The points $ \mathrm{A}(2,9), \mathrm{B}(a, 5) $ and $ \mathrm{C}(5,5) $ are the vertices of a triangle $ \mathrm{ABC} $ right angled at $ \mathrm{B} $. Find the values of $ a $ and hence the area of $ \triangle \mathrm{ABC} $.

AcademicMathematicsNCERTClass 10

Given:

The points $A (2, 9), B (a, 5)$ and $C (5, 5)$ are the vertices of a triangle ABC right angled at B.

To do:

We have to find the values of $a$ and hence the area of $\triangle ABC$.

Solution:

We know that,

The distance between two points \( \mathrm{A}\left(x_{1}, y_{1}\right) \) and \( \mathrm{B}\left(x_{2}, y_{2}\right) \) is \( \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \).

The triangle is right-angled at B.

This implies, by Pythagoras theorem,

\( AB^2+BC^2=AC^2 \).......(i)

Therefore,

\( \mathrm{AB}=\sqrt{(a-2)^{2}+(5-9)^{2}} \)

\( =\sqrt{a^{2}+4-4 a+16} \)

\( =\sqrt{a^{2}-4 a+20} \)

\( \mathrm{BC}=\sqrt{(5-a)^{2}+(5-5)^{2}} \)

\( =\sqrt{(5-a)^{2}+0}=5-a \)

\( \mathrm{CA}=\sqrt{(2-5)^{2}+(9-5)^{2}} \)

\( =\sqrt{(-3)^{2}+(4)^{2}} \)

\( =\sqrt{9+16} \)

\( =\sqrt{25}=5 \)

Substituting the values of \( \mathrm{AB}, \mathrm{BC} \) and \( \mathrm{AC} \) in (i), we get,

\( (5)^{2}=\sqrt{\left(a^{2}-4 a+20\right)^{2}}+(5-a)^{2} \)

\( \Rightarrow 25=a^{2}-4 a+20+25+a^{2}-10 a \)

\( \Rightarrow 2 a^{2}-14 a+20=0 \)

\( \Rightarrow a^{2}-7 a+10=0 \)

\( \Rightarrow a^{2}-2 a-5 a+10=0 \)

\( \Rightarrow a(a-2)-5(a-2)=0 \)

\( \Rightarrow (a-2)(a-5)=0 \)

\( \Rightarrow a=2 \) or \( a=5 \)

If \( a=5, \) the length of \( \mathrm{BC}=0 \) which is not possible because the sides \( \mathrm{AB}, \mathrm{BC} \) and CA from a right angled triangle.

Therefore, \( a=2 \)

The coordinates of \( \mathrm{A}, \mathrm{B} \) and \( \mathrm{C} \) are \( (2,9),(2,5) \) and \( (5,5), \) respectively.

Area of \( \Delta \mathrm{ABC}=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right] \)

\( =\frac{1}{2}[2(5-5)+2(5-9)+5(9-5)] \)

\( =\frac{1}{2}[2 \times 0+2(-4)+5(4)] \)

\( =\frac{1}{2}(0-8+20) \)

\( =\frac{1}{2} \times 12=6 \) 

Hence, the required area of \( \Delta \mathrm{ABC} \) is 6 sq. units.

raja
Updated on 10-Oct-2022 13:28:51

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