Question

abcd is a kite, so ( overline{ac} perp overline{db} ) and ( de = eb ). calculate the length of ( overline{ac} ), to the nearest tenth of a centimeter. answer attempt 1 out of 2 ( ac approx ) cm

Explanation:

Step1: Find the length of \( CE \)

In right - triangle \( CED \), using the Pythagorean theorem \( a^{2}+b^{2}=c^{2} \). Here, \( CD = 7\mathrm{cm}\), \( ED=\sqrt{AD^{2}-AE^{2}}=\sqrt{5^{2}-8^{2}} \) (wait, no, in right - triangle \( AED \), \( AD = 5\mathrm{cm}\), \( AE = 3\mathrm{cm}\) (by Pythagorean theorem in \( AED \): \( ED=\sqrt{AD^{2}-AE^{2}}=\sqrt{5^{2}-3^{2}} = 4\mathrm{cm}\)). In right - triangle \( CED \), \( CD = 7\mathrm{cm}\), \( ED = 4\mathrm{cm}\), then \( CE=\sqrt{CD^{2}-ED^{2}}=\sqrt{7^{2}-4^{2}}=\sqrt{49 - 16}=\sqrt{33}\approx5.7\mathrm{cm}\)

Step2: Calculate the length of \( AC \)

Since \( AC=AE + CE\), and \( AE = 3\mathrm{cm}\), \( CE\approx5.7\mathrm{cm}\), then \( AC=3 + 5.7=8.7\mathrm{cm}\)

Answer:

\(8.7\)