Question

What is the value of m In the figure below in this diagram

Answer 1

Since $\triangle ABD\sim \triangle BCD,$ the corresponding sides are proportional:

$\dfrac{AB}{BC}=\dfrac{BD}{CD}=\dfrac{AD}{BD}.$

You are given that

• $BC=m;$
• $DC=7;$
• $AD=11.$

Thus,

$\dfrac{AB}{m}=\dfrac{BD}{7}=\dfrac{11}{BD}.$

From the second equality $\dfrac{BD}{7}=\dfrac{11}{BD}$ you have that

$BD^2=7\cdot 11=77,\\ \\BD=\sqrt{77}.$

Now consider right triangle BDC (with right angle D). By the Pythagorean theorem,

$BC^2=BD^2+DC^2,\\ \\m^2=(\sqrt{77})^2+7^2=77+49=126,\\ \\m=\sqrt{126}.$

Answer: correct choice is A.

Answer 2

Answer:

(A)$\sqrt{126}$

Step-by-step explanation:

Since, it is given that ΔABD is similar to ΔBCD, then

$\frac{AB}{BC}=\frac{BD}{CD}=\frac{AD}{BD}$

$\frac{AB}{m}=\frac{BD}{7}=\frac{11}{BD}$

Taking second and third equality, we get

$(BD)^{2}=77$

⇒$BD=\sqrt{77}$

Now, from ΔBDC, using the Pythagoras theorem, we get

$(BC)^{2}=(BD)^{2}+(DC)^2$

$m^{2}=(7)^2+(\sqrt{77})^2$

$m^2=49+77$

$m^2=126$

$m=\sqrt{126}$.