Question

A solid is shown in the figure below. The vertices ABCDEFGH form a 4 cm x 2 cm x 2 cm cuboid. The segments AI and DI are both of length 5 cm, and points I, C, G, F, B are coplanar. What is the height (in cm) of point I from the base rectangle EFGH? 

 

Hint:

Using a coordinate system is a powerful technique for solving 3D geometry problems. It transforms geometric properties (like distance and coplanarity) into algebraic equations that are often easier to solve. Always choose a convenient origin to simplify the coordinates.

Answer 1

Correct Answer: 4.8

Step 1: Understanding the Concept: 
This is a 3D coordinate geometry problem. We can solve it by setting up a coordinate system, defining the coordinates of the vertices, and then using distance formulas and plane equations to find the coordinates of point I. The height of I will be its z-coordinate. 
Step 2: Key Formula or Approach: 
1. Distance Formula in 3D: The distance between points \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\) is \(\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}\). 
2. Equation of a Plane: A set of points is coplanar if they satisfy the equation of a single plane. 
Step 3: Detailed Explanation: 
Let's set up a coordinate system with the origin at E(0,0,0). The dimensions are length (x-axis) = 4, width (y-axis) = 2, and height (z-axis) = 2. The coordinates of the relevant vertices are: - E = (0, 0, 0) - A = (0, 0, 2) - D = (0, 2, 2) - B = (4, 0, 2) - C = (4, 2, 2) - F = (4, 0, 0) - G = (4, 2, 0) Let the coordinates of point I be \((x, y, z)\). The height of I from the base EFGH is \(z\). 
Condition 1: Distances AI and DI 
We are given AI = 5 and DI = 5. Using the distance formula: \[ (AI)^2 = (x-0)^2 + (y-0)^2 + (z-2)^2 = 5^2 \implies x^2 + y^2 + (z-2)^2 = 25 \] \[ (DI)^2 = (x-0)^2 + (y-2)^2 + (z-2)^2 = 5^2 \implies x^2 + (y-2)^2 + (z-2)^2 = 25 \] Since both expressions equal 25, we can set them equal to each other: \[ x^2 + y^2 + (z-2)^2 = x^2 + (y-2)^2 + (z-2)^2 \] \[ y^2 = (y-2)^2 = y^2 - 4y + 4 \] \[ 0 = -4y + 4 \implies 4y = 4 \implies y = 1 \] Condition 2: Coplanarity 
The points I, C, G, F, B are coplanar. Let's examine the coordinates of B, C, F, G: - B = (4, 0, 2) - C = (4, 2, 2) - F = (4, 0, 0) - G = (4, 2, 0) All these points have an x-coordinate of 4. This means they all lie on the plane defined by the equation \(x = 4\). For point I\((x, y, z)\) to be on the same plane, its x-coordinate must also be 4. So, \(x = 4\). 
Finding the Height z 
Now we know \(x=4\) and \(y=1\). We can substitute these values back into the distance equation for AI: \[ x^2 + y^2 + (z-2)^2 = 25 \] \[ (4)^2 + (1)^2 + (z-2)^2 = 25 \] \[ 16 + 1 + (z-2)^2 = 25 \] \[ 17 + (z-2)^2 = 25 \] \[ (z-2)^2 = 8 \] \[ z - 2 = \pm\sqrt{8} = \pm 2\sqrt{2} \] Since point I is shown above the cuboid, its height \(z\) must be greater than 2. Thus, we take the positive root: \[ z = 2 + 2\sqrt{2} \] Step 4: Final Answer: 
Using the approximation \(\sqrt{2} \approx 1.4142\): \[ z = 2 + 2(1.4142) = 2 + 2.8284 = 4.8284 \] The height of point I from the base is approximately 4.828 cm. This value is within the specified range of 4.80 - 4.84.