NCERT Solutions For Class 10 Math Chapter – 10 Exercise – 10.2

NCERT Solutions For Class 10 Math Chapter – 10 Exercise – 10.2

 

Q1. Choose the correct option and give justification.

From a point Q , the length of the tangent to a circle is 24 cm and the distance Q from the center is 25 cm. The radius of the circle is

• 7 cm
• 12 cm
• 15 cm
• 5 cm

Solution:-

 

• Is correct option.

 

Q2. Choose the correct option option and give justification.

If TP and TQ are the two tangents to a circle with center O so that ∟POQ = 110⁰ then ∟PTQ is equal to

• 60⁰
• 70⁰
• 80⁰
• 90⁰

Solution:-

 

• Is correct option.

 

Q3. If triangles PA and PB from a point P to a circle with center O are inclined to each other at angle of 80⁰ then ∟POA is equal to

• 50⁰
• 60⁰
• 70⁰
• 80⁰

Solution:-

 

• Is correct option.

 

Q4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Solution:-

 

 

 

 

 

 

 

OB is perpendicular to RS and OA perpendicular to PQ

So , ∟OAP = ∟OAQ = ∟OBR = ∟OBS = 90⁰

∟OBR and ∟OAQ are alternate interior angles.

∟OBR = ∟OAQ and ∟OBS = ∟OAP (Since they also alternate interior angles)

 

It can be said that line PQ and the line RS will be parallel to each  other. ( Hence Proved).

 

Q5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the center.

Solution:-

 

 

 

 

 

 

 

 

PQ intersects CD and AB at R and P respectively.

As , CD // AB,

The line segment PQ is the line of intersection.

 

ORP and RPA are equal as they are alternate interior angles

So , ∟ORP = ∟RPA

And ,

∟RPA = 90⁰

∟ORP = 90⁰

∟ROP + ∟OPA = 180⁰

∟ROP = 90⁰

 

Q6. The length of a tangent from a point A at distance 5 cm from the center of the circle is 4 cm. Find the  radius of the circle.

Solution:-

 

 

 

 

 

 

 

In ∆ABO ,

OA² = AB² + BO² (Using pythagoras theorem )

5² = 4² + BO²

BO = 3 cm

 

Q7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Solution:-

 

 

 

 

Using pythagoras theorem in ∆OPA,

OA² = AP² + OP²

5² = AP² + 3²

AP= 4 cm.

AB = 2AP = 2 X 4 = 8 cm

 

Q8. A quadrilateral ABCD is drawn to circumscribe a circle Prove that  AB + CD = AD + BC.

Solution:-

The figure given is:

 

From this figure we can conclude a few points which are:

(i) DR = DS

(ii) BP = BQ

(iii) AP = AS

(iv) CR = CQ

Since they are tangents on the circle from points D, B, A, and C respectively.

Now, adding the LHS and RHS of the above equations we get,

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

By rearranging them we get,

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

By simplifying,

AD+BC= CD+AB

9. In Fig. 10.13, XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X′Y′ at B. Prove that ∠ AOB = 90°.

Answer:

From the figure given in the textbook, join OC. Now, the diagram will be as-

Now the triangles △OPA and △OCA are similar using SSS congruency as:

(i) OP = OC They are the radii of the same circle

(ii) AO = AO It is the common side

(iii) AP = AC These are the tangents from point A

So, △OPA ≅ △OCA

Similarly,

△OQB ≅ △OCB

So,

∠POA = ∠COA … (Equation i)

And, ∠QOB = ∠COB … (Equation ii)

Since the line POQ is a straight line, it can be considered as a diameter of the circle.

So, ∠POA +∠COA +∠COB +∠QOB = 180°

Now, from equations (i) and equation (ii) we get,

2∠COA+2∠COB = 180°

∠COA+∠COB = 90°

∴∠AOB = 90°

10. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the center.

Answer:

First, draw a circle with centre O. Choose an external point P and draw two tangents PA and PB at point A and point B respectively. Now, join A and B to make AB in a way that it subtends ∠AOB at the center of the circle. The diagram is as follows:

From the above diagram, it is seen that the line segments OA and PA are perpendicular.

So, ∠OAP = 90°

In a similar way, the line segments OB ⊥ PB and so, ∠OBP = 90°

Now, in the quadrilateral OAPB,

∴∠APB+∠OAP +∠PBO +∠BOA = 360° (since the sum of all interior angles will be 360°)

By putting the values we get,

∠APB + 180° + ∠BOA = 360°

So, ∠APB + ∠BOA = 180° (Hence proved).

11. Prove that the parallelogram circumscribing a circle is a rhombus.

Answer:

Consider a parallelogram ABCD which is circumscribing a circle with a center O. Now, since ABCD is a parallelogram, AB = CD and BC = AD.

From the above figure, it is seen that,

(i) DR = DS

(ii) BP = BQ

(iii) CR = CQ

(iv) AP = AS

These are the tangents to the circle at D, B, C, and A, respectively.

Adding all these we get,

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

By rearranging them we get,

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

Again by rearranging them we get,

AB+CD = BC+AD

Now, since AB = CD and BC = AD, the above equation becomes

2AB = 2BC

∴ AB = BC

Since AB = BC = CD = DA, it can be said that ABCD is a rhombus.

12. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.

Answer:

The figure given is as follows:

Consider the triangle ABC,

We know that the length of any two tangents which are drawn from the same point to the circle is equal.

So,

(i) CF = CD = 6 cm

(ii) BE = BD = 8 cm

(iii) AE = AF = x

Now, it can be observed that,

(i) AB = EB+AE = 8+x

(ii) CA = CF+FA = 6+x

(iii) BC = DC+BD = 6+8 = 14

Now the semi perimeter “s” will be calculated as follows

2s = AB+CA+BC

By putting the respective values we get,

2s = 28+2x

s = 14+x

 

By solving this we get,

= √(14+x)48x ……… (i)

Again, the area of △ABC = 2 × area of (△AOF + △COD + △DOB)

= 2×[(½×OF×AF)+(½×CD×OD)+(½×DB×OD)]

= 2×½(4x+24+32) = 56+4x …………..(ii)

Now from (i) and (ii) we get,

√(14+x)48x = 56+4x

Now, square both the sides,

48x(14+x) = (56+4x)2

48x = [4(14+x)]2/(14+x)

48x = 16(14+x)

48x = 224+16x

32x = 224

x = 7 cm

So, AB = 8+x

i.e. AB = 15 cm

And, CA = x+6 =13 cm.

13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Answer:

First draw a quadrilateral ABCD which will circumscribe a circle with its centre O in a way that it touches the circle at point P, Q, R, and S. Now, after joining the vertices of ABCD we get the following figure:

Now, consider the triangles OAP and OAS,

AP = AS (They are the tangents from the same point A)

OA = OA (It is the common side)

OP = OS (They are the radii of the circle)

So, by SSS congruency △OAP ≅ △OAS

So, ∠POA = ∠AOS

Which implies that∠1 = ∠8

Similarly, other angles will be,

∠4 = ∠5

∠2 = ∠3

∠6 = ∠7

Now by adding these angles we get,

∠1+∠2+∠3 +∠4 +∠5+∠6+∠7+∠8 = 360°

Now by rearranging,

(∠1+∠8)+(∠2+∠3)+(∠4+∠5)+(∠6+∠7) = 360°

2∠1+2∠2+2∠5+2∠6 = 360°

Taking 2 as common and solving we get,

(∠1+∠2)+(∠5+∠6) = 180°

Thus, ∠AOB+∠COD = 180°

Similarly, it can be proved that ∠BOC+∠DOA = 180°

Therefore, the opposite sides of any quadrilateral which is circumscribing a given circle will subtend supplementary angles at the center of the circle.