# Finding the length of a side and comparing two areas from a given figure

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.

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

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 $$6$$.

• 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? – rsadhvika Jul 17 at 13:24
• @rsadhvika:$\;$Since $AE:EC=3:5$, it follows that $$AE:AC=AE:(AE+EC)=3:(3+5)=3:8$$ – quasi Jul 17 at 13:29
• 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 :) – rsadhvika Jul 17 at 13:31

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

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.