Question

In the given figure, AB is a chord of length 24 cm (see figure). The tangents At A and B intersectat
point P. If the radius of circle is 13 cm. Find the length of PA,
​

Answer 1

$\huge{\orange{\underline{\red{\mathtt{GIVEN:-}}}}}$

chord length is 24cm.

radius length is 13cm.

$\huge{\orange{\underline{\red{\mathtt{FIND:-}}}}}$

LENGTH OF PA.

$\large{\orange{\underline{\red{\mathtt{APPLYING \: THIS \:THEORAM:-}}}}}$

the perpendicular inserted on the chord from the center of circle bisects that organism.

$\large{\orange{\underline{\red{\mathtt{STEL \: BY \: STEP \:EXPLANATION:- }}}}}$

AP=1/2AB.......(using theorem)

AP=1/$\cancel{2}$×$\cancel{24}$

AP=12cm.

$\huge{\orange{\underline{\red{\mathtt{EXTRA:-}}}}}$

IN ΔOPA.

OA²=OP²+PA².....(Pythagoras theorem)

13²=OP²+12²

169=OP²+144

.°.OP=√25

.°.OP=5cm.

$\huge{\orange{\underline{\red{\mathtt{THANKS.}}}}}$

Answer 2

The length of PA is $\frac{156}{5}$ cm.

To find : Length of PA.

Given :      

AB is a chord.

Length of chord AB = 24 cm.

Tangents A and B intersects point P.

Radius of circle = 13 cm.

From the figure,

Note : Figure is attached below.

OA = Radius ( r ) = 13 cm.

ON is perpendicular to AB.

So, AN = NB = $\frac{AB}{2}$ = $\frac{24}{2}$ = 12 cm.

In ΔONA,

Using Pythagoras theorem,

ON² + NA² = OA²

ON² = OA² - NA²

ON² = 13² - 12²          

ON² = 169 - 144 = 25

ON  = $\sqrt{25}$ = 5

ON = 5 cm.

In ∠AON,

tan ∠AON = $\frac{AN}{ON}$ = $\frac{12}{5}$

In ΔOAP,

tan ∠AON = $\frac{PA}{OA}$

PA = tan ∠AON × OA

     = $\frac{12}{5}\times 13$

PA = $\frac{156}{5}$ cm.

Therefore, the length of PA = $\frac{156}{5}$ cm.

To learn more...

1. Two circles with centres A and B of radius 3 cm and 4 cm respectively intersect at two points C and D such that AC and BC are tangents to the two circles. find the length of the common chord CD

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