It is given that AEFD's area is 738cm². So AEFD cannot be 246cm²

Since GB is twice of GF, and△GFH and △GBH have the same height ,

the area of △GBH is twice that of △GFH because its base is twice the length of △GFH's length.


You have correctly found that the area of △FBH is 369cm².

So area of △ GBH is ⅔ of △FBH

= ⅔ x 369cm²
= 246cm²

And, area of △ GFH = ⅓ x 369cm²
= 123cm²

(△FBH area = △GBH area + △GFH area
If △GFH area = 1 unit,
then △GBH area = 2 units
△FBH area = 3 units)

Next,

△ EFH also = 369cm² because it has the same base length as △FBH and also the same height
(their heights = breadth of rectangle ABCD)


△EFG area = △EFH area - △GFH area
= 369cm² - 123cm²
= 246cm²

Area of △EFG + area of△FBH

= 246cm² + 369cm²
= 615cm²

Alternatively,

Area of △EFG + △FBH

= Area of △EFH + Area of △FBH - area of overlap(which is actually △GFH)

= 369cm² + 369cm² - 123cm²

= 738cm² - 123cm²
= 615cm²

Oops.. lm mistaken AEFD as ABCD.
Thanks @J for highlighted the error.

Glad to help