Question

Solid Mensuration :)

Answer 1

hey harttn you can view the prob I attached just now :D

Answer 2

i was viewing that only

join FC
then, can you find EF ?

Answer 3

because you will know FC and EC

Answer 4

ok i''ll work on it, wait a minute :)

Answer 5

sure, take your time...
ask me if you want to know how to get EF

Answer 6

ok, nah. i know how to ;) thx

Answer 7

using Pythagorean Theorem EF=\[6\sqrt{6}\], am I right?

Answer 8

good :)

Answer 9

now, what you know about FG and DG

Answer 10

yes! what would be the next step?

Answer 11

like you know why FG and DG are equal ?

let them be 'x'

Answer 12

i know why're they're equal :) because they're both tangent lines intersecting at one point, right? :D

Answer 13

good :)
so, now see right triangle EGD
GD = x
hypo = EF+x
ED .....you know

Answer 14

OH! we'll be able to get DG! hahaha it will be like this: \[(6\sqrt{6}+x)^{2}=18^{2}+x ^{2}\] solving for x.....x=3.67in :D right?

Answer 15

i didn't check, i am helping 4 ppl at a time :P
give me some time to verify....

Answer 16

ok ok haha :) its ok take your time!

Answer 17

yes, i get 3.674 too

Answer 18

now we got DG, we need to get the area enclosed by FG and DG, and arc FD

Answer 19

hint :
area of rhombus FCDG - area of sector

Answer 20

***sector FCD

Answer 21

makes sense ? or should i elaborate ?

Answer 22

no! its ok :) no need to ^^