Diagonals of quadrilateral ABCD intersect at O.If area of triangle ACB is 5 sq.units,area of triangle BCD is 9 sq.units,area of triangle CDA is 10 sq.units and area of triangle DAB is 6 sq.units,Find area of triangle AOB.

Im begging you! pls I badly need an answer!! can you pls answer this? Now? As in Now? Thankyouuu :)

Label the areas of the triangles with O as a vertex, in the order named above. If we call their areas a,b,c,d then we have

a+b=5
b+c=9
c+d=10
d+a=6

I find no unique solution to that system of equations. I'll have to think on it some more, I guess. Some other geometric property must place an additional constraint on the values.

Sure, I can help you with that. To find the area of triangle AOB, we can use the fact that the area of a triangle formed by two intersecting diagonals of a quadrilateral is equal to half the product of the lengths of those diagonals.

Let's assume that the diagonals of quadrilateral ABCD intersect at point O. So, we need to find the lengths of the diagonals AC and BD.

From the given information, we know that the area of triangle ACB is 5 sq.units. Since the base of this triangle is AC, we have:
(1) Area of triangle ACB = (1/2) * AC * h, where h is the distance from point C to line AB.

Similarly, the area of triangle CDA is given as 10 sq.units. Since the base of this triangle is CD, we have:
(2) Area of triangle CDA = (1/2) * CD * h', where h' is the distance from point D to line AC.

From equations (1) and (2), we find that:
AC = (2 * area of triangle ACB) / h
CD = (2 * area of triangle CDA) / h'

Next, we know that the area of triangle BCD is 9 sq.units. Since the base of this triangle is BD, we have:
(3) Area of triangle BCD = (1/2) * BD * h'', where h'' is the distance from point B to line CD.

Similarly, the area of triangle DAB is given as 6 sq.units. Since the base of this triangle is AB, we have:
(4) Area of triangle DAB = (1/2) * AB * h''', where h''' is the distance from point A to line BD.

From equations (3) and (4), we find that:
BD = (2 * area of triangle BCD) / h''
AB = (2 * area of triangle DAB) / h'''

Now, the area of triangle AOB will be:
(5) Area of triangle AOB = (1/2) * AC * BD

Substituting the values of AC and BD from above, we can calculate the area of triangle AOB.

Please provide the values of h, h', h'', h''', or any additional information, if available, so that we can proceed with the calculation.

Of course, I'd be happy to help you find the area of triangle AOB! To solve the problem, we can use a property of triangles formed by the intersection of diagonals in a quadrilateral.

First, let's label the areas of the given triangles:
- Area of triangle ACB = 5 sq.units
- Area of triangle BCD = 9 sq.units
- Area of triangle CDA = 10 sq.units
- Area of triangle DAB = 6 sq.units

To find the area of triangle AOB, we need to find the ratio between the areas of triangles AOB and ACB.

Step 1: Find the ratio of areas
Since the diagonals of the quadrilateral intersect at point O, we can use the property that the areas of the triangles formed by the intersection of diagonals are proportional.

From the given areas, we can set up the following ratios:
- (Area of triangle BCD) / (Area of triangle CDA) = (Area of triangle ACB) / (Area of triangle DAB)
- 9 / 10 = 5 / 6

Step 2: Solve the ratio
To solve for the missing ratio, we can cross-multiply the equation:
- 9 * 6 = 10 * 5
- 54 = 50

Step 3: Find the area of triangle AOB
Now that we have the ratio between the areas, we can use it to find the area of triangle AOB.

The ratio tells us that the area of triangle AOB is 54/50 times the area of triangle ACB:
- Area of triangle AOB = (54/50) * (Area of triangle ACB)
- Area of triangle AOB = (54/50) * 5 sq.units
- Area of triangle AOB = 27/10 sq.units

Therefore, the area of triangle AOB is 27/10 square units.

I hope this explanation helps you understand how to find the area of triangle AOB in this problem! Let me know if you have any further questions.