Question

cube B is the image of cube A after dilation by a scale factor of 4. If the surface area of cube A is 114in^2, find the surface area of cube B, the image.

Answer 1

To find the surface area of cube B, which is the image of cube A after dilation by a scale factor of 4, we can use the property of surface area in relation to the scale factor.

When a three-dimensional shape such as a cube is dilated by a scale factor \( k \), the surface area of the new shape increases by a factor of \( k^2 \).

1. **Calculate the scale factor**: The scale factor \( k \) is 4.
2. **Find the square of the scale factor**:
\[
k^2 = 4^2 = 16
\]

3. **Calculate the surface area of cube B**: Since the surface area of cube A is given as 114 in², we can find the surface area of cube B by multiplying the surface area of cube A by \( k^2 \):
\[
\text{Surface area of cube B} = \text{Surface area of cube A} \times k^2
\]
\[
\text{Surface area of cube B} = 114 \, \text{in}^2 \times 16 = 1824 \, \text{in}^2
\]

Therefore, the surface area of cube B is \( 1824 \, \text{in}^2 \).

Answer 2

triangle BCD

E is the midpoint of BD and F is the midpoint of CD. If EF = -6X + 35, and BC = 9x - 35, what is the measure of EF

Answer 3

To find the measure of \( EF \) given that \( EF = -6x + 35 \) and \( BC = 9x - 35 \), we first note that since \( E \) and \( F \) are midpoints, \( EF \) is half the length of \( BC \).

### Step 1: Set up the equation

Since \( EF \) is half of \( BC \), we can express this relationship mathematically:

\[
EF = \frac{1}{2} BC
\]

### Step 2: Substitute the expression for \( BC \)

From the problem, we know that \( BC = 9x - 35 \). Substituting this into the equation, we have:

\[
EF = \frac{1}{2}(9x - 35)
\]

### Step 3: Equate the two expressions for EF

We also have \( EF = -6x + 35 \). Therefore, we can set the two expressions for \( EF \) equal to each other:

\[
-6x + 35 = \frac{1}{2}(9x - 35)
\]

### Step 4: Solve for \( x \)

First, simplify the right side:

\[
-6x + 35 = \frac{9x}{2} - \frac{35}{2}
\]

To eliminate the fraction, multiply the entire equation by 2:

\[
2(-6x + 35) = 9x - 35
\]

This simplifies to:

\[
-12x + 70 = 9x - 35
\]

Now, collect all \( x \) terms on one side and constants on the other:

\[
70 + 35 = 9x + 12x
\]

\[
105 = 21x
\]

Now solve for \( x \):

\[
x = \frac{105}{21} = 5
\]

### Step 5: Find \( EF \)

Substituting \( x = 5 \) back into the expression for \( EF \):

\[
EF = -6(5) + 35
\]

Calculating that gives:

\[
EF = -30 + 35 = 5
\]

### Final Answer

The measure of \( EF \) is \( 5 \) units.

Answer 4

Quadrilateral IJKL is similar to quadrilateral MNOP. Find the measure of side OP. round your answer to the nearest tenth if necessary.

IL = 29.1
KL = 14
MP = 8

Answer 5

To find the measure of side \( OP \) in quadrilateral \( MNOP \) that is similar to quadrilateral \( IJKL \), we can use the property of similar figures that states that the ratios of corresponding sides are equal.

### Step 1: Identify the ratio of corresponding sides

From the given information:
- Side \( IL \) corresponds to side \( MP \).
- Side \( KL \) corresponds to side \( OP \).

We know:
- \( IL = 29.1 \)
- \( KL = 14 \)
- \( MP = 8 \)

### Step 2: Set up the ratio

The ratio of the sides can be set up as follows:

\[
\frac{IL}{MP} = \frac{KL}{OP}
\]

Substituting the known values into the ratio gives:

\[
\frac{29.1}{8} = \frac{14}{OP}
\]

### Step 3: Cross-multiply to solve for \( OP \)

Cross-multiplying gives:

\[
29.1 \cdot OP = 14 \cdot 8
\]

Calculating the right side:

\[
14 \cdot 8 = 112
\]

So the equation simplifies to:

\[
29.1 \cdot OP = 112
\]

### Step 4: Solve for \( OP \)

Now, divide both sides by \( 29.1 \):

\[
OP = \frac{112}{29.1}
\]

Calculating that gives:

\[
OP \approx 3.844
\]

### Step 5: Round to the nearest tenth

Rounding \( 3.844 \) to the nearest tenth gives:

\[
OP \approx 3.8
\]

### Final Answer

The measure of side \( OP \) is approximately \( 3.8 \).