Question

ABCD is a rectangle where side BC is twice side AB. If ΔACQ ~ ΔBAP, find area of ΔBAP : area of ΔACQ.

Answer 1

Given:

ABCD is a rectangle with BC = 2·AB.

ΔACQ ~ ΔBAP.

Step-wise calculation:

1. Let AB = x. Then BC = 2x.

2. Diagonal `AC = sqrt(AB^2 + BC^2)` 

= `sqrt(x^2 + (2x)^2)`

= `x · sqrt(5)`

3. From the similarity ΔACQ ~ ΔBAP the side AC corresponds to BA.

So, the linear scale factor from ΔACQ to ΔBAP is

`(BA)/(AC) = x/(x · sqrt(5))`

= `1/sqrt(5)`

4. The ratio of areas of similar triangles equals the square of the ratio of corresponding sides.

Hence, `(Area (ΔBAP))/(Area (ΔACQ))`

= `(1/sqrt(5))^2`

= `1/5`

Area ΔBAP : Area ΔACQ = 1 : 5.