Circle

Solution of NCERT Exercise 10.2 (part2)

Question (7) 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:

Given, Radius of the bigger circle, OQ = 5 cm

Radius of the smaller circle, OA = 3 cm

Length of chord of the larger circle, PQ = ?

Since, chord of larger circle, PQ is the tangent to the smaller circle

Thus, radius of smaller circle, OA ⊥ PQ

This means, ∠ OAQ = 90⁰

In Δ OAQ

According to Pythagoras Theorem,

OQ² = OA² + AQ²

⇒ 5² = 3² + AQ²

⇒ 25 = 9 + AQ²

⇒ AQ² = 25 – 9

⇒ AQ² = 16

∴ AQ = √ 16 = 4

⇒ AQ = 4 cm

In Δ OAP

OP = 5 cm (because radius of the larger circle)

OA = 3 cm

Thus, as above calculated, PA = 4 cm

Now, PQ = PA + AQ

⇒ PQ = 4 cm + 4 cm = 16 cm

⇒ PQ = 16 cm Answer

Question (8) A quadrilateral ABCD is drawn to circumscribe a circle (see figure). Prove that AB+CD=AD+BC

Solution:

In the given, figure

DR and DS are the tangents to the circle from the same point D

Thus, DR = DS ----------- (i)

And, CR and CQ are tangents on the circle from the same point C

Thus, CR = CQ ------------ (ii)

Similarly, BP and BQ are tangents to the circle from the same point B

Thus, BP = BQ ------------- (iii)

And, AP and AS are tangents to the circle from the same point A

Thus, AP = AS ------------- (iv)

Now, by addition equations (i), (ii), (iii) and (iv), we get

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

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

⇒ CD + AB = AD + BC Proved

Question (9) In figure 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 = 90⁰

Solution:

Now, let join O and C in the given, figure

In Δ OPA and Δ OCA,

OP = OC

[∵ OP and OC are the radii of same circle]

AP and AC are tangents to the circle from the same point A

Thus, AP = AC

AO = AO

[Common sides in both of the triangles.]

Thus According to SSS congruence criterion

Δ OPA ≅ Δ OCA

Thus, ∠ POA = ∠ COA ------------- (i)

Similarly, Δ OQB ≅ Δ OCB

Thus, ∠ QOB = ∠ COB ------------- (ii)

In the given, figure, POQ is the diameter of circle. This means OPQ is a straight line.

Thus, ∠ POA + ∠ COA + ∠ COB + ∠ QOB = 180⁰

⇒ ∠ COA + ∠ COA + ∠ COB + ∠ QOB = 180⁰

[∵ according equation (i) ∠ POA = ∠ COA]

⇒ 2 ∠ COA + + ∠ COB + ∠ QOB = 180⁰

⇒ 2 ∠ COA + + ∠ COB + ∠ COB = 180⁰

[∵ according equation (i) ∠ QOB = ∠ COB]

⇒ 2 ∠ COA + 2 ∠ COB = 180⁰

⇒ ∠ COA + ∠ COB = 180/2 = 90⁰

Now, since, ∠ AOB = ∠ COA + ∠ COB

∴ ∠ AOB = 90⁰ Proved

Question (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 centre.

Solution:

Let there is a circle with centre O.

Let P is an external point from which two tangents PQ and PR has been drawn to the circle.

PQ touches at point Q on the circle.

And PR touches at point R on the circle.

Now, Q and R are joined.

Thus, QR is the line segment which joins points of contact Q and R together such that ∠ AOB at the centre O of the circle.

Now, from figure it is clear that

Radius OQ ⊥ PQ (tangent)

∴ ∠ OQP = 90⁰

Similarly, Radius OR ⊥ PR (tangent)

∴ ∠ ORP = 90⁰

Now, in quadrilateral OQPR

Sum of all interior angles = 360⁰

⇒ ∠ OQP + ∠ QPR + ∠ PRO + ∠ ROQ = 360⁰

S⇒ 90⁰ + ∠ QPR + 90⁰ + ∠ ROQ = 360⁰

⇒ 180⁰ + ∠ QPR + ∠ ROQ = 360⁰

⇒ ∠ QPR + &nag; ROQ = 360⁰ – 180⁰

⇒ ∠ QPR + &nag; ROQ = 180⁰

This means the angle between 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 centre. Proved

Question (11) Prove that the parallelogram circumscribing a circle is a rhombus.

Solution:

Let ABCD is a parallelogram which is circumscribing a circle with centre O.

To prove that, ABCD is a Rhombus

We know that, if all sides of a parallelogram are equal then it is called a Rhombus.

Thus, it is to be proved that, AB = BC = DC = AD

Now, since, ABCD is a parallelogram

∴ AB = DC --------- (i)

And BC = AD ------------- (ii)

Now, DR and DS are tangents to the circle from the same point D

∴ DR = DS ------------- (iii)

Similarly, CR and CQ are tangents to the circle from the same point C

∴ CR = CQ ------------- (iv)

And, BP and BQ are tangents to the circle from same point B

∴ BP = BQ -------------- (v)

And, AP and AS are tangents to the circle from the same point A

∴ AP = AS -------------- (vi)

Now, after adding equation (iii), (iv), (v) and (vi), we get

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

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

⇒ CD + AB = AD + BC

Now, from equation (i) and (ii), after substituting values of CD and AD, we get,

AB + AB = BC + BC

⇒ 2 AB = 2 BC

⇒ AB = BC ------------ (vii)

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

AB = BC = CD = DA

Thus, ABCD is a rhombus. Proved

Question (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 lengths 8 cm and 6 cm respectively (see figure). Find the sides AB and AC

Solution:

Given, CD = 6 cm, and BD = 8 cm

Then, side AB = ? and AC = ?

Let, circle touches the AC side of triangle at point F and AB side of the triangle at point E

Now, point C and E are joined

And B and F are joined

And A and O are joined.

Let, AF = m

Now, CF and CD are tangents to the circle from same point C

Thus, CF = CD = 6 cm

Simiarly, BD and BE are tangents to the circle from same point B

Thus, BE = BD = 8 cm

Similarly, AF and AE are tangents to the circle from

Thus, AE = AF = m

Now, AB = AE + BE = m + 8 cm

BC = BD + CD = 8 cm + 6 cm = 14 cm

CA = AF + CF = m + 6 cm

Now, we know that,

2s = AB + BC + CA

[Where s is the semi-perimeter of triangle]

⇒ 2s = m + 8 + 14 + 6 + m

⇒ 2s = 28 + 2 m

⇒ 2s = 2(14 + m)

⇒ s = 14 + m

Now, according to Heron's Formula, we know that,

Area of Δ AB `=sqrt(s(s-a)(s-b)(s-c))`

`=sqrt((14+m)(m)(8)(6))`

`=sqrt(48(14m+m^2))`

`=sqrt(16xx3(14m+m^2))`

`=4sqrt(3(14m+m^2))`

Now, area of Δ OBC `=1/2xxODxxBC`

`=1/2xx4xx14=28`

And area of Δ OCA `=1/2xxOFxxAC`

`=1/2xx4xx(6+m)`

`=2xx(6+m)`

`=12+2m`

And area of &Dleta; OAB `=1/2xxOExxAB`

`=1/2xx4xx(8+m)`

`=2xx(8+m)`

`=16+2m`

Now, Area of Δ ABC = Area of Δ OBC + Area of Δ OCA + Area of Δ OAB

`=>4sqrt(3(14m+m^2))` = 28 + 12 + 2m + 16 + 2m

`=>4sqrt(3(14m+m^2))` = 56 + 4 m

`=>sqrt(3(14m+m^2))=(4(14+m))/4`

`=>sqrt(3(14m+m^2))=14+m`

Now by squaring both sides

⇒ 3(14m + m²) = (14 + m)²

⇒ 42 m + 3m² = 196 + m² + 28 m

⇒ 42 m – 28 m + 3m² – m² – 196 =0

⇒ 2m² + 14 m – 196 =0

⇒ 2(m² + 7 m – 98) =0

⇒ m² + 7 m – 98 =0

⇒ m² + 14 m – 7 m – 98 = 0

⇒ m (m + 14) – 7 (m + 14) = 0

⇒ (m + 14) (m – 7) = 0

Now, when

m + 14 = 0

∴ m = – 14

And when

m – 7 = 0

∴ m = 7

Now, by ignoring negative value, m = 7 cm

Thus, Now, we have AB = m + 8

⇒ AB = 7 + 8 = 15 cm

[∵ m = 7]

And, AC = 6 + m = 6 + 7 = 13

Thus, AB = 15 cm and AC = 13 cm Answer

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

Solution:

Let ABCD is a quadrilateral which is circumscribing a circle with centre O.

Let quadrilateral touches the circle at point P, Q, R and S/

Let vertices of quadrilateral are joined to the centre of circle.

Now, in Δ OAP and Δ OAS,

AP = AS

[∵ AP and As are tangent from the same point]

OP = OS

[∵ OP and OS are radii of the same cirle]

OA = OA

[Common side of triangles]

∴ According to SSS (side-side-side) criterion

Δ OAP ≅ Δ OAS

Thus,

∠ POA = ∠ AOS

i.e. ∠ a = ∠ 8

Simiarly,

∠ 2 = ∠ 3

∠ 4 = ∠ 5

∠ 6 = ∠ 7

Now,

∠ 1 + ∠ 2 + ∠ 3 + ∠ 4 + ∠ 5 + ∠ 6 + ∠ 8 = 360⁰

⇒ (∠ 1 + ∠ 8) + (∠ 2 + ∠ 3) + (∠ 4 + ∠ 5) + (∠ 6 + ∠ 7) = 360⁰

⇒ 2 ∠ 1 + 2 ∠ 2 + 2 ∠ 5 + 2 ∠ 6 = 360⁰

⇒ 2 (∠ 1 + ∠ 2 + ∠ 5 + ∠ +) = 360⁰

⇒ (∠ 1 + ∠ 2) + (∠ 5 + ∠ 6) = 180⁰

⇒ ∠ AOB + ∠ COD = 180⁰

Simiarly, it can be proved that

∠ BOC + ∠ DOA = 180⁰

Thus, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

MCQs Test

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