IIMP 2025 CLASS 10 MATHS CIRCLE CHAPTER 10

 

Prove that the tangents drawn at the ends of a diameter of a circle are parallel. 

Proof:

∠1 = 90° …(i)
∠2 = 90° …(ii)
∠1 = ∠2----by (1) &(2) (AIA) 
∴PQ || RS

Q 2. Prove that the length of the tangents drawn from an external point to a circle are equal.. 

Answer:

Given :

Let AP and BP be the two tangents drawn from external point P to the circle with centre O.

To Prove : AP = BP

Proof :

In $\Delta$ AOP and $\Delta$BOP

OA = OB (radii) 

$\angle OAP=\angle OBP={90}^{\circ }$ 

(Line drawn from from center to the tangent through the point of contact is perpendicular) 

OP = OP (common)

$\therefore \Delta AOP\cong \Delta BOP$ (R.H.S.) 

$\therefore$ AP = BP ( C P C T ) 

Therefore the length of the

 tangents drawn from an

 external point to a circle are

 equal.

Q 3. Statement: 

BPT (Basic Proportionality Theorem), 

If a line is drawn parallel to one of the  triangle to intersects the other two sides in distinct points, 

then the other two sides of 

the  triangle are divided  into the same ratio. 

Proof:

Given:

In ∆ABC, DE || BC and AB and AC are intersected by DE at points D and E respectively.

To prove:

AD / DB = AE / EC

Construction:

Join BE and CD.and

Draw:

EG⊥AB and DF⊥AC

Proof:

We know that

ar( Δ ADE) = 1 / 2 × AD × EG   

ar( Δ DBE) = 1 / 2 × DB × EG   

So

ar(Δ ADE) / ar(Δ DBE) =

 AD / DB   ------- (1)

Similarly,

ar(Δ ADE) / ar(Δ ECD) =

 AE / EC   ----------(2)

Now, 

Δ DBE and Δ ECD

are the on the same base DE and also between the same parallels i.e. DE and BC,

So

ar(Δ DBE) = ar(Δ ECD)  ---(3)

By (1), (2) , (3) 

AD / DB = AE / EC  

Hence  proved.

Q 4.

Prove √3 is a Irrational number.

Proof:

Let √3 be a rational number.

So, $\sqrt{3}=\frac{\mathrm{p\; \_\_\_\_\; (1)}}{q}$

On squaring both sides

$3=\frac{p^2}{q^2}$
$q^2=\frac{p^2}{3}$

$\Rightarrow \mathrm{3 is\; a\; factor\; of}$$p^2$

$\Rightarrow \mathrm{3\; is\; a\; factor\; of\; p.}$

Now, again let p = 3 c.

So, $\sqrt{3}=\frac{\mathrm{3\; c}}{q}$

On squaring both sides
$3=\frac{\mathrm{9\; c}{}^{}}{q^2}$
$q^2=\mathrm{3\; c}{}^{}$
$c^2=\frac{q^2}{3}$
$\Rightarrow \mathrm{3\; is\; factor\; of}$$q^2$
$\Rightarrow 3$ is a factor of q.

Here 3 is a common factor of p, q  both

 So p, q are not co-prime.

Therefore our assumption is wrong. 

√$\sqrt{3}$ is an irrational number.

Q 5.

Prove that the parallelogram circumscribing a circle is a rhombus. 

 ABCD is a parallelogram  , So opposite sides of a parallelogram are equal. 

$AB=CD.$

$BC=AD.-----++++(1)$

$DR=DS$ --------------(2) 

 (Tangents on the circle from same point $D$)

Similarly

$CR=C\mathrm{Q\; ----------(3)}$

AP = AS ------------(4) 

PB = BQ  -------++(5) 

Adding all these equations we get

$DR+CR+BP+AP=DS+CQ+BQ+AS$

$(DR+CR)+(BP+AP)=(DS+AS)$ +( BQ + CQ) 

$CD+AB=AD+BC$

But   AB =CD   &   BC=AD

     AB + AB = BC + BC

$2AB=2BC$

$\mathrm{Therefore\; A}B=BC-------(6)$

From equation $\left(1\right)$, $\mathrm{\&\; (6)}$we get

$AB=BC=CD=DA$

$\therefore ABCD$ is a Rhombus

Q 6. Find relationship between x and y such that the point P(x, y) is equidistant from the points A (2, 5) and B (-3, 7)

Solution:

Let P (x, y) be equidistant from the points A (2, 5) and B (-3, 7). so

∴ AP = BP

         AP 2 = BP 2 

(x – 2)2 + (y – 5)2 = (x + 3)2 + (y – 7)2

x2 – 4x + 4 + y2 – 10y + 25

                 =

x2 + 6x + 9 + y2 – 14y + 49

⇒ -4x – 10y +29 =  6x - 14y +58 

⇒ -10x + 4y = 29

 is the required relation.

Q 6.

 

 Find relationship between x and y such that the point P(x, y) is equidistant from the points A (2, 5) and B (-3, 7). (2011D)

Solution:

Let P (x, y) be equidistant from the points A (2, 5) and B (-3, 7). so

∴ AP = BP

         AP 2 = BP 2 

(x – 2)2 + (y – 5)2 = (x + 3)2 + (y – 7)2

x2 – 4x + 4 + y2 – 10y + 25

                 =

x2 + 6x + 9 + y2 – 14y + 49

⇒ -4x – 10y +29 =  6x - 14y +58 

⇒ -10x + 4y = 29

 is the required relation.

Q7. 

Draw the graph of 2x + y = 6 and 2x – y + 2 = 0. Shade the region bounded by these lines and x-axis. Find the area of the shaded region. [CBSE 2007]

Solution:

The given system of equation is
2x + y – 6 = 0 ………. (i)
2x – y + 2 = 0 ……… (ii)

 solutions for each equation of the system are:
(i) ⇒ y = 6 – 2x

Table of solutions for 2x + y – 6 = 0

Similarly 

Table of solutions for 2x – y + 2 = 0

Plotting these on the graph

These intersect at (1, 4), so this is the solution. 
In graph, area bounded by the lines and x-axis is

 ∆PAB which is shaded.

Draw PM ⊥ x-axis

PM = y-coordinate of P  (1, 4)    = 4 units

and AB = 1 + 3 = 4 units

∴ Area of shaded region

= Area of ∆PAB = 12 × AB × PM

= 12 × 4 × 4 = 8 sq. units.