# 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.....

1
I suppose you need explanation, right?
yup

2014-11-22T10:34:48+05:30

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1.
5 x = 2y  or  y = 5/2 * x

2.
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

3.
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.

4.
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

5.
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.

thank you so much sir
its ben so kind of you to help me...
its ok. u r welcome. select as best answer.