Embed Size (px)
Citation preview
SAMPLE C
ONTENT
SAMPLE C
ONTENT
Written as per the latest syllabus prescribed by the Maharashtra State Bureau of TextbookProduction and Curriculum Research, Pune.
• In accordance with Q. 5 and Q. 6 of the New Paper Pattern
• Covers Challenging Questions across chapters
• Includes constructions with accurate measurements
Salient Features
Printed at: Print to Print, Navi Mumbai
© Target Publications Pvt. Ltd. No part of this book may be reproduced or transmitted in any form or by any means, C.D. ROM/Audio Video Cassettes or electronic, mechanical
including photocopying; recording or by any information storage and retrieval system without permission in writing from the Publisher.
P.O. No. 164274 Balbharati Registration No.: 2018MH0022TEID: 13421
SAMPLE C
ONTENT
Perfect Std. X Mathematics Challenging Questions is a meticulously compiled handbook for
students of Std. X. It is in accordance to the latest guidelines furnished in the Evaluation Pattern for Std. 10th
by the Maharashtra State Bureau of Textbook Production & Curriculum Research, Pune.
According to the New Paper Pattern of Std. X, students would be asked challenging questions of
Mathematics as they would appear in Q. 5 and Q. 6 of the Board Examination. Taking cognizance of this
change, we have meticulously crafted this book that includes exhaustive yet carefully selected challenging
questions along with solutions spanning across all the chapters of the Text Book.
We are confident that this book will cater to the needs of students and will help them to prepare
extremely well for their SSC Board Examination.
The journey to create a complete book is strewn with triumphs, failures and near misses. If you think
we’ve nearly missed something or want to applaud us for our triumphs, we’d love to hear from you on
From,
Publisher
Edition: First
Disclaimer This reference book is transformative work based on ‘Mathematics; First Edition: 2018’ published by the Maharashtra State Bureau of Textbook Production and Curriculum Research, Pune. We the publishers are making this reference book which constitutes as fair use of textual contents which are transformed by adding and elaborating, with a view to simplify the same to enable the students to understand, memorize and reproduce the same in examinations. This work is purely inspired upon the course work as prescribed by the Maharashtra State Bureau of Textbook Production and Curriculum Research, Pune. Every care has been taken in the publication of this reference book by the Authors while creating the contents. The Authors and the Publishers shall not be responsible for any loss or damages caused to any person on account of errors or omissions which might have crept in or disagreement of any third party on the point of view expressed in the reference book. © reserved with the Publisher for all the contents created by our Authors. No copyright is claimed in the textual contents which are presented as part of fair dealing with a view to provide best supplementary study material for the benefit of students.
PREFACE
SAMPLE C
ONTENT
Question
No. Type of Questions Total Marks Marks with option
1. (A) Solve 4 out of 6 subquestions (1 mark each) 04 06
(B) Solve 2 out of 3 subquestions (2 marks each) 04 06
2. (A) Solve 4 out of 4 MCQ (1 mark each) 04 04
(B) Solve 2 out of 3 subquestions (2 marks each) 04 06
3. (A) Solve activity based 2 subquestions out of 3 (2 marks each) 04 06
(B) Solve 2 out of 3 subquestions (2 marks each) 04 06
4. Solve 3 out of 4 subquestions (3 marks each) 09 12
5. Solve 1 out of 2 subquestions (4 marks each) 04 08
6. Solve 1 out of 2 subquestions (3 marks each) 03 06
Total Marks 40 60 Note: The subquestions in Q. No.1 will be based on syllabus of Std. IX. The division of marks in question papers as per objectives will be as follows.
Mathematics - Part I Mathematics - Part II
Objectives Percentage of marks
Objectives Percentage of
marks
Previous knowledge 20
Previous knowledge 20
Knowledge and understanding 30
Knowledge and understanding 30
Application 40
Application 30
Skill 10
Skill 20
[Reference: ceneje<ì^ jep³e HeeþîeHegmlekeÀ efveefce&leer Je DeY³eeme¬eÀce mebMeesOeve ceb[U, HegCes efveefce&le cetu³eceeHeve DeejeKe[e] [P.S. Scan this Q.R. Code to get a better understanding of the Paper Pattern.]
PAPER PATTERN
SAMPLE C
ONTENT
Textbook Chapter
No. Topic Name Page No.
Part - I
1 Linear Equations in Two Variables 1
2 Quadratic Equations 12
3 Arithmetic Progression 22
4 Financial Planning 25
5 Probability 30
6 Statistics 33
Part - II
1 Similarity 45
2 Pythagoras Theorem 55
3 Circle 61
4 Geometric Constructions 79
5 Co-ordinate Geometry 86
6 Trigonometry 91
7 Mensuration 102
Note: Steps of construction are provided in
Chapter 4 : Geometric constructions for the students’ understanding.
INDEX
SAMPLE C
ONTENT
55
1. In the given figure, if LK = 6 3 , find MK,
ML, KN, MN and the perimeter of MNKL.
Solution: In MLK, MLK = 90 …[Given] MKL = 30 …[Given] LMK = 60
…[Remaining angle of a triangle] MLK is a 306090 triangle.
LK = 32
MK …[Side opposite to 60]
6 3 = 32
MK …[ LK = 6 3 ]
6 3 23 = MK
MK = 12 units
ML = 12
MK …[Side opposite to 30]
ML = 12 12
ML = 6 units In MNK, MKN = 90 …[Given] MNK = 45 …[Given] KMN = 45
…[Remaining angle of a triangle] MNK is a 454590 triangle.
MK = 12
MN …[Side opposite to 45]
12 = 12
MN
MN = 12 2 units
KN = 12 MN …[Side opposite to 45]
KN = 1212 2
KN = 12 units Perimeter of MNKL = LM + LK + KN + MN
= 6 + 6 3 + 12 +12 2 = 18 + 6 3 +12 2 = 6(3 + 3 + 2 2 ) units MK = 12 units, ML = 6 units, KN = 12
units, MN = 12 2 units, The perimeter of MNKL is 6(3 + 3 + 2 2 ) units.
2. In the given figure, PQRV is a trapezium in which seg PQ || seg VR. SR = 4 and PQ = 6. Find VR. Solution: In QSR, QSR = 90 …[Given] SQR = 45 …[Given] SRQ = 45
…[Remaining angle of a triangle] SRQ SQR QS = SR
…[Converse of isosceles triangle theorem] QS = SR = 4 …(i) In PQST, seg PQ || seg TS …[ VTS, TSR] PTS QST …[Each angle is 90] PQST is a rectangle. PT = QS = 4 …(ii)[From (i)] and PQ = TS = 6 …(iii)[ PQ = 6] In PTV, PTV = 90 …[Given] VPT = 60 …[Given] PVT = 30
…[Remaining angle of a triangle] PTV is a 30 60 90 triangle.
PT = 1 PV2
…[Side opposite to 30]
4 = 1 PV2
…[From (ii)]
PV = 8 …(iv)
M 45
30 L K
N
6 3
60 45
P 6 Q
R 4 S T V
LEVEL - 1
Pythagoras Theorem 2
SAMPLE C
ONTENT
56
56
Std. X: Mathematics (Part - II)
VT = 32
PV …[Side opposite to 60]
VT = 32 8 …[From (iv)]
VT = 4 3 …(v) VR = VT + TS + SR …[VTS, TSR] VR = 4 3 + 6 + 4
…[From (v), (iii) and (i)] VR = (10 + 4 3 )
3. In the given figure, PQR = 60 and ray
QT bisects PQR, seg BA ray QP and seg BC ray QR. If BC = 8, find the perimeter of ABCQ.
Solution: Ray QT is the bisector of PQR …(i)[Given] Point B lies on the bisector of PQR BA = BC …[By angle bisector theorem] But, BC = 8 …[Given] BA = BC = 8 …(ii)
BQC = 12PQR …[From (i)]
BQC = 12 60
BQC = 30 …(iii) In BCQ, BCQ = 90 …[Given] BQC = 30 …[From (iii)] QBC = 60
…[Remaining angle of a triangle] BCQ is a 30 60 90 triangle
BC = 12
BQ …[Side opposite to 30]
8 = 12
BQ …[From (ii)]
BQ = 16
QC = 32
BQ …[Side opposite to 60]
QC = 32 16
QC = 8 3
Similarly, QA = 8 3 Perimeter of ABCQ = AB + BC + QC + QA = 8 + 8 + 8 3 + 8 3 = (16 + 16 3 ) = 16(1 + 3 ) units The perimeter of ABCQ is 16(1 + 3 )
units. 4. Prove that the sum of the squares of the
sides of a rhombus is equal to the sum of the squares of diagonals.
Given : ABCD is a rhombus, diagonals AC and BD intersect at O. To Prove : AB2 + BC2 + CD2 + AD2 = AC2 + BD2 Proof: ABCD is a rhombus …[Given] In AOB, AOB = 90
…[Diagonals of a rhombus are perpendicular]
AB2 = OA2 + OB2
…(i)[By Pythagoras theorem]
OA = 12
AC …(ii)
OB = 12
BD …(iii)
AB2 = 21 AC
2
+ 21 BD
2
…[From (i), (ii) and (iii)]
AB2 = 14
AC2 + 14
BD2
AB2 = 2 2AC BD
4
4AB2 = AC2 + BD2
…[Multiplying throughout by 4] AB2 + AB2 + AB2 +AB2 = AC2 + BD2
…(iv) But, AB = BC = CD = AD
…(v)[Sides of a rhombus] AB2 + BC2 + CD2 + AD2 = AC2 + BD2
…[From (iv) and (v)]
P
A B
T
Q C R
8
A
B C
D
O
[Diagonals of a rhombus bisect each other]
SAMPLE C
ONTENT
57
Chapter 2: Pythagoras Theorem
5. ABD is a triangle in which A = 90 and
seg AC seg BD. Show that: i. AB2 = BC BD ii. AD2 = BD CD iii. AC2 = BC CD Proof: In ABD, BAD = 90 …[Given] seg AC hypotenuse BD …[Given] BAD BCA ACD
…(i)[Similarity of right angled triangles] i. BAD BCA …[From (i)]
ABBC
= ADAC
= BDAB
…[Corresponding sides of similar triangles]
ABBC
= BDAB
AB2 = BC BD ii. BAD ACD …[From (i)]
ABAC
= ADCD
= BDAD
…[ Corresponding sides of similar triangles]
ADCD
= BDAD
AD2 = BD CD iii. BCA ACD …[From (i)]
BCAC
= ACCD
= ABAD
…[ Corresponding sides of similar triangles]
BCAC
= ACCD
AC2 = BC CD 6. Prove that three times the square of any
side of an equilateral triangle is equal to four times the square of an altitude.
Given: ABC is an equilateral triangle seg AD side BC …[BDC]
To Prove: 3(AB)2 = 4(AD)2
Proof: In ADB, ADB = 90 AB2 = AD2 + BD2
…(i)[By Pythagoras theorem] ABC is an equilateral triangle. AB = BC = AC
…(ii)[Sides of an equilateral triangle] AB = AC …[From (ii)] Point A is equidistant from B and C. seg AD is bisector of side BC .
…[Perpendicular bisector theorem]
BD = 12
BC …(iii)
AB2 = AD2 + 21 BC
2
…[Substituting the value of (iii) in (i)]
AB2 = AD2 + 2BC
4
4AB2 = 4AD2 + BC2
…[Multiplying both sides by 4] 4AB2 = 4AD2 + AB2 …[From (ii)] 4AB2 AB2 = 4AD2 3AB2 = 4AD2 7. In ABC, AB = AC and D is any point on BC. Prove that: AB2 AD2 = BD CD Construction: Draw seg AE seg BC,
such that BEC. Proof: In AEB, AEB = 90 …[By construction] AB2 = AE2 + BE2
…(i)[By Pythagoras theorem] In AED, AED = 90 …[By construction] AD2 = AE2 + DE2
…(ii)[By Pythagoras theorem] AB2 AD2 = AE2 + BE2 AE2 DE2
…[Subtracting (ii) from (i)] AB2 AD2 = BE2 DE2 AB2 AD2 = (BE DE)(BE + DE) AB2 AD2 = BD(BE + DE) …(iii)[BDE] Also, AB = AC …[Given] Point A is equidistant from points B and C. seg AE is perpendicular bisector of side BC.
…[Perpendiular bisector theorem]
A
B D C
B C
A
D
A
B C D E
LEVEL - 2
SAMPLE C
ONTENT
58
58
Std. X: Mathematics (Part - II)
BE = CE …(iv) AB2 AD2 = BD (CE + DE) …[Substituting the value of (iv) in (iii)] AB2 AD2 = BD CD …[DEC] 8. ABCD is a quadrilateral. M is the
midpoint of diagonal AC and N is the midpoint of diagonal BD. Prove that : AB2 + BC2 + CD2 + DA2
= AC2 + BD2 + 4MN2. Given: ABCD is a quadrilateral.
M and N are the midpoints of diagonal AC and BD respectively.
To prove:AB2 + BC2 + CD2 + DA2 = AC2 + BD2 + 4MN2
Construction: Join seg DM and seg BM. Proof: In ADC, M is the midpoint of side AC. AD2 + CD2 = 2 DM2 + 2 CM2
…[Appollonius theorem]
AD2 + CD2 = 2 DM2 + 221 AC
2
…[M is the midpoint of AC]
AD2 + CD2 = 2DM2 + 2 14
AC2
AD2 + CD2 = 2DM2 + 12
AC2 …(i)
Similarly, in ABC, M is the midpoint of side AC.
AB2 + BC2 = 2BM2 + 12
AC2 …(ii)
AD2 + CD2 + AB2 + BC2
= 2DM2 + 12
AC2 + 2BM2 + 12
AC2
…[Adding (i) and (ii)] AB2 + BC2 + CD2 + DA2 = 2BM2 + 2DM2
+ AC2 …(iii) In DMB, N is the midpoint of side BD. BM2 + DM2 = 2MN2 + 2BN2
…[Appollonius theorem]
BM2 + DM2 = 2MN2 + 221 BD
2
…[N is the midpoint of BD]
BM2 + DM2 = 2MN2 + 2 14
BD2
BM2 + DM2 = 2MN2 + 12
BD2
2BM2 + 2DM2 = 4MN2 + 2 12
BD2
…[Multiplying both sides by 2] 2BM2 + 2DM2 = 4MN2 + BD2
…(iv) AB2 + BC2 + CD2 + DA2 = (4MN2 + BD2) + AC2
…[Substituting (iv) in (iii)] AB2 + BC2 + CD2 + DA2 = AC2 + BD2 + 4MN2 9. In ABC, ACB = 90. seg CD side AB
and seg CE is angle bisector of ACB.
Prove that: 2
2AD AE=BD BE
. [Mar 19]
Given: ACB = 90 seg CD side AB seg CE bisects ACB.
To prove: 2
2
AD AEBD BE
Proof: In ABC, ACB = 90 seg CD side AB …[Given] ADC ACB
…[Similarity of right angled triangles]
ACAB
= ADAC
…[Corresponding sides of similar triangles] AC2 = AD AB …(i) Similarly, CDB ACB
…[Similarity of right angled triangles]
BCAB
= BDBC
…[Corresponding sides of similar triangles] BC2 = BD AB …(ii)
2
2
ACBC
= AD ABBD AB
…[From (i) and (ii)]
2
2
ACBC
= ADBD
…(iii)
Now, seg CE is the bisector of ACB …[Given]
D
A B
M N
C
A
B C
ED
SAMPLE C
ONTENT
59
Chapter 2: Pythagoras Theorem
ACBC
= AEBE
…[Property of angle bisector of a triangle]
2
2
ACBC
= 2
2
AEBE
…(iv)[Squaring both sides]
ADBD
= 2
2AEBE
…[From (iii) and (iv)] 10. In ABC, ABC = 135. Prove that:
AC2 = AB2 + BC2 + 4A(ABC). Construction: Draw seg AD side BC, such that DBC. Proof: In ABC, ABC is an obtuse angle, and seg AD side DC AC2 = AB2 + BC2 + 2BC. BD
…(i)[By application of Pythagoras theorem]
A(ABC) = 12 BC AD
4 A(ABC) = 4 12 BC AD
…[Multiplying both sides by 4] 4A(ABC) = 2BCAD …(ii) ABC + ABD = 180
…[Angles in a linear pair] 135 + ABD = 180 ABD = 45 In ADB, ADB = 90 …[By Construction] ABD = 45 BAD = 45
…[Remaining angle of a triangle] ABD BAD AD = BD
…(iii)[Converse of isosceles triangle theorem]
4A(ABC) = 2BCBD …(iv)[From (ii) and (iii)]
AC2 = AB2 + BC2 + 4A(ABC) …[From (i) and (iv)]
11. PQR is a right angled triangle, right angled at Q such that QR = b and A(PQR) = a. If QN PR, then show that
QN =4 2
2a.bb 4a
.
Proof: Consider, 2ab = 2 A(PQR) QR
= 2 PR QN12
QR
2ab = PR QN QR …(i) Consider, b4 + 4a2
= QR4 + 4[A(PQR)]2
= QR4 + 421 PQ QR
2
= QR4 + 4 1
4 PQ2 QR2
= QR4 + PQ2 QR2 b4 + 4a2 = QR2 [QR2 + PQ2] 4 2 2 2b + 4a QR QR PQ
…(ii)[Taking square root of both sides] In PQR, PQR = 90 QR2 + PQ2 = PR2
…(iii)[By Pythagoras theorem]
4 2
2abb 4a
= 2 2
PR QN QRQR QR PQ
…[Dividing (i) by (ii)]
4 2 2
2ab PR QNb 4a PR
…[From (iii)]
4 2
2ab PR QNPRb 4a
QN = 4 2
2abb + 4a
12. ABC is a triangle, where C = 90. Let BC = a, CA = b, AB = c and let p be the length of the perpendicular from C on AB. Prove that:
i. cp = ab
ii. 21p
= 21a
+ 21b
A
D C B
135
P
Q R
N
b
A
C B
b
a
D c
p
LEVEL - 3
SAMPLE C
ONTENT
60
60
Std. X: Mathematics (Part - II)
Proof: In ACB,
ACB = 90 …[Given] seg CD hypotenuse AB …[Given]
ACB ADC CDB…(i)[Similarity of right angled triangles]
i. ACB ADC …[From (i)] ACAD
= BCCD
= ABAC
…[Corresponding sides of similar triangles]
bAD
= ap
= cb
ap
= cb
cp = ab …(ii)
ii. AB2 = AC2 + BC2
…[By Pythagoras theorem] c2 = b2 + a2
2
2 2
ca b
= 2
2 2
ba b
+ 2
2 2
aa b
…[Dividing throughout by a2b2]
2
2c
a.b = 2
1a
+ 2
1b
…(iii)
2
2c
c.p = 2
1a
+ 2
1b
…[From (ii) and (iii)]
2
2 2
cc p
= 2
1a
+ 2
1b
21p
= 21a
+ 21b
13. In ABC, ACB = 90,seg CD seg AB,seg DE seg CB.Show that: CD2 AC = AD AB DE.
Proof: In ACB,
ACB = 90 …[Given]seg CD hypotenuse AB
CD2 = AD DB…(i)[Theorem of geometric mean]
In DEB and ACB,
DEB ACB …[Each angle is 90]
DBE ABC …[Common angle]
DEB ACB
…[By AA test of similarity]
DEAC
= DBAB
…[Corresponding sides of similar triangles]
AC = DE ABDB - …(ii)
CD2 AC = AD DB DE ABDB
…[Multiplying (i) and (ii)]
CD2 AC = AD AB DE
A
C
D
E
B
![D STD ]STD W T STD WXŒP ST DDDDD ...d ˙˛~q˚std˙˛ tw•p˛]std˙˛w_t˜ std˙˛wxŒp st ddddd (¤ dfid˙˛ƒtw]std˙!ƒstdddddddddddd dddddddddddddddddddddhµµµµµµµ! xstd⁄n"]std#wt˜x](https://img.pdfslide.net/doc/110x75/5f0a52c07e708231d42b1742/d-std-std-w-t-std-wxp-st-ddddd-d-qstd-twapstdwtoe-stdwxp.jpg)
