Draw the Circle Chapter 2
Mathematics Part I Solutions Solutions for Class 10 Math Chapter ii Circles are provided hither with simple step-by-stride explanations. These solutions for Circles are extremely popular among Form 10 students for Math Circles Solutions come up handy for speedily completing your homework and preparing for exams. All questions and answers from the Mathematics Part I Solutions Book of Course 10 Math Chapter 2 are provided here for you for gratis. You will too love the ad-free experience on Meritnation's Mathematics Office I Solutions Solutions. All Mathematics Part I Solutions Solutions for grade Course 10 Math are prepared by experts and are 100% accurate.
Page No 32:
Question 1:
In Δ ABC , we have ∠ A = threescore° and ∠ B = 70°. Is the vertex C inside or exterior the circumvolve with diameter AB?
Reply:
Given: In Δ ABC , ∠ A = 60 ° and ∠ B = 70 °
Using bending sum belongings:
∠ A + ∠ B + ∠ C = 180 °
⇒ lx ° + 70 ° + ∠ C = 180 °
⇒ 130 ° + ∠ C = 180 °
⇒ ∠ C = 180 ° − 130 °
= l °
We know that angle in a semi circle is a right angle.
If AB is the diametre of the circle and then ∠ C should exist equal to xc ° .
However, here ∠ C is less than 90 ° . Therefore, vertex C will lie outside the circumvolve.
Folio No 32:
Question two:
Prove that if a pair of contrary angles of a quadrilateral are right, then a circle can be fatigued through all 4 of its vertices.
Answer:
Consider the figure beneath.
Hither, ABCD is a quadrilateral in which ∠ A = ∠ C = 90 ° .
Nosotros know that angle in a semi circle is a right angle.
Suppose a circle is drawn such that information technology passes through vertex A.
As ∠ BAD = ninety ° , BD acts like the diametre of this circle.
Similarly, suppose a circle is drawn such that information technology passes through vertex C.
As ∠ BCD = 90 ° , BD acts like the diametre of this circle.
Thus, BD is the diametre of the circumvolve passing through vertices A and C.
As the terminate points of the diametre prevarication on the circle, all the 4 vertices of quadrilateral ABCD lie on the circumvolve.
Hence, if a pair of reverse angles of a quadrilateral are right then a circle can exist fatigued through all 4 of its vertices.
Page No 32:
Question 3:
In the quadrilateral ABCD, we have AB = iii cm, BC = 4 cm, Air-conditioning = 5 cm, ∠ A = 120°, ∠ C = seventy°. If we draw the circle with Ac as diameter, which of the four vertices of ABCD would be inside the circumvolve? Which of them would be outside this circle? Is any vertex on the circle? What about the circle with BD as bore.
Answer:
Given: AB = 3 cm, BC = iv cm, AC = 5 cm, ∠ A = 120 ° and ∠ C = seventy °
Consider the post-obit:
AB two = 3 2 = ix
BC 2 = 4 2 = 16
AC 2 = 5 ii = 25
⇒ AC two = AB 2 + BC two
Therefore, past the antipodal of Pythagoras theorem, we go that ∠ B = 90 ° .
Using bending sum property in quadrilateral ABCD:
∠ A + ∠ B + ∠ C + ∠ D = 360 °
⇒ 120 ° + xc ° + 70 ° + ∠ D = 360 °
⇒ 280 ° + ∠ D = 360 °
⇒ ∠ D = 360 ° − 280 ° = eighty °
We know that angle in a semi circle is a right angle.
If we draw a circumvolve with Air conditioning as the diametre and so vertex B will prevarication on the circle.
As the measure of ∠ D is less than 90 ° , vertex D volition lie outside the circle.
Thus, vertices A, B and C volition prevarication on the circle.
Now, let us draw a circle with BD as the diametre.
Every bit ∠ A = 120 ° and ∠ C = 70 ° , vertex A volition lie inside the circle and vertex C will lie outside the circle.
Only vertices B and D will lie on the circumvolve.
Page No 41:
Question 1:
In this figure, what fraction of the circumference of the circle is the length of the arc ADB?
Reply:
Structure: Join points A and B with centre O of the circle.
Given: ∠ ACB = lx °
We know that angle fabricated by an arc at whatsoever indicate on the alternate arc is equal to half the angle fabricated at the center.
We know that measure of an arc is equal to the measure of the respective key angle.
⇒ m (arc ADB) = ∠ AOB
⇒ m (arc ADB) = 120 °
At present,
Thus, the length of the arc ADB is of the circumference of the circumvolve.
Page No 42:
Question ane:
What is the radius of the circumvolve shown below?
Answer:
Given: Δ ABC , with ∠ B = xc ° 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 = three cm (Sides opposite to equal angles are equal in length.)
Using Pythagoras theorem in Δ ABC:
Equally ∠ ABC = 90 ° and nosotros know that angle in a semi circle is a correct angle, Air-conditioning is the diametre of the circle.
Radius of the circle =
Page No 42:
Question 2:
What is the area of the circumvolve shown beneath?
Answer:
Let the given rectangle exist ABCD.
Construction: Bring together BD.
We know that each bending of a rectangle measures 90 ° .
∴ ∠ BAD = ninety °
Nosotros 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 =
Surface area of the circle = π r 2
Page No 42:
Question 3:
How do we draw a triangle with two of the bending 40° and 120° and circumradius three centimetres?
Respond:
To construct the required triangle, we volition use the following property.
Angle made by an arc at any point on the alternate arc is equal to half the bending fabricated at the centre.
The steps of structure are as follows:
(i) Draw a circumvolve with center O and radius 3 cm.
(two) Take a point B on this circle and bring together OB.
(3) Taking O as the centre and OB as the segment, depict an angle of measure out eighty ° such that it intersects the circle at bespeak C.
(Since nosotros need ∠ BAC = 40 ° , we accept fatigued ∠ BOC = 80 ° .)
(4) Taking O equally the centre and OB as the segment, draw an angle of measure 240 ° on the other side of ∠ BOC such that it intersects the circumvolve at point A.
(Since we need ∠ ACB = 120 ° , we have drawn ∠ AOB = 240 ° .)
(five) Join Ac, AB and BC.
Δ ABC is the required triangle.
Page No 42:
Question iv:
How do nosotros draw a angle?
Answer:
To construct the required bending, we volition use the following belongings.
Angle made by an arc at any betoken on the alternate arc is equal to half the angle made at the centre.
The steps of construction are as follows:
(1) Depict a circumvolve of whatever radius with O as the eye.
(2) Take a point B on the circle and join OB.
(3) Taking O as the middle and OB as the segment, describe an bending of measure 45 ° such that information technology intersects the circle at indicate A.
(Since we need ∠ ACB = we have drawn ∠ AOB = 45 ° .)
(iv) Join CA and CB.
∠ ACB is the required angle.
Page No 42:
Question five:
In each of the pictures below, draw a angle, according to the specifications:
(i) At the point A
(2) At the point A with 1 side along OA
(3) At the point A one side forth AB
Answer:
(i)
To construct the required angle at point A, we will employ the following property.
Bending made by an arc at any signal on the alternate arc is equal to half the angle fabricated at the center.
The steps of structure are every bit follows:
(one) Draw a circle of any radius with O equally the centre.
(two) Take a indicate B on the circle and join OB.
(3) Taking O every bit the heart and OB equally the segment, draw an angle of measure 45 ° such that it intersects the circumvolve at point C.
(Since we need ∠ CAB = nosotros have fatigued ∠ COB = 45 ° .)
(4) Join CA and AB.
∠ CAB is the required angle.
(ii)
To construct the required angle at signal A with one side along OA, we volition utilize the following property.
Angle made past an arc at any signal on the alternating arc is equal to half the angle made at the centre.
(1) Describe a circumvolve of whatsoever radius with O as the centre.
(2) Bring together the diametre AB of the circle.
(three) Taking O as the centre and OB as the segment, draw an angle of measure 45 ° such that information technology intersects the circumvolve at point C.
(Since nosotros need ∠ CAB = , we have fatigued ∠ COB = 45 ° .)
(4) Join Air conditioning.
∠ CAB is the required bending.
(3)
To construct the required bending at bespeak A with one side along AB, we volition utilize the following property.
Bending made past an arc at any point on the alternate arc is equal to half the angle made at the eye.
(1) Draw a circle of any radius with O as the heart.
(2) Take 2 points A and B on the upper side of the heart and bring together AB.
(iii) Join OB.
(four) Taking O as the centre and OB as the segment, draw an angle of measure 45 ° such that information technology intersects the circle at point C.
(Since nosotros need ∠ CAB = , we have fatigued ∠ COB = 45 ° .)
(5) Join Air conditioning.
∠ CAB is the required bending.
Folio No 43:
Question 1:
In the picture beneath, 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) … (ane)
We know that angle made by an arc at any point on the alternating arc is equal to half the angle made at the centre.
∴ ∠ AOB = 2 ∠ ACB
Using equation (one):
∠ AOB = 2 ∠ DOB
⇒ OD acts as the bisector of ∠ AOB.
Therefore, OD bisects ∠ AOB.
Yeah, we tin use this concept to draw the bisector of a given angle.
Suppose, in the given figure, nosotros are given only ∠ AOB and CB as the diametre of the circumvolve.
Let usa bring together ane end signal, C of the diametre CB with indicate A of ∠ AOB.
Nosotros get that 2 ∠ ACB = ∠ AOB.
If we draw a line segment passing through indicate O and parallel to Air conditioning such that it intersects the circumvolve at betoken D, and then OD acts as the bisector of ∠ AOB.
Folio No 44:
Question 1:
In the figure below, O is the centre of the circle.
Prove that x + y = 90°
Reply:
Nosotros know that angle made past an arc at any point on the alternate arc is equal to half the angle fabricated at the centre.
∴ ∠ BOC = 2 ∠ BAC
⇒ ∠ BOC = iiy
Now, OB = OC (Radii of the same circumvolve)
∴ ∠ OBC = ∠ OCB = x (Angles reverse to equal sides are equal in mensurate.)
Using angle sum property in Δ OBC:
∠ BOC + ∠ OCB + ∠ CBO = 180 °
⇒ 2y + x + x = 180 °
⇒ twoy + 210 = 180 °
⇒ 2(y + x) = 180 °
⇒ y + 10 =
⇒ x + y = 90 °
Page No 46:
Question ane:
In the figure below, A, B, C, D are points on the circumvolve.
Compute the angles of the quadrilateral ABCD and the angles between its diagonals.
Answer:
We know that angles in a aforementioned segment are equal.
Here, ∠ CBD and ∠ CAD are the angles in the same segment.
⇒ ∠ CBD = ∠ CAD = 30 °
Now, ∠ BAD = ∠ BAC + ∠ CAD = 35 ° + xxx ° = 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 °
At present, ∠ ADC = ∠ ADB + ∠ BDC = l ° + 35 ° = 85 °
We know that angles in the alternate segments are supplementary.
∴ ∠ BAD + ∠ BCD = 180 °
⇒ 65 ° + ∠ BCD = 180 °
⇒ ∠ BCD = 180 ° − 65 ° = 115 °
Likewise, ∠ 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 ° + fifty ° + ∠ CPB = 180 °
⇒ fourscore ° + ∠ CPB = 180 °
⇒ ∠ CPB = 180 ° − eighty ° = 100 °
∠ APD = ∠ BPC = 100 ° (Vertically opposite angles)
Nosotros 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 contrary angles)
Hence, the angles betwixt the diagonals are ∠ CPD = ∠ BPA = 80 ° and ∠ APD = ∠ BPC = 100 ° .
Page No 46:
Question 2:
In the figure below, Δ ABC is equilateral and O is its circumcentre.
Testify 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 aforementioned segment are equal.
⇒ ∠ ADC = ∠ ABC = lx °
Now, OD = OA (Radii of the aforementioned 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 ° + threescore ° + ∠ AOD = 180 °
⇒ 120 ° + ∠ AOD = 180 °
⇒ ∠ AOD = 180 ° − 120 ° = 60 °
⇒ ∠ OAD = ∠ ADO = ∠ AOD = threescore °
Therefore, Δ AOD is an equilateral triangle.
⇒ AO = AD = Practise
Thus, the length of AD is equal to the radius of the circumvolve.
Folio No 46:
Question 3:
In the moving-picture show beneath, Δ PQR is right angled. As well, ∠ A = ∠ P and BC = QR.
Evidence that the diameter of the circumcircle of Δ ABC is equal to the length of PQ.
Answer:
Given: ∠ PRQ = ninety °
We know that angle in a semi circumvolve is a right angle.
If we describe a circle passing through vertex R of Δ PRQ and then PQ acts equally the diametre of the circle.
We have ∠ A = ∠ P and BC = QR.
Nosotros know that angles in the same segment are equal.
BC and QR are equal segments having equal angles, ∠ A and ∠ P . And then, points A, B and C also lie on the same circle on which points P, Q and R lie.
Every bit PQ is the diametre of the circumcircle of Δ PQR , PQ is the diametre of the circumcircle of Δ ABC .
Page No 51:
Question ane:
Show that in a cyclic quadrilateral, the exterior bending at any vertex is equal to the interior angle at the opposite vertex.
Answer:
Consider the cyclic quadrilateral ABCD whose i side DC is extended to betoken E.
We know that reverse angles of a circadian 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 … (ii)
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.
Nosotros know that opposite angles of a parallelogram are equal.
∴ ∠ A = ∠ C … (one)
Let us suppose that ABCD is a cyclic quadrilateral.
Nosotros know that reverse angles of a cyclic quadrilateral are supplementary.
⇒ ∠ A + ∠ C = 180 °
⇒ ∠ A + ∠ A = 180 ° (From equation (ane))
⇒ 2 ∠ A = 180 °
⇒ ∠ C = ninety °
This is a contradiction to the given status that ABCD is a not-rectangular parallelogram.
Thus, our supposition was wrong.
Hence, a not- rectangular parallelogram is not a cyclic quadrilateral.
Page No 51:
Question three:
Prove that non-isosceles trapeziums are not circadian.
Answer:
Consider the non-isosceles trapezium, ABCD.
Let u.s.a. suppose that ABCD is a circadian quadrilateral.
We know that opposite angles of a cyclic quadrilateral are supplementary.
⇒ ∠ B + ∠ D = 180 °
⇒ ∠ B = 180 ° − ∠ D … (1)
ABCD is a not-isosceles trapezium and AB is parallel to CD
∴ ∠ A + ∠ D = 180 ° (Sum of the interior angles on the aforementioned 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 not-isosceles trapezium.
Thus, our supposition was incorrect.
Hence, non-isosceles trapeziums are not circadian.
Page No 51:
Question 4:
In the effigy beneath, ABCD is a square.
How much is ∠ APB ?
Answer:
Given: ABCD is a foursquare.
Construction: Join Air conditioning.
Nosotros know that the diagonals of a square bisect the angles.
∴ (Each angle of a foursquare measures 90 ° )
Now, APBC is a circadian quadrilateral as all its vertices lie on the circumvolve.
Nosotros 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 circadian hexagon ABCDEF every bit shown beneath, ∠ A + ∠ C + ∠ E = ∠ B + ∠ D + ∠ F .
Reply:
Given: A circadian hexagon ABCDEF.
Construction: Join CF.
ABCF is a circadian quadrilateral as all its vertices lie on the circumvolve.
Nosotros know that opposite angles of a cyclic quadrilateral are supplementary.
∴ ∠ B + ∠ CFA = 180 ° … (ane)
∠ A + ∠ BCF = 180 ° … (2)
From equation (one) and (2), nosotros have:
∠ B + ∠ CFA = ∠ A + ∠ BCF … (three)
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), nosotros take:
∠ D + ∠ EFC = ∠ E + ∠ DCF … (half dozen)
Adding equation (3) and equation (half dozen):
∠ B + ∠ CFA + ∠ D + ∠ EFC = ∠ A + ∠ BCF + ∠ East + ∠ DCF
⇒ ∠ B + ∠ D + ∠ EFC + ∠ CFA = ∠ A + ∠ Eastward + ∠ DCF + ∠ BCF
⇒ ∠ B + ∠ D + ∠ F = ∠ A + ∠ East + ∠ C
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