Question

find the area of an equilateral triangle with each side 2 using

∫∫??????

Find area of an equibteral triangle if sive a using polar coordinates such that we are A = dorar that cannot be done wf a single double 2 integral

Answer 1

ar B(1,13) V * */3 >X A(2,0) O 1 (0,0) Let is the We 2 y' OAB is the is the equilateral triangle with each side 2. clearly, OB = AB = 0 A = 2 OA. BP perpendicular line on Then, op = Z OA = { x2 = 1. By pythagorash theorem get OB = op + PB 2 = ?+ PB → PB? = 4-1 = 3 PB : 13 (: PB >0) : Coordinate of B B is (1,13). cartesian equation of AB is y- 43 O-V3 2 → .. = X-1 2-1 y-13 - VE n x- I ? ? y-v3 =-3x + 3 y = -13% +213

Now , x=rcoso, y=rsino gives rsino = -38coso + 253 r sino +38coso = 253 r(sinot J3 (OSO) = 283 213 r sino + 3 coso .: For the triangle : 0505 og and osrs 213 sino tog coso A = Sfx dr do 졸 23 Sinoxiscose r dr НА ( ) do ) ozo r-o 213 sino tus coso T/3 A= [] do ozo Y=0 213 » A = s a do sino tiscoso 020 */3 do 2 > A = Ź • (203)2 S (sing + V3 Coso) O=0 M/3 do > A = 65 (2 (Snø + un cos )) O0

73 do = A = 6 ſ 820 4 (sin Sino + cose cose)? (: sin q = */ cos 7 = do > A = (cos (8-5))" T/ do */ s oco 1/3 A = 3 2 050 [: Cos (0-7) = sint sindt cost coso → A = BS Oso Coj(8-5) > A = 1 sec (0.7%) do [tan (0-1)] C S see? (0-7/6) do » =tan(0-*/6) 6-17] > A = { (tan (5-7) - tan (0-7)) > A = { (tan (273) - tant- 7)) >> A = { (tan (7) + tan (F)) = 3/2 (2 tan 7 fit 3 3a =v3 >> А tan X 2 X = » 7 s N/W wil 33 → A = = A = V3 Area is V3.