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Selina Solutions Class 9 Mathematics Solutions for Exercise 1(D) in Chapter 1 - Chapter 1- Rational and Irrational Numbers
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Question 13 Exercise 1(D)
Using the following figure, show that BD=√𝒙.
Answer:
Let AB=x
BC=1
AC=x+1
Here, AC is diameter and O is the center
OA= OC = OD = radius = \frac{x+1}{2}
And OB = OC – BC = \frac{x+1}{2} -1= \frac{x+1}{2}
Now, using Pythagoras theorem,
OD²= OB²+BD²
\begin{array}{l} \left(\frac{x+1}{2}\right)^{2}=\left(\frac{x-1}{2}\right)^{2}+B D^{2} \\ \Rightarrow B D^{2}=\left(\frac{x+1}{2}\right)^{2}-\left(\frac{x-1}{2}\right)^{2} \\ \Rightarrow \frac{x^{2}+2 x+1-x^{2}+2 x-1}{4} \\ \Rightarrow \frac{4 x}{4}=x \\ \therefore, B D=\sqrt{x} \end{array}
Hence proved.
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