CBSE Class 10th Exam Date:17 Feb' 26 - 17 Feb' 26
Have you ever thought about how accurate measurements are made in engineering projects or how architects design buildings? Construction is the answer! This chapter teaches students a clear understanding of geometric constructions. Major topics such as the division of a line segment, the construction of tangents to a circle, and the creation of geometric shapes under given conditions are covered in this chapter. NCERT exemplar Class 10 Maths chapter 10 solutions are prepared by experts considering the practical nature of this chapter so that students can learn the topics of NCERT Class 10 Maths related to constructions.
This Story also Contains
These NCERT Exemplar Class 10 Maths chapter 10 solutions provide step-by-step solutions to the construction problems to make it easier for students to understand the chapter and its concepts. Notes for Class 10 Mathematics are provided here, the NCERT Notes for Class 10 Maths.
Class 10 Maths Chapter 10 Exemplar Solutions Exercise: 10.1 Page number: 114 Total questions: 6 |
Question:1
To divide a line segment AB in the ratio $5:7$, first, a ray AX is drawn so that$\angle BAX$ is an acute angle, and then at equal distances, points are marked on the ray AX such that the minimum number of these points is
(A) 8 (B) 10 (C) 11 (D) 12
Answer(D) 12
Solution
Given: $\angle BAX$ is an acute angle.
The required ratio is $5:7$
Let m = 5, n = 7
Steps of construction
1. Draw any ray AX making an acute angle with AB.
2. Locate 12 points on AX at equal distances. (Because here $m+n= 12$)
3. Join $A_{12}B$
4. Through the point $A_{5}$ draw a line parallel to $A_{12}B$ intersecting AB at the point P.
Then $AP:PB= 5:7$
$\because A_{5}P\parallel A_{12}B$
$\therefore \frac{AA_{5}}{A_{5}A_{12}}= \frac{AP}{PB}$ (By Basic Proportionality theorem)
By construction $\frac{AA_{5}}{A_{5}A_{12}}= \frac{5}{7}$
$\therefore \frac{AP}{PB}= \frac{5}{7}$
Hence, the number of points is 12.
Question:2
To divide a line segment AB in the ratio $4:7$, a ray AX is drawn first such that $\angle BAX$ is an acute angle and then points $A_{1}, A_{2 }, A_{3}\cdots$ are located at equal distances on the ray AX and the point B is joined to:
(A) A12 (B) A11 (C) A10 (D) A9
Answer(B) A11
Solution
Given: $\angle BAX$ is an acute angle
The required ratio is $4:7$
Let m = 4, n = 7
$m+n= 4+7= 11$
Steps of construction
1. Draw any ray AX making an acute angle with AB.
2. Locate 11 points on AX at equal distances (because m + n = 11)
3. Join $A_{11}B$
4. Through the point A4 draw a line parallel to $A_{11}B$ intersecting AB at the point P.
Then $AP:PB= 4:7$
Hence, point B is joined to A11.
Question:3
To divide a line segment AB in the ratio $5:6$, draw a ray AX such that $\angle BAX$ is an acute angle, then draw a ray BY parallel to AX and the points $A_{ 1}, A_ {2}, A_{3}\cdots$and $B _{1}, B_{2}, B_{3}\cdots$are located at equal distances on ray AX and BY, respectively. Then the points joined are
(A) A5 and B6 (B) A and B5 (C) A4 and B5 (D) A5 and B4
Answer(A) A5 and B6
Solution
Given: $\angle BAX$ and $\angle ABY$ both are acute angles and AX parallel to BY
The required ratio is $5:6$
Let m = 5 , n = 6
Steps of construction
1. Draw any ray AX making an acute angle with AB.
2. Draw a ray $BY\parallel AX$.
3. Locate the points $A_{1},A_{2},A_{3},A_{4},A_{5}$ on AX at equal distances
4. Locate the points $B_{1}, B_{2}, B_{3 }, B_{ 4 }, B_{5}$ on BY at a distance equal to the distance between points on the AX line.
5. Join $A_{5}B_{6}$.
Let it intersect AB at a point C in the figure.
Then $AC:CB= 5:6$
Here $\bigtriangleup AA_{5}C$ is similar to $\bigtriangleup BB_{6}C$
Then $\frac{AA_{5}}{BB_{6}}= \frac{5}{6}= \frac{AC}{BC}$
$\therefore$by construction $\frac{AA_{5}}{BB_{6}}= \frac{5}{6}$
$\therefore \frac{AC}{BC}= \frac{5}{6}$
$\therefore$ Points joined one A5 and B6.
Question:4
To construct a triangle similar to a given $\bigtriangleup ABC$ with its sides of $\frac{3}{7}$ the corresponding sides of $\bigtriangleup ABC$, first, draw a ray BX such that CBX is an acute angle and X lies on the opposite side of A Aconcerningo BC. Then locate points B1, B2, B3, ... on BX at equal distances, and the next step is to join
(A) B10 to C (B) B3 to C (C) B7 to C (D) B4 to C
Answer(C) B7 to C
Solution Given: $\angle CBX$ is an acute angle.
Steps of construction
1. Draw any ray BX making an angle with BC on the side opposite vertex A.
2. Locate 7 points on BX in equidistant
3. Now join B7 to C
4. Draw a line through B3${C}'$ parallel to B7C .
Question:5
Answer (B) 8
Solution
To construct a triangle similar to a triangle, with its sides $\frac{8}{5}$ of the corresponding sides of a given triangle, the minimum number of points to be located at an equal distance is equal to the greater of 8 and 5 in .$\frac{8}{5}$ Here $8> 5$
So, the minimum number of points to be located at equal distances on ray BX is 8.
Question:6
To draw a pair of tangents to a circle which are inclined to each other at an Angle of $60^{\circ}$, it is required to draw tangents at the endpoints of those two radii of the circle, and the angle between them should be
(A) $135^{\circ}$ (B) $90^{\circ}$ (C) $60^{\circ}$ (D) $120^{\circ}$
Answer(D) 120°
Solution
According to question:-
Given :$\angle QPR= 60^{\circ}$
Let $\angle QOR= x$
As we know, the angle between the agent and the radius of a circle is 90 degrees
$\angle PQO= \angle PRO= 90^{\circ}$
We know that $\angle PQO+ \angle PRO+ \angle QPR+\angle QOR= 360^{\circ}$
[$\because$ sum of interior angles of quadrilateral is $360^{\circ}$ ]
$90^{\circ}+90^{\circ}+x+60^{\circ}= 360^{\circ}$|
$240+x= 360^{\circ}$
$x= 120^{\circ}$
Class 10 Maths Chapter 10 exemplar solutions Exercise: 10.2 Page number: 115 Total questions: 4 |
Question:1
Write True or False and give reasons for your answer in each of the following:
By geometrical construction, it is possible to divide a line segment in the ratio $\sqrt{3}:\frac{1}{\sqrt{3}}$
Answer [True]
Solution
We need both positive integers to divide a line segment in the ratio.
So, we can simplify it by multiplying both the terms by .$\sqrt{3}$
We obtain $\sqrt{3}\times \sqrt{3} : \sqrt{3}\times \frac{1}{\sqrt{3}}$
$\Rightarrow 3:1$
So, the required ratio is $3:1$
In geometrical construction is possible to divide a line segment in the ratio $3:1$
Question:2
Write True or False and give reasons for your answer in each of the following:
To construct a triangle similar to a given $\bigtriangleup ABC$ with its sides $\frac{7}{3}$ of the corresponding sides of $\bigtriangleup ABC$, draw a ray BX making an acute angle with BCand X lies on the opposite side of A concerning BC. The points B1, B2,..., B7 are located at equal distances on BX, B3 is joined to C and then a line segment $B_6{C}'$ is drawn parallel to B3C where ${C}'$ lies on BC produced. Finally, line segment ${A}'{C}'$ is drawn parallel to AC.
Answer: False
Solution
According to the question:-
To construct a triangle similar to a given $\bigtriangleup ABC$ with its sides $\frac{7}{3}$ of the corresponding sides of $\bigtriangleup ABC$
1. Draw a line segment BC
2. Taking B and C as centres, draw two arcs of suitable radii intersecting each other at A.
3. Join BA and CA?ABC is the required triangle.
4. From B draw any ray BX downwards making an acute angle CBX.
5. Locate seven points B1, B2, b3, …. B7 on BX such that BB1 = B1B2 = B1B3 = B3B4 = B4B5 = B5B6 = B6B7.
6. Join B3C and from B7 draw a line B7C’ ? B3C intersecting the extended line segment BC at C’.
7. From point C’ draw C’A’? CAintersectsg the extended line segment BA at A
But as given, if we join B3C and from B6 draw a line B6C’? B3C intersecting the extended line segment BC at C’.
$BB_{3}/BB_{6}={BC}/{{BC}'} = 3/6$
${BC}/{{BC}'} = 1/2$
$BC:{BC}'= 1:2$
Hence, the sides are not in the ratio of 7:3
So, the required triangle can not be constructed in this way.
Hence, the given statement is false.
Question:3
Write True or False and give reasons for your answer in each of the following:
A pair of tangents can be constructed from a point P to a circle of radius 3.5 cm situated at a distance of 3 cm from the centre
Answer [False]
Solution
Radius, r = 3.5 cm
Point distance from centre = 3 cm
But, $r> d\Rightarrow 3.5 cm> 3cm$
So, the point P lies inside the circle.
So, a pair of tangents cannot be drawn to point P to a circle.
Question:4
Write True or False and give reasons for your answer in each of the following:
A pair of tangents can be constructed to a circle inclined at an angle of $ 170^ 70^{\circ}$.
Answer [True]
Solution
Tangent: - It is a straight line that touches the curve but does not cross it. A pair of tangents can be constructed to a circle inclined at an angle greater than $0^{\circ}$ but less than $180^{\circ}$. Here, the inclination angle is $170^{\circ}$. Hence, it is possible.
Class 10 Maths Chapter 10 exemplar solutions Exercise: 10.3 Page number: 116 Total questions: 4 |
Question:1
Draw a line segment of length 7 cm. Find a point P on it which divides it in the ratio $3:5$.
Solution
Given: AB = 7cm
The required ratio is $3:5$
Let m = 3, n = 5
$m+n= 3+5= 8$
Steps of construction
1. Draw a line segment AB = 7cm.
2. Draw a ray AX making an acute angle with AB
3. Locate 8 points $A_{1},A_{2},A_{3}\cdots A_{8}$ on AX on equidistant. (because $m+n= 8$ )
4. Join $BA_{8}$
5. Through the point $A_{3}$ draw a line parallel to $BA_{8}$ which intersect line AB at P..
Here triangle AA3P is similar to triangle $AA_{8}B$
$AA_{3}/AA_{8}= AP/AB= 3/5$ (by construction)
Therefore $AP:BP= 3:5$
Question:2
Solution
Given BC = 12cm, AB = 5 cm, $\angle B= 90^{\circ}$
Steps of construction
1. Draw a line $B C=12 \mathrm{~cm}$
2. From $B$ draw $A B=5 \mathrm{~cm}$ which makes an angle of $90^{\circ}$ at $B$.
3. Join AC
4. Make an acute angle at B as $<C B X$
5. On BX mark 3 points at equal distance $X_1, X_2, X_3$.
6. Join $ X_3C$
7. From $X_2$ draw $X_2 C^{\prime} \| X_3 C$ intersect AB at $C^{\prime}$
8. From point $C^{\prime}$ draw $C^{\prime} A^{\prime} \| C A$ intersect $\mathrm{AB} A^{\prime}$
Now $\triangle A^{\prime} B C^{\prime}$ is the required triangle and $\triangle A^{\prime} B C^{\prime}$ is also a right triangle.
Question:3
Solution
Given : BC = 6cm, CA = 5cm, AB = 4cm
Scale factor = 5/3
Let m = 5 and n = 3
Steps of construction
1. Draw a line BC = 6c cm
2. Taking B and C as centre mark arcs of length 4cm and 5cm respectively,y intersects each other at point A.
3. Join BA and CA
4 . Draw a ray BX making a g acute angle with BC.
5.Locate 5 Points on BX in equidistant as B1, B2, B3, B4, B5.
6.Join B3 to C and draw a line through B5 parallel to B3C to intersect at BC extended at ${C}'$.
7. Draw a line through ${C}'$ parallel to AC intersect AB extended at ${A}'$ .
Now ${A}'B{C}'$ is required triangle.
Question:4
Solution
Given : Radius = 4cm
Steps of construction
1. Draw a circle at radius r = 4 cm at point O.
2. Take a point P at a distance of 6cm from point O and join PO
3. Draw a perpendicular bisector of line PO. M is the midpoint of PO.
4. Taking M as the centre, we draw another circle of radius equal to MO, and it intersects the given circle at points Q and R
5. Now, join PQ and PR.
Then PQ and PR one the required two tangents.
Class 10 Maths Chapter 10 exemplar solutions Exercise: 10.4 Page number: 117-118 Total questions: 7 |
Question:1
Solution
Given :
AB = 5 cm and AC = 7 cm
$AP= \frac{3}{4}AB\: \: \cdots 1$
$AQ= \frac{1}{4}AC\: \: \cdots 2$
From equation 1
$AP= \frac{3}{4}AB$
$AP= \frac{3}{4}\times 5= \frac{15}{4}\: \: \left [ \because AB= 5cm \right ]$
P is any point on B
$\therefore PB= AB-AP= 5-\frac{15}{4}= \frac{20-15}{4}= \frac{5}{4}cm$
$\frac{AP}{AB}= \frac{15}{4}\times \frac{4}{5}= \frac{1}{3}$
$AP:AB= 1:3$
$\therefore$scale of a line segment AB is $\frac{1}{3}$
Steps of construction
1. Draw a line segment AB = 5 cm
2. Now draw ray AO which makes an angle,
3. Which A as centre and radius equal to 7 cm, draw an arc cutting line AO at C
4. Draw ray AP with acute angle BAP
5. Along AP make 4 points $A_{1},A_{2},A_{3},A_{4}$ with equal distance.
6. Join $A_{4}B$
7. From $A_{3}$ draw $A_{3}P$ which is parallel to $A_{4}B$ which meet AB at point P.
Then P is a point which divides AB in a ratio 3: 1
AP : PB = 3 : 1
8. Now draw ray AQ, with an acute angle CAQ.
9. Along AQ mark 4 points $B_{1},B_{2},B_{3},B_{4}$ with equal distance.
10. Join $B_{4}C$
11. From $B_{1}$ draw $B_{1}Q$ which is parallel to $B_{4}C$ which meet AC at point Q.
Then Q is a point which divides AC in a ratio 1 : 3
AQ : QC = 1 : 3
12. Finally, you join PQ, and its measurement is 3.25 cm.
Question:2
Solution
Steps of construction
1. Draw, A-line AB = 3 cm
2 Draw a ray by making $\angle ABP= 60^{\circ}$
3. Taking B as the centre and radius equal to 5 cm. Draw an arc which cuts BP at point C
4. Again draw ray AX making $\angle {Q}'AX= 60^{\circ}$
5. With A as the centre and radius equal to 5 cm, draw an arc which cuts AX at point D
6. Join C and D Here ABCD is a parallelogram
7. Join BD , BD is a diagonal of parallelogram ABCD
8. From B draw a ray BQ with any acute angle at po, int B, i.e., $\angle CBQ$ is the acute angle
9. Locate 4 points $B_{1},B_{2},B_{3},B_{4}$ on BQ with equal distance.
10. Join $B_{3}C$ and from $B_{4},{C}'$ parallel to $B_{3}C$ which intersect at point ${C}'$
11. From point ${C}'$ draw line ${C}'{D}'$ which is parallel to CD
12. Now draw a line segment ${D}'{A}'$ parallel to DA
Note : Here ${A}',{C}'$ and ${D}'$ are the extended sides.
13. ${A}'B{C}'{D}'$ is a parallelogram in which ${A}'{D}'= 6\cdot 5\, cm$ and ${A}'{B}= 4\, cm$ and ${A}'B{D}'= 60^{\circ}$ divide it into triangles $B{C}'{D}'$ and ${A}'{BD}'$ by the diagonal ${BD}'$
Question:3
Solution
Steps of construction
1. Draw two concentric circles with centre O and radii 3 cm and 5 cm
2. Taking any point P on the outer cijoin, join P and O
3. Draw a perpendicular bisector of OP let M be the midpoint of OP
4. Taking M as the centre and OM as the radius, draw a circle which cuts the inner circle at Q and R
5. Join PQ and PR. Thu, PQ and PR are required tangents
On measuring PQ and PR, we find that PQ = PR = 4 cm
Calculations
$\bigtriangleup OQP, \angle OQP= 90^{\circ}$
[usiPythagoras' theorem]
$\left ( 5 \right )^{2}= \left ( 3 \right )^{2}+\left ( PQ \right )^{2}$
$25-9= PQ^{2}$
$16= PQ^{2}$
$\sqrt{16}= PQ$
$4cm= PQ$
H, the length of both tangents is 4 cm.
Question:4
Solution
Steps of construction
1. a Draw a line BC = 5 cm
2. Taking B and C as centres, draw two arcs of equal radius 6 cm intersecting each other at point A.
3. Join AB and AC DABC is required isosceles triangle
4 From B draw ray $B_{X}$ with an acute angle $CB{B}'$
6. draw $B_{1},B_{2},B_{3},B_{4}$ at $BX$ with equal distance
7. Join $B_{3}C$ and from $B_{4}$ draw line $B_{4}D\parallel B_{3}C,$ , intersect extended segment BC at point D.
8. From point D draw $DE\parallel CA$ meting BA produced at E.
Then EBD is required triangle. We can name it PQR.
Justification
$\because B_{4}D\parallel B_{3}C$
$\therefore \frac{BC}{CD}= \frac{3}{1}\Rightarrow \frac{CD}{BC}= \frac{1}{3}$
$Now\, \therefore \frac{BD}{BC}= \frac{BC+CD}{BC}= 1+\frac{CD}{BC}= 1+\frac{1}{3}= \frac{4}{3}$
$Also\, DE\parallel CA$
$\therefore \bigtriangleup ABC\sim \bigtriangleup DBE$
$\frac{EB}{AB}= \frac{DE}{CA}= \frac{BD}{BC}= \frac{4}{3}$
Question:5
Solution
Given : AB = 5 cm, BC = 6 cm
Steps of construction
1. Draw a line segment AB = 5 cm
2.draw $< ABO= 60^{\circ}$ B taking as a centre draw an arc of radius BC=6cm
3. Join AC, DABC is the required triangle
4.From point A draw any ray $A{A}'$ with acute angle $\angle BA{A}'$
5.Mark 7 points $B_{1},B_{2},B_{3},B_{4},B_{5},B_{6},B_{7}$ with equal distance.
6.Join $B_{7}B$ and form $B_{5}$ draw $B_{5}X\parallel B_{7}B\, \, BY$making the angle equal From point X draw $XY\parallel BC$ intersecting AC at Y. Then, DAMN is the required triangle whose sides are equal to $\frac{5}{7}$ of the corresponding sides of the $\bigtriangleup ABC.
Justification: Here, $B_{5X}\parallel B_{7}B$ [by construction]
$\therefore \frac{AX}{XB}= \frac{5}{2}\Rightarrow \frac{XB}{AX}= \frac{2}{5}$
Now $\frac{AB}{AX}= \frac{AX+XB}{AX}$
$1+\frac{XB}{AX}= 1+\frac{2}{5}= \frac{7}{5}$
Also, $XY\parallel BC$
$\therefore \bigtriangleup AXY\sim \bigtriangleup ABC$
$\frac{AX}{AB}= \frac{AY}{AC}= \frac{YX}{BC}= \frac{5}{7}$
Question:6
Solution
Steps of construction
1. Draw a circle of radius OA = 4 cm with centre O
2. Produce OA to P such that $OA= AP=4 cm$
3. Draw a perpendicular bisector of OP = 8 =8cm
4. Now, taking A as the centre, we draw the circle of radius AP = OA = 4 cm
5Which intersectsst the circle at x and y
6. Join PX and PY
7. PX and PY are the tangents to the circle
Justification
In $\bigtriangleup OAX$ we have
$OA= OP= 4\, cm$ (Radius)
$AX= 4\, cm$ (Radius of circle with centre A)
$\therefore OAX$ is equilateral triangle
$\angle OAX= 60^{\circ}$
$\Rightarrow \angle XAP= 120^{\circ}$
In $\bigtriangleup PAX$ we have
$PA= AX= 4cm$
$\angle XAP= 120^{\circ}$
$\angle APX= 30^{\circ}$
$\Rightarrow APY= 30^{\circ}$
Hence $\angle XPY= 60^{\circ}$
Question:7
Solution
Steps of construction
1 . Draw a line segment BC = 6 cm
2 . Taking B and C as centres, draw an arc of radius AB 4cm and AC = 9 cm
3. Join AB and AC
4 . Triangle ABC is a required triangle. From Bd raw ray BM with acute angle $\angle XBM$
5.Make 3 points $B_{1},B_{2},B_{3}$ on BM with equal distance
6.Join $B_{2}C$ and $B_{3}$ draw $B_{3}X\parallel B_{2}C$ intersecting BC at X From point X draw XY||CA intersecting the extended line segment BA to Y Then $\bigtriangleup BXY$ is the required triangle whose sides are equal to$\frac{3}{2}$ of the $\bigtriangleup ABC$
Justification :
Here $B_{3}X\parallel B_{2}C$
$\therefore \frac{BC}{CX}= \frac{2}{1}$
$\therefore \frac{BX}{BC}= \frac{BC+CX}{BC}= 1+\frac{1}{2}= \frac{3}{2}$
Also $XY\parallel CA$
$\bigtriangleup ABC\sim \bigtriangleup YBX$
$\therefore \frac{YB}{AB}= \frac{YX}{AC}= \frac{BX}{BC}= \frac{3}{2}$.
Here are three angles that are the same, but the three sides are not the same.
$\therefore$The two triangles are not congruent because, if two triangles are congruent, then they have the same shape and size.
These Class 10 Maths NCERT Exemplar Chapter 10 solutions provide a basic knowledge of Construction, which has great importance in higher classes.
The questions based on Construction can be practised in a better way, along with these solutions.
These Class 10 Maths NCERT exemplar solutions in Chapter 10 Constructions are appropriate to solve other books such as NCERT Class 10 Maths, A Textbook of Mathematics by Monica Kapoor, RD Sharma Class 10 Maths, and RS Aggarwal Class 10 Maths.
Here are the subject-wise links for the NCERT solutions of class 10:
Given below are the subject-wise NCERT Notes of class 10 :
Here are some useful links for NCERT books and the NCERT syllabus for class 10:
Given below are the subject-wise exemplar solutions of class 10 NCERT:
Frequently Asked Questions (FAQs)
Yes, we can divide any line segment into any number of parts with any length ratio.
Yes, we can draw a common tangent for two given circles with the help of construction.
The chapter Constructions is vital for Board examinations as it holds around 2-3% weightage of the whole paper.
Generally, you can expect to get either a Long answer or a Very Long answer question in the board examination. A thorough study from NCERT exemplar Class 10 Maths solutions chapter 10 can help you ace the questions on Constructions.
On Question asked by student community
Hello! If you are looking for the CM Shri School admission result, here is the link provided by Careers360. You can check your result directly and stay updated about the admission status. I’ll be attaching it for your reference.
https://school.careers360.com/articles/cm-shri-school-admission-test-2025
https://school.careers360.com/articles/cm-shri-school-admission-test-result-2025
Hello Rajesh!!!
I have question paper of science of 9th standard . Here is your question paper link where you have click to open and wrap it up.
https://school.careers360.com/boards/dge-tamil-nadu/tn-class-9-quarterly-exam-question-paper-2025-26
Hello,
For your information, the class 7 2025 result of the CM Shri School is expected to be published on September 20, 2025. So, if you want to see and download the results, then you can visit the official website of the CM Shri School: edudel.nic.in.
I hope it will clear your query!!
Hello
Visit the official website of the Rajasthan Education Department or the Shala Darpan portal.
Look for the “Latest News” or “Examination” section. Check school notice boards or ask your class teacher for updates.
Some district education office websites may also upload the key. Avoid unofficial websites as they may provide incorrect or fake keys.
Hrllo,
If you want to view the CM Shri School 2025-26 admission test result , visit the official website, click on the "CM Shri Schools Admission Test - 2025." then select the "CM Shri Admission Test Result 2025 - Merit List" link. Here you need to log in with your credentials and view or download the merit list pdf.
I hope it will clear your query!!
This ebook serves as a valuable study guide for NEET 2025 exam.
This e-book offers NEET PYQ and serves as an indispensable NEET study material.
As per latest syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
As per latest syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters
As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE
As per latest syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters