Question

1. (45) find the volume of this right solid. dimensions are in meters. first find the area of a triangular end.
2. (49) find the surface area of the right solid.

Explanation:

Step1: Find the base of the triangle

Assume the triangular base is a right triangle? Wait, no, the height of the triangle is 12, one side is 13, so by Pythagoras, the base \( b \) of the triangle: \( b=\sqrt{13^{2}-12^{2}}=\sqrt{169 - 144}=\sqrt{25} = 5\)? Wait, no, maybe the length of the base of the triangle (the side perpendicular to the height 12) – wait, maybe the triangular end is a triangle with height 12, and the other side? Wait, the solid is a right prism (since it's a right solid) with triangular base. Wait, the length of the prism (the distance along the direction perpendicular to the triangular end) is 11 (from the diagram, the 11 is the length of the prism). Wait, first, for problem 7: volume of a right prism is \( V= \text{Area of base} \times \text{length of prism} \).

First, find the area of the triangular end. Let's assume the triangular end is a triangle with height 12 (the vertical height) and base \( b \). Wait, maybe the triangle is a right triangle? Wait, the side is 13, height 12, so the base of the triangle (the horizontal side) is \( \sqrt{13^{2}-12^{2}} = 5 \)? Wait, no, maybe the base of the triangle is, say, let's check: if the triangle has height 12, and the hypotenuse 13, then the base is 5 (since \( 5 - 12 - 13 \) triangle). Then the area of the triangular end is \( \frac{1}{2} \times \text{base} \times \text{height} \). Wait, but what's the base? Wait, maybe the base of the triangle is, let's see, the length of the prism is 11. Wait, maybe the triangular end is a triangle with base, say, let's re-examine. Wait, the diagram shows a triangular prism: the triangular face has height 12, one side 13, and the length of the prism (the distance between the two triangular faces) is 11. Also, maybe the base of the triangle (the side of the triangle that's the base) is, let's compute the base of the triangle: using Pythagoras, \( b = \sqrt{13^2 - 12^2} = 5 \)? Wait, no, that would be if it's a right triangle with legs 12 and \( b \), hypotenuse 13. So \( b = 5 \). Then the area of the triangular end is \( \frac{1}{2} \times 5 \times 12 = 30 \) square meters. Then the volume of the right solid (prism) is \( \text{Area of base} \times \text{length of prism} = 30 \times 11 = 330 \) cubic meters. Wait, but maybe I made a mistake. Wait, maybe the base of the triangle is not 5. Wait, maybe the triangle is not a right triangle? Wait, no, the dashed line is the height (12) of the triangle, so the triangle is a triangle with height 12, and the base is, say, let's see, the other side? Wait, maybe the base of the triangle is, for example, if the triangle has height 12, and the length of the base (the side opposite the height) is, let's check the diagram again. Wait, the problem says "First find the area of a triangular end." So let's do that.

Step1: Find the base of the triangular end

The triangular end is a triangle with height 12 m and hypotenuse 13 m (the side). Using the Pythagorean theorem, the base \( b \) of the right triangle is:
\( b = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5 \) m.

Step2: Calculate the area of the triangular end

The area \( A \) of a triangle is \( \frac{1}{2} \times \text{base} \times \text{height} \). Substituting \( b = 5 \) m and height \( h = 12 \) m:
\( A = \frac{1}{2} \times 5 \times 12 = 30 \) m².

Step3: Calculate the volume of the right solid (prism)

The volume \( V \) of a right prism is \( V = \text{Area of base} \times \text{length of prism} \). The length of the prism (the distance between the two triangular faces) is 11 m. So:
\( V = 30 \times 11 = 330 \) m³.

For problem 8: Surface area of the right solid (triangular prism). The surface area is the sum of the areas of all faces. A triangular prism has two triangular faces and three rectangular faces.

Step1: Area of the two triangular faces

We already know the area of one triangular face is 30 m², so two of them is \( 2 \times 30 = 60 \) m².

Step2: Area of the three rectangular faces

- First rectangular face: base of triangle (5 m) × length of prism (11 m): \( 5 \times 11 = 55 \) m².
- Second rectangular face: height of triangle (12 m) × length of prism (11 m): \( 12 \times 11 = 132 \) m².
- Third rectangular face: hypotenuse of triangle (13 m) × length of prism (11 m): \( 13 \times 11 = 143 \) m².

Step3: Sum all areas

Total surface area \( = 60 + 55 + 132 + 143 = 60 + 55 = 115; 115 + 132 = 247; 247 + 143 = 390 \) m².

Answer (for problem 7):

The volume of the right solid is \( \boxed{330} \) cubic meters.

Answer (for problem 8):

The surface area of the right solid is \( \boxed{390} \) square meters.

Answer:

(for problem 7):
The volume of the right solid is \( \boxed{330} \) cubic meters.