Question

In square ABCD, shown here, point F is on side AB such that
AF:FB = 2:1, and point G is on side AD such that AG:GD = 3:1.
What is the ratio of the area of triangle AFG to the area of
pentagon FBCDG? Express your answer as a common fraction​

Answer 1

Given:

In square ABCD, shown here, point F is on side AB such that

AF:FB = 2:1, and point G is on side AD such that AG:GD = 3:1.

To Find:

Ratio of the area of triangle AFG to the area of

pentagon FBCDG.

Solution:

It is given that, AF:FB = 2:1 & AG:GD = 3:1

=> AF:AB = 2:3 & AG:AD = 3:1

Now, Let the side of square be s.

Area of square = $s^{2}$

Since, AF:AB = 2:3 & AG:AD = 3:1

=> AF = $\frac{2}{3} s$ & AG = $\frac{3}{4} s$

Now, Area of triangle AFG = $\frac{1}{2}$ × b × h

Area of triangle AFG = $\frac{1}{2}$ × $\frac{2}{3} s$ × $\frac{3}{4} s$ = $\frac{s}{4} ^{2}$

Now, it is clear from figure that,

Area of pentagon FBCDG = Area of square ABCD - Area of triangle AFG

Area of pentagon FBCDG = $s^{2} - \frac{s^{2} }{4}$ = $\frac{3s^{2} }{4}$

Now,

Ar. AFG : Ar. FBCDG = $\frac{\frac{S^{2} }{4} }{\frac{3S^{2} }{4} } = \frac{1}{3}$

Hence,  ratio of the area of triangle AFG to the area of

pentagon FBCDG is 1:3.