Kolkata board math's chapter 3 solution

Theorems Related To Circle

Class 10^th Mathematics West Bengal Board Solution

Let Us Work Out 3.1

1. Let us see the adjoining figure of the circle with centre O and write the radii which…
2. Let us write in the following by understanding it. i. In a circle, there are number of…
3. With the help of scale and pencil compass let us draw a circle and indicate centre,…
4. Let us write true or false: i. The circle is a plane figure. ii. The segment is a plane…

Let Us Work Out 3.2

1. The length of a radius of a circle with its centre O is 5 cm. and the length of its…
2. The length of a diameter of a circle with its centre at O is 26 cm. The distance of the…
3. The length of a chord PQ of a circle with its centre O is 4 cm. and the distance of PQ…
4. The lengths of two chords of a circle with its centre O are 6 cm. and 8 cm. If the…
5. If the length of a chord of a circle is 48 cm and the distance of it from the centre is…
6. In the circle of adjoining figure with its center at O, OP ⊥ AB; if AB = 6 cm. and PC =…
7. A straight line intersects one of the two concentric circles at the points A and B and…
8. I prove that, the two intersecting chords of any circle cannot bisect each other unless…
9. The two circles with centers X and Y intersect each other at the points A and B. A is…
10. The two parallel chords AB and CD with the lengths of 10 cm and 24 cm in a circle are…
11. The centers of two circles are P and Q; they intersect at the points A and B. The…
12. The two chords AB and AC of a circle are equal. I prove that, the bisector of ∠BAC…
13. If the angle-bisector of two intersecting chords of a circle passes through its…
14. I prove that, among two chords of a circle the length of the nearer to centre is…
15. Let us write by proving the chord with the least length through any point in a circle.…
16. Q16A1 The lengths of two chords of a circle with centre O are equal. If ∠AOB = 60°, then…
17. Q16A2 The length of a radius of a circle is 13 cm. and the length of a chord of a circle…
18. Q16A3 Ab and CD are two equal chords of a circle with its centre O. If the distance of the…
19. Q16A4 The length of each of two parallel chord is 16 cm. If the length of the radius of…
20. Q16A5 The centre of two concentric circle is O; a straight line intersects a circle at the…
21. Let us write True/False: i. Only one circle can be drawn through three collinear…
22. Let us fill in the blanks: i. If the ratio of two chords PQ and RS of a circle with…
23. Two equal circles of radius 10 cm. intersect each other and the length of their…
24. AB and AC are two equal chords of a circle having the radius of 5 cm. The centre of…
25. The lengths of two chords AB and CD of a circle with its centre O are equal. If ∠AOB…
26. P is any point in a circle with its centre O. If the length of the radius is 5 cm.…
27. The two circles with their centres at P and Q intersect each other at the points A…

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Let Us Work Out 3.1

Question 1.

Let us see the adjoining figure of the circle with centre O and write the radii which are situated in the segment PAQ.

Answer:

Radius is the line segment that joins center and circumference of a circle.

So, OA, OC, OP, OQ and OB are the radii.

Question 2.

Let us write in the following by understanding it.

i. In a circle, there are number of points.

ii. The greatest chord of the circle is.

iii. The chord divides the circular region into two.

iv. All diameters of the circle pass through.

v. If two segments are equal, then their two arcs are in length.

vi. The sector of the circular region is the region enclosed by the arc and the two.

vii. The length of the line segment joining the point outside the circle and the center is than the length of radius.

Answer:

(i) Infinite

The circle is a collection of infinite number of points lying on a plane, each of the points is equidistant from a fixed point on that plane.

(ii) Diameter

A diameter is the greatest chord of any circle.

(iiii) Segment

Any chord of a circle divide the circle in two parts, the one with major area is known as major segment and the other part is known as minor segment.

(iv) Origin

As you can see in this image, Diameter has to pass through origin.

(v) equal

Equal arcs subtend equal segments and sectors.

(vi) radii

A sector is the region enclosed by an arc and two radii as shown below.

OAB is a sector.

(vii) Greater than radius

Question 3.

With the help of scale and pencil compass let us draw a circle and indicate centre, chord, diameter, radius, major arc, minor arc on it

Answer:

Question 4.

Let us write true or false:

i. The circle is a plane figure.

ii. The segment is a plane region.

iii. The Sector is a plane region.

iv. The chord is a line segment

v. The arc is a line segment.

vi. There are finite number of chords of same length in a circle

vii. One and only one circle can be drawn by taking a fixed point as its centre.

viii. The lengths of the radii of two congruent circles are equal.

Answer:

(i) True

Plane Figure, A figure drawn on a 2-D plane.

(ii) True

The segment of a circle is the region bounded by a chord and the arc subtended by the chord.

As you can see in the figure the segment is a plane figure.

(iii) True

A sector is the part of a circle enclosed by two radii of a circle and their intercepted arc.

As you can see in the figure the Sector is a plane figure.

(iv) True

A line segment is a distance between two points.

A chord is a distance between two points on the circumference.

(v) False

Arc is a segment of differential Curve.

(vi) False

There are infinite number of Chords of same length.

(vii) False

Concentric circles are circles with a common center. As you can see in this image We can draw infinite circles.

(viii) True

Two circles are congruent if they have the same size. The size can be measured as the radius, diameter or circumference. They can overlap

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Let Us Work Out 3.2

Question 1.

The length of a radius of a circle with its centre O is 5 cm. and the length of its chord AB is 8 cm. Let us write by calculating, the distance of the chord AB from the centre O.

Answer:

Given radius = 5cm, AB = 5cm

Perpendicular from the center of the circle to any Chord bisects it in two line segments

So, AD = BD

In ODB

OB = 5cm

BD = 4cm

Using Pythagoras Theorem

base² + Height² = Hypotenuse²

⇒ OD² + DB² = OB²

⇒ OD² = 25-16

⇒ OD² = 9

⇒ OD = 3cm

Question 2.

The length of a diameter of a circle with its centre at O is 26 cm. The distance of the chord PQ from the point O is 5 cm. Let us write by calculating, the length of the chord PQ.

Answer:

Given Length of the diameter = 26cm, Distance of Chord PQ From center = 5cm

Radius

Radius = 13cm

In O PD , Using Pythagoras Theorem

Hypotenuse² = base² + Height²

⇒ OP² = OD² + PD²

⇒ 13² = 5² + PD²

⇒ PD² = 169-25

⇒ PD² = 144

⇒ PD = 12cm

⇒ PQ = 2 PD

⇒ PQ = 24cm

Question 3.

The length of a chord PQ of a circle with its centre O is 4 cm. and the distance of PQ from the point O is 2.1 cm. Let us write by calculating, the length of its diameter.

Answer:

Given PQ = 4cm, Distance from the Center is 2,1cm

PQ = 2 PD

Perpendicular from the center of the circle to any Chord bisects it in two line segments

⇒ PD

⇒ PD = 2cm

In O PD , Using Pythagoras Theorem

⇒ OD² + PD ² = OP²

⇒ OP² = 2.1² + 2²

⇒ OP² = 8.41

⇒ OP = 2.9cm

Question 4.

The lengths of two chords of a circle with its centre O are 6 cm. and 8 cm. If the distance of smaller chord from centre is 4 cm, then let us write by calculating, the distance of other chord from the centre.

Answer:

Let Smaller Chord is EF, the other chord is BC, distance from the centre o to smaller chord is OG.

⇒ BC = 8cm

⇒ CD

Perpendicular from the center of the circle to any Chord bisects it in two line segments

⇒ CD = 4cm

⇒ EF = 6cm

⇒ EG

⇒ EG = 3cm

⇒ OG = 4cm

In OEG, Using Pythagoras Theorem

⇒ OE² = OG² + EG²

⇒ OE² = 16 + 9

⇒ OE² = 25

⇒ OE = 5cm

In OCD,Using Pythagoras Theorem

⇒ OC² = CD² + OD²

⇒ 5² = 4² + OD²

⇒ 25 = 16 + OD²

⇒ OD² = 9

⇒ OD = 3cm

Question 5.

If the length of a chord of a circle is 48 cm and the distance of it from the centre is 7 cm. then let us write by calculating, the length of radius of the circle.

Answer:

Given AB = 48cm, OP = 7cm

AP

Perpendicular from the center of the circle to any Chord bisects it in two line segments

⇒ AP = 24cm

In OAP, Using Pythagoras Theorem

⇒ OA² = AP² + OP²

⇒ OA² = 24² + 7²

⇒ OA² = 576 + 49

⇒ OA² = 625

⇒ OA = 25cm

Question 6.

In the circle of adjoining figure with its center at O, OP ⊥ AB; if AB = 6 cm. and PC = 2cm, then let us write by calculating, the length of radius of the circle.

Answer:

Given, AB = 6cm, PC = 2cm

AP

Perpendicular from the center of the circle to any Chord bisects it in two line segments

⇒ AP

⇒ AP = 3cm

⇒ OA = OC

⇒ OA = OP + CP ………..(1)

In OAP, Using Pythagoras Theorem

OA² = OP² + AP²

⇒ (OP + PC)² = OP² + 9 (from eq.(1))

⇒ OP² + PC² + 2(OP)(PC) = OP² + 9

⇒ PC² + 2(OP)(PC) = 9

⇒ 4 + 2(OP)(2) = 9

⇒ 2(OP)(2) = 5

⇒ OP

Radius = OC = OP + PC

⇒ OC = 2 +

⇒ OC = 2.25cm

Question 7.

A straight line intersects one of the two concentric circles at the points A and B and the other at the point C and D. I prove with reason that AC = DB.

Answer:

Given, Concentric Circles, CD and AB chords.

To prove: AC = DB

Construction: OP is a perpendicular bisector of CD and AB.

In OAP and OBP, We have

AP = BP

Perpendicular from the center of the circle to any Chord bisects it in two line segments

In OCP and ODP

CP = DP

⇒ CA + AP = BP + DB

⇒ CA = DB + BP-AP

⇒ CA = DB + 0

⇒ CA = DB

Question 8.

I prove that, the two intersecting chords of any circle cannot bisect each other unless both of them are diameters of the circle.

Answer:

Given, AB and CD are Diameters.

To prove: OA = OB, OC = OD

Construction: Point D joined with B, Point A joined with C.

DAC = 90⁰ (Rectangle)

ACB = 90⁰ (Rectangle)

CBD = 90⁰ (Rectangle)

BDA = 90⁰ (Rectangle)

ACBD is Rectangle, So AD = CB, AD BCand BD = AC,BD AC

In OCB and ODA

OCB = ODA (Interior angles, BC AD)

BC = AD (Rectangle)

OBC = OAD (Interior angles, BC AD)

BCA SA Congruency

In OCB ODA

Hence Using CPCT, OA = OB, OD = OC

Hence Proved.

Question 9.

The two circles with centers X and Y intersect each other at the points A and B. A is joined with the mid-point ‘S’ of XY and the perpendicular on SA through the point A is drawn which intersects the two circles at the points P and Q. Let us prove that PA = AQ.

Answer:

Given, Two Circles with Centers X and Y intersect each other at point A and B. S is a mid-point of XY, AS is perpendicular to PQ.

To Prove: AP = AQ

Construction: XM perpendicular to chord PA. And YN perpendicular to chord AQ of respective circles.

Since SA is also perpendicular to PQ (given)

So, all these perpendiculars to the same line are parallel to each other.

SX = SY

AM = AN

So,

AP = AQ

Question 10.

The two parallel chords AB and CD with the lengths of 10 cm and 24 cm in a circle are situated on the opposite sides of the centre. If the distance between two chords AB and CD is 17 cm. then let us write by calculating, the length of the radius of the circle.

Answer:

Given, AB = 10cm, CD = 24cm, PQ = 17cm

AP

Perpendicular from the center of the circle to any Chord bisects it in two line segments

AP

AP = 5cm

CQ

⇒ CQ

⇒ CQ = 12cm

In OAP, Using Pythagoras Theorem

OA² = AP² + OP²

⇒ OA² = 5² + OP²

⇒ OA² = 25 + OP² ………… (1)

⇒ In OCQ, Using Pythagoras Theorem

⇒ OC² = CQ² + OQ²

⇒ OC² = 12² + OQ²

⇒ OC² = 144 + OQ² ………. (2)

⇒ OC = OA

⇒ 144 + OQ² = 25 + OP²

⇒ OP²-OQ² = 119

⇒ (OP-OQ)(OP + OQ) = 119 (using (A² + B² = (A + B)(A-B))

⇒ OQ + OP = 17 ………. (3)

⇒ (OP-OQ)17 = 119

⇒ OP-OQ = 119/17

⇒ OP-OQ = 7 …….. (4)

Eq.3 + Eq.4

⇒ 2OP = 24

⇒ OP = 12cm

⇒ OP-OQ = 7

⇒ OQ = OP-7

⇒ OQ = 12-7

⇒ OQ = 5cm

In OAP, Using Pythagoras Theorem

⇒ OC² = 144 + OQ²

⇒ OC² = 144 + 25

⇒ OC = 13cm

Question 11.

The centers of two circles are P and Q; they intersect at the points A and B. The straight line parallel to the line-segment PQ through the point A intersects the two circles at the points C and D. I prove that, CD = 2PQ.

Answer:

Given, CD is parallel to PQ, Two Circles with centers P and Q intersect at point A and B.

To Prove: 2PQ = CD

Construction: PE and QE are the Perpendiculars on the Chord CD From centers P and Q respectively.

⇒ DE = EA

⇒ CF = AF

⇒ DE + EA + AF + FC = DC

⇒ 2EA + 2AF = DC

⇒ 2(EA + AF) = DC ……..(1)

PQ and CD are parallel lines and PE and QE are the perpendiculars.

So, AEP = 90⁰,AFQ = 90⁰,EPQ = 90⁰,AFQ = 90⁰

EPQF is a rectangle

So, EF = PQ ……… (2)

⇒ 2(EA + AF) = DC

⇒ 2EF = DC

⇒ 2PQ = DC

Question 12.

The two chords AB and AC of a circle are equal. I prove that, the bisector of ∠BAC passes through the centre.

Answer:

Given, AC = AB, BAF = CAF

To prove: FB = FC

In ABF and CAF

AC = AB (Given)

BAF = CAF (Given)

AF = AF (Common)

ABF CAF

BFA = CFA (CPCT)

FB = FC (CPCT)

AE is a perpendicular bisector of chord BC, So, It has to pass through the center of the circle.

Question 13.

If the angle-bisector of two intersecting chords of a circle passes through its centre, then let me prove that the two chords are equal.

Answer:

Given, OF is angle bisector of AFC.

Construction: OQ ⊥ AB and OP ⊥ CD

In ΔOFQ and ΔOFP
∠OFQ = ∠OFP (given)
OF = OF(Common)
∠OQF = ∠OPF(Construction)

AAS Congruency.
ΔOPR ≅ΔOPQ.

∴ OR = OQ (C.P.C.T)

Hence AB = CD

Question 14.

I prove that, among two chords of a circle the length of the nearer to centre is greater than the length of the other.

Answer:

To Prove: DH>EG

Construction: B is the center of the circle, CD and EF are the chords, BH and BG are the perpendicular bisector of CD and EF respectively.

Given BH<BG As it given, CD is nearer to center than EF.

In BDH

⇒ BD² = BH² + DH² …………… (1)

In BEG

⇒ BE² = BG² + EG² ………… (2)

⇒ BD = BE

From Eq1 and Eq2

⇒ BH² + DH² = BG² + EG²

⇒ BH<BG

⇒ BH²<BG²

⇒ BH²-BG²<0

From Eq.3

⇒ BH² + DH² = BG² + EG²

⇒ BH²-BG² = EG²-DH²

⇒ EG²-DH²<0

⇒ EG²<DH²

⇒ EG<DH

Question 15.

Let us write by proving the chord with the least length through any point in a circle.

Answer:

To Prove CD<BG

Let, Center is O, Two chords are BG and CD, OM and OE are perpendicular bisector of BG and CD respectively.

In OME,

⇒ OE>OM

In OMB, Using Pythagoras Theorem

⇒ OB² = OM² + BM² ………. (1)

In OCE

⇒ OC² = OE² + CE² …………….. (2)

⇒ OC = OB

⇒ OM² + BM² = OE² + CE²

⇒ OE>OM

So, BM>CE

⇒ BG>CD

As Chord goes near, its length increases.

Question 16.

The lengths of two chords of a circle with centre O are equal. If ∠AOB = 60°, then the value of ∠COD is
A. 40^o

B. 30^o

C. 60^o

D. 90^o

Answer:

In AOB and COD

AB = CD

OA = OC = OB = OD

SSS Congruency.

So each and every angles and sides should be equal.

COD = 60⁰

Question 17.

The length of a radius of a circle is 13 cm. and the length of a chord of a circle is 10 cm, the distance of the chord from the centre of the circle is
A. 12.5 cm

B. 12 cm

C.

D. 24 cm

Answer:

Given radius = 13cm, PQ = 10cm

⇒ PD

⇒ PD

⇒ PD = 5cm

Using Pythagrus theorem in O PD

⇒ OP² = OD² + PD ²

⇒ 13² = OD² + 5²

⇒ OD² = 169-25

⇒ OD² = 144

⇒ OD = 12cm

Question 18.

Ab and CD are two equal chords of a circle with its centre O. If the distance of the chord AB from the point O is 4 cm. Then the distance of the chord from the centre O of the circle is
A. 2 cm

B. 4 cm

C. 6 cm

D. 8 cm

Answer:

In AOB and COD

AB = CD

OA = OC = OB = OD

SSS Congruency.

So each and every angles and sides should be equal.

OAB = OCD = OBA = ODC

⇒ EA

⇒ FC

⇒ AB = CD

So, EA = FC

In AOE and COE

⇒ EA = FC

⇒ EAO = FCO

⇒ OA = OC

SAS Congruency.

So, By CPCT OE = OF = 4cm

Question 19.

The length of each of two parallel chord is 16 cm. If the length of the radius of the circle is 10 cm, then the distance between two chords is
A. 12 cm

B. 16 cm

C. 20 cm

D. 5 cm

Answer:

Let Parallel Chords are AB and CD. OE and OF are perpendicular bisector of AB and CD respectively.

Given, OB = OA = OD = OC = 10cm, AB = CD = 16cm

⇒ DF

⇒ DF

⇒ DF = 8cm

⇒ EB

⇒ EB

⇒ EB = 8cm

In AOE

⇒ OA² = AE² + OE²

⇒ OE² = OA²-AE²

⇒ OE² = 10²-8²

⇒ OE² = 100-64

⇒ OE² = 36

⇒ OE = 6cm

As we can see from the previous question both the triangles are congruent.

So, OE = OF

Hence Distance between parallel lines = OE + OF = 6 + 6 = 12cm

Question 20.

The centre of two concentric circle is O; a straight line intersects a circle at the points A and B and other circle at the points C and D. If AC = 5 cm. then the length of BD is
A. 2.5 cm

B. 5 cm

C. 10 cm

D. none of these

Answer:

In OAP and OBP

⇒ AP = BP

⇒ OA = OB

⇒ OP = OP

In OCP and ODP

⇒ CP = DP

⇒ CA + AP = BP + DB

⇒ CA = DB + BP-AP

⇒ CA = DB + 0

⇒ CA = DB

So, BD = 5cm

Question 21.

Let us write True/False:

i. Only one circle can be drawn through three collinear points

ii. The two circles ABCDA and ABCEA are same circle.

iii. If two chord AB and AC of a circle with its centre O are situated on the Opposite side of the radius OA, then ∠OAB=∠OAC.

Answer:

(i) False

There is no Circle passes through the 3 Collinear points.

(ii) True

Only one Circle Passes through three Collinear Points.

(iii) False

You can see in this image, AB and AC are Situated on the Opposite side of the radius OA, Angles OAB and OCA are not Equal. It holds only for Equal chords.

Question 22.

Let us fill in the blanks:

i. If the ratio of two chords PQ and RS of a circle with its centre O is 1 : 1, then ∠POQ: ∠ROS=_______________

ii. The perpendicular bisector of any chord of a circle is __________ of that circle.

Answer:

(i) 1:1

In AOB and COD

AB = CD

OA = OC = OB = OD

SSS Congruency.

So each and every angles and sides should be equal.

OAB = OCD = OBA = ODC

(ii) Passes through the origin of that Circle.

As you can see in this image The perpendicular bisector passes through the centre.

Question 23.

Two equal circles of radius 10 cm. intersect each other and the length of their common chord is 12 cm. Let us determine the distance between the two centers of two circle

Answer:

Let Center of the Circles are A and B. CD is a common Chord of the circle. AB is the perpendicular bisector of the chord CD.

If AB is a perpendicular bisector of CD then it should passes through both the centers.

So, AB is the distance that we need to calculate.

Given, AC = 10cm, CD = 12cm

⇒ CM

⇒ CM

⇒ CM = 6cm

In ACM

⇒ AC² = AM² + CM²

⇒ AM² = AC²-CM²

⇒ AM² = 100-36

⇒ AM² = 64

⇒ AM = 8cm

In BCM

⇒ BC² = BM² + CM²

⇒ BM² = BC²-CM²

⇒ BM² = 100-36

BM² = 64

BM = 8cm

AB = 16cm

Question 24.

AB and AC are two equal chords of a circle having the radius of 5 cm. The centre of the circle

is situated at the outside of the triangle ABC. If AB = AC = 6 cm. then let us calculate the length of the chord BC.

Answer:

Given, AC = AB, BAF = CAF, AB and AC are two equal Chords of a circle, therefore the centre of the circle lies on the bisector of ∠BAC.

In ABF and CAF

⇒ AC = AB

⇒ BAF = CAF

⇒ AF = AF

⇒ ABF CAF

⇒ BFA = CFA

⇒ FB = FC

AE is a perpendicular bisector of chord BC.

In ΔABF, by Pythagoras theorem,

⇒ AB² = AF² + BF²

⇒ BF² = 6² - AF² .............(1)

In OBF

⇒ OB² = OF² + BF²

⇒ 5² = (5 - AF)² + BF²

⇒ BF² = 25 - (5 - AP)² ...........(2)

Equating (1) and (2), we get

⇒ 6² - AF² = 25 - (5 - AF)²

⇒ 11 - AF² = -25 - AF² + 10AF

⇒ 36 = 10AF

⇒ AF = 3.6 cm

Putting AF in (1), we get

⇒ BF² = 6² - (3.6)² = 23.04

⇒ BF = 4.8 cm

⇒ BC = 2BF = 2 × 4.8 = 9.6 cm

Question 25.

The lengths of two chords AB and CD of a circle with its centre O are equal. If ∠AOB = 60° and CD = 6 cm. then let us calculate the length of the radius of the circle.

Answer:

In AOB and COD

⇒ AB = CD

⇒ OA = OC = OB = OD

SSS Congruency.

So each and every angles and sides should be equal.

AB = 6cm

⇒ COD = 60⁰

⇒ AE

⇒ AE

⇒ AE = 3cm

In AOE and BOE

⇒ OA = OB

⇒ OE = OE

⇒ AE = BE

Hence Using SSS congruency

⇒ AOE BOE

⇒ AOE = BOE

⇒ AOE = 30⁰

⇒ Sin30

⇒

⇒ OA = 2AE

⇒ OA = 2(3)

⇒ OA = 6cm

Question 26.

P is any point in a circle with its centre O. If the length of the radius is 5 cm. and OP = 3 cm., then let us determine the least of the chord passing through the point P.

Answer:

Given: OP = 3cm, Radius = 5cm

Let OA, OC radius of the circle, OM is perpendicular Bisector Passes through the centre O of Chord CD. OP is a Perpendicular Bisector of Chord AB passes through the Centre O.

In OPM

OMP = 90⁰

Using, Pythagoras Theorem

⇒ OP² = OM² + PM²

⇒ OP>OM

We have proved in Question 14, That Nearer chord is greater than the other.

It shows us that

AB<CD

Hence The Least chord passes through point P is AB, OP is Perpendicular bisector of the chord.

In OPA

Using, Pythagoras Theorem

⇒ OA² = OP² + AP²

⇒ 5² = 3² + AP²

⇒ AP² = 25-9

⇒ AP² = 16

⇒ AP = 4cm

Using above Theorem

AB = 8cm

Question 27.

The two circles with their centres at P and Q intersect each other at the points A and B. Through the point A, a straight line parallel to PQ intersects the two circles at the points C and D respectively. If PQ = 5 cm., then let us determine the length of CD.

Answer:

Given: PQ = 5cm, CD is parallel to PQ, Two Circles with centers P and Q intersect at point A and B.

Construction: EF and FQ are Perpendicular bisector drawn from P and Q Respectively.

The Perpendicular from The Centre to the chord, bisects the chord.

⇒ DE = EA

⇒ CF = AF

⇒ DE + EA + AF + FC = DC

⇒ 2EA + 2AF = DC

⇒ 2(EA + AF) = DC ………..(1)

EFPQ (Given)

PEA = 90^o (Construction)

QFA = 90^o (Construction)

Using Interior Angle Theorem,

PEA + BPE = 180

BPE = 90^o

PQFE is a rectangle.

So, EF = PQ …………… (2)

⇒ 2(EA + AF) = DC

⇒ 2EF = DC

⇒ 2PQ = CD

CD = 10cm