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NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals
Edited By Ramraj Saini | Updated on Sep 27, 2023 10:40 PM IST
NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals
Quadrilaterals class 9 questions and answers are discussed here. These NCERT solutions are developed by the expert at Careers360 taking care of latest CBSE syllabus 2023. the solutions are provided for all the NCERT problems in simple and easy to understand language covering step by step all the concepts which ultimately help during the exam.
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This Story also Contains
- NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals
- Quadrilaterals Class 9 Questions And Answers PDF Free Download
- Quadrilaterals Class 9 Solutions - Important Formulae And Points
- Quadrilaterals Class 9 NCERT Solutions (Intext Questions and Exercise)
- Summary Of Class 9 Quadrilaterals NCERT Solutions
- Important Concepts Learned in NCERT Maths Chapter 8 Class 9 – Quadrilaterals
- Quadrilaterals Class 9 Solutions - Exercise Wise
- NCERT Solutions For Class 9 Maths - Chapter Wise
- NCERT Solutions For Class 9 - Subject Wise
In class 9 maths chapter 8 question answer, you will get to know about angles sum property, rhombus, parallelogram, square, rectangle, trapezium, kite, etc. The chapter is starting with recollecting the angle sum property of a quadrilateral. In total there are 2 exercises that consist of 37 questions. NCERT solutions for class 9 maths chapter 8 Quadrilaterals has the solutions to all the 37 questions. Here you will get NCERT solutions for class 9 Maths also.
Also Read :
- NCERT Exemplar Solutions for Class 9 Maths Chapter Quadrilaterals
- Quadrilaterals Class 9 Maths Chapter Notes
Quadrilaterals Class 9 Questions And Answers PDF Free Download
Quadrilaterals Class 9 Solutions - Important Formulae And Points
Quadrilateral:
Sum of all angles = 360°
Parallelogram:
A diagonal of a parallelogram divides it into two congruent triangles.
In a parallelogram, diagonals bisect each other.
In a parallelogram, opposite angles are equal.
In a parallelogram, opposite sides are equal.
Square:
Diagonals of a square bisect each other at right angles and are equal, and vice-versa.
Triangle:
A line through the midpoint of a side of a triangle parallel to another side bisects the third side (Midpoint theorem).
The line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to half the third side.
Parallelogram Angle Bisectors:
In a parallelogram, the bisectors of any two consecutive angles intersect at a right angle.
If a diagonal of a parallelogram bisects one of the angles of a parallelogram, it also bisects the second angle.
Rectangle:
The angle bisectors of a parallelogram form a rectangle.
Each of the four angles of a rectangle is a right angle.
Rhombus:
The diagonals of a rhombus are perpendicular to each other.
Free download NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals for CBSE Exam.
Quadrilaterals Class 9 NCERT Solutions (Intext Questions and Exercise)
Class 9 maths chapter 8 NCERT solutions - Exercise: 8.1
Q1 The angles of quadrilateral are in the ratio . Find all the angles of the quadrilateral.
Answer:
Given : The angles of a quadrilateral are in the ratio .
Let the angles of quadrilateral be .
Sum of all angles is 360.
Thus,
All four angles are :
Q2 If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Answer:
Given: ABCD is a parallelogram with AC=BD.
To prove: ABCD is a rectangle.
Proof : In ABC and
BAD,
BC= AD (Opposite sides of parallelogram)
AC=BD (Given)
AB=AB (common)
ABC
BAD (By SSS)
(CPCT)
and (co - interior angles)
Hence, it is a rectangle.
Answer:
Given: the diagonals of a quadrilateral bisect each other at right angles i.e.BO=OD.
To prove: ABCD is a rhombus.
Proof : In AOB and
AOD,
(Each
)
BO=OD (Given )
AO=AO (common)
AOB
AOD (By SAS)
AB=AD (CPCT)
Similarly, AB=BC and BC=CD
Hence, it is a rhombus.
Q4 Show that the diagonals of a square are equal and bisect each other at right angles.
Answer:
Given : ABCD is a square i.e. AB=BC=CD=DA.
To prove : the diagonals of a square are equal and bisect each other at right angles i.e. AC=BD,AO=CO,BO=DO and
Proof : In BAD and
ABC,
(Each
)
AD=BC (Given )
AB=AB (common)
BAD
ABC (By SAS)
BD=AC (CPCT)
In AOB and
COD,
OAB=
OCD (Alternate angles)
AB=CD (Given )
OBA=
ODC (Alternate angles)
AOB
COD (By AAS)
AO=OC ,BO=OD (CPCT)
In AOB and
AOD,
OB=OD (proved above)
AB=AD (Given )
OA=OA (COMMON)
AOB
AOD (By SSS)
AOB=
AOD (CPCT)
AOB+
AOD =
2. AOB =
AOB =
Hence, the diagonals of a square are equal and bisect each other at right angles.
Answer:
Given : ABCD is a quadrilateral with AC=BD,AO=CO,BO=DO, COD =
To prove: ABCD is a square.
Proof: Since the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a rhombus.
Thus, AB=BC=CD=DA
In BAD and
ABC,
AD=BC (proved above )
AB=AB (common)
BD=AC
BAD
ABC (By SSS)
(CPCT)
BAD+
ABC =
(Co-interior angles)
2. ABC =
ABC =
Hence, the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Q6 (i) Diagonal AC of a parallelogram ABCD bisects (see Fig.
). Show that
it bisects also.
Answer:
Given: DAC=
BAC ................1
DAC=
BCA.................2 (Alternate angles)
BAC=
ACD .................3 (Alternate angles)
From equation 1,2 and 3, we get
ACD=
BCA...................4
Hence, diagonal AC bisect angle C also.
Q6 (ii) Diagonal AC of a parallelogram ABCD bisects (see Fig.
). Show that
ABCD is a rhombus.
Answer:
Given: DAC=
BAC ................1
DAC=
BCA.................2 (Alternate angles)
BAC=
ACD .................3 (Alternate angles)
From equation 1,2 and 3, we get
ACD=
BCA...................4
From 2 and 4, we get
ACD=
DAC
In ADC,
ACD=
DAC (proved above )
AD=DC (In a triangle,sides opposite to equal angle are equal)
A parallelogram whose adjacent sides are equal , is a rhombus.
Thus, ABCD is a rhombus.
Q7 ABCD is a rhombus. Show that diagonal AC bisects as well as
and diagonal BD bisects
as well as
.
Answer:
In ADC,
AD = CD (ABCD is a rhombus)
3=
1.................1(angles opposite to equal sides are equal )
3=
2.................2 (alternate angles)
From 1 and 2, we have
1=
2.................3
and 1=
4.................4 (alternate angles)
From 1 and 4, we get
3=
4.................5
Hence, from 3 and 5, diagonal AC bisects angles A as well as angle C.
In ADB,
AD = AB (ABCD is a rhombus)
5=
7.................6(angles opposite to equal sides are equal )
7=
6.................7 (alternate angles)
From 6 and 7, we have
5=
6.................8
and 5=
8.................9(alternate angles)
From 6 and 9, we get
7=
8.................10
Hence, from 8 and 10, diagonal BD bisects angles B as well as angle D.
Q8 (i) ABCD is a rectangle in which diagonal AC bisects as well as
. Show that:
ABCD is a square
Answer:
Given: ABCD is a rectangle with AB=CD and BC=AD 1=
2 and
3=
4.
To prove: ABCD is a square.
Proof : 1=
4 .............1(alternate angles)
3=
4 ................2(given )
From 1 and 2, 1=
3.....................................3
In ADC,
1=
3 (from 3 )
DC=AD (In a triangle, sides opposite to equal angle are equal )
A rectangle whose adjacent sides are equal is a square.
Hence, ABCD is a square.
Q8 (ii) ABCD is a rectangle in which diagonal AC bisects as well as
. Show that:
diagonal BD bisects as well as
.
Answer: 
In ADB,
AD = AB (ABCD is a square)
5=
7.................1(angles opposite to equal sides are equal )
5=
8.................2 (alternate angles)
From 1 and 2, we have
7=
8.................3
and 7 =
6.................4(alternate angles)
From 1 and 4, we get
5=
6.................5
Hence, from 3 and 5, diagonal BD bisects angles B as well as angle D.
Q9 (i) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that (see Fig.
). Show that:
Answer:
Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that .
To prove :
Proof :
In
DP=BQ (Given )
ADP=
CBQ (alternate angles)
AD=BC (opposite sides of a parallelogram)
(By SAS)
Q9 (ii) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that (see Fig.
). Show that:
Answer:
Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that .
To prove :
Proof :
In
DP=BQ (Given )
ADP=
CBQ (alternate angles)
AD=BC (opposite sides of a parallelogram)
(By SAS)
(CPCT)
Q9 (iii) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that (see Fig.
). Show that:
Answer:
Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that .
To prove :
Proof :
In
DP=BQ (Given )
ABQ=
CDP (alternate angles)
AB=CD (opposite sides of a parallelogram)
(By SAS)
Q9 (iv) In parallelogram ABCD, two points P and Q are taken on diagonal BD such that (see Fig.
). Show that:
Answer:
Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that .
To prove :
Proof :
In
DP=BQ (Given )
ABQ=
CDP (alternate angles)
AB=CD (opposite sides of a parallelogram)
(By SAS)
(CPCT)
Answer:
Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that .
To prove: APCQ is a parallelogram
Proof :
In
DP=BQ (Given )
ADP=
CBQ (alternate angles)
AD=BC (opposite sides of a parallelogram)
(By SAS)
(CPCT)...............................................................1
Also,
In
DP=BQ (Given )
ABQ=
CDP (alternate angles)
AB=CD (opposite sides of a parallelogram)
(By SAS)
(CPCT)........................................2
From equation 1 and 2, we get
Thus, opposite sides of quadrilateral APCQ are equal so APCQ is a parallelogram.
Answer:
Given: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD.
To prove :
Proof: In ,
APB=
CQD (Each
)
ABP=
CDQ (Alternate angles)
AB=CD (Opposite sides of a parallelogram )
Thus, (By SAS)
Answer:
Given: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD.
To prove :
Proof: In ,
APB=
CQD (Each
)
ABP=
CDQ (Alternate angles)
AB=CD (Opposite sides of a parallelogram )
Thus, (By SAS)
(CPCT)
Answer:
Given : In and
,
and
.
To prove : quadrilateral ABED is a parallelogram
Proof : In ABED,
AB=DE (Given)
AB||DE (Given )
Hence, quadrilateral ABED is a parallelogram.
Answer:
Given: In and
,
and
.
To prove: quadrilateral BEFC is a parallelogram
Proof: In BEFC,
BC=EF (Given)
BC||EF (Given )
Hence, quadrilateral BEFC is a parallelogram.
Answer:
To prove: and
Proof :
In ABED,
AD=BE.................1(ABED is a parallelogram)
AD||BE .................2(ABED is a parallelogram)
In BEFC,
BE=CF.................3(BEFC is a parallelogram)
BE||CF .................4(BEFC is a parallelogram)
From 2 and 4 , we get
AD||CF
From 1 and 3, we get
AD=CF
Answer:
To prove : quadrilateral ACFD is a parallelogram
Proof :
In ABED,
AD=BE.................1(ABED is a parallelogram)
AD||BE .................2(ABED is a parallelogram)
In BEFC,
BE=CF.................3(BEFC is a parallelogram)
BE||CF .................4(BEFC is a parallelogram)
From 2 and 4 , we get
AD||CF...........................5
From 1 and 3, we get
AD=CF...........................6
From 5 and 6, we get
AD||CF and AD=CF
Thus, quadrilateral ACFD is a parallelogram
Answer:
In ACFD,
AC=DF (Since, ACFD is a parallelogram which is prooved in part (iv) of the question)
Answer:
In and
,
AB=DE (Given )
BC=EF (Given )
AC=DF ( proved in (v) part)
(By SSS rule)
Q12 (i) ABCD is a trapezium in which and
(see Fig.
). Show that
[ Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]
Answer:
Given: ABCD is a trapezium in which and
To prove :
Proof: Let A be
1,
ABC be
2,
EBC be
3,
BEC be
4.
In AECD,
AE||DC (Given)
AD||CE (By cnstruction)
Hence, AECD is a parallelogram.
AD=CE...............1(opposite sides of a parallelogram)
AD=BC.................2(Given)
From 1 and 2, we get
CE=BC
In BCE,
.................3 (opposite angles of equal sides)
...................4(linear pairs)
.....................5(Co-interior angles)
From 4 and 5, we get
(Since,
)
Q12 (ii) ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that
[ Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]
Answer:
Given: ABCD is a trapezium in which and
To prove :
Proof: Let A be
1,
ABC be
2,
EBC be
3,
BEC be
4.
(Co-interior angles)
(Co-interior angles)
Thus, (Since ,
)
Q12 (iii) ABCD is a trapezium in which and
(see Fig.
). Show that
[ Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]
Answer:
Given: ABCD is a trapezium in which and
To prove :
Proof: In ,
BC=AD (Given )
AB=AB (Common )
(proved in (i) )
Thus, (By SAS rule)
Q12 (iv) ABCD is a trapezium in which and
(see Fig.
). Show that diagonal AC
diagonal BD
[ Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]
Answer:
Given: ABCD is a trapezium in which and
To prove: diagonal AC diagonal BD
Proof: In ,
BC=AD (Given )
AB=AB (Common )
(proved in (i) )
Thus, (By SAS rule)
diagonal AC diagonal BD (CPCT)
Class 9 maths chapter 8 question answer - Exercise : 8.2
Answer:
Given: ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig ). AC is a diagonal.
To prove :
and
Proof: In ACD,
S is the midpoint of DA. (Given)
R is the midpoint of DC. (Given)
By midpoint theorem,
and
Answer:
Given: ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig ). AC is a diagonal.
To prove :
Proof : In ACD,
S is mid point of DA. (Given)
R is mid point of DC. (Given)
By mid point theorem,
and
...................................1
In ABC,
P is mid point of AB. (Given)
Q is mid point of BC. (Given)
By mid point theorem,
and
.................................2
From 1 and 2,we get
and
Thus,
Answer:
Given : ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig ). AC is a diagonal.
To prove : PQRS is a parallelogram.
Proof : In PQRS,
Since,
and
.
So,PQRS is a parallelogram.
Answer:
Given: ABCD is a rhombus in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA . AC, BD are diagonals.
To prove: the quadrilateral PQRS is a rectangle.
Proof: In ACD,
S is midpoint of DA. (Given)
R is midpoint of DC. (Given)
By midpoint theorem,
and
...................................1
In ABC,
P is midpoint of AB. (Given)
Q is mid point of BC. (Given)
By mid point theorem,
and
.................................2
From 1 and 2,we get
and
Thus, and
So,the quadrilateral PQRS is a parallelogram.
Similarly, in BCD,
Q is mid point of BC. (Given)
R is mid point of DC. (Given)
By mid point theorem,
So, QN || LM ...........5
LQ || MN ..........6 (Since, PQ || AC)
From 5 and 6, we get
LMPQ is a parallelogram.
Hence, LMN=
LQN (opposite angles of the parallelogram)
But, LMN= 90 (Diagonals of a rhombus are perpendicular)
so, LQN=90
Thus, a parallelogram whose one angle is right angle,ia a rectangle.Hence,PQRS is a rectangle.
Answer:
Given: ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively.
To prove: the quadrilateral PQRS is a rhombus.
Proof :
In ACD,
S is the midpoint of DA. (Given)
R is the midpoint of DC. (Given)
By midpoint theorem,
and
...................................1
In ABC,
P is the midpoint of AB. (Given)
Q is the midpoint of BC. (Given)
By midpoint theorem,
and
.................................2
From 1 and 2, we get
and
Thus, and
So, the quadrilateral PQRS is a parallelogram.
Similarly, in BCD,
Q is the midpoint of BC. (Given)
R is the midpoint of DC. (Given)
By midpoint theorem,
and
...................5
AC = BD.......................6(diagonals )
From 2, 5 and 6, we get
PQ=QR
Thus, a parallelogram whose adjacent sides are equal is a rhombus. Hence, PQRS is a rhombus.
Answer:
Given: ABCD is a trapezium in which , BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig.
).
To prove: F is the mid-point of BC.
In ABD,
E is the midpoint of AD. (Given)
EG || AB (Given)
By converse of midpoint theorem,
G is the midpoint of BD.
In BCD,
G is mid point of BD. (Proved above)
FG || DC (Given)
By converse of midpoint theorem,
F is the midpoint of BC.
Answer:
Given: In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively
To prove: the line segments AF and EC trisect the diagonal BD.
Proof : In quadrilatweral ABCD,
AB=CD (Given)
(E and F are midpoints of AB and CD)
In quadrilateral AECF,
AE=CF (Given)
AE || CF (Opposite sides of a parallelogram)
Hence, AECF is a parallelogram.
In DCQ,
F is the midpoint of DC. (given )
FP || CQ (AECF is a parallelogram)
By converse of midpoint theorem,
P is the mid point of DQ.
DP= PQ....................1
Similarly,
In ABP,
E is the midpoint of AB. (given )
EQ || AP (AECF is a parallelogram)
By converse of midpoint theorem,
Q is the midpoint of PB.
OQ= QB....................2
From 1 and 2, we have
DP = PQ = QB.
Hence, the line segments AF and EC trisect the diagonal BD.
Answer:
Given: ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA. AC, BD are diagonals.
To prove: the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.
Proof: In ACD,
S is the midpoint of DA. (Given)
R is midpoint of DC. (Given)
By midpoint theorem,
and
...................................1
In ABC,
P is the midpoint of AB. (Given)
Q is the midpoint of BC. (Given)
By midpoint theorem,
and
.................................2
From 1 and 2, we get
and
Thus, and
So, the quadrilateral PQRS is a parallelogram and diagonals of a parallelogram bisect each other.
Thus, SQ and PR bisect each other.
Answer:
Given: ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D.
To prove :D is mid point of AC.
Proof: In ABC,
M is mid point of AB. (Given)
DM || BC (Given)
By converse of mid point theorem,
D is the mid point of AC.
Answer:
Given: ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallels to BC intersects AC at D.
To prove :
Proof : ADM =
ACB (Corresponding angles)
ADM=
. (
ACB =
)
Hence, .
Answer:
Given: ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D.
To prove :
Proof : In ABC,
M is the midpoint of AB. (Given)
DM || BC (Given)
By converse of midpoint theorem,
D is the midpoint of AC i.e. AD = DC.
In AMD and
CMD,
AD = DC (proved above)
ADM =
CDM (Each right angle)
DM = DM (Common)
AMD
CMD (By SAS)
AM = CM (CPCT)
But ,
Hence, .
Summary Of Class 9 Quadrilaterals NCERT Solutions
Following topics are covered in the chapter:
- Angle Sum Property
- Types of Quadrilaterals
- Properties of a Parallelogram
- Condition for a Quadrilateral to be a Parallelogram
- The Mid-point Theorem
Important Concepts Learned in NCERT Maths Chapter 8 Class 9 – Quadrilaterals
- A quadrilateral is a polygon with four sides and four angles.
- The sum of the angles in a quadrilateral is 360 degrees.
- The opposite sides of a parallelogram are parallel and congruent.
- The opposite angles of a parallelogram are congruent.
- The diagonals of a parallelogram bisect each other.
- A rectangle is a parallelogram with four right angles.
- The diagonals of a rectangle are congruent and bisect each other.
- A square is a rectangle with four congruent sides.
- The diagonals of a square are congruent and bisect each other at right angles.
- A rhombus is a parallelogram with four congruent sides.
- The diagonals of a rhombus bisect each other at right angles.
- The area of a quadrilateral can be calculated by dividing it into triangles and using the formula for the area of a triangle.
- The perimeter of a quadrilateral is the sum of the lengths of its sides.
- The properties of quadrilaterals can be used to solve problems related to real-life situations.
Quadrilaterals Class 9 Solutions - Exercise Wise
Students can practice class 9 maths ch 8 question answer using the exercise line given below.
NCERT Solutions For Class 9 Maths - Chapter Wise
Chapter No. | Chapter Name |
Chapter 1 | |
Chapter 2 | |
Chapter 3 | |
Chapter 4 | |
Chapter 5 | |
Chapter 6 | |
Chapter 7 | |
Chapter 8 | Quadrilaterals |
Chapter 9 | |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | |
Chapter 13 | |
Chapter 14 | |
Chapter 15 |
NCERT Solutions For Class 9 - Subject Wise
How To Use NCERT Solutions For Class 9 Maths Chapter 8 Quadrilaterals
- Read and memorize the properties related to all kinds of quadrilaterals.
- Have a glance through some solved to understand the pattern of the solution to that particular question.
- Now check your learning on the practice exercises problems.
- If you stuck in any question then you can assist yourself using NCERT solutions for class 9 maths chapter 8 Quadrilaterals.
- Following the above-written steps, you can get 100% out of the chapter.
Also Check NCERT Books and NCERT Syllabus here:
Frequently Asked Questions (FAQs)
1. What are the important topics in Quadrilaterals ch 8 maths class 9?
Angle sum property of a quadrilateral, properties of the parallelogram, another condition for a quadrilateral to be a parallelogram, and mid-point theorem are the important topics covered in class 9th quadrilateral solution. students can practice these NCERT solutions for class 9 to command the concepts.
2. How does the NCERT solutions are helpful ?
NCERT solutions are not only helpful for the students if they stuck while solving NCERT problems but also it will provide them new ways to solve the problems. These solutions are provided in a very detailed manner which will give them conceptual clarity.
3. In NCERT Solutions for Class 9 Maths Chapter 8, how does the concept of "quadrilaterals" get defined?
Quadrilateral class 9 solutions define a quadrilateral as a two-dimensional shape that has four sides or edges and four corners or vertices. Quadrilaterals are commonly recognized by their standard shapes, such as rectangle, square, trapezoid, and kite, but they can also have irregular and undefined shapes.
4. Where can I find the complete solutions of class 9 chapter 8 maths ?
Here you will get the detailed NCERT solutions for class 9 by clicking on the link. for ease students can study quadrilateral class 9 pdf both online and offline mode.
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Option 2)
3.125 × 10-2 |
| Option 3)
1.25 × 10-2 |
Option 4)
2.5 × 10-2 |
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