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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. Please kindly answer all the questions with explanation. It is very urgently needed.......... Plzzzzz help..........

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See diagram. Let BF = x

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

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.