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abcd is a kite, so \\( \\overline { a c } \\perp \\overline { d b } \\)…

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.

Explanation:

Step1: Find the length of \( CE \)

In right - triangle \( CDE \), use the Pythagorean theorem \( a^{2}+b^{2}=c^{2} \). Here, \( CD = 3\mathrm{cm}\), \( DE=\frac{4}{2}=2\mathrm{cm}\). Let \( CE = x \), then \(x=\sqrt{CD^{2}-DE^{2}}\).

$$x=\sqrt{3^{2}-2^{2}}=\sqrt{9 - 4}=\sqrt{5}\approx2.2\mathrm{cm}$$

Step2: Find the length of \( AE \)

In right - triangle \( ADE \), use the Pythagorean theorem. Here, \( AD = 4\mathrm{cm}\), \( DE = 2\mathrm{cm}\). Let \( AE=y\), then \(y=\sqrt{AD^{2}-DE^{2}}\).

$$y=\sqrt{4^{2}-2^{2}}=\sqrt{16 - 4}=\sqrt{12}\approx3.5\mathrm{cm}$$

Step3: Calculate the length of \( AC \)

Since \( AC=AE + CE \), substitute the values of \( AE \) and \( CE \).

$$AC=\sqrt{5}+\sqrt{12}\approx2.2+3.5 = 5.7\mathrm{cm}$$

Answer:

\(5.7\)