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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

Guest Feb 23, 2015

Best Answer 

 #1
avatar+889 
+10

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.

Bertie  Feb 23, 2015
 #1
avatar+889 
+10
Best Answer

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.

Bertie  Feb 23, 2015

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