Question

Kindly see the picture given in the attachment​

Answer 1

Topic

① Circles- Tangent

A tangent of a circle forms a right angle with the diameter.

② Triangles- Similarity of Three Right Triangles

In the following attachment of three similar right triangles, the formulas are gotten. (Refer to the attachment please.)

Solution

$\overline{AB}$ is a tangent, and $\overline{BC}$ is the diameter. Hence $\angle ABC=90^{\circ}$ and $\triangle ABC$ is a right triangle, right-angled at $\angle {B}$.

Let's construct a line segment $\overline{BD}$. $\overline{BC}$ is the diameter, and $D$ is a point on the circle. The measure of circumference angle is half the angle at the center. $\triangle BCD$ is a right triangle, right-angled at $\angle {D}$.

Now, $\triangle ABC\sim \triangle BCD\ \text{(AA similarity)}$.

$\implies \overline{AC} :\overline{BC} =\overline{BC}: \overline{CD} \ \text{(The Ratio of Corresponding Sides)}$

$\implies \overline{BC}^{2}=\overline{AC} \cdot \overline{CD}$

We proved it using the similarity of the right triangle. Now we can solve our problem.

$\implies \overline{AC} \cdot \overline{CD}=8^{2}$

$\implies \overline{AC} \cdot \overline{CD}=64$

The rest of the formulas can also be proved in the same manner.

This is the required answer.

Answer 2

$\large\underline{\sf{Given \:Question - }}$

In the given figure, BC is diameter of a circle. AB is tangent to a circle at point B. If radius of circle is 4 cm. Find the value of AC × DC.

(a) 192 cm

(b) 128 cm

(c) 64 cm

(d) cannot be determined

$\green{\large\underline{\sf{Solution-}}}$

Given that,

BC is diameter of circle and AB is tangent to a circle at point B.

Also, radius of circle is 4 cm, it implies BC = 8 cm

We know, Radius is perpendicular to tangent.

So,

$\bf\implies \:AB \perp \: BC$

$\bf\implies \: \angle \: ABC = 90 \degree$

Now, Join DB

Since, BC is diameter and we know angle in semi-circle is right angle.

$\bf\implies \: \angle \: BDC = 90 \degree$

Now, Consider In triangle CDA and triangle CBA

$\rm :\longmapsto\: \angle \: CDB \: = \: \angle \: ABC \: \: \{ \: each \: 90 \degree \: \}$

$\rm :\longmapsto\: \angle \: DCB \: = \: \angle \: ACB \: \: \{ \: common \: \}$

$\rm :\longmapsto\: \triangle \: CDB \: = \: \triangle \: CBA \: \: \{ \: By \: AA \: similarity \: \}$

$\rm :\longmapsto\:\dfrac{CD}{CB} = \dfrac{BC}{AC}$

$\rm :\longmapsto\: {BC}^{2} = AC \times CD$

$\rm :\longmapsto\: {8}^{2} = AC \times CD$

$\bf\implies \: AC \times CD = 64 \: cm$

Hence, Option (c) is correct

Additional Information :

1. Length of tangent drawn to a circle from external point are equal.

2. Tangents are equally inclined to the line segment joining the centre and external point.

3. Angle is same segment are equal.

4. Angle subtended by an arc at center is double the angle subtended by the same arc on circumference.

5. Sum of angles of opposite pair of cyclic quadrilateral is supplementary.

6. Radius is perpendicular to tangent.