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use parallelogram hijk at the right. find each measure. 9. ( hi = ) 10.…

Question

use parallelogram hijk at the right. find each measure.

  1. ( hi = )
  2. ( gh = )
  3. ( mangle kih = )
  4. ( mangle kji = )
  5. ( kh = )
  6. ( hj = )
  7. ( mangle jih = )
  8. ( mangle hki = )

Explanation:

Step1: Property of parallelogram (opposite sides equal)

In a parallelogram \(HIJK\), \(HI = KJ\). Given \(KJ = 16\), so \(HI=16\)

Step2: Diagonals bisect each other

In parallelogram \(HIJK\), diagonals \(HJ\) and \(KI\) bisect each other. Given \(GI = 7\), so \(GH=GI = 7\)

Step3: Alternate - interior angles

Since \(HK\parallel IJ\), \(\angle KIH=\angle GKI\). Given \(\angle GKI = 28^{\circ}\), so \(m\angle KIH = 28^{\circ}\)

Step4: Alternate - interior angles

Since \(HK\parallel IJ\), \(\angle KJH=\angle GHJ\). In \(\triangle HJI\), \(HI = 16\), \(IJ = 10\), \(HJ\) is a side. But for \(\angle KJI\), using the property of parallelogram and angle - sum. Wait, another approach: \(\angle KJI=\angle HKI\) (alternate - interior angles for \(HK\parallel IJ\) and transversal \(KJ\)). But no, better: In parallelogram \(HIJK\), \(\angle KJI=\angle HKI\) (not correct). Wait, using the fact that \(HK\parallel IJ\) and \(KJ\) is a transversal. \(\angle KJH=\angle GHJ\). Wait, no. Wait, \(\angle KJI\): Since \(HK\parallel IJ\), \(\angle KJH=\angle GHJ\). But from the given \(\angle HKI = 28^{\circ}\) (alternate - interior angles for \(HK\parallel IJ\) and transversal \(KJ\)) is wrong. Wait, correct: \(\angle KJI=\angle HKI\) (alternate - interior angles for \(HK\parallel IJ\) and transversal \(KJ\)) is wrong. Wait, \(HK\parallel IJ\), \(KI\) is a transversal. \(\angle HKI=\angle JIK\) (alternate - interior angles). But for \(\angle KJI\), since \(HI = KJ = 16\), \(IJ = 10\), \(KI\) is a diagonal. Wait, no, property of parallelogram: \(KH = IJ\). Given \(IJ = 10\), so \(KH = 10\)

Step5: Diagonals bisect each other

\(HJ = 2\times HG\). Since \(HG = 7\), \(HJ=14\)

Step6: Angle - calculation

In \(\triangle HJI\), \(HI = 16\), \(IJ = 10\), \(HJ = 14\). Using the law of cosines in \(\triangle HJI\) to find \(\angle JIH\). Wait, no, using the property of parallelogram: \(\angle HKI+\angle KHI+\angle HIK = 180^{\circ}\). Given \(\angle KHI = 84^{\circ}\), \(\angle HIK = 28^{\circ}\), so \(\angle HKI=68^{\circ}\). Then \(\angle JIH=\angle HKI = 68^{\circ}\) (alternate - interior angles for \(HK\parallel IJ\) and transversal \(HI\))

Step7: Angle - sum

In \(\triangle HKI\), \(\angle HKI=180^{\circ}-\angle KHI-\angle KIH\). Given \(\angle KHI = 84^{\circ}\), \(\angle KIH = 28^{\circ}\), so \(m\angle HKI=68^{\circ}\)

Answer:

  1. \(16\)
  2. \(7\)
  3. \(28^{\circ}\)
  4. \(68^{\circ}\)
  5. \(10\)
  6. \(14\)
  7. \(68^{\circ}\)
  8. \(68^{\circ}\)