Question

In figure O is the centre of a circle of radius 5 cm.T is a point such that OT=13CM and OT intersects the circle at E.if AB is the tangent to the circle at E.find AB. sorry no figure i hope somebody can find it out

Answer 1

Hey ! Buddy . Below is your answer. Hope it helps!!

Answer 2

(The figure missing in the question is attached below)

Given:

Radius of the circle = 5 cm

Length of OT = 13 cm

To find:

The length of AB

Calculation:

The tangents AP, AE are drawn to the circle from the same point A. Therefore the lengths of the tangents will be equal

AP = AE = a(say)

Also from the figure we have

=> PA + AT = PT

=> AT = (12-x) cm

In the given figure, ΔOPT is a right angle triangle. By applying pythagoras theorem, we have

$\Rightarrow (OT)^2=(OP)^2+(PT)^2 \\ \\\Rightarrow 13^2=5^2+ (PT)^2 \\ \\\Rightarrow 169=25+ (PT)^2 \\ \\\Rightarrow (PT)^2 = 144 \\ \\\Rightarrow PT=12\ cm$

ΔAET is a right angle triangle. By applying pythagoras theorem, we have

$\Rightarrow (AT)^2=(AE)^2+(ET)^2 \\ \\\Rightarrow (12-x)^2=x^2+ (13-5)^2 \\ \\\Rightarrow (12-x)^2 =x^2+ 64 \\ \\\Rightarrow 144+x^2-24x=x^2+64 \\ \\\Rightarrow 24x=80\\ \\\Rightarrow x=\frac{10}{3} \\ \\\Rightarrow AE=BE= \frac{10}{3}$

From the given figure,

$\Rightarrow AB=AE+EB\\\\\Rightarrow AB=\frac{10}{3}+ \frac{10}{3}\\\\\Rightarrow AB= \frac{20}{3}$

$\boxed{The\ value\ of\ AB\ is\ \frac{20}{3}\ cm}$