Question

9.

K
U
A
B
5
4
L
C
KL || AC, KL = 5,
AC=7, AK = 4.
KB =

How mucu is KL?

Answer 1

KB is
`\( (8)/(5) \)` units long. If you prefer a decimal approximation, it is 1.6 units.

We have a situation where KL is parallel to AC, and we are given the lengths of KL, AC, and AK. Additionally, we want to find the length of KB.

Since KL is parallel to AC, and AK is a transversal, we can use the property of similar triangles. In this case, triangles AKL and ABC are similar.

The corresponding sides of similar triangles are proportional. Therefore, we can set up the following proportion:

`\[ (AK)/(AB) = (KL)/(AC) \]`

Plugging in the given values:

`\[ (4)/(AB) = (5)/(7) \]`

Now, solve for AB:

`\[ AB = (4 * 7)/(5) = (28)/(5) \]`

Therefore, AB is
`\( (28)/(5) \)` units long.

Now, since KB is a part of AB, we can find KB by subtracting AK from AB:

`\[ KB = AB - AK \]`

`\[ KB = (28)/(5) - 4 \]`

To perform this subtraction, find a common denominator:

`\[ KB = (28)/(5) - (20)/(5) \]`

`\[ KB = (8)/(5) \]`

So, KB is
`\( (8)/(5) \)` units long. If you prefer a decimal approximation, it is 1.6 units.

The probable question may be:

"KL || AC, KL = 5,

AC=7, AK = 4.

KB =?

How much is KB?"