QUESTION IMAGE
Question
find the indicated measure in ▱hijk. explain.
- hi
- kh
- gh
- hj
- m∠kih
- m∠jih
- m∠kji
- m∠hki
Step1: Recall properties of parallelogram
In parallelogram $HIJK$, opposite - sides are equal.
Step2: Find $HI$
Since $KJ$ and $HI$ are opposite - sides, and $KJ = 16$, then $HI=16$.
Step3: Find $KH$
Since $IJ$ and $KH$ are opposite - sides, and $IJ = 10$, then $KH = 10$.
Step4: Use triangle properties for $GH$
In $\triangle HKI$, we don't have enough information to find $GH$ directly from the given parallelogram properties. Assume $G$ is the mid - point of $HI$ and $KJ$ (if not given otherwise, we can't solve). If $G$ is the mid - point of $HI$, and $HI = 16$, then $GH=\frac{HI}{2}=8$.
Step5: Find $HJ$
We can use the law of cosines in $\triangle HKI$. In parallelogram $HIJK$, $\angle KHI=84^{\circ}$, $KH = 10$, $HI = 16$. In $\triangle HKI$, by the law of cosines, $HJ^{2}=KH^{2}+HI^{2}-2\cdot KH\cdot HI\cdot\cos\angle KHI$.
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Step6: Find $\angle KIH$
In $\triangle HKI$, we know $\angle KHI = 84^{\circ}$ and $\angle HKI=28^{\circ}$. Using the angle - sum property of a triangle ($\angle KHI+\angle HKI+\angle KIH = 180^{\circ}$), then $\angle KIH=180^{\circ}-84^{\circ}-28^{\circ}=68^{\circ}$.
Step7: Find $\angle JIH$
Since $\angle KIH = 68^{\circ}$ and $\angle KIJ$ is a straight - line angle with $\angle KIH$, and in a parallelogram adjacent angles are supplementary, $\angle JIH = 180^{\circ}-\angle KIH=112^{\circ}$.
Step8: Find $\angle KJI$
In parallelogram $HIJK$, $\angle KJI$ and $\angle KHI$ are opposite angles, so $\angle KJI = 84^{\circ}$.
Step9: Find $\angle HKI$
Given $\angle HKI = 28^{\circ}$.
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- $HI = 16$
- $KH = 10$
- If $G$ is the mid - point of $HI$, $GH = 8$
- $HJ\approx17.96$
- $m\angle KIH = 68^{\circ}$
- $m\angle JIH = 112^{\circ}$
- $m\angle KJI = 84^{\circ}$
- $m\angle HKI = 28^{\circ}$