AB is a tangent to the circle with centre 'O' and 'A' is the point of contact. If OB = 5 sqrt2 cm, then prove that riangle AOB is an isosceles triangle. Find the radius of the circle.

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IV. Answer the following questions:

24. AB is a tangent to the circle with centre 'O' and 'A' is the point of contact. If OB = $5\sqrt{2}$ cm, then prove that $\triangle AOB$ is an isosceles triangle. Find the radius of the circle.

Step 1: Prove $\triangle AOB$ is an isosceles triangle.
A tangent to a circle is perpendicular to the radius at the point of contact. Therefore, $\angle OAB = 90^\circ$.
In $\triangle AOB$, the sum of angles is $180^\circ$.
We are given $\angle OBA = 45^\circ$ (from the diagram).

$$\angle AOB + \angle OBA + \angle OAB = 180^\circ$$ $$\angle AOB + 45^\circ + 90^\circ = 180^\circ$$ $$\angle AOB + 135^\circ = 180^\circ$$ $$\angle AOB = 180^\circ - 135^\circ$$ $$\angle AOB = 45^\circ$$

Since $\angle OBA = 45^\circ$ and $\angle AOB = 45^\circ$, two angles of $\triangle AOB$ are equal. Therefore, the sides opposite to these equal angles must be equal, which means $OA = AB$.
Hence, $\triangle AOB$ is an isosceles triangle.

Step 2: Find the radius of the circle.
In the right-angled triangle $\triangle AOB$, we can use trigonometric ratios. We know $\angle OBA = 45^\circ$ and $OB = 5\sqrt{2}$ cm.
The radius is $OA$.

$$\sin(\angle OBA) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{OA}{OB}$$ $$\sin(45^\circ) = \frac{OA}{5\sqrt{2} \text{ cm}}$$ $$\frac{1}{\sqrt{2}} = \frac{OA}{5\sqrt{2} \text{ cm}}$$ $$OA = \frac{5\sqrt{2}}{\sqrt{2}} \text{ cm}$$ $$OA = 5 \text{ cm}$$

The radius of the circle is $\boxed{\text{5 cm}}$.

25. Find the zeroes of the quadratic polynomial $p(x) = x^2 - 2x - 8$ and verify the relationship between the zeroes and the coefficients.

Step 1: Find the zeroes of the polynomial.
To find the zeroes, set $p(x) = 0$:

$$x^2 - 2x - 8 = 0$$

Factor the quadratic expression. We need two numbers that multiply to $-8$ and add to $-2$. These numbers are $-4$ and 2.

$$(x-4)(x+2) = 0$$

Set each factor to zero:

$$x-4 = 0 \Rightarrow x = 4$$ $$x+2 = 0 \Rightarrow x = -2$$

The zeroes of the polynomial are $\boxed{4 \text{ and } -2}$.

Step 2: Verify the relationship between the zeroes and the coefficients.
For a quadratic polynomial $ax^2 + bx + c$, the sum of zeroes is $-\frac{b}{a}$ and the product of zeroes is $\frac{c}{a}$.
For $p(x) = x^2 - 2x - 8$, we have $a=1$, $b=-2$, $c=-8$.
Let the zeroes be $\alpha = 4$ and $\beta = -2$.

• Sum of zeroes:

  $$\alpha + \beta = 4 + (-2) = 2$$

  From coefficients:

  $$-\frac{b}{a} = -\frac{(-2)}{1} = 2$$

  The sum of zeroes matches the formula.

• Product of zeroes:

  $$\alpha \beta = 4 \times (-2) = -8$$

  From coefficients:

  $$\frac{c}{a} = \frac{-8}{1} = -8$$

  The product of zeroes matches the formula. The relationship between the zeroes and coefficients is verified.

26. Prove that "The tangent at any point of a circle is perpendicular to the radius through the point of contact".

Step 1: Set up the proof.
Let a circle have center O and radius $r$. Let AB be a tangent to the circle at point P. We need to prove that the radius OP is perpendicular to the tangent AB, i.e., $OP \perp AB$.

Step 2: Consider a point on the tangent.
Take any point Q on the line AB, other than P. Since AB is a tangent, it touches the circle at only one point, P. This means that any other point Q on the line AB must lie outside the circle.

Step 3: Compare distances.
If Q lies outside the circle, then the distance OQ must be greater than the radius OP.

$$OQ > OP$$

Step 4: Conclude the proof.
This implies that OP is the shortest distance from the center O to any point on the line AB. In geometry, the shortest distance from a point to a line is always the perpendicular distance.
Therefore, $OP \perp AB$.
Hence, the tangent at any point of a circle is perpendicular to the radius through the point of contact.

27. $\triangle ABC \sim \triangle PQR$. In $\triangle ABC$, $\angle A = 55^\circ$, $\angle C = 35^\circ$ and the sides of $\triangle ABC$ are in the ratio $3:4:5$. If the perimeter of $\triangle PQR$ is $60 \text{ cm}$, then find the measure of sides and angles of $\triangle PQR$.

Step 1: Find the angles of $\triangle ABC$.
The sum of angles in a triangle is $180^\circ$.

$$\angle A + \angle B + \angle C = 180^\circ$$ $$55^\circ + \angle B + 35^\circ = 180^\circ$$ $$90^\circ + \angle B = 180^\circ$$ $$\angle B = 180^\circ - 90^\circ$$ $$\angle B = 90^\circ$$

Step 2: Find the angles of $\triangle PQR$.
Since $\triangle ABC \sim \triangle PQR$, their corresponding angles are equal.

$$\angle P = \angle A = 55^\circ$$ $$\angle Q = \angle B = 90^\circ$$ $$\angle R = \angle C = 35^\circ$$

The angles of $\triangle PQR$ are $\boxed{\angle P = 55^\circ, \angle Q = 90^\circ, \angle R = 35^\circ}$.

Step 3: Find the sides of $\triangle PQR$.
The sides of $\triangle ABC$ are in the ratio $3:4:5$. Let the sides be $3k, 4k, 5k$.
Since $\triangle ABC \sim \triangle PQR$, the sides of $\triangle PQR$ will also be in the same ratio $3:4:5$. Let the sides of $\triangle PQR$ be $3x, 4x, 5x$.
The perimeter of $\triangle PQR$ is given as $60 \text{ cm}$.

$$\text{Perimeter of } \triangle PQR = 3x + 4x + 5x$$ $$60 \text{ cm} = 12x$$ $$$$