2-2BD.ED+DC^2+ED^2+2CD.ED

= 2AD^2+BD^2+CD^2

= 2(AD^2+ BD^2)

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We know,

LCM × HCF = Product of numbers

18 × LCM = 612 × 1314

LCM = (612× 1314)/18

LCM = 44,676

There are 50 brackets. So for making x^49, we will multiply x 49 times and mutiply the constant term 1 time. Since the no. of possible constant terms which may be taken are 50

Sn=(-1–3–5….-99)x^49

=50*100/2(-1)x^49 = -2500x^ 49

Coefficient of x^49= -2500option b

1210000 is the answer

-454 bhai bahut easy hai ye to..

315-428+612-953315+184-953499-953=-454