In the adjoining figure ADE and ABC are two similar triangles, point D divides AB in the raito 2:1 and point E divides AC in the ratio 1:2 . If the area of `triangleADE` is 23 square units, then find the ar(`squareDECB)`.

Explanation

We have
`(AD)/(DB)= 2/1 and (AE)/(EC)= 1/2 `
Let AD = 2x `Rightarrow DB=x and "let" AE = y Rightarrow EC= 2y Rightarrow AC= 3y`
`(AD)/(DB) ne (AE)/(EC)`
` DEcancel||BC`.
` angleD ne angleB and angleE ne angleC`.
But since `triangleADE and triangleABC` are similar and `angleA` is common to both the triangles.
`triangleADE and triangleABC`
`angleA= angleA` (common)
and `angleD= angleC` (obvious)
`triangleADE~triangleACB` (AA corollary)
`(AD)/(AC) = (DE)/(CB) = (AE)/(AB) Rightarrow (2x)/(3y)= y/(3x) Rightarrow 6x^(2)= 3xy Rightarrow 2x= y`
`(ar(triangleADE))/(ar(triangle(ACB))= (AD^(2))/(AC^(2))`
if two triangles are similar then ratio of their areas is equal to the ratio of square of their corresponding sides)
`(4x^(2))/(9y^(2))= ((2x)^(2))/(9y^(2))= (y^(2))/(9y^(2))= 1/9`
`23/(ar(triangleACB))= 1/9`
` ar (triangleACB)= 23xx9 = 209` sq units
`ar ("square "DECB)= ar(triangleABC)-ar(triangleADE)`
207-23= 184 sq units.