Mathematics Part ii Solutions Solutions for Class 10 Math Chapter 2 - Tangents

Answer:

Given: Two circles with centre A and B, which cut each other at point P.

AP is a tangent to the circle centred at B.

We know that any tangent to circle is perpendicular to the radius to the point of contact.

AP is a tangent to the circle centred at B and BP is the radius of the circle.

∴ ∠BPA = 90° 

BT is a straight line passing through point P.

We know that sum of angles forming a linear pair is 180°.

∴ ∠BPA + ∠APT = 180°

⇒ 90° + ∠APT = 180° 

⇒ ∠APT = 90°

Therefore, AP is perpendicular to BP.

We know that a line drawn through any point of a circle, perpendicular to the radius through that point is a tangent to the circle.

Thus, BP is a tangent to the circle centred at A.

Answer:

(1)

The steps of construction to construct the required figure are as follows:

1) Draw a circle with centre O and radius OA = 2 cm.

2) Extend line segment OA to a point P such that OP = 6 cm (= 4 cm + 2 cm).

3) Mark a point B on line segment OP such that PB = BO = = 3 cm.

4) Draw a circle with B as the centre and BO as the radius. Name the points where it intersects the previous circle at S and T.

5) Join PS and PT.

(2)

The given figure is:

We know that any tangent to circle is perpendicular to the radius to the point of contact.

PS is the tangent to the given circle and SO is the radius of the circle.

∴ ∠OSP = 90°

Applying Pythagoras theorem in ΔOSP:

OP² = PS² + SO²

Now, we can construct the required figure. The steps of construction are as follows:

1) Draw a circle with centre O and radius OA = 2.5 cm.

2) Extend line segment OA to a point P such that OP = 6.5 cm.

3) Mark a point B on line segment OP such that PB = BO = = 3.25 cm.

4) Draw a circle with B as the centre and BO as the radius. Name the points where it intersects the previous circle at S and T.

5) Join PS, PT and OS.

(3)

We have joined the centre of the circle given in the figure with the point where the tangent intersects the circle.

The given figure is:

We know that any tangent to circle is perpendicular to the radius to the point of contact.

PS is the tangent to the given circle and SO is the radius of the circle.

∴ ∠OSP = 90°

Applying Pythagoras theorem in ΔOSP:

OP² = PS² + SO²

⇒ SO² = OP² − PS²

Now, we can construct the required figure. The steps of construction are as follows:

1) Draw a circle with centre O and radius OA = 3.32 cm.

2) Extend line segment OA to a point P such that OP = 5.2 cm.

3) Mark a point B on line segment OP such that PB = BO = = 2.6 cm.

4) Draw a circle with B as the centre and BO as the radius. Name the points where it intersects the previous circle at S and T.

5) Join PS and PT.

Answer:

Given: OR = OQ = 15 cm and ∠ROQ = 120°

Construction: Join OP.

In ΔOPR and ΔOPQ:

OR = OQ (Radii of the same circle)

OP = OP (Common side)

PR = PQ (Length of tangents drawn from a point outside the circle are equal)

If all the sides of one triangle are equal to the corresponding sides of the other triangle then the two triangles are congruent.

∴ ΔOPR ≅ ΔOPQ 

⇒ ∠POR = ∠POQ (Congruent parts of congruent triangles are congruent)

⇒ ∠POR = ∠POQ =  

In ΔPOR: 

= tan 60°

⇒ RP = OR tan 60°

= 15 cm ×

= 15 × 1.732 cm

= 25.98 cm

We know that lengths of tangents drawn from an external point are equal.

∴ PQ = PR = 25.98 cm

Answer:

Given: A circle with centre O

Tangents at points A and B intersect at point P.

(i)

Construction: Join AO, OB and OP.

We know that any tangent to circle is perpendicular to the radius to the point of contact.

AP is a tangent to the circle and OA is the radius of the circle.

∴∠OAP = 90°

Applying Pythagoras theorem in ΔOAP:

OP² = OA² + AP²

_⇒OA² = OP² − AP² … (1)

Similarly, we have OB² = OP² − BP² …(2)

OA = OB (Radii of the same circle) 

From equations (1) and (2), we have:

OP² − AP² = OP² − BP²  

⇒ AP² = BP²

⇒ AP = BP

Thus, point P is equidistant from A and B.

(ii)

Construction: Join AB.

In ΔOAP and ΔOBP:

PA = PB (Proved in part (1))

PO = PO (Common side)

OA = OB (Radii of the same circle)

As all sides of one triangle are equal to the corresponding sides of the other triangle, ΔOAP ≅ ΔOBP. 

Corresponding parts of congruent triangles are congruent.

⇒∠AOP = ∠BOP …(1)

∠APO = ∠BPO  

Thus, line segment OP bisects ∠APB. 

In ΔAQO and ΔBQO:

OA = OB (Radii of the same circle) 

∠AOP = ∠BOP (From equation (1))

OQ = OQ (Common side)

As two sides and included angle of one triangle are equal to the two sides and included angle of the other triangle, ΔAQO ≅ ΔBQO. 

Corresponding parts of congruent triangles are congruent.

⇒ AQ = BQ 

Thus, line segment OP bisects AB.

(iii)

We have shown in part (2) that ΔAQO ≅ ΔBQO. 

∴ ∠AQO = ∠BQO (Corresponding parts of congruent triangles are congruent)

We know that sum of angles forming a linear pair are equal.

∴ ∠AQO + ∠BQO = 180°

⇒ ∠AQO = ∠BQO = 90° 

In ΔAQO and ΔPBO:

∠AQO = ∠PBO = 90°

∠AOP = ∠BOP (Proved in part (1))

∴ ΔAQO ∼ ΔPBO (By angle angle criterion)

Answer:

Construction: Joining the centre O of the circle with the point at which the tangents touch the circle.

After naming, we have the following figure.

We know that any tangent to circle is perpendicular to the radius to the point of contact.

AE is the tangent to the circle and OE is the radius of the circle.

∴ ∠OEA = 90°

Similarly, ∠AFO = 90°

In quadrilateral AFOE: 

∠A = ∠OEA = ∠OFA = 90°

Thus, by angle sum property of quadrilaterals:

∠EOF = 90°

Also, OE = OF = r (say) (Radii of the same circle) 

As all angles of quadrilateral OEAF are of measure 90° and adjacent sides are equal, quadrilateral OEAF is a square. 

We know that lengths of tangents drawn from a point outside a circle are equal.

⇒ AE = AF  

⇒ AE = AF = OE = OF = r (All sides of a square are equal in length) 

Now, AE + AF = r + r 

= 2r 

= d (where d is the diametre of the circle) …. (1)

Also, DE = DC and BC = BF (Lengths of tangents drawn from a point outside a circle) …(2)

We know that perimeter of a triangle is equal to the sum of the lengths of its sides.

∴ Perimeter of ΔDAB = AD + DB + BA

= AD + DC + CB + BA 

= AD + DE + BF + BA (From equation (2))

= AE + AF  

= d (From equation (1))

Therefore, the perimeter of the right-angled triangle is equal to the diametre of the circle. 

Answer:

We know that perimeter of a triangle is equal to the sum of the lengths of its sides.

Perimeter of ΔABC = AB + BC + CA

= (AP + PB) + (BQ + QC) + (AR + RC) …(1) 

We know that lengths of tangents drawn from a point outside a circle are equal.

⇒ AP = AR ... (2)

BP = BQ ... (3)

CR = CQ ... (4)

Substituting the values from equations (2), (3) and (4) in equation (1):

Perimeter of ΔABC = AP + BQ + BQ + CR + AP + CR

= 2(AP + BQ + CR)

Answer:

The rough sketch of the required figure can be drawn as follows:

We know that the central angle of the smaller arc between the two points on a circle and the angle between the tangents at these points are supplementary.

∠ECF + ∠EOF = 180°

⇒ ∠EOF = 180° − 40° = 140°

We also know that any tangent to circle is perpendicular to the radius to the point of contact.

∴ ∠OFC = ∠OGB = ∠OHD = ∠OEC = 90°

We will use these measurements to construct the required figure.

The steps of construction are as follows:

1) Draw a circle with centre O and radius, OF = 3 cm.

2) Draw ∠OFX of measure 90° at point F by taking OF as the base. Extend ray XF downwards.

3) Draw ∠EOF of measure 140° at point O by taking FO as the base. 

4) Draw ∠OEC of measure 90° at point E by taking OE as the base. Extend CE downwards.

5) Extend the line segments EO and OF such that they intersect the circle at points G and H respectively.

6) Draw ∠OGB of measure 90° at point G by taking OG as the base, where B is a point on line FX. Extend BG downwards.

7) Draw ∠OHD of measure 90° at point H by taking OH as the base, where D is a point on ray CE.

8) Extend line segment DH to intersect ray BG at point A.

ABCD is the required rhombus with all the four sides touching the circle.

Answer:

In a polygon, sum of all the sides is given by (n − 2) × 180°, where n is the number of sides.

For a regular pentagon, n = 5

Sum of all the angles = (5 − 2) × 180°

= 3 × 180° 

= 540° 

Since all the angles of a regular pentagon are equal, measure of each angle = = 108°.

The rough sketch of the required figure can be drawn as follows:

We know that the central angle of the smaller arc between the two points on a circle and the angle between the tangents at these points are supplementary.

∴ ∠PBQ + ∠POQ = 180°

⇒ 108° + ∠POQ = 180°

⇒ ∠POQ = 180° − 108° = 72° 

We also know that any tangent to circle is perpendicular to the radius to the point of contact.

∴ ∠OPB = ∠OQB = ∠OTA = ∠OSE = ∠ORD = 90°

We will use these measurements to construct the required figure.

The steps of construction are as follows:

1) Draw a circle with centre O and radius, OQ = 4 cm.

2) Draw ∠OQX of measure 90° at point Q by taking OQ as the base. Extend ray QX upwards.

3) Draw ∠QOP of measure 72° at point O by taking QO as the base such that point P lies on the circle. 

4) Draw ∠OPB of measure 90° at point P by taking OP as the base such that B is a point on line QX. Extend ray PB on the left side.

5) Draw ∠POT of measure 72° at point O by taking PO as the base such that point T lies on the circle.

6) Draw ∠OTA of measure 90° at point T by taking OT as the base such that A is a point on line PB. Extend ray TA upwards.

7) Draw ∠TOS of measure 72° at point O by taking TO as the base such that point S lies on the circle.

8) Draw ∠OSE of measure 90° at point S by taking OS as the base such that E is a point on line TA. Extend ray ES upwards.

9) Draw ∠SOR of measure 72° at point O by taking SO as the base such that point R lies on the circle.

10) Draw ∠ORD of measure 90° at point R by taking OR as the base such that D is a point on line SE. 

11) Extend ray RD downwards to intersect line QX at point C.

ABCDE is the required pentagon with all the sides touching the circle.

Answer:

Given: A circle with centre O

PA and PB are tangents to the circle through a point P and AB is the chord.

We know that lengths of tangents drawn from a point outside the circle are equal.

∴ PA = PB

In ΔPAB, PA = PB 

We know that angles opposite to equal sides are equal in measure.

∴ ∠PBA = ∠PAB

Thus, in a circle, the tangents at two points make equal angles with the chord joining the points of contact. 

Answer:

Using angle sum property in ΔABC:

∠ABC + ∠BCA + ∠CAB = 180°

⇒ 30° + 70° + ∠CAB = 180°

⇒ 100° + ∠CAB = 180°

⇒ ∠CAB = 180° −100° = 80°

We know that lengths of tangents drawn from a point outside the circle are equal.

∴ BQ = BR

AQ = AP

CP = CR

In ΔBQR, BQ = BR

We know that angles opposite to equal sides are equal in measure.

∠BRQ = ∠BQR 

By angle sum property:

∠BQR + ∠QRB + ∠RBQ = 180°

⇒ 2∠BQR + 30° = 180° 

⇒ 2∠BQR = 150°

⇒ ∠BQR = ∠BRQ = 75°

Similarly, in ΔRPC:

∠CRP = ∠CPR (As CP = CR)

By angle sum property:

∠CRP + ∠RPC + ∠PCR = 180°

⇒ 2∠CRP + 70° = 180°

⇒ 2∠CRP = 110°

⇒ ∠CRP = ∠CPR = 55°

In ΔAQP:

∠APQ = ∠AQP (As AQ = AP)

By angle sum property:

∠APQ + ∠PQA + ∠QAP = 180°

⇒ 2∠APQ + 80° = 180°

⇒ 2∠APQ = 100°

⇒ ∠AQP =∠APQ = 50°

We know that each angle between a chord and the tangent at one of its ends in a circle is equal to the angle in the segment on the other side of the chord. 

∴ ∠BQR = ∠RPQ = 75°

∠RPC = ∠RQP = 55°

∠AQP = ∠QRP = 50°

Hence, the measures of all the angles of smaller triangle are 75°, 55° and 50°.

Answer:

In a circle, each angle between a chord and the tangent at one of its ends is equal to the angle in the segment on the other side of the chord.

So, the pair of equal angles are:

∠BAC = ∠UCB …(1) 

∠BAC = ∠SBC …(2)

∠ABC = ∠TCA …(3)

∠ABC = ∠PAC …(4)

∠ACB = ∠QAB …(5)

∠ACB = ∠RBA …(6)

From (1) and (2): 

∠UCB = ∠SBC 

From (3) and (4):

∠TCA = ∠PAC  

From (5) and (6): 

∠QAB = ∠RBA  

So, there are nine pairs of equal angles in the given figure.

Answer:

The incircle for triangle with sides 4 cm, 5 cm and 6 cm as can be constructed as follows:

1) Draw a line segment AB of length 5 cm.

2) Taking A as the centre and 5 cm as the radius, draw an arc on the upper side of AB.

3) Taking B as the centre and 6 cm as the radius, draw an arc cutting the previously drawn arc at point C. 

4) Join AC and BC.

ΔABC is the required triangle.

The incircle for the given equilateral triangle can be constructed as follows:

1) Draw angle bisectors of ∠A and ∠B. Let them intersect at point O. 

2) With O as the centre, draw OK perpendicular to AC. Name the point where it intersects AC as D.

3) With O as the centre and OD as the radius, draw a circle. The circle so drawn touches all the sides of the triangle.

The given figure shows the required incircle of the given triangle.

Answer:

An equilateral triangle with side 6 cm can be constructed as follows:

1) Draw a line segment AB of length 6 cm.

2) Taking A as the centre and 6 cm as the radius, draw an arc on the upper side of AB.

3) Taking B as the centre and 6 cm as the radius, draw an arc cutting the previously drawn arc at point C. 

4) Join AC and BC.

ΔABC is the required equilateral triangle.

The incircle for the given equilateral triangle can be constructed as follows:

1) Draw angle bisectors of ∠A and ∠B. Let them intersect at point O. 

2) With O as the centre, draw OK perpendicular to AC. Name the point where it intersects AC as D.

3) With O as the centre and OD as the radius, draw a circle. The circle so drawn touches all the sides of the triangle.

The given figure shows the required incircle of the given triangle.

The circumcircle for the given equilateral triangle can be constructed as follows:

1) Draw perpendicular bisectors of AB and AC. Let these perpendicular bisectors intersect at point O.

2) With O as the centre and radius OA = OB = OC, draw a circle passing through the points A, B and C.

The given figure shows the required circumcircle of the given triangle.

Answer:

A square with side 5 cm as length can be drawn as follows:

1) Draw a line segment AB of length 5 cm.

2) At point A and B, draw ∠XAB = ∠YBA = 90°.

3) Taking A and B as the centres and 5 cm as the radius, draw arcs on rays AX and BY. Name the point of intersection as D and C.

4) Join DC.

Quadrilateral ABCD is the required square.

The incircle of this square can be drawn as follows: 

1) Draw angle bisectors of ∠A and ∠B. Let them intersect at point O. 

2) With O as the centre, draw OK perpendicular to AB. Name the point where it intersects AB as P.

3) With O as centre and OP as the radius, draw a circle. The circle so drawn touches all the sides of the triangle.

The given figure shows the required incircle of the given square.

The circumcircle for the given square can be constructed as follows:

1) Draw perpendicular bisectors of AB and BC. Let them intersect at point O.

2) With O as the centre and radius OA = OB = OC = OD, draw a circle passing through the points A, B, C and D.

The given figure shows the required circumcircle of the given square.

Answer:

The given figure is:

We know that sum of interior angles on the same side of the transversal are supplementary.

∴∠A + ∠B = 180°

⇒ 50° + ∠B = 180°

⇒ ∠B = 180° − 50° = 130° 

The given rhombus can be drawn as follows:

1) Draw a line segment AB of length 5 cm.

2) At point A, draw ∠TAB = 50°.

3) Taking A as the centre and 5 cm as the radius, draw an arc on ray AT. Name the point of intersection as D.

4) At point B, draw ∠UBA = 130°.

5) Taking B as the centre and 5 cm as the radius, draw an arc on ray BU. Name the point of intersection as C.

6) Join CD.

Quadrilateral ABCD is the required rhombus.

The incircle of this rhombus can be made as follows:

1) Draw angle bisectors of ∠A and ∠B. Let them intersect at point O.

2) With O as the centre, draw OK perpendicular to AB. Name the point where it intersects AB as P.

3) With O as centre and OP as the radius, draw a circle. The circle so drawn touches all the sides of the triangle.

The given figure shows the required incircle of the given rhombus.

Answer:

Case I.

Consider an equilateral triangle ABC. 

Let O′ be its incentre.

We know that incentre of a circle is obtained by the intersection of the angles bisectors of a triangle.

BO′ is the angle bisector of ∠ABC.

∴ ∠ABO′ = ∠O′BC =

Similarly, we have ∠O′CA = ∠O′AC = ∠O′AB = ∠O′BA = 30°

Case II.

Now, again consider the same triangle with angles marked on it.

Let O be the circumcentre of the given triangle.

As O is the circumcentre of ΔABC, OA = OB = OC

In ΔOBC, OB = OC

⇒ ∠2 = ∠1 …(1)

Also, ∠1 + ∠6 = ∠2 + ∠3 (As each angle of an equilateral triangle is equal)

⇒ ∠6 = ∠3 …(2)

Similarly, in ΔOAB:

∠5 = ∠6 …(3)

In ΔOAC:

∠3 = ∠4 …(4)

From equations (2) and (4), we have:

∠4 = ∠6 …(5)

From equations (3) and (5), we have:

∠4 = ∠5

As ∠4 + ∠5 = 60°

⇒ 2∠4 = 60°

⇒ ∠4 = 30°

Similarly, we can show that ∠1 = ∠2 = ∠3 = ∠4 = ∠5 = ∠6 = 30°, i.e., ∠ABO = ∠OBC = ∠OCA = ∠OAC = ∠OAB = ∠OBA = 30°.

Therefore, BO, AO and CO is the angle bisectors of the angles of the given triangle ABC.

However, BO′, AO′ and CO′ is the angle bisectors of the angles of the given triangle ABC according to our assumption in case I.

Thus, O and O′ coincide.

Hence, the circumcentre and incentre of an equilateral triangle are same. 

We know that centroid of an equilateral triangle coincides with circumcentre and orthocentre.

Also, centroid divides the median in the ratio 2:1.

Here, AD is the median. 

⇒ AO = and OD =

⇒ AD = 3 OD

⇒ AO = 2 OD

⇒ AO : OD = 2 : 1 

Thus, the ratio of the circumradius to inradius is 2:1.