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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.....

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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.