Question

figure cdef is a parallelogram. 10r - 20, 2r + 12, 6r. what is the value of r? 2 3 4 5

Explanation:

Step1: Recall parallelogram property

In a parallelogram, opposite sides are equal. So, \( CF = DE \) and \( CD = FE \). Wait, looking at the sides: \( CF = 10r - 20 \), \( DE = 6r \)? Wait no, maybe \( CD = FE \)? Wait, no, let's check the labels. The sides: \( CD = 2r + 12 \), \( FE = 6r \)? Wait, no, maybe \( CF = DE \)? Wait, the figure is CDEF, so sides: \( CD \) and \( FE \) are opposite? Wait, maybe \( CF \) and \( DE \) are opposite, and \( CD \) and \( FE \) are opposite? Wait, no, let's re-express. Wait, in parallelogram CDEF, opposite sides are equal. So \( CF = DE \) and \( CD = FE \). Wait, \( CF = 10r - 20 \), \( DE = 6r \)? No, maybe \( CD = FE \)? Wait, \( CD = 2r + 12 \), \( FE = 6r \)? No, that doesn't make sense. Wait, maybe \( CF = DE \) and \( CD = FE \)? Wait, no, let's check the other pair. Wait, maybe \( CF = DE \): \( 10r - 20 = 6r \)? No, solving that: \( 10r - 6r = 20 \), \( 4r = 20 \), \( r = 5 \). Wait, but let's check the other pair. If \( r = 5 \), then \( CD = 2(5) + 12 = 22 \), \( FE = 6(5) = 30 \). No, that's not equal. Wait, maybe I mixed up the sides. Wait, maybe \( CD = DE \)? No, that's not a property. Wait, no, in a parallelogram, opposite sides are equal. So \( CD \) is opposite to \( FE \), and \( CF \) is opposite to \( DE \). Wait, the labels: C, D, E, F. So the sides are CD, DE, EF, FC. So CD || EF and CD = EF; CF || DE and CF = DE. So CD = EF: \( 2r + 12 = 6r \)? Wait, \( 2r + 12 = 6r \), then \( 12 = 4r \), \( r = 3 \). Wait, but then CF = 10r - 20 = 10(3) - 20 = 10, DE = 6r = 18? No, that's not equal. Wait, maybe CF = DE: \( 10r - 20 = 6r \), so \( 4r = 20 \), \( r = 5 \). Then CD = 2(5) + 12 = 22, EF = 6(5) = 30. No, that's not equal. Wait, maybe I got the sides wrong. Wait, the figure: C to D is \( 2r + 12 \), D to E is \( 6r \), E to F is \( 10r - 20 \), F to C is... Wait, no, maybe the sides are CD and EF, and CF and DE. So CD = EF: \( 2r + 12 = 10r - 20 \)? Wait, \( 2r + 12 = 10r - 20 \), then \( 12 + 20 = 8r \), \( 32 = 8r \), \( r = 4 \). Then DE = 6r = 24, CF = 10r - 20 = 20? No, that's not equal. Wait, I'm confused. Wait, maybe the sides are CF and DE, and CD and FE. Wait, CF is \( 10r - 20 \), DE is \( 6r \), so \( 10r - 20 = 6r \) → \( 4r = 20 \) → \( r = 5 \). Then CD is \( 2r + 12 = 22 \), FE is \( 6r = 30 \). No, that's not equal. Wait, maybe the other pair: CD = FE. CD is \( 2r + 12 \), FE is \( 6r \). So \( 2r + 12 = 6r \) → \( 12 = 4r \) → \( r = 3 \). Then CF is \( 10r - 20 = 10(3) - 20 = 10 \), DE is \( 6r = 18 \). No, that's not equal. Wait, maybe the problem has a typo, but let's check the answer choices. The options are 2,3,4,5. Let's test r=5: CF=10*5-20=30, DE=6*5=30. Oh! Wait, maybe CF and DE are opposite sides. So CF = DE: \( 10r - 20 = 6r \). Then 10r - 6r = 20 → 4r=20 → r=5. Then CD=2*5+12=22, EF=6*5=30? No, that's not equal. Wait, no, maybe CD and EF are opposite. Wait, CD=2r+12, EF=10r-20? Wait, the figure: C to D is 2r+12, D to E is 6r, E to F is 10r-20, F to C is... Wait, maybe the sides are CD and EF, and DE and CF. So CD=EF: 2r+12=10r-20 → 8r=32 → r=4. Then DE=6r=24, CF=10r-20=20. No, that's not equal. Wait, maybe I misread the sides. Let me check again. The figure: C to D: 2r+12, D to E: 6r, E to F: 10r-20, F to C:? Wait, maybe CF is 10r-20, DE is 6r, so CF=DE: 10r-20=6r → r=5. Then CD=2*5+12=22, EF=6*5=30? No, that's not equal. Wait, maybe the problem is that CD and FE are equal, and CF and DE are equal. Wait, CD=2r+12, FE=6r; CF=10r-20, DE=6r? No, DE is 6r? Wait, no, DE is 6r, CF is 10r-20. So CF=DE: 10r-20=6r → r=5. Then CD=2*5+12=22, FE=6*5=30. No, that's not equal. Wait, maybe the sides are CD=FE and CF=DE. So CD=2r+12, FE=6r; CF=10r-20, DE=6r? No, DE is 6r? Wait, maybe DE is 10r-20? No, the label is DE as 6r? Wait, the user's figure: CDEF is a parallelogram. Sides: CF is 10r - 20, CD is 2r + 12, DE is 6r. So in a parallelogram, opposite sides are equal. So CF (from C to F) and DE (from D to E) are opposite? Wait, no, in parallelogram CDEF, the vertices are C, D, E, F in order. So the sides are CD, DE, EF, FC. So CD is adjacent to DE, DE to EF, EF to FC, FC to CD. So opposite sides: CD and EF, DE and FC. Therefore, CD = EF and DE = FC. So DE = FC: 6r = 10r - 20. Then 10r - 6r = 20 → 4r = 20 → r = 5. And CD = EF: 2r + 12 = EF. If r=5, CD=22, so EF=22. Wait, but EF is labeled as... Wait, maybe EF is 6r? No, the label for DE is 6r, EF is 10r - 20? No, the user's figure: F to E is 10r - 20? Wait, no, the figure shows: C to F is 10r - 20, C to D is 2r + 12, D to E is 6r. So FC = 10r - 20, DE = 6r. So DE = FC: 6r = 10r - 20 → 4r = 20 → r = 5. Then CD = 2r + 12 = 22, and EF should be equal to CD, so EF = 22. If EF is 6r, then 6r = 22 → r = 22/6 ≈ 3.666, which is not an option. Wait, maybe I got the opposite sides wrong. Maybe CD and DE are not adjacent? No, in a parallelogram, consecutive vertices are connected by sides. So C to D to E to F to C. So CD and DE are adjacent, DE and EF are adjacent, etc. So opposite sides: CD and EF, DE and CF. Therefore, CD = EF and DE = CF. So DE = CF: 6r = 10r - 20 → 4r = 20 → r = 5. Then CD = 2r + 12 = 22, so EF = 22. But EF is labeled as... Wait, maybe EF is 2r + 12? No, the figure shows EF as... Wait, maybe the problem has a typo, but among the options, r=5 makes DE=CF (6*5=30, 10*5-20=30). Ah! Wait, 10r -20 when r=5 is 30, and 6r when r=5 is 30. So DE=6r=30, CF=10r-20=30. So they are equal. Then CD=2r+12=22, and EF should be equal to CD, so EF=22. Maybe EF is labeled as something else, but the key is that DE and CF are opposite sides, so they must be equal. So solving 6r = 10r -20 gives r=5.

Step1: Identify parallelogram property

In a parallelogram, opposite sides are equal. So \( DE = CF \).

Step2: Set up the equation

Given \( DE = 6r \) and \( CF = 10r - 20 \), we set \( 6r = 10r - 20 \).

Step3: Solve for \( r \)

Subtract \( 6r \) from both sides: \( 0 = 4r - 20 \).
Add 20 to both sides: \( 20 = 4r \).
Divide by 4: \( r = 5 \).

Answer:

5