In the figure let #B(20, 0)# and #C(0, 30) #lie in x and y axis respectivelly. The angle, #/_ACB= 90°#. A rectangle DEFG is inscribed in triangle ABC. Given that the area of triangle CGF is 351, calculate the area of the rectangle DEFG?

enter image source here

1 Answer
Oct 8, 2016

#468#

Explanation:

#hat(ACB) = pi/2# so

#bar(AO)cdot bar(OB)=bar(OC)^2# so

#abs(Delta(ACB)) = 1/2(bar(AO)+bar(OB))bar(OC)#

We know that #abs(Delta(GCF)) = 351# and

#abs(bar(OB))=20# and #bar(OC)=30# then

#abs(bar(AO))=30^2/20=45# so

#abs(Delta(ACB)) =1/2(45+20)30=975#

We have also

#abs(Delta(ACB))/bar(AB)^2=abs(Delta(GCF))/bar(GF)^2# then

#bar(GF) = bar(AB)sqrt(abs(Delta(GCF))/abs(Delta(ACB)))=65sqrt(351/975)=65(3/5)#

Calling now #h_1# the height of #Delta GCF# we have

#h_1/bar(GF)=bar(OC)/bar(AB)# and also

#h_2 = bar(OC)-h_1# so the sought area is

#square = bar(GF) cdot h_2 = bar(GF) cdot bar(OC)(1-bar(GF)/bar(AB)) = 468#