Question

réponse:
cd 2
10 le bloc de béton ① doit être concassé et recyclé pour fabriquer le bloc de béton ②, qui a la forme
d’un prisme droit à base trapézoïdale et dont la longueur est d’au plus 2 m.
est - il possible de respecter cette contrainte de longueur ?
bloc de béton ①
bloc de béton ②
45 cm
100°
80 cm
75 cm
36 cm
20 cm
15 cm
?
30 cm
réponse:

Explanation:

Step1: Calculate Volume of Bloc ①

The figure for Bloc ① seems to be a quadrilateral (maybe a kite or composed of triangles). Let's break it into two triangles. First triangle: base 45 cm, height 36 cm. Area $A_1 = \frac{1}{2} \times 45 \times 36 = 810$ $cm^2$. Second triangle: angle 100°, sides 45 cm and 80 cm? Wait, no, maybe the other part. Wait, maybe Bloc ① is a quadrilateral with area calculated as sum of two triangles. Wait, maybe the first triangle is with base 45, height 36, and the second triangle with sides 80 and 45? No, better: maybe Bloc ① is a parallelogram? No, the diagram shows a triangle with base 45, height 36, and another triangle with angle 100°, sides 80 and 45? Wait, no, let's re-express. Wait, the first part: triangle with base 45 cm, height 36 cm: area $A_1 = \frac{1}{2} \times 45 \times 36 = 810$ $cm^2$. The second triangle: using the formula for area with two sides and included angle: $A_2 = \frac{1}{2} \times 45 \times 80 \times \sin(100^\circ)$. $\sin(100^\circ) \approx 0.9848$, so $A_2 = \frac{1}{2} \times 45 \times 80 \times 0.9848 \approx 1772.64$ $cm^2$. Wait, no, maybe Bloc ① is a quadrilateral with length 75 cm? Wait, maybe I misread. Wait, Bloc ①: the sides are 45, 75, 80, and the height 36, angle 100°. Wait, perhaps Bloc ① is a trapezoid? No, better to calculate volume. Wait, Bloc ①: let's assume it's a prism? Wait, no, the problem is about recycling, so volume should be equal (since material is recycled, volume of Bloc ① = volume of Bloc ②). Wait, Bloc ② is a right prism with trapezoidal base. Volume of Bloc ② = Area of trapezoidal base × length (let's call length $L$). Area of trapezoidal base: $\frac{(20 + 30)}{2} \times 15 = \frac{50}{2} \times 15 = 375$ $cm^2$. So Volume of Bloc ② = $375 \times L$ $cm^3$. Now Volume of Bloc ①: let's calculate its area (assuming it's a 2D figure, then volume is area × length, but maybe Bloc ① has length 75 cm? Wait, the diagram shows Bloc ① with sides 45, 75, 80, and the height 36, angle 100°. Wait, maybe Bloc ① is a quadrilateral with area: first triangle (45, 36) and second triangle (80, 45, angle 100°). Wait, no, maybe Bloc ① is a parallelogram? No, let's recast. Wait, the first triangle: base 45, height 36: area 810. The second triangle: sides 80 and 45, included angle 100°: area $\frac{1}{2} \times 80 \times 45 \times \sin(100°) \approx \frac{1}{2} \times 80 \times 45 \times 0.9848 \approx 1772.64$. Then total area of Bloc ① (2D) is $810 + 1772.64 = 2582.64$ $cm^2$. Then volume of Bloc ① (if length is 75 cm, since the side is 75 cm) is $2582.64 \times 75 = 193698$ $cm^3$. Now Volume of Bloc ②: trapezoidal base area is 375 $cm^2$, length $L$ (in cm). So $375 \times L = 193698$. Then $L = \frac{193698}{375} \approx 516.53$ cm. Convert to meters: $516.53$ cm = $5.1653$ m. Wait, but the constraint is length at most 2 m (200 cm). Wait, that can't be. Wait, maybe I messed up the volume of Bloc ①. Wait, maybe Bloc ① is a triangle? No, the diagram shows a quadrilateral. Wait, maybe Bloc ① is a kite with diagonals? No, let's check again. Wait, the first part: triangle with base 45, height 36: area 810. The other part: a triangle with sides 80 and 75? No, the side is 75 cm. Wait, maybe Bloc ① is a parallelogram with base 75 and height? No, the angle is 100°. Wait, perhaps the correct way is: Bloc ① is a quadrilateral with area calculated as the sum of two triangles: one with base 45, height 36, and another with base 80, height? Wait, no, the problem is in French: "Le bloc de béton ① doit être concassé et recyclé pour fabriquer le bloc de béton ②, qui a la forme d’un prisme droit à base trapézoïdale et dont la longueur est d’au plus 2 m." So volume of ① = volume of ②. Let's recalculate Volume of ②: base trapezoid: (20 + 30)/2 * 15 = 375 cm². Let length be L (cm). Volume of ②: 375L cm³. Volume of ①: let's look at Bloc ①: the figure has a triangle with base 45, height 36, and another triangle with sides 80 and 45? Wait, no, maybe Bloc ① is a quadrilateral with area = (45×36)/2 + (80×45×sin(100°))/2? Wait, 45×36/2 = 810, 80×45×sin(100°)/2 ≈ (80×45×0.9848)/2 ≈ 1772.64, total area ≈ 2582.64 cm². Then if Bloc ① is a prism with length 75 cm (the side 75 cm), volume of ① is 2582.64×75 = 193698 cm³. Then Volume of ②: 375L = 193698 → L = 193698 / 375 ≈ 516.53 cm = 5.1653 m. But the constraint is length at most 2 m (200 cm). Wait, that's more than 2 m. But that can't be. Wait, maybe I made a mistake in the volume of Bloc ①. Wait, maybe Bloc ① is a triangle? No, the diagram shows a quadrilateral. Wait, maybe the first triangle is with base 45, height 36, and the second triangle is with base 80, height 36? No, the angle is 100°. Wait, perhaps the correct area of Bloc ① is (45 + 80)/2 × 36? No, that's a trapezoid. Wait, 45 and 80 as the two bases, height 36, angle 100°? No, the height is 36, so maybe it's a trapezoid with bases 45 and 80, height 36, and then length 75? Wait, volume would be area × length = ((45 + 80)/2 × 36) × 75. Let's calculate that: (125/2 × 36) × 75 = (2250) × 75 = 168750 cm³. Then Volume of ②: 375L = 168750 → L = 168750 / 375 = 450 cm = 4.5 m. Wait, still more than 2 m. Wait, maybe the first triangle is 45×36/2, and the second is 80×36/2? No, that would be (45 + 80)/2 × 36 = 2250, same as before. Wait, maybe the angle 100° is a red herring, or I misread the diagram. Wait, the diagram for Bloc ①: left side 45, height 36, right side 80, angle 100°, bottom side 75. So maybe it's a quadrilateral with area = area of triangle (45, 36) + area of triangle (80, 75, 100°)? No, using the formula for area with two sides and included angle: $\frac{1}{2}ab\sin\theta$. So for the second triangle: sides 80 and 75, angle 100°? Wait, 45 and 80? No, the left side is 45, right side 80, bottom 75. Maybe the correct area is $\frac{1}{2}×45×36 + \frac{1}{2}×80×75×\sin(100°)$? Let's calculate: first part: 810, second part: 0.5×80×75×0.9848 ≈ 0.5×6000×0.9848 ≈ 2954.4. Total area: 810 + 2954.4 = 3764.4 cm². Then volume of ①: 3764.4 × 75 (length) = 282330 cm³. Volume of ②: 375L = 282330 → L = 282330 / 375 = 752.88 cm = 7.5288 m. Still more than 2 m. Wait, this can't be. Maybe I messed up the volume of ②. Wait, the base is trapezoidal: 20 cm and 30 cm are the two parallel sides, height 15 cm. So area is (20 + 30)/2 × 15 = 375 cm². Correct. Then length of ② is volume / 375. Volume of ①: let's check the diagram again. Maybe Bloc ① is a triangle with base 45 + 80 = 125, height 36, and length 75? No, that's a prism. Wait, the problem says "bloc de béton", so it's a 3D block. So Bloc ①: maybe a prism with base a quadrilateral. The quadrilateral has area: one triangle (45, 36) and another triangle (80, 45) with angle 100°? No, the key is that when we calculate the volume, we find that the length of Bloc ② is more than 2 m? But the question is "Est-il possible de respecter cette contrainte de longueur?" (Is it possible to respect the length constraint of at most 2 m?). Wait, maybe my calculation is wrong. Let's do it again. Volume of Bloc ①: suppose it's a prism with base area A and length 75 cm. Base area A: from the diagram, the base is a quadrilateral with a triangle (45, 36) and a triangle (80, 45) with angle 100°. Wait, no, the first triangle: base 45, height 36: area = 0.5×45×36 = 810 cm². The second triangle: sides 45 and 80, angle 100°: area = 0.5×45×80×sin(100°) ≈ 0.5×45×80×0.9848 ≈ 1772.64 cm². Total base area A = 810 + 1772.64 = 2582.64 cm². Volume of Bloc ①: 2582.64 × 75 = 193698 cm³. Volume of Bloc ②: base trapezoid area 375 cm², length L: 375L = 193698 → L = 193698 / 375 ≈ 516.53 cm = 5.1653 m. Which is more than 2 m (200 cm). Wait, but maybe the length of Bloc ① is not 75 cm? Maybe the 75 cm is the height? No, the diagram shows 75 as the bottom side. Wait, maybe the problem is that I misinterpret the shape of Bloc ①. Maybe Bloc ① is a triangle with base 45 + 80 = 125, height 36, and length 75? No, that would be volume 0.5×125×36×75 = 168750 cm³, then L = 168750 / 375 = 450 cm = 4.5 m. Still more than 2 m. Wait, maybe the angle 100° is the angle between 45 and 80, so the area is 0.5×45×80×sin(100°) ≈ 1772.64, and the other part is 0.5×45×36 = 810, total area 2582.64, length 75, volume 193698, length of ② is ~5.16 m > 2 m. So the answer would be no? But that seems odd. Wait, maybe I made a mistake in the volume of Bloc ①. Let's check the units: 2 m = 200 cm. If L = 516 cm > 200 cm, then it's not possible. Wait, but maybe the length of Bloc ① is 45 cm? No, the diagram shows 75. Wait, maybe the problem is that Bloc ① is a parallelogram with base 75 and height 36, and another part? No, I think the key is that when we calculate the length of Bloc ②, it's more than 2 m, so the answer is no. But let's confirm.

Step2: Compare Length with 2 m

We found that the length of Bloc ② is approximately 516.53 cm (or 5.1653 m), which is greater than 2 m (200 cm). So it is not possible to respect the length constraint.

Answer:

No, it is not possible to respect the length constraint (the length of Bloc ② would be approximately 5.17 m, which is more than 2 m).