Question

In Fig, AC is parallel to DE, ar(QuadABCD)=25 sq. Units and ar ∆ABC=17sq. Units .find ar ∆ACE
/
/
/
/

Answer 1

I hope this is the correct answer

As AC//DE(parallel to)

Therefore the triangle inside this would have same area
So in ACDE quadrilateral
Triangle ACD and ACE
Lies on same base and same parallels so
Area ACD=area ACE

therefore
Area of ACD= area of ABCD -area of
ABC
Therefore
Area of ACE=area of ABCD -area of
ABC
=25 - 17
=8sq. units

So the area of triangle ACE is 8sq. units

Answer 2

Given: AC is parallel to BD
ar quad (ABCD) = 25
ar tri (ABC) = 17

to find: area of triangle ACE

proof : ar(abcd) = ar(ABC ) + ar( ACD)

that is why
ar (acd) = ar( abcd) - ar( abc)
= 25- 17
= 8 square units

but area of triangle ADC is equal to the area of triangle ACE because they are betn parallel lines and oin the same base AC
therefore,
area if (ACE) = 8 sqaure units