0
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ Now, I have edited the question. Sorry for that major mistake @expiTTp1z0 $\endgroup$
    – onlymaths
    Jun 24 '17 at 17:04
  • $\begingroup$ Maybe we need to find $\frac{P_{\Delta CDF}}{P_{\Delta ABC}}$? $\endgroup$ Jun 24 '17 at 17:07
  • $\begingroup$ Yes, you are right @MichaelRozenberg $\endgroup$
    – onlymaths
    Jun 24 '17 at 17:09
  • $\begingroup$ $BE:EA=(2\cdot36):60=6:5$. So $BD:DC=6:5$ and hence $AF:FC=6:5$. The ratio of the perimeter of $\triangle CDF$ to the ratio of the perimeter of $\triangle ABC$ is $5:11$ $\endgroup$
    – CY Aries
    Jun 24 '17 at 17:11
  • $\begingroup$ How did you find BE:EA and BD:DC? @CYAries Please elaborate more $\endgroup$
    – onlymaths
    Jun 24 '17 at 17:13
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

$$S_{\Delta EDA}=30,$$ 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: $$\frac{P_{\Delta DFC}}{P_{\Delta ABC}}=\sqrt{\frac{25}{36+60+25}}=\frac{5}{11}$$

$\endgroup$
2
  • $\begingroup$ How did you find BE:EA=6:5? $\endgroup$
    – onlymaths
    Jun 24 '17 at 17:33
  • $\begingroup$ @onlymaths Because $\frac{BE}{EA}=\frac{S_{\Delta BED}}{S_{\Delta AED}}=\frac{36}{\frac{1}{2}\cdot60}=\frac{6}{5}$. $\endgroup$ Jun 24 '17 at 20:34
0
$\begingroup$

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 $5:11$.

$\endgroup$
4
  • $\begingroup$ How triangle ADE and triangle will have the same height? @CYAries $\endgroup$
    – onlymaths
    Jun 24 '17 at 17:24
  • $\begingroup$ For $\triangle ADE$, if you take $AE$ as the base, the height is the perpendicular from $D$ to $AE$. For $\triangle BDE$, if you take $BE$ as the base, the height is the perpendicular from $D$ to $BE$. Since $A$,$E$ and $B$ are collinear, the two triangles share the same height. $\endgroup$
    – CY Aries
    Jun 24 '17 at 17:27
  • $\begingroup$ But it is not given that D will make an angle of 90 degree? How did you reach this conclusion? I am really sorry for bothering you so much but please help me out $\endgroup$
    – onlymaths
    Jun 24 '17 at 17:28
  • $\begingroup$ You don't need $DE\perp AB$. Drop a perpendicular from $D$ to $AB$ and meet $AB$ at $K$. $DK$ is the common height. $K$ is not necessarily the same point as $E$. $\endgroup$
    – CY Aries
    Jun 24 '17 at 17:31
0
$\begingroup$

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

Triangle 1

$$\therefore \ \text{Area of }\triangle AEF = \text{Area of }\triangle DEF = \frac12 \text{ Area of $AEDF$ parallelogram} = \frac12 \cdot 60 = 30$$ 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.