Question

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?

Answer 1

`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`