We call ABC be the original triangle, and A'B'C' be the new triagngle, We have:

\(\dfrac{A'B'}{AB}=\dfrac{B'C'}{BC}=\dfrac{C'A'}{CA}=1.2\)

=> \(\Delta ABC\) is similar to \(\Delta A'B'C'\)

=> \(\dfrac{h}{h'}=1.2\) (with h, h' are the heights of them corespondsively)

=> \(\dfrac{area\left(ABC\right)}{area\left(A'B'C'\right)}=\dfrac{\dfrac{1}{2}a.h}{\dfrac{1}{2}a'h'}=1.2\times1.2=1.44\)

=> the area is increased by 44%.

There are 29−3 = 26 people in the room each one of whom speaks at least one of French or English. Since 11 speak French and 24 speak English, there are exactly 11+24−26 = 9 that speak both languages.

From 1 to 1000 there are \(\dfrac{995-5}{10}+1=100\) digits 5 at the unit digits (5, 15, 25, ..., 995).

From 50 to 59 there are 10 digits 5 at the tenth digits (50, 51, ..., 55).

=> From 1 to 1000 there are 10x10 = 100 digits 5 at the tenth digits (50, 51, ..., 55, 150, 151,..., 195, ....).

From 500 to 599 there are 100 digits 5 at the hundredth.

So, there are 100 + 100 + 100 = 300 digits 5 in the list of numbers 1, 2, 3, .., 1000. 

At the first time that the hands are perpendicular after 5:00, let θ in radians be the smallest angle that the minute hand makes with a line \(l\) passing through the center of the hand’s position at noon. Then the smallest angle between \(l\) and the hour hand is

   2π(5/12) + (θ/12).

It follows that

(5π/6) + (θ/12) − θ = π/2

so that

 π/3 = 11θ/12.

Hence, θ = 4π/11. The number of hours elapsed is (4π/11)/(2π) = 2/11.