QUESTION IMAGE
Question
figure w x y z is a kite. the diagonals of the kite meet at point p.
given: ( w p = 8 ) units, ( x p = 6 ) units, ( y p = 12 ) units, ( z p = 6 ) units.
drag and drop the values to reflect the measures in the diagram.
note: some values may not be used.
drag & drop the answer
note: use ctrl+d to drag the option via keyboard
Step1: Calculate the length of \(WZ\) and \(WX\)
Use the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\) in right - triangle \(WPZ\) and \(WPX\).
For \(WZ\): \(WP = 8\), \(ZP=6\). Then \(WZ=\sqrt{WP^{2}+ZP^{2}}=\sqrt{8^{2}+6^{2}}=\sqrt{64 + 36}=\sqrt{100}=10\)
For \(WX\): \(WP = 8\), \(XP = 6\). Then \(WX=\sqrt{WP^{2}+XP^{2}}=\sqrt{8^{2}+6^{2}}=\sqrt{64 + 36}=\sqrt{100}=10\)
Step2: Calculate the length of \(YZ\) and \(YX\)
Use the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\) in right - triangle \(YPZ\) and \(YPX\).
For \(YZ\): \(YP = 12\), \(ZP = 6\). Then \(YZ=\sqrt{YP^{2}+ZP^{2}}=\sqrt{12^{2}+6^{2}}=\sqrt{144+36}=\sqrt{180}\approx13.4\)
For \(YX\): \(YP = 12\), \(XP = 6\). Then \(YX=\sqrt{YP^{2}+XP^{2}}=\sqrt{12^{2}+6^{2}}=\sqrt{144 + 36}=\sqrt{180}\approx13.4\)
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The values for \(WZ\) and \(WX\) are \(10\), the values for \(YZ\) and \(YX\) are approximately \(13.4\)