The following figure came in two national-exam paper (different question in each paper).

enter image description here

First Question:

What is the length of $AE$?


A) $\frac{5}{4}$ cm

B) $\frac{\sqrt{41}}{5}$ cm

C) $\frac{7}{5}$ cm

D) $\frac{15}{8}$ cm

Second Question


First value: area of $\triangle ABC$

Second value: area of $\triangle CDE$


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.

  • $\begingroup$ $AB\perp BC$? $AC\perp BD$? $AB||DC$? $\endgroup$ – Michael Rozenberg Jul 17 at 13:03
  • $\begingroup$ @MichaelRozenberg Yes. $\endgroup$ – Hussain-Alqatari Jul 17 at 13:05
  • $\begingroup$ If so, this quadrilateral does not exist because we need $4^2=3\cdot5,$ which is wrong. $\endgroup$ – Michael Rozenberg Jul 17 at 13:07
  • $\begingroup$ @MichaelRozenberg $AB\perp BC$, $AB || DC$. But $AB$ is not $\perp BC$ $\endgroup$ – Hussain-Alqatari Jul 17 at 13:10
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    $\begingroup$ @MichaelRozenberg Yes - No - Yes. $\endgroup$ – Hussain-Alqatari Jul 17 at 13:19

enter image description here

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 $6$.

  • $\begingroup$ quasi your way looks very elegant. if I may ask how did you figure out $AE = \frac{3}{8}AC$ directly with out setting up a proportion ? Is there some nice trick here? $\endgroup$ – rsadhvika Jul 17 at 13:24
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    $\begingroup$ @rsadhvika:$\;$Since $AE:EC=3:5$, it follows that $$AE:AC=AE:(AE+EC)=3:(3+5)=3:8$$ $\endgroup$ – quasi Jul 17 at 13:29
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    $\begingroup$ Oh nice nice! With that in mind we can directly do: $AE = \frac{3}{8}AC$ and $EC = \frac{5}{8}AC$. I see now.. Thank you so much for explaining this you're awesome :) $\endgroup$ – rsadhvika Jul 17 at 13:31

For first question you may use similar triangles - $\triangle AEB$ and $\triangle CED$: $$\dfrac{3}{5} = \dfrac{x}{5-x}$$

enter image description here


For the first part use similarity of triangles to simplify computation.

You have similar triangles $ABE$ and $CDE$ with ratio of sides being $3$ to $5$

You also know $AC=5$ so you find $AE=5\times \frac {3}{8}=\frac {15}{8}$ which is choice $D$

Your have solved the second part correctly.


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