Find the cracked area

Question

You have a square of side 1m, consider that both triangles have vertices at the midpoint of the square, find the cracked area

$S=2 S\triangle - S_ \boxed{LMNP}=2. \frac{1}{2}.\frac{1}{2}.1 - S \boxed{}=\frac{1}{2}- S_\boxed{LMNP}$

Need to find the area $S_\boxed{LMNP}??$

Answer 1

Here's your diagram, with several perpendicular lines, points and line lengths added:

With $\lvert LS\rvert = x$, since $\measuredangle LES = \measuredangle LEQ = \frac{\pi}{4}$, we have $\lvert LQ\rvert = x$. Also, $\triangle ILQ \sim \triangle IJE$, with $\lvert IE\rvert = 2\lvert EJ\rvert$, so $\lvert IQ\rvert = 2\lvert LQ| = 2x$. Thus, $\lvert QE\rvert = 1 - 2x$. Since $\lvert LQ\rvert = \lvert QE\rvert = x$, then

$$x = 1 - 2x \;\;\to\;\; x = \frac{1}{3} \tag{1}\label{eq1A}$$

Similarly, and by symmetry, we get

$$z = \frac{1}{3} \tag{2}\label{eq2A}$$

Using that $\triangle EPT \sim \triangle EKF$, then $\lvert ET\rvert = 2\lvert PT\rvert = 2y$. Also, from $\lvert PR\rvert = \lvert ET\rvert = 2y$ and $\triangle IPR \sim \triangle IJE$, then $\lvert IR\rvert = 2\lvert PR\rvert = 4y$. Thus,

$$1 = \lvert IE| = \lvert IR| + \lvert RE| = 4y + y = 5y \;\;\to\;\; y = \frac{1}{5} \tag{3}\label{eq3A}$$

From \eqref{eq1A}, since $\lvert EJ\rvert = \frac{1}{2}$, using the half base times height formula for triangle area results in

$$S_{ELJ} = \frac{1}{12} \tag{4}\label{eq4A}$$

Similarly, from \eqref{eq3A}, we have

$$S_{EPJ} = \frac{1}{20} \tag{5}\label{eq5A}$$

Thus, by the difference in triangle areas, we get

$$S_{EPL} = S_{ELJ} - S_{EPJ} = \frac{1}{12} - \frac{1}{20} = \frac{5}{60} - \frac{3}{60} = \frac{1}{30} \tag{6}\label{eq6A}$$

With $\lvert EF\rvert = 1$, using \eqref{eq2A} and the standard triangle area formula again, gives that

$$S_{ENF} = \frac{1}{6} \tag{7}\label{eq7A}$$

Using the difference of areas results in

$$\begin{equation}\begin{aligned} S_{LMNP} & = S_{EMF} - S_{ENF} - S_{EPL} \\ & = \frac{1}{4} - \frac{1}{6} - \frac{1}{30} \\ & = \frac{15 - 10 - 2}{60} \\ & = \frac{1}{20} \end{aligned}\end{equation}\tag{8}\label{eq8A}$$

Answer 2

• $\;L$ is the centroid of $\triangle EFI$ and $N$ the centroid of $\triangle EFG$ so $ML = MN = \dfrac{EM}{3} = \dfrac{\sqrt{2}}{6}$ and $LJ = NK = \dfrac{EK}{3} = \dfrac{\sqrt{5}}{6}$.

• $\triangle EPJ \sim \triangle EFK$ so $EP = \dfrac{EJ \cdot EF}{EK} = \dfrac{\sqrt{5}}{5}$ then $PN = EK - EP - NK = \dfrac{2 \sqrt{5}}{15}$.

• $\triangle LPN$ and $\triangle LMN$ are right triangles so $LP^2 = LN^2 - PN^2 = 2\, LM^2 - PN^2 = \dfrac{1}{45}.$

The area in question is the sum of the areas of the two right triangles $\,\dfrac{LM^2 + LP \cdot PN}{2} = \dfrac{1}{20}$.

Answer 3

Here we can do analytic geometry. Define $E$ to be the origin. Define the x-axis to be rightward and the y-axis to be upward in the provided diagram.

The equations for each of the drawn segments can be as follows: $$\begin{align*} \overline{EG}&: y=x \\ \overline{EK}&: y=\frac{x}{2} \\ \overline{IF}&: y=1-x \\ \overline{IJ}&: y=1-2x \end{align*}$$ We can solve the system of equations for intersecting lines. These will give us the coordinates of the points $L, M, N,$ and $P$.

Point $L$ is the intersection of lines $EG$ and $IJ$. Solving the system, we get that $L$'s coordinates are $\left(\dfrac{1}{3}, \dfrac{1}{3}\right).$

Point $M$ is the intersection of lines $EG$ and $IF$. Solving the system, we get that $M$'s coordinates are $\left(\dfrac{1}{2}, \dfrac{1}{2}\right).$

Point $N$ is the intersection of lines $EK$ and $IF$. Solving the system, we get that $N$'s coordinates are $\left(\dfrac{2}{3}, \dfrac{1}{3}\right).$

Point $P$ is the intersection of lines $EK$ and $IJ$. Solving the system, we get that $P$'s coordinates are $\left(\dfrac{2}{5}, \dfrac{1}{5}\right).$

Notice the y-coordinates of $L$ and $N$ are the same. Therefore, the distance between them is $\left|\frac{2}{3} - \frac{1}{3}\right| = \frac{1}{3}.$ We can then find out that the length of the altitude of $M$ to $\overline{LN}$ would be $\left|\frac{1}{2} - \frac{1}{3}\right| = \frac{1}{6}.$ Similarly, the length of the height from $P$ to $LN$ would be $\left|\frac{1}{5} - \frac{1}{3}\right| = \frac{2}{15}.$

Adding the length of the heights together, we get $\frac{3}{10}$. Multiply this by the length of $LN$ and dividing by $2$, we get that the answer is $\boxed{\frac{1}{20}}$

Answer 4

Locate all midpoints of the given square and connect them as shown.

All color-shaded quadrilaterals are equal in area.

The required = $0.25$ of the square $Pqrs$ and the area of the latter can be found in many texts.

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Square $Pqrs$ is surrounded by $4$ congruent triangles (e.g. $\triangle IEP$).

By ‘midpoint theorem’, $Is = sP = x$, say. Note that $[\triangle IEP] = 0.5(x)(2x) = x^2 = [\square Pqrs].$

Therefore, the required area $= (1/4)[\square Pqrs] = (1/4)(1/5)[\square EFGI] = 1/20.$

Answer 5

Here's an easy, purely geometric approach:

1.) Extend a perpendicular from $N$ to line $IE$ at $O$. It is obvious (and easy to show) that $MN=ML$, therefore line $NO$ passes through $L$ as well. Notice that $\angle NIO=45^\circ$ and, therefore $NO=OI$. If we set $LO=x$, then due to similarity of triangles ($\triangle IOL$ is similar to $\triangle IEJ$), $IO=2x$. This means that $LO=NL=x$.

2.)Since $\angle OEL=45^\circ$ as well, this means $LO=NL=OE=x$. We know that $IE=1$ and also that $IE=3x$, therefore $x=\frac{1}{3}$. Notice that $\triangle EPJ$ is similar to $\triangle IEJ$, since $NO \parallel EF$, this implies that $\triangle LPN$ is similar to $\triangle EPJ$.

3.) Since we know $x=\frac{1}{3}$, we can compute, via similarity of triangles, that $PL=\frac{1}{3\sqrt{5}}$ and $PN=\frac{2}{3\sqrt{5}}$. Since we know that $\angle LMN=90^\circ$ and that $MN=ML$, we find that $MN=\frac{1}{3\sqrt{2}}$. With this information, we can easily find the areas of $\triangle LPN$ and $\triangle MLN$. We get $\frac{1}{45}$ and $\frac{1}{36}$. Adding that, we get the area of $\quad PLMN=\frac{1}{20}$

Answer 6

Set $$ E=(0,0),F=(1,0),G=(1,1),I=(0,1) $$ in the following diagram

It is given that $$ J=\left(\frac12,0\right),K=\left(1,\frac12\right) $$ Solving the equations in red simultaneously, we get $$ L=\left(\frac13,\frac13\right),M=\left(\frac12,\frac12\right),N=\left(\frac23,\frac13\right),P=\left(\frac25,\frac15\right) $$ Then, using the cross product restricted to $2$ dimensions: $(x_1,y_1)\times(x_2,y_2)=x_1y_2-x_2y_1$ (which is the dot product of the unit normal with the $3$-dimensional cross product), we get the area to be $$ \begin{align} |NMLP| &=\frac12(N\times M\color{#A0A0A0}{+\overbrace{M\times L}^{M\parallel L}}+L\times P\color{#A0A0A0}{+\overbrace{P\times N}^{P\parallel N}})\\ &=\frac12\left(\left(\frac23,\frac13\right)\times\left(\frac12,\frac12\right)\color{#A0A0A0}{+0}+\left(\frac13,\frac13\right)\times\left(\frac25,\frac15\right)\color{#A0A0A0}{+0}\right)\\ &=\frac12\left(\frac16-\frac1{15}\right)\\ &=\frac1{20} \end{align} $$