## Subject IV.

Given the cube ABCDEFGH, where $B G \cap F C=\{P\}$ and the area of triangle DBP is equal to $25 \sqrt{3} \mathrm{~cm}^{2}$.

a) Calculate the length of the edge of the cube.

b) Determine $\tan u^{0}$, where $u^{0}$ is the measure of the angle between the lines DP and EB.

c) Calculate the value of a trigonometric function for the measure of the angle determined by the planes (EDG) and (EBG).

## Note:

All subjects are mandatory.

Working time 3 hours.

Each subject is graded from 0 to 7 points.

Any other solution leading to the problem's solution is accepted.

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## GRADE 8



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## Subject IV.

![](https://cdn.mathpix.com/cropped/2024_06_07_725beade569d86b4c0a5g-4.jpg?height=821&width=857&top_left_y=378&top_left_x=691)

a) If we denote the edge of the cube with 2a, we get:

$\mathrm{BD}=2 \mathrm{a} \sqrt{2}$ (diagonal in the square $\mathrm{ABCD}$);

$B P=a \sqrt{2}$ (radius of the circumscribed circle of the square $\mathrm{FBCG}$);

$D P=a \sqrt{6}$ (Pythagorean theorem in $\triangle D P T$, where T is the midpoint of $[B C]$).

We obtain $\triangle B P D$ right-angled at $\mathrm{P}$ (converse of the Pythagorean theorem).......1p

Equating the area: $5 \sqrt{3}=\frac{B P \cdot D P}{2}$, we get $\mathrm{a}=5$. Therefore, the edge of the cube is $10 \mathrm{~cm} \ldots \ldots \ldots \ldots .1 \mathrm{p}$

b) The lines DP and EB are skew. Draw a line through P parallel to EB.

[O'P] is the midline in $\triangle E G B$, $\mathrm{O}^{\prime}$ is the center of the square $\mathrm{EFGH}$.

$u^{0}=m(\widehat{D ; E} B)=m\left(D \widehat{P ; O^{\prime} P}\right)=m\left(\widehat{D P O^{\prime}}\right)$

$D P=5 \sqrt{6} \mathrm{~cm} ; \mathrm{O}^{\prime} \mathrm{P}=5 \sqrt{2} \mathrm{~cm} ; \mathrm{DO}^{\prime}=5 \sqrt{6} \mathrm{~cm}$ $.1 \mathrm{p}$

We will obtain $\operatorname{tg} \mathrm{u}^{0}=\sqrt{11}$. $1 p$

c)(EDG) $\cap(E D B)=E G$

$\triangle E B G$ equilateral $\Rightarrow \mathrm{BO}^{\prime} \perp \mathrm{EG}, \mathrm{BO}^{\prime}=5 \sqrt{6} \mathrm{~cm}$.

$\triangle E D G$ equilateral $\Rightarrow \mathrm{DO}^{\prime} \perp \mathrm{EG}, \mathrm{DO}^{\prime}=5 \sqrt{6} \mathrm{~cm}$.

$m\left[(E D \widehat{G) ;(E B G)}]=m\left(D \widehat{O^{\prime} ; B} O^{\prime}\right)=m\left(\widehat{D O^{\prime} B}\right)=a^{0}\right.$ $.1 p$

We obtain $\cos a^{0}=\frac{1}{3}$ or $\sin a^{0}=\frac{2 \sqrt{2}}{3}$.