Question

Geometry question

Answer 1

HIT ME WITH IT!! :D

Answer 2

(((:))))

Answer 3

Um sweetheart that isnt a question :)

Answer 4

or Mister

Answer 5

Tag me when you finished writing your paragraph 00.00

Answer 6

\(BN=15,MC=18\) and \(\angle BKC=90^{\circ}\), then find the area of \(\triangle ABC\)

Answer 7

@Here_to_Help15

Answer 8

THey forgot to tag you

Answer 9

Are you missing some given condition? I got the area BMNC = 135
But no way to go further. ha!!

Answer 10

lol i forget to state that \(BN\) and \(CN\) are medians

Answer 11

BN and CM

Answer 12

ha!! it is a big missing point.

Answer 13

Then we have MN// BC, right?

Answer 14

ok, let me carefully draw it out and we discuss later. kid @xapproachesinfinity

Answer 15

http://www.math10.com/en/geometry/trapezium-trapezoid-central-median.html

Answer 16

@xapproachesinfinity That is what I said. If M, N are middle points of AB, AC, then MN?// BC

Answer 17

I give up. hehehe.... old man need eating something. Young boy try to help him, ok?

Answer 18

let me do paper and pen hehhe

Answer 19

?

Answer 20

@amistre64

Answer 21

is the area 180?
since BN is a median there for BK=2/3BN=2/3 (15)=10
similarly CK=2/3 (18)=12
and since ck is the altitude of the triangle CKB ( angle bkc=90)
therefore Area of BKC=12x10/2=60
and since all triangle created by the centroid are equal in area
now i create another median coming from the vertex A cutting BC at H
then i have CKH=1/2 BKC=30 (As a result of K being a Centroid)
now all it follow that all angles have the same area if K is the centroid
therefore Total area=6(30)=180

Answer 22

this makes sense now !