Question :

In the given figure, $\angle ABC=90°$ and BD$\perp$AC. If $AB=5.7$cm, $BD=3.8$cm and $CD=5.4$cm, find $BC$. Given Figure

By similarity in $\triangle ABC \sim \triangle BDC$, (By $\frac{AB}{BD}=\frac{BC}{DC}$ ) I got $BC=8.1$cm but by Pythagorean theorem I got $BC=6.6$.

Which one is right?

  • 1
    $\begingroup$ There is no correct answer. The given information is inconsistent. $\endgroup$
    – Blue
    Sep 3, 2020 at 6:55
  • $\begingroup$ You are getting two different answers as the dimensions given are incorrect for a right angled triangle. $\endgroup$
    – Math Lover
    Sep 3, 2020 at 7:01
  • $\begingroup$ Thanks. When I constructed triangel ABC with BC=8.1,I found CD 6.5cm and when I constructed ABC with BC=6.6cm, I found CD=4.9cm. $\endgroup$
    – Wolgwang
    Sep 3, 2020 at 7:16

1 Answer 1


Because this triangle does not exist!

Indeed, by the Pythagorean theorem we obtain: $$AD=\sqrt{5.7^2-3.8^2}.$$

Since $$\measuredangle BAD=90^{\circ}-\measuredangle ABD=\measuredangle CBD,$$ we see that: $\Delta ABD\sim\Delta BCD$ and we obtain: $$BD^2=AD\cdot DC,$$ which gives $$AD=\frac{3.8^2}{5.4}=\frac{361}{135}$$ and easy to see that $$\sqrt{5.7^2-3.8^2}\neq\frac{361}{135}$$

  • $\begingroup$ Can you explain the first step? $\endgroup$
    – Wolgwang
    Sep 3, 2020 at 8:01
  • $\begingroup$ Which similarity criterion did you use? $\endgroup$
    – Wolgwang
    Sep 3, 2020 at 8:05
  • $\begingroup$ @Mayank angle-angle $\endgroup$ Sep 3, 2020 at 8:05
  • $\begingroup$ But in the $\triangle$ABD and $\triangle$BCD , only one angle is equal to other triangle i.e. 90°. $\endgroup$
    – Wolgwang
    Sep 3, 2020 at 8:12
  • $\begingroup$ @Mayank Also, $\measuredangle BAD=\measuredangle CBD.$ $\endgroup$ Sep 3, 2020 at 8:15

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