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abcd is a kite, so ac ⊥ db and de = eb. calculate the length of ac, to …

Question

abcd is a kite, so ac ⊥ db and de = eb. calculate the length of ac, to the nearest tenth of a centimeter.

Explanation:

Step1: Use Pythagorean theorem for \(\triangle CED\)

In right - triangle \(CED\), by the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\). Let \(CE\) be one leg, \(DE = 11\) be the other leg, and \(CD=40\) be the hypotenuse. Then \(CE=\sqrt{CD^{2}-DE^{2}}\).

$$CE=\sqrt{40^{2}-11^{2}}=\sqrt{(40 + 11)(40 - 11)}=\sqrt{51\times29}=\sqrt{1479}\approx38.5$$

Step2: Use Pythagorean theorem for \(\triangle AED\)

In right - triangle \(AED\), with \(AD = 60\) (hypotenuse) and \(DE = 11\) (one leg). Let \(AE\) be the other leg. Then \(AE=\sqrt{AD^{2}-DE^{2}}\).

$$AE=\sqrt{60^{2}-11^{2}}=\sqrt{(60 + 11)(60 - 11)}=\sqrt{71\times49}=\sqrt{3479}\approx59.0$$

Step3: Calculate the length of \(AC\)

Since \(AC=AE + CE\), then \(AC\approx38.5+59.0 = 97.5\)

Answer:

\(97.5\)