Question

question 2 of 10 what is the value of m in the figure below? in this diagram, △abd ~ △bcd.

Explanation:

Step1: Use property of similar triangles

Since $\triangle ABD\sim\triangle BCD$, the ratios of corresponding sides are equal. That is, $\frac{AD}{BD}=\frac{BD}{CD}$. Given $AD = 14$ and $CD = 5$, and let $BD$ be the common - side ratio. Cross - multiplying gives $BD^{2}=AD\times CD$.

Step2: Calculate $BD^{2}$

$BD^{2}=14\times5=70$.

Step3: Apply Pythagorean theorem in $\triangle BCD$

In right - triangle $\triangle BCD$, by the Pythagorean theorem $BC^{2}=BD^{2}+CD^{2}$. Since $BD^{2}=70$ and $CD = 5$, then $BC^{2}=70 + 25=95$. So, $m=\sqrt{95}$.

Answer:

D. $\sqrt{95}$