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2. Given square CANE with diagonals

intersecting at B. mZCNE = 3a + 2b,
AC = 35, and CE = 6a + 5.
Find the value of a +b.

2. Given square CANE with diagonals intersecting at B. mZCNE = 3a + 2b, AC = 35, and-example-1
User Cain
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2 Answers

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What’s something goes up but never comes down?
User Remko Jansen
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The value of a + b is 15. Hence, the option a) is correct.

To find the value of a + b, we need to use the information given in the problem.

First, let's look at the angle CNE. We are told that m\angle CNE is equal to 3a + 2b degrees.

Next, we know that AC is one side of the square and has a length of 35 units. This means that all sides of the square are equal to 35 units.

We are also given that CE is another side of the square and has a length of 6a + 5 units.

Since all sides of a square are equal, AC = CE.

So we can set up the equation:

35 = 6a + 5

Now, let's solve for a.

35 - 5 = 6a

30 = 6a

Divide both sides by 6:

5 = a

Now that we have the value of a, we can substitute it back into the equation to find the value of b.

35 = 6a + 5

35 = 6(5) + 5

35 = 30 + 5

35 = 35

This confirms that the value of a is indeed 5.

Finally, we can find the value of b by substituting the value of a back into the equation for the angle:

m\angle CNE = 3a + 2b

3(5) + 2b = 35

15 + 2b = 35

Subtract 15 from both sides:

2b = 20

Divide both sides by 2:

b = 10

Now we can find the value of a + b:

a + b = 5 + 10

a + b = 15

User Psalaets
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