Question based on geometry(area)

Question

In triangle ABC, three points D, E and F are chosen on sides BC, AB and AC respectively such that DE is parallel to AC and DF is parallel to AB. If the area of triangle BDE is 36 and that of quadrilateral AEDF is 60, then find the ratio of the perimeter of triangle CDE to that of triangle ABC. Only I could think of is that quadrilateral AEDF is a parallelogram. I will be really grateful if anyone could help me with this question.

Answer 1

Let $h$ be the height of trapazoid $DFAB$ which will also be the height of triangles $BDE$ and $AFE$. Then,

$$\frac{1}{2} \times h \times (DF + AB) = 96$$

$$\frac{1}{2} \times h \times BE = 36$$

$$\frac{1}{2} \times h \times AE = 30$$

$$\frac{1}{2} \times h \times (AE + BE) = \frac{1}{2} \times h \times AB = 66$$

$$\implies \frac{DF}{AB} = \frac{5}{11}$$

Since $\Delta CDF \sim \Delta ABC \implies \frac{P_{\Delta CDF}}{P_{\Delta ABC}} = \frac{DF}{AB} = \frac{5}{11}$

Note: for similar triangles the ratio of length of similar edges are equal.

Answer 2

$$S_{\Delta EDA}=30,$$ which gives $$\frac{BE}{EA}=\frac{36}{30}=\frac{6}{5}=\frac{BD}{DC}=\frac{AF}{FC}.$$ Thus, $$S_{\Delta DFC}=\frac{5}{6}\cdot30=25$$ and we get the answer: $$\frac{P_{\Delta DFC}}{P_{\Delta ABC}}=\sqrt{\frac{25}{36+60+25}}=\frac{5}{11}$$

Answer 3

The area of $\triangle AED$ is $60\div2=30$.

Since $\triangle ADE$ and $\triangle BDE$ have the same height from $D$,

$BE:EA=\text{area of }\triangle BDE:\text{area of }\triangle ADE=36:30=6:5$

By intercept theorem, $BD:DC=BE:EA=6:5$

So, $DC:BC=5:11$.

As $\triangle CDF\sim\triangle CBA$, the ratio of the perimeters is $5:11$.

Answer 4

The quadrilateral $AEDF$ is a parallelogram, and therefore $AE=DF=a$, $AF=DE=b$, and $\triangle AEF \equiv \triangle DEF$.

$$\therefore \ \text{Area of }\triangle AEF = \text{Area of }\triangle DEF = \frac12 \text{ Area of $AEDF$ parallelogram} = \frac12 \cdot 60 = 30$$ If $h$ is the height of parallelogram $AEDF$, $h$ will also be the height of $\triangle BDE$ and $\triangle AEF$. $$\therefore \ \text{Area of }\triangle AEF = \frac12 \cdot ah = 30 \qquad \text{and} \qquad \text{Area of }\triangle BDE = \frac12 \cdot ch = 36 $$ Thus, $\frac{c}{a} = \frac{36}{30}= \frac{6}{5}$. Therefore, $c=\frac{6}{5}a$.

As shown in the diagram (use your knowledge on parallelograms), you may find $\triangle BDE$ and $\triangle CDF$ are similar. Thus, their corresponding side lengths have the same ratio:

$$ \frac{BE}{DF} = \frac{BD}{DC} = \frac{DE}{CF} = \frac{c}{a} = \frac{d}{e} = \frac{b}{f} = \frac{6}{5}$$

Thus, $\frac{d}{e} = \frac{6}{5}$ and $\frac{b}{f} = \frac{6}{5}$ as well.

If perimeters of $\triangle ABC$ and $\triangle CDF$ are $P_{\triangle ABC}$ and $P_{\triangle CDF}$, respectively:

$$P_{\triangle ABC} = (a + c) + (d+e) + (f+b) = \left(a + \frac{6}{5}a\right) + \left(e + \frac{6}{5}e\right) + \left(f + \frac{6}{5}f\right)\\ = \frac{11}{5}\left(a + e + f\right) = \frac{11}{5}P_{\triangle CDF} $$

Simillarly, if perimeters of $\triangle BDE$ and $\triangle CDF$ are $P_{\triangle BDE}$ and $P_{\triangle CDF}$, respectively: $P_{\triangle BDE}= c+d+b = \frac{6}{5}a + \frac{6}{5}e + \frac{6}{5}f = \frac{6}{5}\left(a+e+f\right) = \frac{6}{5}P_{\triangle CDF} $

Yet, the qusetion is asling the ratio of $P_{\triangle ABC}$ and $P_{\triangle CDE}$, which cannot be found using the given data.