The following figure came in two national-exam paper (different question in each paper).
First Question:
What is the length of $AE$?
Choices:
A) $\frac{5}{4}$ cm
B) $\frac{\sqrt{41}}{5}$ cm
C) $\frac{7}{5}$ cm
D) $\frac{15}{8}$ cm
Second Question
Compare:
First value: area of $\triangle ABC$
Second value: area of $\triangle CDE$
Choices:
A) First value > Second value
B) First value < Second value
C) First value = Second value
D) Given information is not enough
My Attempt:
Finding the equation of $AC$ and $BD$ in order to find the coordinates of E:
We can assume that $C$ is the origin $(0,0)$ and $A$ is $(3,4)$, then the equation of $AC$ is $y=\frac{4}{3}x$, and the equation of $BD$ is $y=-\frac{4}{5}x+4$.
Next, solve the system of equations for $x$ and $y$, we end up with the coordinates of $E(\frac{15}{8},\frac{5}{2})$.
To solve the first question, we can use the distance between two points formula,
therefore $AE=\sqrt{(3-\frac{15}{8})^2+(4-\frac{5}{2})^2}=\frac{15}{8}$ cm.
Hence D is the correct choice.
To solve the second question, we can use the area of triangle formula,
area of $\triangle ABC=\frac{1}{2}\times 3\times 4=6$ cm$^2$.
area of $\triangle CDE=\frac{1}{2}\times 5\times \frac{5}{2}=6.25$ cm$^2$.
Hence B is the correct choice.
In the exam, calculators are not allowed. It is simple to determine to coordinates of $E$ without a calculator, but it will take time!
For the first question, I think there is a good way to solve, maybe $BD$ divides $AC$ into two parts ($AE$ and $CE$) in a ratio, this ratio can be calculated somehow.
For the second question, I think also a good way to think, like moving $D$ will make $\triangle CDE$ bigger or smaller, while $\triangle ABC$ will be unchanged. But this has a limitation since the areas $6$ and $6.25$ cm$^2$ are close, it will not be obvious to us.
If we have $75$ seconds (on average) to solve each of these questions, then how can we solve them? [remember that in some questions in the national exam, we do not need to be accurate $100$%].
Any help will be appreciated. Thanks.
Triangles $ABE$ and $CDE$ are similar, with ratio $3:5$.
Hence $$AE=\frac{3}{8}AC=\frac{3}{8}\cdot 5=\frac{15}{8}$$
By the same similarity, if $h$ is the height of triangle $CDE$ from $E$ to $CD$, then $$h=\frac{5}{8}BC=\frac{5}{8}\cdot 4=\frac{5}{2}$$ hence the area of triangle $CDE$ is $$\frac{1}{2}\cdot CD\cdot h=\frac{1}{2}\cdot 5\cdot \frac{5}{2}=\frac{25}{4}=6.25$$ whereas the area of triangle $ABC$ is $6$.