Question

abcd is a kite, so \\( \overline { a c } \perp \overline { d b } \\) and \\( d e = e b \\). calculate the length of \\( \overline { a c } \\), to the nearest tenth of a centimeter.
answer attempt 2 out of 2
\\( a c \approx \\) cm submit answer

Explanation:

Step1: Find the length of \( DE \) and \( EB \)

Since \( DE = EB \) and \( DB=8\) cm, then \( DE=\frac{DB}{2}=\frac{8}{2} = 4\) cm.

Step2: Use the Pythagorean theorem to find \( CE \)

In right - triangle \( CDE \), by the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\) (where \(c = CD = 6\) cm and \(a = DE=4\) cm, and \(b = CE\)).

$$CE=\sqrt{CD^{2}-DE^{2}}=\sqrt{6^{2}-4^{2}}=\sqrt{36 - 16}=\sqrt{20}=2\sqrt{5}\approx4.5$$

Step3: Use the Pythagorean theorem to find \( AE \)

In right - triangle \( ADE \), by the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\) (where \(c = AD = 5\) cm and \(a = DE = 4\) cm, and \(b = AE\)).

$$AE=\sqrt{AD^{2}-DE^{2}}=\sqrt{5^{2}-4^{2}}=\sqrt{25 - 16}=\sqrt{9}=3$$

Step4: Calculate the length of \( AC \)

Since \(AC=CE + AE\), then \(AC\approx4.5+3=7.5\)

Answer:

\(7.5\)