∆ADC is comprised of 2 regions: ∆ADE and ∆CDE
We're told that ∆ADE and ∆CDE have areas 16 ans 12 respectively.
So, the area of ∆ADC = 16 + 12 = 28



First recognize that ∆ADC and ∆BCD both have the same base.
Also, if we let h = the height of ∆ADC, then ∆BCD also has height h


area of triangle = (base)(height)/2
So, if ∆ADC and ∆BCD have the same base and the same height, then they must have the same area.

So, if the area of ∆ADC is 28, then the area of ∆BCD must also be 28


Answer: 28

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In the figure, what is the area of triangle ADB given that the area of triangle ACE is 4 and the area of triangle CDE is 3?
Give your answer as a fraction.

can you please solve it for me

Sir, I kindly request you to please go through the question

as inverting the figure changes a lot and it's asked for another triangle that too infraction.

and the answer is supposed to be 28/3, would really kind of you if you could explain to me

It is correct indeed. \(28\) is the answer.

The base of both the triangle is same. Hence the formula used is correct.

Can you please post your way of reaching \(\frac{28}{3}\)? Perhaps the mistake can be evaluated better.

Regards

Sir, I kindly request you to please go through the question

as inverting the figure changes a lot and it's asked for another triangle that too infraction.

and the answer is supposed to be 28/3, would really kind of you if you could explain to me

In the figure, what is the area of triangle ADB given that the area of triangle ACE is 4 and the area of triangle CDE is 3?
Give your answer as a fraction.

can you please solve it for me as the question is a bit different would request you to through and figure and values in the question are different