However, for the benefit of those poor Sec Three students who've yet to learn trigo at this stage, here's another method utilizing the ratios obtained in Parts 2 and 3:

From Part 3,

Let Δ ABE = 4 units^{2}

Let Δ CDE = 9 units^{2}

From Part 2,

⇒ Δ BCE = 2/3 x 9 = 6 units^{2}

Also, since Δ ABE and Δ ADE has the same base lengths (6cm and 9cm), and given they too share a common height `h` (similar to the case in Part 2), one can deduce that:

⇒ Δ ADE = 3/2 x 4 = 6 units^{2}

Looking at the diagram above,

哗……数学与 Ratios 真的好神奇啊！

]]>Area of △BCE = ½(6)(35-14)sin∠BEC = 63sin∠BEC cm²

Area of △ADE = ½(9)(14)sin∠DEA = 63sin∠DEA cm²

∴The ratio = 63sin∠BEC : 63sin∠DEA = 1:1. ]]>

In △ABE & △CDE,

∠ABE=∠CDE (alt ∠s, AB//CD)

∠EAB=∠ECD (alt ∠s, AB//CD)

∠BEA=∠DEC (vert opp ∠s)

∴△ABE ~ △CDE (AAA)

Great job in proving that △ABE & △CDE are *similar* using the all angles equal case (AAA - refer to this chart for the cases of similar triangles)! Note that you don't really need to show the workings for the proof if the question didn't explicitly ask for it in your exam.

However, you do need to convince yourself and state that △ABE & △CDE are *similar* first, and then obtain the ratios of the corresponding sides (see working below) in order to solve this part 😉

∴BE:DE = AE:CE (corr ∠s, ~△s)

6:9 = AE:(35-AE)

2(35-AE) = 3AE

AE = 14 cm.

2. Another diagram to brighten things up 🙂

The altitudes of △BCE & △CDE are the same and let this altitude be h; the bases are different.

∴The ratio = (6h/2): (9h/2) = 2:3.

Yes though △BCE & △CDE are NOWHERE near similar they both share a common height `h` when calculating their respective areas - which then makes it easy to cancel `h` out when their areas are expressed as a ratio.

3. ∵△ABE ~ △CDE (from 1.)∴the ratio: 62:92 = 4:9. 62:92??? Fatty finger typo again???

This part tests your understanding of **areas of similar figures** i.e.

So since it's already proven in part 1 that △ABE & △CDE are similar, you can simply get the ratio of their areas by squaring the ratio of any two corresponding lengths of the triangles in one step BAM!

P.S. Make sure you're clear which is the numerator/denominator though!

4. I can tell that this appears more difficult because students would more likely focus on the six corners given. However they may ignore the angles, in which the road to success lies. Stare at this: ∠BEC=∠DEA (vert opp ∠s)

Area of △BCE = ½(6)(35-14) = 63 cm2

Area of △ADE = ½(9)(14) = 63 cm2

∴The ratio = 63:63 = 1:1.

哗……好神奇。难怪数学这么有趣！