QUESTION IMAGE
Question
given $parallelogram luck$, use the figure and the theorems in items 1, 3, and 4 to find the following.
- solve for $x$ if $mangle kcu = 10x - 15$ and $mangle k = 6x + 3$.
$x=$
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Step1: Use the property of parallelogram
In parallelogram \( LUCK\), \( LU\parallel KC\). Then \(\angle KCU=\angle LKC\) (alternate - interior angles). Also, in \(\triangle LKC\), \(LK = UC\) and \(LU\parallel KC\), so \(\angle KCU=\angle LKC\). Since \(LUCK\) is a parallelogram, \(LK\parallel UC\). By the property of parallelogram (alternate - interior angles), \(\angle KCU=\angle LKC\). And in a parallelogram, \(LK\parallel UC\), so \(\angle KCU=\angle LKC\). Also, \(m\angle KCU = 10x-15\) and \(m\angle K=6x + 3\).
Step2: Set up the equation
Since \(\angle KCU\) and \(\angle K\) are related by the property of the parallelogram (alternate - interior angles in a parallelogram), we set \(10x-15=6x + 3\).
Subtract \(6x\) from both sides:
\(10x-6x-15=6x-6x + 3\)
\(4x-15=3\)
Add \(15\) to both sides:
\(4x-15 + 15=3+15\)
\(4x=18\)
Divide both sides by \(4\):
\(x=\frac{18}{4}=\frac{9}{2}=4.5\)
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\(x = 4.5\)