Mathematics Part I Solutions Solutions for Class 10 Math Chapter 1 Arithemetic Sequences are provided here with simple step-by-step explanations. These solutions for Arithemetic Sequences are extremely popular among Class 10 students for Math Arithemetic Sequences Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part I Solutions Book of Class 10 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Mathematics Part I Solutions Solutions. All Mathematics Part I Solutions Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

#### Page No 32:

#### Question 1:

In ΔABC, we have ∠A = 60° and ∠B = 70°. Is the vertex C inside or outside the circle with diameter AB?

#### Answer:

Given: In Δ*ABC*, ∠*A* = 60° and ∠*B* = 70°

Using angle sum property:

∠*A* + ∠*B* + ∠*C* = 180°

⇒ 60° + 70° + ∠*C* = 180°

⇒ 130° + ∠*C* = 180°

⇒ ∠*C* = 180° − 130°

= 50°

We know that angle in a semi circle is a right angle.

If AB is the diametre of the circle then ∠*C* should be equal to 90°.

However, here ∠*C* is less than 90°. Therefore, vertex *C* will lie outside the circle.

#### Page No 32:

#### Question 2:

Prove that if a pair of opposite angles of a quadrilateral are right, then a circle can be drawn through all four of its vertices.

#### Answer:

Consider the figure below.

Here, *ABCD* is a quadrilateral in which ∠*A* = ∠*C* = 90°.

We know that angle in a semi circle is a right angle.

Suppose a circle is drawn such that it passes through vertex* A*.

As ∠*BAD** *= 90°, *BD* acts like the diametre of this circle.

Similarly, suppose a circle is drawn such that it passes through vertex *C*.

As ∠*BCD* = 90°, *BD* acts like the diametre of this circle.

Thus, *BD* is the diametre of the circle passing through vertices *A* and *C*.

As the end points of the diametre lie on the circle, all the four vertices of quadrilateral *ABCD* lie on the circle.

Hence, if a pair of opposite angles of a quadrilateral are right then a circle can be drawn through all four of its vertices.

#### Page No 32:

#### Question 3:

In the quadrilateral ABCD, we have AB = 3 cm, BC = 4 cm, AC = 5 cm, ∠A = 120°, ∠C = 70°. If we draw the circle with AC as diameter, which of the four vertices of ABCD would be inside the circle? Which of them would be outside this circle? Is any vertex on the circle? What about the circle with BD as diameter.

#### Answer:

Given: *AB* = 3 cm, *BC* = 4 cm, *AC* = 5 cm, ∠*A* = 120° and ∠*C* = 70°

Consider the following:

*AB*^{2} = 3^{2} = 9

*BC*^{2} = 4^{2} = 16

*AC*^{2} = 5^{2} = 25

⇒ *AC*^{2} = *AB*^{2} + *BC*^{2}

Therefore, by the converse of Pythagoras theorem, we get that ∠*B* = 90°.

Using angle sum property in quadrilateral *ABCD*:

∠*A* + ∠*B* + ∠*C* + ∠*D* = 360°

⇒ 120° + 90° + 70° + ∠*D* = 360°

⇒ 280° + ∠*D* = 360°

⇒ ∠*D* = 360° − 280° = 80°

We know that angle in a semi circle is a right angle.

If we draw a circle with *AC* as the diametre then vertex *B* will lie on the circle.

As the measure of ∠*D* is less than 90°, vertex *D* will lie outside the circle.

Thus, vertices *A*, *B* and *C* will lie on the circle.

Now, let us draw a circle with *BD* as the diametre.

As ∠*A* = 120° and ∠*C* = 70°, vertex *A* will lie inside the circle and vertex *C *will lie outside the circle.

Only vertices *B *and *D* will lie on the circle.

#### Page No 41:

#### Question 1:

In this figure, what fraction of the circumference of the circle is the length of the arc ADB?

#### Answer:

**Construction: **Join points *A* and *B *with centre *O* of the circle.

Given: ∠*ACB* = 60°

We know that angle made by an arc at any point on the alternate arc is equal to half the angle made at the centre.

We know that measure of an arc is equal to the measure of the corresponding central angle.

⇒ *m* (arc *ADB*) = ∠*AOB*

⇒ *m* (arc *ADB*) = 120°

Now,

Thus, the length of the arc *ADB* is of the circumference of the circle.

#### Page No 42:

#### Question 1:

What is the radius of the circle shown below?

#### Answer:

Given: Δ*ABC*, with ∠*B* = 90° and ∠*C* = 45°

Using angle sum property:

∠*A* + ∠*B* + ∠*C* = 180°

⇒ ∠*A* + 90° + 45° = 180°

⇒ ∠*A* + 135° = 180°

⇒ ∠*A* = 180° − 135°

= 45°

∠*A* = ∠*C*

∴ *BC* = *AB* = 3 cm (Sides opposite to equal angles are equal in length.)

Using Pythagoras theorem in Δ*ABC*:

As ∠*ABC* = 90° and we know that angle in a semi circle is a right angle, *AC* is the diametre of the circle.

Radius of the circle =

#### Page No 42:

#### Question 2:

What is the area of the circle shown below?

#### Answer:

Let the given rectangle be *ABCD*.

**Construction: **Join *BD*.

We know that each angle of a rectangle measures 90°.

∴ ∠*BAD** *= 90°

We know that angle in a semi circle is a right angle.

Therefore, *BD* is the diametre of the circle.

Using Pythagoras theorem in Δ*ABD*:

Radius of the circle, *r* =

Area of the circle = *π**r*^{2}

#### Page No 42:

#### Question 3:

How do we draw a triangle with two of the angle 40° and 120° and circumradius 3 centimetres?

#### Answer:

To construct the required triangle, we will use the following property.

Angle made by an arc at any point on the alternate arc is equal to half the angle made at the centre.

The steps of construction are as follows:

(1) Draw a circle with centre *O* and radius 3 cm.

(2) Take a point *B* on this circle and join* OB*.

(3) Taking *O* as the centre and *OB* as the segment, draw an angle of measure 80° such that it intersects the circle at point *C*.

(Since we need ∠*BAC* = 40° , we have drawn ∠*BOC* = 80°.)

(4) Taking *O* as the centre and *OB* as the segment, draw an angle of measure 240° on the other side of ∠*BOC** *such that it intersects the circle at point* A*.

(Since we need ∠*ACB* = 120°, we have drawn ∠*AOB* = 240°.)

(5) Join *AC*, *AB* and *BC*.

Δ*ABC* is the required triangle.

#### Page No 42:

#### Question 4:

How do we draw aangle?

#### Answer:

To construct the required angle, we will use the following property.

Angle made by an arc at any point on the alternate arc is equal to half the angle made at the centre.

The steps of construction are as follows:

(1) Draw a circle of any radius with *O *as the centre.

(2) Take a point *B* on the circle and join *OB*.

(3) Taking *O* as the centre and *OB* as the segment, draw an angle of measure 45° such that it intersects the circle at point *A*.

(Since we need ∠*ACB* =we have drawn ∠*AOB* = 45°.)

(4) Join *CA* and *CB*.

∠*ACB* is the required angle.

#### Page No 42:

#### Question 5:

In each of the pictures below, draw aangle, according to the specifications:

(i) At the point A

(ii) At the point A with one side along OA

(iii) At the point A one side along AB

#### Answer:

(i)

To construct the required angle at point *A*, we will use the following property.

Angle made by an arc at any point on the alternate arc is equal to half the angle made at the centre.

The steps of construction are as follows:

(1) Draw a circle of any radius with *O *as the centre.

(2) Take a point *B* on the circle and join *OB*.

(3) Taking *O* as the centre and *OB* as the segment, draw an angle of measure 45° such that it intersects the circle at point *C*.

(Since we need ∠*CAB* =we have drawn ∠*COB* = 45°.)

(4) Join *CA* and *AB*.

∠*CAB* is the required angle.

(ii)

To construct the required angle at point *A *with one side along* OA*, we will use the following property.

(1) Draw a circle of any radius with *O *as the centre.

(2) Join the diametre *AB* of the circle.

(3) Taking *O* as the centre and *OB* as the segment, draw an angle of measure 45° such that it intersects the circle at point *C*.

(Since we need ∠*CAB* =, we have drawn ∠*COB* = 45°.)

(4) Join AC.

∠*CAB* is the required angle.

(iii)

To construct the required angle at point *A *with one side along* AB*, we will use the following property.

(1) Draw a circle of any radius with *O *as the centre.

(2) Take two points *A* and *B* on the upper side of the centre and join *AB*.

(3) Join *OB*.

(4) Taking *O* as the centre and *OB* as the segment, draw an angle of measure 45° such that it intersects the circle at point *C*.

(Since we need ∠*CAB* =, we have drawn ∠*COB** *= 45°.)

(5) Join *AC*.

∠*CAB* is the required angle.

#### Page No 43:

#### Question 1:

In the picture below, O is the centre of the circle and the line OD is parallel to the line CA.

Prove that OD bisects ∠AOB.

#### Answer:

Given: *CA* is parallel to* OD*.

*CB* acts as the transversal.

∴ ∠*ACO* = ∠*DOB* (Corresponding angles) … (1)

We know that angle made by an arc at any point on the alternate arc is equal to half the angle made at the centre.

∴ ∠*AOB *=** **2*∠**ACB*

Using equation (1):

∠*AOB *=** **2∠*DOB*

⇒ *OD *acts as the bisector of ∠*AOB*.

Therefore, *OD* bisects ∠*AOB*.

Yes, we can use this concept to draw the bisector of a given angle.

Suppose, in the given figure, we are given only ∠*AOB* and *CB *as the diametre of the circle.

Let us join one end point, *C* of the diametre *CB* with point *A* of ∠*AOB*.

We get that 2∠*ACB* = ∠*AOB.*

If we draw a line segment passing through point *O* and parallel to* AC* such that it intersects the circle at point *D*, then *OD* acts as the bisector of ∠*AOB*.

#### Page No 44:

#### Question 1:

In the figure below, O is the centre of the circle.

Prove that *x* + *y* = 90°

#### Answer:

We know that angle made by an arc at any point on the alternate arc is equal to half the angle made at the centre.

∴ ∠*BOC* = 2∠*BAC*

⇒ ∠*BOC* = 2*y*

Now, *OB* =* OC* (Radii of the same circle)

∴ ∠*OBC* = ∠*OCB** =* *x * (Angles opposite to equal sides are equal in measure.)

Using angle sum property in Δ*OBC*:

∠*BOC* + ∠*OCB** *+ ∠*CBO** *= 180°

⇒ 2*y* + *x* + *x* = 180°

⇒ 2*y* + 2*x* = 180°

⇒ 2(*y* + *x*) = 180°

⇒ *y* + *x* =

⇒ *x *+ *y* = 90°

#### Page No 46:

#### Question 1:

In the figure below, A, B, C, D are points on the circle.

Compute the angles of the quadrilateral ABCD and the angles between its diagonals.

#### Answer:

We know that angles in a same segment are equal.

Here, ∠*CBD* and ∠*CAD* are the angles in the same segment.

⇒ ∠*CBD* = ∠*CAD* = 30°

Now, ∠*BAD* = ∠*BAC* + ∠*CAD* = 35° + 30° = 65°

∠*ADB* and ∠*ACB* are the angles in a same segment.

⇒ ∠*ADB* = ∠*ACB** *= 50°

∠*BDC* and ∠*BAC* are the angles in a same segment.

⇒ ∠*BDC* = ∠*BAC* = 35°

Now, ∠*ADC* = ∠*ADB* + ∠*BDC* = 50° + 35° = 85°

We know that angles in the alternate segments are supplementary.

∴ ∠*BAD* + ∠*BCD* = 180°

⇒ 65° + ∠*BCD* = 180°

⇒ ∠*BCD* = 180° − 65° = 115°

Also, ∠*ABC* + ∠*ADC* = 180°

⇒ ∠*ABC* + 85° = 180°

⇒ ∠*ABC* = 180° − 85° = 95°

∴ ∠*BAD* = 65°, ∠*ABC* = 95°, ∠*BCD* = 115° and ∠*ADC* = 85°

Using angle sum property in Δ*PBC*:

∠*PBC* + ∠*BCP* + ∠*CPB** *= 180°

⇒ 30° + 50° + ∠*CPB* = 180°

⇒ 80° + ∠*CPB* = 180°

⇒ ∠*CPB* = 180° − 80° = 100°

∠*APD** *= ∠*BPC* = 100° (Vertically opposite angles)

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

∴ ∠*BPC* + ∠*BPA** *= 180°

⇒ 100° + ∠*BPA* = 180°

⇒ ∠*BPA* = 180° − 100° = 80°

∠*CPD* = ∠*BPA* = 80° (Vertically opposite angles)

Hence, the angles between the diagonals are ∠*CPD* = ∠*BPA* = 80° and ∠*APD** *= ∠*BPC* = 100°.

#### Page No 46:

#### Question 2:

In the figure below, ΔABC is equilateral and O is its circumcentre.

Prove that the length of AD is equal to the radius of the circle.

#### Answer:

**Construction:** Join AO.

Given: Δ*ABC** *is an equilateral triangle.

⇒ ∠ABC = 60°

We know that angles in the same segment are equal.

⇒ ∠*ADC* = ∠*ABC* = 60°

Now,* OD *=* OA * (Radii of the same circle)

∴ ∠*OAD* = ∠*ADO* = 60° (Angle opposite to equal sides are equal in measure.)

Using angle sum property in Δ*AOD*:

∠*OAD* + ∠*ADO* + ∠*AOD* = 180°

⇒ 60° + 60° + ∠*AOD* = 180°

⇒ 120° + ∠*AOD* = 180°

⇒ ∠*AOD** *= 180° − 120° = 60°

⇒ ∠*OAD* = ∠*ADO* = ∠*AOD** *= 60°

Therefore, Δ*AOD* is an equilateral triangle.

⇒ *AO* = *AD* = *DO *

Thus, the length of *AD* is equal to the radius of the circle.

#### Page No 46:

#### Question 3:

In the picture below, ΔPQR is right angled. Also, ∠A = ∠P and BC = QR.

Prove that the diameter of the circumcircle of ΔABC is equal to the length of PQ.

#### Answer:

Given: ∠*PRQ* = 90°

We know that angle in a semi circle is a right angle.

If we draw a circle passing through vertex *R* of Δ*PRQ* then* PQ *acts as the diametre of the circle.

We have ∠*A* = ∠*P* and *BC* =* QR.*

We know that angles in the same segment are equal.

*BC* and *QR* are equal segments having equal angles, ∠*A* and ∠*P*. So, points *A*, *B* and *C* also lie on the same circle on which points *P*, *Q* and *R* lie.

As *PQ* is the diametre of the circumcircle of Δ*PQR*, *PQ* is the diametre of the circumcircle of Δ*ABC*.

#### Page No 51:

#### Question 1:

Prove that in a cyclic quadrilateral, the exterior angle at any vertex is equal to the interior angle at the opposite vertex.

#### Answer:

Consider the cyclic quadrilateral ABCD whose one side DC is extended to point E.

We know that opposite angles of a cyclic quadrilateral are supplementary.

∴ ∠*DAB* + ∠*DCB* = 180°

⇒ ∠*DCB* = 180° − ∠*DAB* … (1)

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

∴ ∠*BCE** *+ ∠*BCD** *= 180°

⇒ ∠*BCD** *= 180° − ∠*BCE* … (2)

From equation (1) and equation (2), we get:

180° − ∠*DAB* = 180° − ∠*BCE*

⇒ −∠*DAB* = −∠*BCE*

⇒ ∠*DAB* = ∠*BCE*

Hence, in a cyclic quadrilateral, the exterior angle at any vertex is equal to the interior angle at the opposite vertex.

#### Page No 51:

#### Question 2:

Prove that the non-rectangular parallelogram is not a cyclic quadrilateral.

#### Answer:

Consider the non-rectangular parallelogram,** **ABCD.

We know that opposite angles of a parallelogram are equal.

∴ ∠*A* = ∠*C* … (1)

Let us suppose that ABCD is a cyclic quadrilateral.

We know that opposite angles of a cyclic quadrilateral are supplementary.

⇒ ∠*A* + ∠*C* = 180°

⇒ ∠*A* + ∠*A* = 180° (From equation (1))

⇒ 2∠*A* = 180°

⇒ ∠C = 90°

This is a contradiction to the given condition that* ABCD* is a non-rectangular parallelogram.

Thus, our supposition was wrong.

Hence, a non- rectangular parallelogram is not a cyclic quadrilateral.

#### Page No 51:

#### Question 3:

Prove that non-isosceles trapeziums are not cyclic.

#### Answer:

Consider the non-isosceles trapezium,** **ABCD.

Let us suppose that ABCD is a cyclic quadrilateral.

We know that opposite angles of a cyclic quadrilateral are supplementary.

⇒ ∠*B* + ∠*D* = 180°

⇒ ∠*B* = 180° − ∠*D* … (1)

*ABCD* is a non-isosceles trapezium and *AB *is parallel to *CD*

∴ ∠*A* + ∠*D* = 180° (Sum of the interior angles on the same side of the transversal is 180°.)

⇒ ∠*A* = 180° − ∠*D* … (2)

From equation (1) and equation (2), we get:

∠*A* = ∠*B*

This is not possible as *ABCD* is a non-isosceles trapezium.

Thus, our supposition was wrong.

Hence, non-isosceles trapeziums are not cyclic.

#### Page No 51:

#### Question 4:

In the figure below, ABCD is a square.

How much is ∠APB?

#### Answer:

Given: *ABCD* is a square.

**Construction:** Join AC.

We know that the diagonals of a square bisect the angles.

∴ (Each angle of a square measures 90°)

Now,* APBC* is a cyclic quadrilateral as all its vertices lie on the circle.

We know that opposite angles of a cyclic quadrilateral are supplementary.

∴ ∠*APB* + ∠*ACB* = 180°

⇒ ∠*APB *+ 45° = 180°

⇒ ∠*APB* = 180° − 45°

⇒ ∠*APB* = 135°

#### Page No 51:

#### Question 5:

Prove that in a cyclic hexagon ABCDEF as shown below, ∠A + ∠C + ∠E = ∠B + ∠D + ∠F.

#### Answer:

Given: A cyclic hexagon *ABCDEF*.

**Construction: **Join CF.

*ABCF* is a cyclic quadrilateral as all its vertices lie on the circle.

We know that opposite angles of a cyclic quadrilateral are supplementary.

∴ ∠*B* + ∠*CFA* = 180° … (1)

∠*A* + ∠*BCF* = 180° … (2)

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

∠*B* + ∠*CFA* = ∠*A* + ∠*BCF* … (3)

*CDEF* is a cyclic quadrilateral as all its vertices lie on the circle.

∴ ∠*D* + ∠*EFC* = 180° … (4)

∠*E* + ∠*DCF* = 180° … (5)

From equation (4) and (5), we have:

∠*D* + ∠*EFC* = ∠*E* + ∠*DCF* … (6)

Adding equation (3) and equation (6):

∠*B* + ∠*CFA* + ∠*D* + ∠*EFC* = ∠*A* + ∠*BCF* + ∠*E* + ∠*DCF*

⇒ ∠*B* + ∠*D* + ∠*EFC* + ∠*CFA* = ∠*A* + ∠ *E* + ∠ *DCF* + ∠*BCF*

⇒ ∠*B* + ∠*D* + ∠*F* = ∠*A* + ∠*E** *+ ∠*C*

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