why will it not require first statement 
i hate data sufficiency 😞
@sauravd2001 said:why will it not require first statement i hate data sufficiency
First term to cancel hi ho jayega
a+5d=1/4 + a+3d
a+5d=1/4 + a+3d
@Torque024 said:First term to cancel hi ho jayega a+5d=1/4 + a+3dFor second x=sqrt(3)/2?
isnt it a+5d = 5/4(a+3d) ?
Find all integers a, b, c, d, and e, such that
a^2 = a + b - 2c + 2d + e - 8;
b^2 = -a - 2b - c + 2d + 2e - 6;
c^2 = 3a + 2b + c + 2d + 2e - 31;
d^2 = 2a + b + c + 2d + 2e - 2;
e^2 = a + 2b + 3c + 2d + e - 8:
a^2 = a + b - 2c + 2d + e - 8;
b^2 = -a - 2b - c + 2d + 2e - 6;
c^2 = 3a + 2b + c + 2d + 2e - 31;
d^2 = 2a + b + c + 2d + 2e - 2;
e^2 = a + 2b + 3c + 2d + e - 8:
The one who will get the answer at first will get the chance to burn::
:mg:
@Torque024 second to i get it why 30 degree is required because smallest angle has smallest side opposite to it but first one mein
if i take initial height as
1 and continus height as 2 then
1,3,5,7,9,11,13........toh 13*1/4=toh kuch alag hi ayega
aur 4 ke sath toh option fit baithta hai
@catter2011 said:1) both a & b2) neither ??
neither kaise.. agar
if we take hypotenuse that is AC as rt 3
AB= 3/2 and
BC = rt 3 /2
then BD = rt3/4
then AD is => rt39/ 4.
correct me if i am wrong..
@albiesriram said:neither kaise.. agar if we take hypotenuse that is AC as rt 3AB= 3/2 and BC = rt 3 /2 then BD = rt3/4 then AD is => rt39/ 4.correct me if i am wrong..
D hawa mein hein.. D is nt fixed
@sauravd2001 said:@Torque024 bhai i support u in question second that both are required u get height of ad as 1.5?
@Torque024 said:sqrt(3) aa rha mera ruk check krta hu
In a right angle triangle ABC right angled at B. There is a point D such that BD = DC. Find the length of AD.
location of D is not mentioned, how come u are calculating the length of AD?
sorry if i'm misinterpreting your approach
location of D is not mentioned, how come u are calculating the length of AD?
sorry if i'm misinterpreting your approach
@all 1/4 more means 1/4 times more not 1/4 feet more .@sauravd2001 @catter2011 @Torque024 . Sorry for wrong tag if any..
@albiesriram said:neither kaise.. agar if we take hypotenuse that is AC as rt 3AB= 3/2 and BC = rt 3 /2 then BD = rt3/4 then AD is => rt39/ 4.correct me if i am wrong..
Yes M too getting sqrt(39)/4 
@krum What a thought process sir !! of course its cannot be determined.. and my wrong answered list goes +1.
.
@krum bhai dekh hypotnuese root3 toh
perpendicular aur base root2 aur 1 ho skate hain
ab dekh figure ke hisab se angle a =30 toh 1 will be opposite it as smaller angle has smaller side opposite to it
so bd=dc hue .5 and .5 and perpendicular hai hi hamare pass root 2 toh
ad hue hypotneuse
toh ad^2=(1/2)^2+(root2)^2=1/4+2=9/4
ad=root9/4
ad=3/2.....kuch galti hai kya ?:P
how many squares between 1 and 1000 (both inclusive) can be expressed as the diff. of the squares of 2 non negative integers. A.750 B. 748 C.300 D.250
please show approach as well.
@krum said:In a right angle triangle ABC right angled at B. There is a point D such that BD = DC. Find the length of AD.location of D is not mentioned, how come u are calculating the length of AD?sorry if i'm misinterpreting your approach
u r8 bro...should be cdb
...mai hi utna nhi soch pata
...mai hi utna nhi soch pata@krum said:In a right angle triangle ABC right angled at B. There is a point D such that BD = DC. Find the length of AD.location of D is not mentioned, how come u are calculating the length of AD?sorry if i'm misinterpreting your approach
AB^2+BC^2=3
AB^2+(2*BD)^2=3 -----eq 1
Sin 60 = AB/AC
AB= Sin60*Sqrt(3)
From AB we can find BD from eq 1
AD= AB^2+ BD^2
AB^2+(2*BD)^2=3 -----eq 1
Sin 60 = AB/AC
AB= Sin60*Sqrt(3)
From AB we can find BD from eq 1
AD= AB^2+ BD^2
@sauravd2001 said:@krum bhai dekh hypotnuese root3 toh perpendicular aur base root2 aur 1 ho skate hain ab dekh figure ke hisab se angle a =30 toh 1 will be opposite it as smaller angle has smaller side opposite to it so bd=dc hue .5 and .5 and perpendicular hai hi hamare pass root 2 toh ad hue hypotneuse toh ad^2=(1/2)^2+(root2)^2=1/4+2=9/4ad=root9/4ad=3/2.....kuch galti hai kya ?
yaar itna nai padha jata, tu bas ye bata D point kahan par hai, sur soch kar batana
Ya D hawa mai ho sakta hai Fas gaye :/
@realslimshady said:how many squares between 1 and 1000 (both inclusive) can be expressed as the diff. of the squares of 2 non negative integers. A.750 B. 748 C.300 D.250 please show approach as well.
750..