iN THE FIGURE AB:BC=4:3 AND BD//CE and area of triangle ADB is 24cm^2 and the area of the trapezium BCED is 60 cm^2, find Area of triangle ADE
+893 Start with some construction.
Draw a line through A parallel to DB and EC and a line through C to meet this line in a right angle. Call the point of intersection F. Extend DB to meet the new line through C at G.
Since AB:BC = 4:3, by similar triangles FG:GC = 4:3, so let FG = 4a and GC = 3a, for some constant a.
Let the length of CE = p and the length of BD =q.
The area of the parallelogram is 60, so (p + q).3a/2 = 60, so, pa + qa = 40................(1)
The area of the triangle ABD is 24, so, q.4a/2 = 24, so qa = 12................(2)
From (1) and (2), pa = 28.
The area of the big triangle ACE = p.7a/2 = 28*7/2 = 98,
in which case the area of the triangle ADE = 98 -24 - 60 = 14.
+893 Start with some construction.
Draw a line through A parallel to DB and EC and a line through C to meet this line in a right angle. Call the point of intersection F. Extend DB to meet the new line through C at G.
Since AB:BC = 4:3, by similar triangles FG:GC = 4:3, so let FG = 4a and GC = 3a, for some constant a.
Let the length of CE = p and the length of BD =q.
The area of the parallelogram is 60, so (p + q).3a/2 = 60, so, pa + qa = 40................(1)
The area of the triangle ABD is 24, so, q.4a/2 = 24, so qa = 12................(2)
From (1) and (2), pa = 28.
The area of the big triangle ACE = p.7a/2 = 28*7/2 = 98,
in which case the area of the triangle ADE = 98 -24 - 60 = 14.
