Question

as shown in the figure, square abcd has a side length of 4. if $overline{be}=5$, then the length of $ce=___$ a $sqrt{5}$ b 5 c $sqrt{15}$ d $sqrt{17}$

Explanation:

Step1: Find length of AE in right - triangle ABE

In square ABCD, AB = 4. In right - triangle ABE, by the Pythagorean theorem \(AE=\sqrt{BE^{2}-AB^{2}}\). Since BE = 5 and AB = 4, then \(AE=\sqrt{5^{2}-4^{2}}=\sqrt{25 - 16}=\sqrt{9}=3\).

Step2: Find length of DE

Since AD = 4 and AE = 3, then \(DE=AD - AE=4 - 3 = 1\).

Step3: Find length of CE in right - triangle CDE

In right - triangle CDE, CD = 4 and DE = 1. By the Pythagorean theorem \(CE=\sqrt{CD^{2}+DE^{2}}\). Substitute CD = 4 and DE = 1 into the formula, we get \(CE=\sqrt{4^{2}+1^{2}}=\sqrt{16 + 1}=\sqrt{17}\).

Answer:

D. \(\sqrt{17}\)