My observation

Three consecutive terms of an arithmetic progression can represent three sides of a triangle provided that their smallest term is greater than the common difference.

Trick to draw a trapezium

A TR A P eziuM (trapezium) is a TRiangle And a ParallelograM.

How to draw 90 degrees angle quickly without using protractor?

Draw 90 degrees angle:

1) Take any point B.

2) Draw a horizontal line.

3) Take any point D say to right of B, say at a distance 5 cm from B.

4) Take any ​radius more than half 

​of​ BD say 3 cm and draw arcs with center B & D on upper part of seg BD.

5) Name the point of intersection of arcs as A.

6) Draw a ray from D to A so that it extends beyond A, above B and to the left of B. 

7) With point A as center and radius equal to AD, draw an arc PQ to cut the ray DA at point C.

8) Join C & B.

9) Angle CBD = 90 degrees. i.e. CB is perpendicular to BD.

This construction is based on following properties and theorem respectively:
(1) In right angled triangle, median on hypotenuse is half the hypotenuse.
(2) Angle inscribed in a semi-circle is 90 degrees

My sum HOTS quadratic equations

Quadratic equation HOTS

If 3 cm & 5 cm are two sides of triangle whose area is 6 sq.cm. then find it's third side.

Angle bisector sum's solution

Angle bisector sum

Cube root by long division method

Method of cube root by long division method is similar to method of square root by long division method except:

1) in stead of adding the latest quotient part we have to add take full quotient

2) in stead of adding the number before multiplication we have to:

2a) find 300 x quotient square

2b) 30 x quotient

2c) take a judgement from step 2a) whiat will be next number

2d) multiply 2b) by that number

2e) take square of that number

2f) Add answers of 2a) 2 d) and 2e) and

2g) multiply 2f) by that number ( step 2c)'s judgement )

Any way the steps are re-written in order:

(1) Make groups of 3 digits starting from right-most (unit place digit) number.

(2) take the left-most group. It may have 1,2 or 3 numbers or digits

(3) find the largest cube less than this number.

(4) Subtract it from step (2)'s number. Write in quotient's place cube root of the number whose cube you have just now subtracted.

(5) Bring the next group of 3 numbers down i.e. write it besides the subtraction(remainder)  of the last step.

(6) Next number b you have to find such that [300 x a2 + 30a x b + b2] x b should be less than (as we have been doing in normal division method) (not much less than) this number.

Here "a"  is the quotient (answer of cube root) you have got till now.

(7a) Continue the procees by taking the steps (5) and (6) . If the given number is known to be a perfect cube or is a perfect cube, your process will stop when you come to last group of 3 digits.

In other words, continue the process till your last number lets you continue.

(7b) If the given number is not a perfect cube then you have to continue the process till you get 1 more place than the number of decimal points upto which you want accuracy.

Hope I have explained the method clearly. 2 3 1 1 2 Working for for Step 1 2 12 345 678 910 111 Working for Step 2 0 0 0 Step 1 8 1 1 1 4 345 a= 2 2 8 2 4345 b= 3 3 27 3 Step 2 4167 300a2 1200 3.620833 4 64 4 178 678 30a 60 30ax b 180 5 125 5 178678 b2 9 6 216 6 Step 3 159391 300a2+30ab+b2 Total 1389 7 343 7 19287 910 (300a2+30ab+b2) x b 4167 8 512 8 19287910 9 729 9 Step 4 16015231 3272679 111 Working for Step 3 Step 5 3272679111 23 3204709928 a= 23 67969183 b= 1 300a2 158700 1.125885 30a 690 30ax b 690 b2 1 300a2+30ab+b2 Total 159391 (300a2+30ab+b2) x b 159391 Working for Step 4 a= 231 b= 1 300a2 16008300 1.204869 30a 6930 30ax b 6930 b2 1 300a2+30ab+b2 Total 16015231 (300a2+30ab+b2) x b 16015231 Working for Step 5 a= 2311 b= 2 300a2 1602216300 2.042595 30a 69330 30ax b 138660 b2 4 300a2+30ab+b2 Total 1602354964 (300a2+30ab+b2) x b 3204709928 12345678910111 Cube root  23112.04241

Hemant's unique method

Quadratic Equations by Completing Square Method

Sums on Geometry chapter Circle

Squares of 1001 to 1031

Trick to remember squares of selected 4 digit numbers starting from 1000 till 1031.

Write "10", to its right double of last 2 digits from  given number say ex. 1024. Then 2x24=48.

to the right of this "1048" write square of 24 i.e. 576.
your answer is 1048576.

Triangle Similarity Pythagoras Theorem HOTS

HOTS

If ABC is a triangle, P, Q & R are midpoints of sides AB, AC and BC respectively. If AS is perpendicular to BC then prove that PQRS is a cyclic quadrilateral.

Hint:
The proof is based on:
1) midpoint theorem
2) In right angled triangle, median on hypotenuse is half the hypotenuse.
3) Two triangles with common base and equal opposite angles form cyclic quadrilateral.

How to decide dimensions of cyclic quadrilateral?

Circle with center O		PQ = a	QR=b	RS=c	SP=d
Radius of circle	R	5			Sin	R Sin theta	Cos	R Cos theta	Height of isosceles triangle	Angle between Radius & side
Angle bet P & Q		60	30	0.523599	0.5	2.5	0.866025	4.330127	4.330127019	60	Angle OPQ
Angle bet Q & R		120	60	1.047198	0.866025404	4.330127	0.5	2.5	2.5	30	Angle OQR
Angle bet R & S		90	45	0.785398	0.707106781	3.5355339	0.707107	3.535534	3.535533906	45	Angle ORS
Angle bet S & P		90	45	0.785398	0.707106781	3.5355339	0.707107	3.535534	3.535533906	45	Angle OSP
PQ		5
QR		8.660254
RS		7.071068
SP		7.071068
Area of Triangle OPQ		10.82532	10.825
Area of Triangle OQR		10.82532
Area of Triangle OPQ		12.5
Area of Triangle OPQ		12.5
Are of Quad PQRS		46.65064
Angle bet PQ & QR		90
Angle bet QR & RS		75
Angle bet RS & SP		90
Angle bet SP & PQ		105
Angle Q + Angle S		180	Because it is a cyclic quadrilateral
Angle P + Angle R		180	Because it is a cyclic quadrilateral
Sum ofangles of quad.		360				rcumfewrence
Circumference		31.41593
Perimeter of quad		27.80239	It is always less than circumference of the circle in which it is inscribed
Diagonal PR					Sin	R Sin Theta
Angle bet P and R	Angle OPQ + Angle OQR	180	90	1.570796	1	5
	(1/2PR)/R= Sin (1/2*Angle OPR)
Diagonal PR =	10
Diagonal QS					Sin	R Sin
Angle bet Q and S	Angle OQR + Angle ORS	210	105	1.832596	0.965925826	4.8296291
	(1/2QS)/R= Sin (1/2*Angle OQS)
Diagonal QS =	9.659258263
Let E bepoint of intersection of the diagonals
PE/ER= da /cb	0.577350269			By Similarity of triagles
PR/ER =	1.577350269
ER =	6.339745962
PE = PR-ER	3.660254038
SE/EQ= cd/ba	1.154700538
SQ/EQ=	2.154700538
EQ=	4.482877361
SE=SQ-EQ	5.176380902
PE*ER=	23.20508076
SE*EQ=	23.20508076
Internal intersecting chord property proved.

In a cyclic quadrilateral, given 3 sides and included angle between two of these sides, how to find fourth side?

Given 3 sides and angle included between 2 sides how to find fourth side of cyclic quadrilateral
PQ=a	2
QR=b	3	Radians	Cos	Sin
Angle PQR	135	2.356194	-0.70711	0.707107
RS=c	4	RS can not be greater than 2R
Find PR by cosine rule	21.48528	4.635222
2R = PR / Sin (angle PQR)	6.555194	So R=	3.277597
Find angle ROS by (1/2*RS)/R= Sin (1/2*Angle ROS)	0.610203	0.656317	37.6042	0.656317	0.792245	0.610203
1/2*Angle ROS=Angle RPS	37.6042
AngleRSP= 180 deg- Angle PQR	(Reason- Oppo. Angle of cyclic quad. Are supplementary						45	82.6042	97.3958	97.3958
Find angle SRP by anglesum property of triangle	97.3958	1.699877		0.991681
Angle SOR = 2 x Angle SRP
1/2*SP/R= Sin(Angle SRP)
Thus SP can be found, in otherwordsSP can befoundby Sine Rule	SP=	6.500658
Remember the constant of sine rule is 2xR