Please solve q 87 ?

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3 Answers
May 18, 2018

#AB=6# cm

Explanation:

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Let #|ABC|# denote area of #DeltaABC#,
let #|FBD|=2a#,
as #BD=DC, => |FDC|=2a, => |FBC|=2a+2a=4a#,
given #AE:ED=1:2#,
#=> |FAC|:|FDC|=1:2, => |FAC|=a#
#=> |ABC|=|FAC|+|FBC|=a+4a=5a#
#=> AF:AB=|FAC|:|ABC|=1:5#
#=> AB=5*AF=5xx1.2=6# cm

May 18, 2018

# |AB|=6 # cm

Explanation:

There's probably a cool trick I'm missing, but I didn't peek at the other answer. When I'm stuck I find putting it on the Cartesian grid is a way to make progress.

Let's say #B(0,0)=O# at the origin, #C(6c,0)#, and #A(6a,6b)#. Out of experience I put a factor of 6 on to try to avoid fractions; we'll see how that goes.

#D# is the foot of the median, so the midpoint of BC,

#D=1/2(B+C) = D(3c,0)#

#E# is one third the way along AD from #A# to #D.# If it was #2/3# it would be the centroid.

# E = A +1/3(D-A) = 2/3 A + 1/3 D = E(4a+c, 4b) #

#F# is the meet of CE and AB.

# F = C + t (E -C) = B+ u(A-B) = uA quad # because #B=O#

#(6c,0) +t(\ ( 4a+c,4b) -(6c,0) \ ) = u (6a,6b) #

Y coordinate first, it's easier: # 4bt = 6 b u #

#2t = 3u#

X: # 6c + ( 4a+c-6c ) t = 6 au = 2a (3u) = 2a(2t)=4at #

#6 c = 5 c t #

# t = 6/5 , quad quad u = 2/3 t = 4/5#

#|AF|= |F-A|=|uA-A| = |u-1||A| = (1-u)|AB|#

# |AB| = |AF|/{1-u} = |AF|/{1-4/5}= 5 |AF|#

We're given # |AF|=1.2 # cm # =6/5 # cm so

# |AB|=6 # cm

May 18, 2018

#6#

Explanation:

Second answer for this one.

The statement is claimed for any triangle, so we only need to work it out for one. Let's forget about the 1.2 for a bit and work this out for one of the usual suspects, say 45/45/90 with legs length six.

We'll put the right angle at the origin and the sides along the axes, #A(6,0), B(0,6), C(0,0) #.

AD is the median to BC so D is the midpoint of BC, #D(0,3).#

E on AD such that #AE=1/3 AD#. We go 2/3 along the way from D toward A for each coordinate, so E=#(4,1)#

CE #C(0,0),E(4,1)# intersects AB #A(6,0),B(0,6)# at F.

CE is #x=4y# and AB is #y=6-x# so #y=6-4y# or #5y=6# or #y=6/5#. #quad x=4y=24/5#

# F(24/5, 6/5)#

# |AF|^2 = (6-24/5)^2+(6/5)^2 = 2(6/5)^2 #

# |AB|^2=6^2+6^2=2(6^2)#

#|AB|^2/|AF|^2= 5^2 #

So in the original question

#|AB|=5 |AF| = 5(1.2)=6#, choice (1)