Question

In the figure, ABCD is a rectangle. G is a point on CD. The area of

the triangle ∆ AGH is 7 square centimetres and the area of the

triangle ∆ BGH is 4 square centimetres.

a) What is the area of the triangle ∆ AGD?

b) What is the area of the triangle ∆ BGC?

c) What is the area of the triangle ∆ ABG?

d) What is the area of the rectangle ABCD?

e) What is the relation between he area of the triangle ABG

and that of the rectangle ABCD​​

please give answer with explanation

Answer 1

Answer:

Draw line FZ, where Z is the midpoint of AE; FZ divides triangle AEF into two equal-area triangles — their bases are equal, and altitude is equal. Call the area of each of them x.

Draw line ZY, where Y is the midpoint of AB; that line divides the area-4 triangle into area 1 below ZY and area 3 above it.

So above FZY we have area 3 + x + 3; below FZY we have area 1 +x + 5. So FZY divides the rectangle in half — it’s a straight line midway between the rectangle’s edges and parallel to them.

So the length of FZ is midway between the lengths of CE and AD, which means x is midway between 3 and 5, since the area-3 triangle and the area-5 triangle and the two area-x triangles all have the same altitude.