1.In a parallelogram ABCD, AB= 8cm. The altitudes corresponding to sides AB and AD are respectively 4cm and 5cm. Find the measure of AC.
2.In a parallelogram ABCD,P is the mid point of BC and P is connected to A and D.If the area of the parallelogram is 80cm², then find the area of ΔADP.
3.ΔABC and ΔDBC are two triangles on the same base BC and between the parallel lines m and n at the top and bottom respectively.If AB=3cm, BC=5cm, angle A= 90 degree, find the area of Δ DBC.
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See diagram.
Let  BF = x

BC = root(x^2 + 16) = AD
DE = root(64-25) = root(39)

AE = AD+DE = root(x^2+16) + root(39)
AC^2 = AF^2 + FC^2 = (8+x)^2 + 4^2    -- eq 1
AC^2 = AE^2 + EC^2
         = (x^2 +16 + 39 + 2 root[39 x^2 + 39*16] + 5^2   -- eq 2

Equating equations 1 and 2  and solving for x, we get

   x = root(39)*4/5

Now find AC  using equation 1. 

Area of parallelogram is base * height.  Area of triangle ADP is 1/2 * base * height.   As both base and height are same, we have the area of triangle is 1/2 of area of parallelogram = 40 cm^2

see figure 2.

triangle ABC is a right angle triangle.  AC^2 =  BC^2 - AB^2 = 25 - 9 = 16
 AC = 4 cm
area of ABC = 1/2 * AC * AB = 3*4/2 = 6 cm^2
area of ABD is same as that of ABC as  both triangles have the same altitude and base.

2 5 2
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