Geometry 2

Pls explain this step again-

My method is, area of tri(ADE)/area of tri(ABC) = DE^2 / BC^2 since tri(ADE) ~ tri(ABC)

ok...understood triangles are similar..but confusion over how u reached DE^2/ BC^2...pls elaborate this step

It is 9/16

I approached via algebra (this is for the ones not familiar with the properties of similar triangles ):

Let the height of ADE = x, height of BCDE = y

Then:

S(ADE) = 3x/2
S(BCDE) = (3+5)/2 *y = 4y
S(ABC) = 5*(x+y)/2

Since S(ABC)=S(ADE)+S(BCDE)
3x/2 + 4y = 5(x+y)/2
3x+8y=5x+5y
x=1.5y

Now the ratio of ADE to BCDE = 3x/2 / 4y
3*1.5y/2 * 1/4y
4.5y/8y
4.5/8
9/16

I agree that the answer is B: 9:16 .

Please note though that the triangles DO NOT have to be similar. However, since this case applies to all trapezium, triangle pairs, you can simplify the problem by assuming ADE is similar to ABC. I took a slightly different approach. Since there are no restrictions given on Triangle ABC, I decided to create one that simplifies the problem (see attached image).

I assume that ABC is a right triangle, and that DE is parallel to BC so that ADE is similar to ABC. Therefore,

Area(Triangle) = Area(ADE)
Area(Trapezium) = Area(ABC) - Area(ADE)

Area(ADE) = 3 * 3 / 2 = 9/2
Area(ABC) = 5 * 5 / 2 = 25/2

Area(Triangle) = 9/2
Area(Trapezium) = 25/2 - 9/2 = 16/2 = 8

Area(Triangle) : Area(Trapezium) = 9/2 : 8 = 9 : 16

[quote="virtualanimosity"]Pls explain this step again-

My method is, area of tri(ADE)/area of tri(ABC) = DE^2 / BC^2 since tri(ADE) ~ tri(ABC)

ok...understood triangles are similar..but confusion over how u reached DE^2/ BC^2...pls elaborate this step




ADK & ABH are similar triangle too:
AD/AB=AK/AH

Tri ADE & ABC are similar triangle:
AD/AB=AE/AC=DE/BC

SADE/SABC=(AK*DE/2)/(AH*BC/2)
=AK*DE/AH*BC
=AK/AH * DE/BC
=AD/AB * DE/BC
=DE/BC * DE/BC
=DE^2/BC^2

Is it clear to U now Virtualanimosity?

Thanks man...crystal clear
great detailed explanation..really helpful!!