1. Write a linear equation for x=2,y=5.
2. If a triangle and a rhombus are on the same base and between the parallels,find the ratio of area of triangle and area of triangle and area of rhombus.
3. Diagonal AC and BD of a quadrilateral ABCD intersect each other at P. Show that
ar (ΔAPB)×ar (ΔCPD) = ar (ΔAPD)×ar (ΔBPC).
4. Express y in terms of x if it being given that 3x+y-9=0. Check the points (3,0) and (2,2) lie on the equation.
5.P and Q are any points lying on the sides DC and AD of a parallelogram ABCD.Show that ar (ΔAPB) = ar (ΔBQC).
Plzzz answer each and every question. Don't leave any question. Plzz do carefuly. Its a bit urgent.Plzzz helppp.....

I suppose you need explanation, right?


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    5 x = 2y  or  y = 5/2 * x

    base AD.  Triangle = AQD.    Q is on BC.  
    Draw QR perpendicular from Q on to AD meeting it at R.
    Let Altitude QR  = distance between AD & BC

       Area of Rhombus ABCD =  AD * QR 
       area of triangle APD = 1/2 AD * QR
             ratio = 1/2

    Draw ABCD.   Draw P as intersection of AC and BD.  Draw perpendicular from B to AC meeting AC at R.  Draw perpendicular from D on to AC to meet it at  Q.

            Area triangle APB * area of triangle CPD
                  =  1/2 AP * BR  *  1/2  CP * QD
                   = 1/2  AP * QD   *  1/2  CP * BR
                  = area of triangle APD  *  Area of triangle CPB.

       3x + y - 9 = 0    =>        y = 9 - 3 x
             (3,0)  =>    0 = 9 - 3 * 3 = 0   TRUE  => It is on the line
             (2,2)    =>   2y = 9 - 3 * 2     FALSE => not on the line

         Proof is similar to that in question 2 above.
           Draw a perpendicular from P onto AB meeting it at S.
             Area of triangle APB = 1/2* AB * PS = 1/2 * Area of parallelogram
         Draw a perpendicular from Q on to BC meeting it at T.
            Area of triangle BQC = 1/2 * BC * QT = 1/2 * area of parallelogram
               hence areas of triangles APB and BQC  are both equal.

2 5 2
thank you so much sir
its ben so kind of you to help me...
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