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frika

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.

boffeemadrid

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

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

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

\(m^2=49+77\)

\(m^2=126\)

\(m=\sqrt{126}\).

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