Q. 1 In figure, 𝑃𝑄 is a chord of length 16 𝑐𝑚, of a circle of radius 10 𝑐𝑚. The tangents at 𝑃 and 𝑄 intersect at a point 𝑇. Find the length of 𝑇𝑃.

 

(OR)

 

Q.2 Draw triangle 𝐴𝐵𝐶 such that 𝐵𝐶 = 5 𝑐𝑚, ∠𝐴𝐵𝐶 = 60°.∠𝐴𝐶𝐵 = 30º. Now construct Δ𝐴'𝐵𝐶' corresponds to Δ𝐴𝐵𝐶 with 𝐴'𝐵 : 𝐴𝐵 = 3 : 2

Given that,

 𝑃𝑄 is a chord of length 16 𝑐𝑚 

Also, radius  10 𝑐𝑚 

The tangents at 𝑃 and 𝑄 intersect at a point 𝑇. 

We have to find the length of 𝑇𝑃.

 Here, 𝑂𝑃 = 10 𝑐𝑚 

𝑃𝑀 = 𝑀Q

$=\frac{8}{2}$

= 4 𝑐𝑚 

In ∆𝑂𝑀𝑃,

By using Pythagoras Theorem 

𝑂𝑃² = 𝑃𝑀² + 𝑂𝑀² 

10² = 8² + 𝑂𝑀² 

𝑂𝑀² = 100 − 64 

𝑂𝑀² = 36 

𝑂𝑀 = 6 𝑐𝑚 

Let ∠𝑃𝑂𝑀 be θ

$\begin{array}{l}\tan \theta =\frac{PM}{MO}\\ =\frac{8}{6}\end{array}$

In ∆𝑂𝑇𝑃,

$$\begin{gathered} 𝑐𝑜𝑡 \theta = \frac{𝑂𝑃}{ 𝑇𝑃} \\ = \frac{10 }{ 𝑇𝑃} \end{gathered}$$

We know that $𝑂𝑃\perp 𝑇𝑃$ 

Radius is always perpendicular to Tangent.

$Now, we know that: \cot \theta =\frac{1}{\tan \theta }$

$\Rightarrow 𝑐𝑜𝑡 \theta =\frac{ 6 }{8} (\because 𝑡𝑎𝑛 \theta = 8 6 )$

$\Rightarrow \frac{ 6 }{8} = \frac{10}{ 𝑇𝑃}$

$\Rightarrow 𝑇𝑃 = \frac{80 }{6}$

$\Rightarrow 𝑇𝑃 = 13. 33 𝑐𝑚$

$Hence, the length of 𝑇𝑃 is 13.33 cm$

(OR)

1. Draw a line segment BC  of length 5 cm. 

2. Draw the angles of 60°  and 30⁰  on the points B  and C  respectively which intersect each other at A.  

3. ΔABC is the given triangle. 

4. Draw a ray BX  making an acute angle with BC. 

5. Locate three points B₁, B₂ and B₃ on line segment BX. Such that BB₁ = B₁B₂ = B₂B₃

6. Join B₂C. 

7. Draw B₃C' || B₂C to intersect the extended line BC' at C'.

8. Through C draw a line parallel to AC intersecting extended line segment BA at A'.

ΔA'BC' is the required triangle.