3 Marks Question

Question 513 Marks

Find the length of altitude AD of an isosceles $\triangle\text{ABC}$ in which AB = AC = 2a units and BC = a units.

Answer

Given: $\triangle\text{ABC}$ in which AB = AC = 2a units and BC = a units. Const: Draw $\text{AD}\perp\text{BC}$ then D is the midpoint of BC.

In $\triangle\text{ABC}$ $\text{BC}=\text{a}$ and $\text{BD}=\frac{\text{BC}}{2}=\frac{\text{a}}{2}$ In $\triangle\text{ADB},$ $(\text{AB})^2=\text{AD}^2+\text{BD}^2$ $\text{AD}^2=\Big(\text{AB}^2-\text{BD}^2\Big)$ $\text{AD}^2=\bigg[(\text{2a})^2-\Big(\frac{\text{a}}{2}\Big)^2\bigg]$ $\text{AD}^2=\bigg[4\text{a}^2-\frac{\text{a}^2}{4}\bigg]=\frac{15\text{a}^2}{4}$ $\Rightarrow\text{AD}=\frac{\text{a}\sqrt{15}}{2}\text{units}$

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Question 523 Marks

If the lengths of the sides BC, CA and AB of a $\triangle\text{ABC}$ are a, b and c respectively and AD is the bisectore of $\angle\text{A}$ then find the lengths of BD and DC.

Answer

Given $\triangle\text{ABC}$ in which AD, the bisector of $\angle\text{A}$ meets BC in D.
Let = x ⇒ DC = (a - x)
Then by the angle Bisector theorem,
$\frac{\text{BD}}{\text{DC}}=\frac{\text{AB}}{\text{AC}}$
$\Rightarrow\frac{\text{x}}{\text{a}-\text{x}}=\frac{\text{c}}{\text{b}}$
$\Rightarrow\text{xb}=\text{c}(\text{a}-\text{x})$
$\Rightarrow\text{xb}=\text{ac}-\text{xc}$
$\Rightarrow\text{xb}+\text{xc}=\text{ac}$
$\Rightarrow\text{x}=\frac{\text{ac}}{\text{b+c}}$
So, $\text{BD}=\frac{\text{ac}}{\text{b+c}}$
$\text{DC}=\text{a}-\text{x}\Rightarrow\text{DC}=\text{a}-\frac{\text{ac}}{\text{b+c}}$
$=\frac{\text{ab}+\text{ac}-\text{ac}}{\text{b}+\text{c}}=\frac{\text{ab}}{\text{b}+\text{c}}$
Hence, $\text{BD}=\frac{\text{ac}}{\text{b}+\text{c}}$ and $\text{DC}=\frac{\text{ab}}{\text{b}+\text{c}}.$

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Maths 3 Marks Question