Let ∆ABC be any triangle, where points $E, F, C$ are collinear, $D, E, B$ are collinear, $A, D, F$ are collinear and such that $| CF | = | F E |, | DE | = | EB |, | AD | = | DF |$, as in the following figure:
If $| AC | = | CB |=\sqrt{14}$ dm and Area (∆DEF) = 0.01 $m^{2}$ , so calculate $| AB |$ in cm.
Hello, can you help me with this Geometry problem? Thanks.
On
It's better late than never. Vectors approach.
The only benefit of vectors approach I see -- it's fast and requires no thinking. "0. Choose convinient basis. 1. Express all what's given and what's needed via basis vectors. 2 Solve system of equations. 3. Don't think".
However, the choosing basis step requires some thinking or some experience.
Here we could choose the basis of $\overrightarrow{AB},\,
\overrightarrow{AC}$ but then expressing the area of $\triangle DEF$ would be troblesome. Instead, we choose $e:=\overrightarrow{DE},\,f:=\overrightarrow{DF}$ as basis vectors (and $D$ as the origin, meaning $X=x\mathbf{e}+y\mathbf{f}$ reads $\overrightarrow{DX}=x\mathbf{e}+y\mathbf{f}$).
I think you don't mind if I borrow your image
$$A=-f,\,C=f+\overrightarrow{EF}=2f-e,\,B=2e$$ We need $|AC|$ and $|CB|$: $$\overrightarrow{CA}=A-C=-3f+e,\, \overrightarrow{CB}=B-C=3e-2f$$ We result in a linear transform: $$\begin{pmatrix} \overrightarrow{CA}\\ \overrightarrow{CB} \end{pmatrix}= \begin{pmatrix} 1&-3\\ 3&-2 \end{pmatrix} \begin{pmatrix} \mathbf{e}\\ \mathbf{f} \end{pmatrix}$$ It's known that a linear transform $x\to Ax$ scales areas by $|A|$ times, thus $$\mathrm{Area}_{ABC}=\operatorname{abs}\left(\left| \begin{array}{cc}1&-3\\ 3&-2\end{array}\right|\right)\mathrm{Area}_{DEF}=7\mathrm{Area}_{DEF}$$ then we can $$\mathrm{Area}_{ABC}=\frac12\cdot|CA|\cdot|CB|\cdot\sin\angle ACB$$ hence we find $$\sin\angle ACB=1$$ and then apply cosine rule to find $AB$. However, it requires some thinking.
Let's go the straight way: $$\cos \angle EDF=\frac{e\cdot f}{|e|\cdot|f|},$$ $$\cos^2 \angle EDF=\frac{(e\cdot f)^2}{e^2\cdot f^2},$$ $$\sin^2 \angle EDF=1-\frac{(e\cdot f)^2}{e^2\cdot f^2},$$ but $$\left(\mathrm{Area}_{DEF}\right)^2= \frac14 e^2 f^2 \sin^2 \angle EDF= \frac14\left(e^2\cdot f^2-(e\cdot f)^2\right)$$ Hence we have (I'll have decimeter as an unit): $$\begin{cases} CA^2=9f^2-6ef+e^2=14\\ CB^2=9e^2-12ef+4f^2=14\\ \frac14\left(e^2\cdot f^2-(e\cdot f)^2\right)=1 \end{cases}$$ solvig this for $e^2,\,f^2,\,ef$ will result in a quadratic equation. However let's feed it to wolframalpha, obtaining $$ \begin{cases} e^2 = \frac{26}{7}\\ f^2 = \frac{20}{7}\\ ef = \frac{18}{7} \end{cases} $$ and $$AB^2=(B-A)^2=(2e+f)^2=4e^2+4ef+f^2=28.$$
First, you should notice that:
$$P_{ABD}=2P_{DEF}\tag{1}$$
Why is that so? Obviously $AD=DF$. But the height of triangle $ABD$ drawn from point $B$ perpendicular to the side $AD$ is twice the hight of triangle $DEF$ drawn from point $E$ perpendicular to the side $DF$ (because $DB=2DE$). Strict proof is elementary and I leave it up to you.
In the same way you can show that:
$$P_{AFC}=P_{AEB}=2P_{DEF}\tag{2}$$
From (1) and (2) it follows that the area $P$ of triangle $ABC$ is:
$$P=7P_{DEF}=0.07\text{m}^2=7\text{dm}^2$$
Triangle $ABC$ is isosceles with $AC=b=\sqrt{14}\text{dm}$. Denote with $a$ the length of basis $AB$ and with $h$ the corresponding height drawn from point $C$ perpendicular to segment $AB$.
We have the following equations:
$$\frac{ah}2=P\tag{3}$$
$$\left(\frac a2\right)^2+h^2=AC^2=b^2\tag{4}$$
From (3):
$$h=\frac{2P}a\tag{5}$$
Replace that into (4):
$$\frac{a^2}{4}+\frac{4P^2}{a^2}=b^2$$
$$a^4-4b^2a^2+16P^2=0$$
$$a^2=\frac{4b^2\pm\sqrt{16b^4-64P^2}}{2}$$
$$a^2=2b^2\pm2\sqrt{b^4-4P^2}$$
$$a^2=2\cdot14\pm\sqrt{14^2-4\cdot7^2}=28$$
$$a=2\sqrt{7}\text{dm}$$