Question

select the correct answers.
if sides bd and dc have the same length, what is the length of side bc?
a. 1 unit
b. \\( \sqrt { 2 } \\) units
c. 2 units
d. \\( \sqrt { 3 } \\) units
e. 4 units

Explanation:

Step1: Find length of BD

In right triangle \(ABD\), \(AB = 1\), \(AD = 1\). By Pythagorean theorem, \(BD=\sqrt{AB^{2}+AD^{2}}=\sqrt{1^{2}+1^{2}}=\sqrt{2}\).

Step2: Find length of BC

Given \(BD = DC=\sqrt{2}\), and \(\angle BDC = 90^{\circ}\). By Pythagorean theorem, \(BC=\sqrt{BD^{2}+DC^{2}}=\sqrt{(\sqrt{2})^{2}+(\sqrt{2})^{2}}=\sqrt{2 + 2}=\sqrt{4}=2\)? Wait, no, wait. Wait, no, \(BD = DC\), and \(\angle BDC\) is right angle. Wait, no, first, \(BD\) length: in triangle \(ABD\), right - angled at \(A\), so \(BD=\sqrt{1^{2}+1^{2}}=\sqrt{2}\). Then, triangle \(BDC\) is right - angled at \(D\), and \(BD = DC=\sqrt{2}\)? Wait, no, the problem says \(BD\) and \(DC\) have the same length. Wait, no, maybe I made a mistake. Wait, \(AB = 1\), \(AD = 1\), so \(BD=\sqrt{1 + 1}=\sqrt{2}\). Then \(DC = BD=\sqrt{2}\), and triangle \(BDC\) is right - angled at \(D\), so \(BC=\sqrt{BD^{2}+DC^{2}}=\sqrt{(\sqrt{2})^{2}+(\sqrt{2})^{2}}=\sqrt{2 + 2}=\sqrt{4}=2\)? But wait, the options have C as 2 units. Wait, but let's re - check. Wait, \(AB = 1\), \(AD = 1\), right triangle \(ABD\), so \(BD=\sqrt{1^{2}+1^{2}}=\sqrt{2}\). Then \(DC = BD=\sqrt{2}\), and \(\angle BDC = 90^{\circ}\), so \(BC=\sqrt{BD^{2}+DC^{2}}=\sqrt{(\sqrt{2})^{2}+(\sqrt{2})^{2}}=\sqrt{2 + 2}=\sqrt{4}=2\). So \(BC = 2\) units.

Answer:

C. 2 units