There is a circle with a centre T and a radius of 8 cm. Two tangents KE and KI are drawn from point K, which is 17 cm from the centre. The area of the quadrilateral TEKI is:

The correct answer is 120 sq.cm.

Step-by-Step Explanation:

1. Understand the given properties:

We are given a circle centered at point T with radius $r=8\text{cm}$.

Point K is outside the circle such that the distance from the centre T to K is $TK=17\text{cm}$.

Two tangents, KE and KI, are drawn from point K to touch the circle at points E and I, respectively. Thus, $TE=TI=8\text{cm}$.

2. Properties of Tangents to a Circle:

A tangent to a circle is perpendicular to the radius drawn to the point of contact.

Therefore, $\angle TEK=90°$ and $\angle TIK=90°$.

This means that $△TEK$ and $△TIK$ are right-angled triangles.

3. Calculate the length of the tangents:

In the right-angled triangle $△TEK$, applying the Pythagorean theorem:

$T{K}^2=T{E}^2+K{E}^2$

${17}^2=8^2+K{E}^2$

$289=64+K{E}^2$

$K{E}^2=289-64=225$

$KE=\sqrt{225}=15\text{cm}$

4. Calculate the total area of quadrilateral TEKI:

The quadrilateral TEKI is composed of two congruent right-angled triangles: $△TEK$ and $△TIK$.

The area of one right triangle is given by:

$\text{Area of}△TEK=\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}\times 8\times 15=60\text{sq.cm}$

Since quadrilateral TEKI consists of two such triangles:

$\text{Total Area}=2\times 60=120\text{sq.cm}$