# Triangles - Class 10 - Mathematics

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## Extra Questions - Triangles | NCERT | Mathematics | Class 10

In the given figure, $\triangle ABC$ is a triangle right-angled at $B$ and $BD \perp AC$. If $AD=4$ cm and $CD=5$ cm, then find $BD$ and $AB$.

Given the configuration in $\triangle ABC$, where $\angle B$ is $90^\circ$ and $BD \perp AC$, the lengths of $AD$ and $CD$ are $4 \text{ cm}$ and $5 \text{ cm}$, respectively.

**Step 1: Similar Triangles**The triangles $\triangle ADB$ and $\triangle CDB$ are similar by AAA similarity because $\angle ADB = \angle CDB = 90^\circ$ and $\angle DAB = \angle DBC$. Similarity criteria yield:
$$
\frac{BD}{AD} = \frac{DC}{BD}
$$

**Step 2: Calculating $BD$**From the similarity relation and substituting the given segment lengths, we get:
$$
BD^2 = AD \times DC = 4 \times 5
$$
Thus, solving for $BD$, we find:
$$
BD = \sqrt{20} = 2 \sqrt{5} \text{ cm}
$$

**Step 3: Calculating $BC$ Using Pythagoras' Theorem**In the right triangle $\triangle BDC$, using Pythagoras' theorem:
$$
BC^2 = BD^2 + CD^2 = (2 \sqrt{5})^2 + 5^2 = 20 + 25 = 45
$$
Then:
$$
BC = \sqrt{45} = 3\sqrt{5} \text{ cm}
$$

**Step 4: Calculating $AB$**Using proportionality from the triangles' similarity:
$$
\frac{BD}{DC} = \frac{BA}{BC}
$$
Substituting known values and solving for $BA$:
$$
\frac{2 \sqrt{5}}{5} = \frac{BA}{3 \sqrt{5}}
$$
Solving for $BA$:
$$
BA = \frac{2 \sqrt{5} \times 3 \sqrt{5}}{5} = 6 \text{ cm}
$$

### Conclusion

The length of $BD$ is

**$2 \sqrt{5} \text{ cm}$**The length of $AB$ is

**$6 \text{ cm}$**.

In the figure, the bisector of $\angle B$ and $\angle C$ meet at $O$. Then $\angle BOC$ is:

A) $90^{\circ} + \frac{1}{2} \angle A$

B) $90^{\circ} + \frac{1}{2} \angle B$

C) $90^{\circ} + \frac{1}{2} \angle C$

D) $90^{\circ} - \frac{1}{2} \angle A$

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