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Saturday, August 30, 2014

DAV Question Papers for class 9 and 10 SA-1 September 2014 - 2015 [U-Like and Oswal]

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Std. IX Practice Paper  2014

9th English DAV SA-1 Question paper
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Std. X  Practice Paper  2014

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Std. VIII Practice Paper  2014

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Summative Assessment-I (Sample Papers 2014-15) | Delhi Public School         Download

Thursday, August 7, 2014

weather climate and adaptation CBSE class 7

Q1. What are the elements of weather?
Answer: The elements of weather are humidity, temperature, rainfall, wind speed, etc.

Q2. What do you understand by the climate of the place?

Answer: The average weather [pattern taken over a long time is called climate of the place.

Q3. How are all changes in the weather caused?

Answer: All changes in the weather are caused by the sun.
Q4. Why does the tropical region have a hot climate?

Answer:  It has a hot climate because of its location around the equator. Even in the coldest month the temperature is generally higher than about 15degree C. During the hot  summers, the temperature may cross 40 degrees.

Q5. What do you mean by adaptations?

Answer: It is the phenomenon of acquiring certain features and habits that ensure better survival in given surroundings.

Q6. Why are the animals in the tropical rainforests exhibit variety of adaptations?

Answer: The climatic conditions in rainforests are highly suitable for supporting an enormous number and variety of animals. Since the number is large, there is intense competition for food and shelter. Adaptations ensure survival.

Q7. What properties do make polar bears and penguins good swimmers?

Answer: Their bodies are streamlined and their feet have webs, making them good swimmers.

Q8. By which body part, elephant can tear the bark of trees that it loves to eat?

Answer:  With the help of modified teeth called tusk.

Q9. What do you understand by weather of the place?

Answer: The day to day condition of the atmosphere at a place with respect to the temperature, humidity, rainfall, wind speed, etc, is called the weather at that place.

Q10. What is the function of sticky pads in red eyed frog?

Answer:  Sticky pads on feet help the red-eyed frog to climb trees on which it lives.

Q11. Give reasons;-

a. Polar bears have 2 thick layers of fur.

Answer: 2 layers of fur are to keep the body of polar bear warm in cold weather.

b. Penguins have a thick skin and a lot of fats.

Answer:  To protect them from cold.

c. Polar bears are white in color.

Answer:  The white fur is not easily visible in the snowy white background and this helps them to hide away from predators and prey.

d. Penguins can’t fly but still have streamlined body.

Answer:  Streamlined body enables the penguins to swim in water.

Q12. Why do birds migrate during winters?

Answer:  Birds migrate to warmer places when winter sets in, they come back after the winter is over.

Q13. Define habitat.

Answer:  The natural living place of an organism.

Q14. Explain, with examples, why we find animals of certain kind living in particular climatic conditions.

Answer: Animals are adapted to survive in the conditions in which they live. Animals living in very cold and hot climate must possess special features to protect  themselves against the extreme cold or heat. For e.g.,
1. In Polar Regions, Polar bear- has two thick layers of fur, white coloured fur, very strong sense of smell, long curved and sharp claws
2. In tropical regions, many animals have sensitive hearing, sharp eyesight, thick skin and a skin colour which helps them to camouflage by blending with the surroundings. E.g. big cats

Q15 How do elephant living in the tropical rainforest adapt itself?

Answer:  It has trunk which it uses for as a nose and also used for picking up the food.
 Tusks are modified teeth .it can tear the bark of the trees with it.
 Large ears of the elephant help it to hear even very soft sounds.

CBSE Class-7th Science Weather, Climate and Adaptations of Animal and Climate
Weather describes the condition of the atmosphere. It might be sunny, hot, windy or cloudy, raining or snowing. 
The weather depends on the temperature, precipitation, humidity and atmospheric pressure of the part of atmosphere (air) closest to the surface of the earth. The weather is constantly changing as temperature and humidity change in the atmosphere. 
Climate is the average weather conditions - temperature, pressure, precipitation and humidity - expected for a certain place. Climate is based on the average weather experienced over 30 years or more. Read full post:


Searches related to weather and climate class 7

Tuesday, July 29, 2014

CBSE IX Congruence of Triangle Solved Questions

CBSE Exam  Congruence of  Triangle Solved Questions
Q. 1. Prove that Sum of Two Sides of a triangle is greater than twice the length of median drawn to third side.
Given: Δ ABC in which AD is a median.
To prove: AB + AC > 2AD.
Construction: Produce AD to E, such that AD = DE. Join EC.
Proof: In ΔADB and ΔEDC,
AD = DE              (Construction)
BD = BD             (D is the mid point of BC)
ADB = EDC       (Vertically opposite angles)
ΔADB      ΔEDC   (SAS congruence criterion)
AB = ED               (CPCT)
In ΔAEC,
AC + ED > AE           (Sum of any two sides of a triangles is greater than the third side)
AC + AB > 2AD      (AE = AD + DE = AD + AD = 2AD & ED = AB)

Q. 2. ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see the given figure). Show that BCD is a right angle.
In ΔABC,
AB = AC (Given)
⇒ ∠ACB = ∠ABC (Angles opposite to equal sides of a triangle are also equal)
In ΔACD,      AC = AD
⇒ ∠ADC = ∠ACD (Angles opposite to equal sides of a triangle are also equal)
In ΔBCD,
∠ABC + ∠BCD + ∠ADC = 180º (Angle sum property of a triangle)
⇒ ∠ACB + ∠ACB +∠ACD + ∠ACD = 180º
⇒ 2(∠ACB + ∠ACD) = 180º         
⇒ 2(∠BCD) = 180º            
⇒ ∠BCD = 90º
  
Q.3.Given: two triangles ABC and PQR in which AB=PQ, BC=QR , median AM =median PN prove that triangle ABC is congruent to triangle PQR.

In ∆ ABM  and ∆ PQN 
AB   =  PQ                           ( Given )
AM  =  PN                           ( Given )
And  BM   =  QN   (  As M and N are the midpoint of sides BC and QR  respectively and given BC=  QR ) ∆ ABM 
 ∆ PQN             ( By SSS rule )
SO,
 ABM   =   PQN             ( by  CPCT )
Now  In ∆ ABC  and ∆ PQR
AB   =  PQ                           ( Given )
BC   =  QR                           ( Given )
And
 ABC   =   PQR              ( As we proved ) 
 ∆ ABC    ∆ PQR            ( By SAS  rule )                                       ( Hence proved )

Q.4. The vertex angle of an isosceles triangle is twice the sum of its base angles. Find the measure of all the angles.
Let ABC be an isosceles ∆.Let the measure of each of the base angles = x
Let B = C = x
Now, vertex angle = A = 2x
Now,A + B + C = 180°   [angle sum property]
2x + x + x = 180°4x = 180x = 180/4=450
So, measure of each of the base angles = 45°
Now, measure of the vertex angle = 90°

Q. 5. Prove that the triangle formed by joining the midpoints of the sides of an equilateral triangle is also equilateral.
Let DEF be the midpoints of sides of a triangle ABC( with D on BC, E on AB and F on AC ).
 Now, considering triangles AEF and ABC, angles
EAF = BAC and AE / AB = 1/2 and AF/AC = 1/2. 
Hence, both triangles are similar by the SAS ( Side - Angle - Side ) criterion and correspondingly as AE/AB=AF/AC=EF/BC ( similar triangle properties ), EF =BC/2.
The cases DF=AC/2 and DE=AB/2 can be proved in the same way.
So, AB=BC=AC (from the given data)
2DF=2EF=2DE
DE=EF=DF
So triangle DEF is also Equilateral Triangle
The triangle formed by joining the mid-points of the equilateral triangle is also an equilateral triangle

Q. 6. In triangle PQR, PQ> PR. QS and RS are the bisectors of angle Q and angle R. Prove that SQ> SR
In PQR, we have,       
PQ > PR               [given]
 PRQ > PQR    [angle opposite to longer side of a  is greater]
12PRQ > 12PQR     ........(1)
Since, SR bisects R, thenSRQ = 1/2PRQ      ........(2)
Since SQ bisects P, thenSQR = 1/2PQR   .......(3)
Now, from (1), we have     1/2PRQ > 1/2PQR
⇒∠SRQ > SQR     [using (2) and (3)]
Now, in SQR, we have    SRQ > SQR       [proved above]
 SQ > SR           [side opposite to greater angle of a  is longer

Q.7. In triangle ABC (A at the top) , D is any point on the side BC. Prove that AB+BC+CA 2AD
In triangle ABD,
AB+BD >AD (Sum of two sides of a triangle is greater than the third side) ... (1)
In triangle ACD,
AC+CD>AD (Sum of two sides of a triangle is greater than the third side)  ...(2)
Adding eq. (1) and (2)
AB+(BD+CD)+AC> AD+AD
AB+BC+AC> 2AD

Q.8. In triangle ABC, if AB is the greatest side, then prove that angle c is greater than 60 degrees
It is given that, AB is the longest side of the ∆ABC.
 AB > BC   and  AB > AC.Now,    AB > BC⇒∠C > A    (angle opposite to longer side is greater)  ....(1)
Also,AB > AC⇒∠C > B    (angle opposite to longer side is greater)   ....(2)
adding (1) and (2) , 
we getC + C > A + B
2C > A + B2C + C > A + B + C3C > 180°⇒∠C > 60°

Q.9. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see the given figure). Show that A > C and B > D.
Let us join AC.
In Δ ABC,
AB < BC (AB is the smallest side of quadrilateral ABCD)
∴ ∠2 < ∠1 (Angle opposite to the smaller side is smaller) ... (1)
In ΔADC,
AD < CD (CD is the largest side of quadrilateral ABCD)
∴ ∠4 < ∠3 (Angle opposite to the smaller side is smaller) ... (2)
On adding equations (1) and (2), we obtain
∠2 + ∠4 < ∠1 + ∠3
⇒ ∠C < ∠A
⇒ ∠A > ∠C

Let us join BD.
In ΔABD,
AB < AD (AB is the smallest side of quadrilateral ABCD)
∴ ∠8 < ∠5 (Angle opposite to the smaller side is smaller) ... (3)
In ΔBDC,
BC < CD (CD is the largest side of quadrilateral ABCD)
∴ ∠7 < ∠6 (Angle opposite to the smaller side is smaller) ... (4)
On adding equations (3) and (4), we obtain
∠8 + ∠7 < ∠5 + ∠6
⇒ ∠D < ∠B            ⇒ ∠B > ∠D

Q.10.  If S. is any point on the side QR of triangle PQR, prove that PQ+QR+RP> 2PS
 In   ΔPQS,
PQ + QS > PS   (i) ……………..(Sum of two sides of a triangle is greater than the third side)
In   ΔPSR,
PR + SR > PS  ……(ii)… Sum of two sides of a triangle is greater than the third side)
Adding (i) and (ii), we get
PQ + QS + PR + SR > 2PS
PQ + QR + PR > 2PS  (QS + SR = QR) Hence proved.

Q.11. Prove that the difference of any two sides of a triangle is less than the third side.
Construction: Take a Point D on AB such that AD = AC and join CD
Prove that : AB – AC < BC , AB – BC < AC and BC-AC <AB
Proof: In Δ ACD, Ext <4 > <2
but ,  AD = AC => <1 =  <2
So , < 4  > < 1 ----------------(i)
Now , In Δ BCD, ext <1 > <3 -------------(ii)
Then from (i)  and  (ii) 
< 4  > <3      =>       BC > BD
But, BD = AB – AD and AD = AC         => BD = AB – AC
So, BC > AB – AC

Q.12. that Sum of any two sides of  triangle is greater than third side .
Solution:.
Construction: Extend BA to D Such that AD = AC
Proof : In Δ DACD,  DA=CA.
Therefore, ADC=ACD [ isosceles triangle have two equal angles]
ADC + <1  > ACD 
Thus, BCD >BDC [by Euclid's fifth common notion.]
In  DCB 
BCD >  BDC, So, BD>BC.
But  BD=BA+AD, and AD=AC.
Thus,  BA+AC>BC.
A similar argument shows that AC+BC>BA and BA+BC>AC.

OR, Another way to prove
Draw a triangle,  ABC and line perpendicular to AC passing through vertex B.
Prove that BA + BC > AC

From the diagram, AM is the shortest distance from vertex A to BM. and CM is the shortest distance from vertex C to BM.
i.e. AM < BA and CM < BC
By adding these inequalities, we have
AM + CM < BA + BC
=> AC < BA + BC (
 AM + CM = AC)
BA + BC > AC (Hence Proved)

Q.13. if one acute angle in a right angled triangle is double the other then prove that the hypotenuse is double the shortest side
Given: In Δ ABC , <B = 900 and <ACB = 2 <CAB
Prove that AC = 2BC
Construction: Produce CB to D such that BC  = BD Join  to AD
Proof :  In Δ ABD, and ABC
BD = BC ; AB = AB and <B = <B = 900
By SAS congruency ,    D  ABD ≅ ABC
By CPCT, AD = AC
<DAB = <BAC = X0
So, < DAC =  2X0  
=> <ACB = <ACD
Now in Triangle Δ ADC, <DAC = <ACD= 2X0
So, AD = DC
=> AC = DC = 2BC Proved

Q. 14. Prove that in a triangle the side opposite to the largest angle is the longest.
Solution:
Given , in Δ ABC,  <ABC < <ACB
There is a triangle ABC, with angle ABC > ACB.      
Assume line AB = AC
Then angle ABC = ACB, This is a contradiction       
Assume line AB > AC
Then angle ABC < ACB, This also contradiction our hypothesis
So we are left with only one possibility ,AC> AB, which must be true
Hence proved:  AB < AC       

Q. 15. Prove that in a triangle the angle opposite to the longer side is the longest.
Solution:
Given, in Δ ABC,  AC > AB.
Construction: Take a point D on AC such that AB = AD
Proof: Angle ADB > DCB      
< ADB = <ABD          
So < ABD > <DCB (or ACB) 
< ABC >  <ABD, so < ABC > <ACB 

Q. 16.In a Δ ABC ,<B = 2<C. D is a point on BXC such that AD bisect < BAC and AB = CD. Prove that < BAC = 72 degree
In ΔABC, we have
∠B = 2∠C or, ∠B = 2y, where ∠C =  y
AD is the bisector of ∠BAC. So, let ∠BAD = ∠CAD =  x
Let BP be the bisector of ∠ABC. Join PD.
In ΔBPC, we have
∠CBP = ∠BCP =  y  ⇒ BP = PC ... (1)
Now, in ΔABP and ΔDCP, we have
∠ABP = ∠DCP =  y
AB = DC  [Given]
and, BP = PC  [Using (1)]
So, by SAS congruence criterion, we have
Δ ABP  Δ DCP
<BAP = < CPD and AP = DP
<CDP = 2x  then <ADP = < DAP = x    [<A = 2x]
In ΔABD, we have
∠ADC = ∠ABD + BAD ⇒  x  + 2x   = 2y  +  x  ⇒  x  =  y
In ΔABC, we have
∠A + ∠B + ∠C = 180°
⇒ 2x  + 2y  +  y  = 180°
⇒ 5x  = 180°
⇒  x  = 36°
Hence, ∠BAC = 2x  = 72°

You may also use this way:

Q.17,  If o is any point in the interior of triangle ABC .Prove that  
(a)  AB + AC > OB + OC
(b) AB + BC + CA > OA + OB + OC
(c )OA +OB+OC>1/2(AB+BC+CA)
Construction: Produce BO to meet AC at D
In D ABD, AB + AD > BD => AB + AD > OB + OD   ------(i)
In D OCD, OD + DC > OC    ------(ii)
Adding (i) and (ii) we get,
AB + AD + OD + DC  > OB + OD + OC    
=> AB + AC > OB + OC    --------- (iii)                   Hence prove (a)
Similarly we get ,
BC + BA > OA + OC                ---------(iv)
and , CA + CB > OA + OB       ---------(v)
Adding (iii),(iv)and (v) we get,
2(AB + BC + CA) > 2(OA + OB + OC)
AB + BC + CA > OA + OB + OC                       Hence prove (b)
In D  OAB , D OBC and D OCA
[OA + OB > AB ] + [OB + OC>BC] + [ OC + AO > AC]
2[OA + OB + OC]  > AB + BC + CA
[OA + OB + OC]  > ½ [AB + BC + CA]          
Hence prove (c)
Check more stuff on CBSE IX  Congruence of  Triangle
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CBSE IX Congruence of Triangle Solved Questions          Download File