Question

In the figure given below, CD is the diameter of the circle which meets the chord AB at P such that AP = BP = 12 cm. If DP = 8cm , find the radius of the circle.
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Answer 1

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Step-by-step explanation:

Answer 2

The radius of the circle is 13 cm.

Step-by-step explanation:

We are given in the figure that CD is the diameter of the circle which meets the chord AB at P such that AP = BP = 12 cm. Also, DP = 8 cm.

As, we know that OA = OB = OD = radius of the circle.

Let OD = r ; which means OA = OB = r

We are given that DP = 8cm.

Since, OD = OP + DP

              r  =  OP + 8

So, OP = (r - 8) cm.

Now, it is also provided that AP = BP = 12 cm which means P is the mid-point of chord AB and which further means that;

$\angle$OPA = $\angle$OPB = 90°  {right angle traingles}

So, In $\triangle$OPA we will use Pythagoras Theorem, i.e;

    $(\text{Hypotenuse})^{2} = (\text{Base})^{2} +(\text{Perpendicular})^{2}$

          $(\text{OA})^{2} = (\text{AP})^{2} +(\text{OP})^{2}$

           $\text{r}^{2} = (\text{12})^{2} +(\text{r-8})^{2}$

           $\text{r}^{2} = 144 +\text{r}^{2} +64-16\text{r}$      

            $16\text{r}=144+64$

              r  =  $\frac{208}{16}$  =  13 cm

Hence, the radius of the circle is 13 cm.