A triangular prism ABC-DEF has an isosceles triangle as its base with $AB = AC = 9$ cm, $BC = 6$ cm, and $AD = 3$ cm. The prism is cut by a plane passing through points B, C, and D, resulting in two solids, "A" and "I". (1) Find the ratio of the volumes of solids "A" and "I". (2) By how many $cm^2$ is the surface area of one of the solids "A" and "I" larger than the other?

1. Problem Description

A triangular prism ABC-DEF has an isosceles triangle as its base with $AB = AC = 9$ cm, $BC = 6$ cm, and $AD = 3$ cm. The prism is cut by a plane passing through points B, C, and D, resulting in two solids, "A" and "I".

(1) Find the ratio of the volumes of solids "A" and "I".

(2) By how many $cm^2$ is the surface area of one of the solids "A" and "I" larger than the other?

2. Solution Steps

(1) Volume Ratio:

The volume of the original prism is given by:

$V_{prism} = Area_{triangle} \times height$

The area of triangle ABC is:

$Area_{ABC} = \frac{1}{2} \times BC \times h$

where $h$ is the height of the triangle ABC from vertex A to BC.

Let $M$ be the midpoint of BC. Then $BM = MC = 3$.

$AM = \sqrt{AB^2 - BM^2} = \sqrt{9^2 - 3^2} = \sqrt{81 - 9} = \sqrt{72} = 6\sqrt{2}$.

$Area_{ABC} = \frac{1}{2} \times 6 \times 6\sqrt{2} = 18\sqrt{2}$.

$V_{prism} = 18\sqrt{2} \times 3 = 54\sqrt{2}$.

Solid "A" is the solid ABCD. It is a triangular prism with the base being the triangle ABC, and the height along AD is

3. Solid "I" is the solid BCFED.

Consider the volume of solid "I". This solid is the triangular prism ABC-DEF minus the solid ABCD (solid "A"). So, $V_I = V_{prism} - V_A$.

To compute the volume of solid "A", we can view it as a pyramid DBC, A. Let's consider the triangle BCD. The area of the triangle BCD is half of the area of the rectangle BCFE. The volume of Solid A (ABCD) = Volume of pyramid A-BCD can be computed as (1/3) * (Area BCD) * height (AM).

Also, note that the volume of solid A can be regarded as the prism ABC-DEF subtracting solid I. Consider point D to be at height 0 and point A is at height

3. So that the plane passing through DBC cut off a plane.

Consider the volume of solid "A" as $\frac{1}{2} V_{prism}$.

So, $V_A = \frac{1}{2} V_{prism} = \frac{1}{2} \times 54\sqrt{2} = 27\sqrt{2}$.

Therefore, $V_I = V_{prism} - V_A = 54\sqrt{2} - 27\sqrt{2} = 27\sqrt{2}$.

$V_A : V_I = 27\sqrt{2} : 27\sqrt{2} = 1:1$.

(2) Surface Area:

Surface area of prism = $2*Area_{triangle} + perimeter * height = 2(18\sqrt{2}) + (9+9+6)*3 = 36\sqrt{2} + 24*3 = 36\sqrt{2} + 72$.

The cut introduces the surface area of triangle BCD on both A and I.

Solid A: Surface area = $AB \cdot AD + AC \cdot AD + Area_{ABC} + Area_{BCD}$

Solid I: Surface area = $BE \cdot BC + CE \cdot BC + Area_{DEF} + Area_{BCD}$.

$Area_{BCD} = \frac{1}{2} BC \cdot BD = \frac{1}{2} 6 \cdot 3 \sqrt{2} = 9 \sqrt{2}$

Area A = $9*3 + 9*3 + 18\sqrt{2} + 9 \sqrt{2} = 27 + 27 + 18\sqrt{2} + 9\sqrt{2} = 54 + 27 \sqrt{2}$

Surface area I = $6*3 + 6*3 + 18 \sqrt{2} + 9 \sqrt{2} = 18+18 +18 \sqrt{2} + 9 \sqrt{2} = 36 + 27 \sqrt{2}$

The difference is $|54+27 \sqrt{2} - (36+27\sqrt{2})|=18$

3. Final Answer

(1) 1:1

(2) Solid A is larger by 18 $cm^2$.