higly simple

okk
Thx

tan 30°=1/√3=2r/a...r is inradius

now area of the square 2r²

2r²=a²/6

oh...tx subhash.....

yipee[54]

cheer up this time it is B [1]

again i did it with out the help of st-II[2]

Second one

Option A

Stat 1 is true and stat - 2 is also true
stat 2 is explaining stat - 1

both true...but not a correct explanation

but without it ...st-I can't be explained[12]

try the second one

Yes but can be proved by the method given therefore it is true

but why does statement 2 explain statement 1? we can even prove it without the help of statement 2.!!

here's a solution to counter option A.

1/3<log₂₀3<1/2

=> 20^1/3<3<20^1/2

2³<20<3³ => 2<20^1/3<3 ; 4²<20<5² => 4<20^1/2<5 => 3<20^1/2

thus we can prove it without the help of statement 2 right.!

let me know if my solution is violating some condition of logarithms !

The option has to be A stat2 explains stat 1

See let e = 3

log₂₀3 = log₃3/ log₃20
2<log₃ 20 < 3
Therefore it lies between 1/2 and 1/3

Statement 1: A circle is inscribed in an equilateral triangle of side a. The area of any square inscribed in it is
and
Statement 2: In an equilateral triangle, the inradius is half the circumradius.

try this

i seem to get this wrong everytime

if i had answered A then answer would be B saying stat2 doesnt completely explain stat1