Area of a triangle problem.

[AndyB]

Hi Friends,

I am facing difficulty in solving the below problem, Could you please help me with the solution.


In triangle ABC shown above, DE || BC and DE/BC = 1/4. If area of triangle ADE is 10, find the area of the trapezium BCED and the area of triangle CED.1

[sanju09]

Hi Friends,

I am facing difficulty in solving the below problem, Could you please help me with the solution.


In triangle ABC shown above, DE || BC and DE/BC = 1/4. If area of triangle ADE is 10, find the area of the trapezium BCED and the area of triangle CED.1

Please refer to the Basic Proportionality Theorem in triangles, which states that "If a line is drawn parallel to one side of a triangle, to intersect the other two sides in distinct points, the other two sides are divided in the same ratio."

In other words triangle ABC ~ triangle ADE, and since, DE/BC = 1/4, the areas of triangle ADE and triangle ABC will be in the ratio 1/16. That means when the area of triangle ADE is 10, the area of triangle ABC will be 160. Now, area of trapezium BCED is 160 - 10 = 150.

The line segment DC will divide the trapezium BCED into two triangles viz. triangle CED and triangle CBD, in which

Area of triangle CED = ½ × DE × distance between the parallel lines DE and BC, and

Area of triangle CBD = ½ × BC × distance between the parallel lines DE and BC, and

But DE = ¼ BC, hence DC will divide the trapezium BCED is divided in the ratio 1:4 and Area of triangle CED = 1/5 × area of the trapezium BCED = 1/5 × 150 = 30.

[AndyB]

Hi Sanju,

Thank you very much for taking the time to give a detailed explanation.
I am not that good at the basics of triangles.

I am confused at the following point:

The line segment DC will divide the trapezium BCED into two triangles
viz. ΔCED and ΔCBD (Yes it does)


Area of ΔCED = ½ × DE × distance between the parallel lines DE and BC

Why have you taken the "distance between the parallel lines DE and BC" in the area calculation, when we are considering the ΔCED.
Please help me in understanding the logic behind it.

[GMATGuruNY]

In triangle ABC shown above, DE || BC and DE/BC = 1/4. If area of triangle ADE is 10, find the area of the trapezium BCED and the area of triangle CED.1

Plug in values that satisfy the given conditions:



In the figure above, DE =1.
Since DE/BC = 1/4, BC = 4.

Since the area of ∆ADE is 10:
(1/2)(DE)(AF) = 10
(1/2)(1)(AF) = 10
AF = 20.

Since DE||BC, ∆ADE is similar to ∆ABC.
Since ∆ADE is similar to ∆ABC, and DE/BC = 1/4, all corresponding lengths must yield a ratio of 1:4.
Thus, AF/AG = 1/4:
20/AG = 1/4
AG = 80.

Since AG = 80 and AF = 20, FG = 80-20 = 60. This is the height both of trapezoid BCED and of ∆CED.

Trapezoid BCED:
The area of a trapezoid = (b1 + b2)(h)/2.
Thus, the area of trapezoid BCED = (1+4)(60)/2 = 150.

∆CED:


If we consider DE the base, then the area = (1/2)bh = (1/2)(1)(60) = 30.

[AndyB]

Hi Sanju/GmatGuru,

Thanks again, for the very well explained answer.
I have understood the solution.

The important point here was the height of the scalene triangle CED is the distance between parallel lines.

[ArunangsuSahu]

Area of the trapezium=1/2*(DE+BC)*h
A = 1/2*(5*DE)*h

Area of the triangle CED = 1/2*DE*h =B

B:A=1:5

Area of B=150/5=30

[malikbasha03]

Sir hw the triangles are similar please let me know................