In a triangle ABC, PQ is the midpoint of AB and AC. PQ = 2cm, BC =3cm. The area of trapezium PQCB = 10cm^2.

If the area of triangle APQ = xcm^2. Calculate the value of x.

Sorry. If PQ joins the midpoints of AB and AC, then it is parallel to BC and half its length.

You cannot have PQ=2 and BC=3.

To find the value of x (the area of triangle APQ), we need to use the formula for the area of a trapezium.

The area of a trapezium can be calculated using the formula:
Area = (1/2)(sum of parallel sides)(distance between the parallel sides)

In this case, the parallel sides of the trapezium are PQ and BC, and the distance between them is equal to the height of the trapezium. Let's call the height h.

Given information:
PQ = AC = 2 cm
BC = 3 cm
Area of trapezium PQCB = 10 cm^2

Using the formula for the area of a trapezium, we have:
10 = (1/2)(PQ + BC)(h)

Substituting the given values:
10 = (1/2)(2 + 3)(h)
10 = (1/2)(5)(h)
10 = (5/2)(h)

To solve for h, we can multiply both sides of the equation by 2/5:
(2/5)(10) = h
4 = h

Now that we know the height of the trapezium is 4 cm, we can calculate the area of triangle APQ by using the formula for the area of a triangle:

Area = (1/2)(base)(height)

In triangle APQ, the base is equal to PQ (which is given as 2 cm) and the height is equal to the height of the trapezium (which we found to be 4 cm).

Substituting the values into the formula:
Area = (1/2)(2)(4)
Area = 4 cm^2

Therefore, the value of x (the area of triangle APQ) is 4 cm^2.