ICSE Maths Important Formulas

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.. IN TERNET '~.tl' ~_ : F O R S ti [ '~e l l l , teIn te ,rn e t f ro m $50 (1'~-I-;; U S A R M Y S h ~ p p lr lg t o lr a q a rid A f g l im J a l ii li :s t a n


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...- _ . . . . . . . . . .GUESS PAIPERS

ICSE C:BSE IGCSE ALEVEL IB lIT AIEEE CA TYbcomlIn )) N ote : W ben yo u pro ve tw o trian gles a re s im ila r in th e g iv en fig ure , th en ta ke tw o ro ug h sk etc hes , a nd M a rk th e v alu es e xa ctl'l.E x am p l e In th e g iv en E gu re , f1 _ = a. (A lte rn ate a ng le s a re e qu al)

o = fi _ ( Al te rn at e a ng le s a re e ql Ja l) @ ~3 BB yAA s i m i l a r i t y 1 '1 ABC~ 1 '1 EDC CVIm p Ste p 1: d r aw tw o sm a ll r o u g h sk e tch e s w h ich (m u s t lo o k lik e ex a c t ly s am e )h a s to ~

/\ /\ reprm rJ tL l ABC an d fl EDC~ ~ D64 ') 2 E

A i~.B E~DStep 5: N ow easily YOI l ca n i de ntfy th e th ird le tt er o f t he b ot h th e tria ng le sC CA U B E~D there fore ~~ = ~~ ABE D

Ste p 2 T he n m ark e qu iv al en t p arts jJ _ an d f .1_ a t th e s am e p os itio n o f t he tw o t ria ng le s/ \ / \ / \ / \ H en c e n am e W I th a p p ro p ia te le t te r s ll_a s A an d f1 _ a s E~ ~ = i> A~E~

Ste p 3 s i m i l a r l y m a rk e qu iv ale nt p ar ts a. a n d c l_ a t th e s am e p os it io n o f t he tw o tr ia ng le sH e n c e name W I t h app rop i a t e le tte rs fl _a s B a nd f A . as D/\ /\~ /\ /\~ LA ~BL...i1D

Step 4 : a s a resu lt the f i n a l p ic tu re lo ok s l ik e t he fo llo win g o ne , th en e asily y ou c an a ss um e th e t hird le tte r o f t he e ac h tria ng le

T ho ug h tt t ak es 'M o re ' t im e to u nd ers ta nd , Do n't g iv e-u p h op e As th is p oin t h elp s a lo t to so lv e m os t o f t he p ro ble ms in s imi l ar i t yw it h g re a t e as e.

Size transformation'Length of the model = 'k' times the actual length, [Here' k' is to be taken as 1 /10000]Area of the model = 'k2' times the actual area. [Here ,~, is to be taken as (1 /10000) 2 ]Volume of the model = 'k3, times the actual volume. [Here ' 1 2 ' is to be taken as (1 /l0000) 3]

14.Symmetry.A line which divides the given figure into two identical parts is known as line of'S ymmetry'

1. An angle has (Jlle line of symmetry.2 . A Square has 4 hnes of symmetry.3. A Rectangle has 2 lines of symmetry.4. A Parallelogram has No lines of symmetry.5. A Rhombus has 2 T I i 1 e S of symmetry.6. An Isosceles Triangle has One line of symmetry.

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TC'.sF CR~F TGC~F AI FVFI TR TTT A'ilillFFFCA liVhr:nm


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7. An Equilateral Triangle has 3 lines of symmetry.8. A Circle has Infinite l ines of symmetry.9. A Regular Polygon with 'n' sides has 'n' lines of symmetry.

For ex: A Regular pentagon (5 sides) has 5 lines of symmetryA Regular hexagon (6 sides) has 6 lines of symmetry.

Nnte;_ Angle of a regular polygon =( 2n - 4 )x 90n, Here' n' refers number of sides of a polygon(This formula may be used to construct a regular Polygon)

15.Loci. The :I.ocus; ofa line segment is it's Pljeroendlcu[ar bIsector. TIle LOCUS 0 an angie-is Its Angle tsector,

For solving most of the' Locus' problems, the above two points are good enough. In addition tothese points, You should have the basic knowledge of geometrical constructions. Also look at thegiven figure in terms of either 'line segments' or 'angles'16.Circles. &Tangents.

T he p e rp e n d ic u la r d raw nf ro rn th e c e n tre o f a c irc le , b is e c ts th e c h o rd o f 'th e C i rc leO M j_A B, therefore A11 =1ffi

Equal chords of a circle are equi distant from the center.T h e a ng ie s u b t e nd ed a t th e c en te r b y a n a rc = D o u b le t h e a n g le at a n y pa r t o f t h e ci rcunfererce o f t h e C i r c l e .

A n g l e s in t h e s am e s e gm e nt a r e e qu al( lD O K a t t h e f ee t o f t h e b o t h t h e w . g le s, t h ey a re r es ti n g a t t h e s am e por t s , B u t t he r m d d l e l e t te rs a re d i ff e re n t)I . e / l d r n = ! A Y E .

A i t e ll l a te s e gm ~ D t t h eo re m , T a c i r c l e , If a t a n g e n t i s d m ' l l , < ln d a c h o r d is drai\tl f r o m th e p o e t o f c o n t ac t t h e nA n ~e m ad e b efw ee n th e c bo rd a n d f i le t a n g e n t = A l 1 ~ e m a de i n t h e a lt er na te s eg m en ti l . . ; ; f 1 _

The sum of opposite angles of a cyclic quadrilateral is always 180.

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_ _ J !II I II I II I~ _~ - ... ...- _.I!!!!!I!II_ -GUESS PAIPERSICSE CBSE IGCSE ALEVEL IS lIT AilillEEECA lfYbcom

17.Circumference &Area of a Circle. Area of a Circle = 1tr. Perimeter of a Circle =2 1t r Area of sector = .1L1t r.

360 Length of an arc = .1L21t r. Area of ring = 1t (iq~r) Distance moved by a wheel in one revolution = Circumference of the wheel. Number of revolutions = Teal disaHee ffl6, edCircumference of the wheel.Area of an equilateral triangle = J . ] _ Side',~ While solving 'Mensuratio~' problems, take care of the following.

1. If diameter of a circle is given, then find the radius first2. em'6\H'Ret~edfllli~~1fM\libP.)f~\l; H i ' e ' ~n~I1fh~RI


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I

Volnme of a Sphereical shell = Outer Vol - Inner. Vol.!T, R' - - ' ! IT r 33 ~Volume ofhemi sphere = 7S7t~[Take halfthe volume ofa sphere]

[Take half the SA of a sphere]T A 8 f H em ~ j J R e f e ~ ~ t f p + 7tr =>37tr2

/W h en c om bin atio n o f s olid ; are ~en , ~' r ~ ~ : } y ~ O : e a sk ed t o I if Jd t lI .e TSA o f ! h e g iv en s ol id , Y O l lreVo l o f Cane G SA o f Cane supposed to t ak e CSA of cone + CSA o ff ue m ii ph er e t o g et T SA

'+ Vol o f Hemis phe r e + CSA onIem isphere Don 'nake TSA of COl le + TSA ofhemisob . e re 10 . et T SAT o !c l V o l o f t h e ; .o li d T SA of t he w en s o l i d

While solving the combination of solids it would be better if you take common: If a solid is melted and zecast intonumber of ather small solids, thenVOlume 01me larger WUO=-No x voiume 0 me smaller SOfIa

For Ex: A cylinder is melted and cast into smaller spheres. Find the number of spheres If an 'Ice cre'hWt1.ffiffitP{;.,hl!life


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GUESS PAIPERSICSE CSSE IGCSE ALEVEL IS IIT AIIIEEECA liYbcom Tan ( 90 - e ) = Cot e : Cot ( 90 - e ) = Tan e

20&iehts & Distances\ V h i J e ~ o t \ > i ! l gH e i g h t , & D i ,t a tl c es , id en ti f y , w h eth er a ny o ne s id e o f t he tw o 1 ria n~ es a re g iv en o r ! lo tIf a n y o l l e s i d e o f a t r i : l . .~ e is g iv enC o ol I I I , i f y ou k n o w o ne 5 id e o f t h e t r ia ng le th en , y ou c an f i n d t he o th e r s id e, w h o se v a l u e m a y b e u se d to s o l v e th eo t ~ e r t ri a n g l e ,

If l l l lY s i d e of t h e t ri a ng l e is n o t l 1 i v e nIn th e a djo m in g f ig ur e ,a ~e as t a ny on e s id e o f th e tw o t r ia ng le s a re n ot@ 7 e n ( th ou gh D C is g re e n, i t IS a p a r t o f s id e B DIfl tr u s e as e I de nt i f y th e c om m o n s id e, a nd m a ke it s ub je ct c f t he f or m ul at h e n equate t h e s u bj ec t In b o t h th e cam . [i n t h is e x a m p l e , ' h is t h e c om m on s id e to b oth th e t r ia ng le s, s o m a ke 'h ' as

th e s ub J e c t In b o th t he t ri an g le s ]

AADLk_d:2 0 C II

21.Graphical Representation. Don't forget to write the scale on x-axis, and on y-axis. To find the 'Lower quartile' take N/4 [Here N is L f] then take the corresponding point on X-axis T S f i n s t l 1 e ~ ~ a t 9 1 .1 f a 1 lW ~ l ~ ~ i ~ P m ~diRIm~I}'iffi~~!R~ X-axis

22.Measures of Central Tendency.F"or llD-2[OIlP.ed data . Arithmetic Mean = Sum of observatIOns

No of observations: Mpffitd=thI~~lWl! W & U 8 ! f f l m\\TeqhwlPalafnthM~~narJi or' Descending' order, then

Median = (N+I)12 term value of the given data, in case of the data is having odd no of observations.

Median = [(N/2) + (N+I )12)]/2 term value of the given data, in case of the data is havingeven number of observations.For grnuped dataArithmetic Mean = I . . . . f X (Direct method)LfArithmetic Mean = a + f (short cut method)Arithmetic Mean = a + I..fu x C (step-deviation method)

Probability of an event: P(event)23.Probability.

Number of favorable outcomes

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ICSE CSSE IGCSE ALEVEL IS lIT AIIIEEECA lrYbcom


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Total number of outcomesIn a deck of playing cards, there are four symbols~ (Spades i n .B l ack colour) having A, 2,3,4,5,6,7,8,9,10,J,K, and Q total 13 cards

.. (Clubs in Black colour) having A, 2,3,4,5,6,7,8,9,10,J,K, and Q total 13 cards

., (Hearts in Red colour) having A, 2,3,4,5,6,7,8,9,1O,J,K,and Q total ~ ~Jack, King and Queen are known as 'Face Cards' , As these cards are having some pictures on

~~~9lte~tlgatwa~~3lji9,,~~ ~~~\1!.genHktks for them, but, they willgive a clear cut idea to solve the problemTo solve ffiMt of the problems, I ~trongly sugge s t the fol lowing way.

xtract the detail, !ata from the que stionand then copy it

W r ite th e re la te d fo rm u la

roce ed with the calculations

Prostate Cancer Email Prostate Cancer Expert/Author Larry Pope www.nutrition2000info.comFreshers - Job in 99 Days A Job in an IT Company is 99 Days Away, Register Now for more Details www.niiteducation.Play Chess vs Computer Play Chess against Computer Free registration to get a rating, www.chesstempo.com

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rcsnMaths Important Formulas
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Down l o a d th is D oc um e n t fu r Fre ePr in tM o b i1e Co lle ct io ns Re po rt D oc um e n tTh i s is a p riv ate d oc um e n t. 8Info and RatingNo t e ss t n m y s~ s s p a p e r si n d i a n h i s t o r ye xa m p ap ersth e e x am p ap er slC se g ue ss p ap er(m o re ta gs )c h i r a g a g a rw a l

Uke Be the f irst of your f riends to l ikethis.

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