Wednesday 27 December 2017
Monday 25 December 2017
Friday 22 December 2017
Wednesday 20 December 2017
CLASS-9 - MATHS
WINTER VACATION HOMEWORK / Ch- HERON’S
FORMULA
Q.1) A field is in the form of
parallelogram has sides 60 m and 40 m and one of its diagonals is 80 m long. Find
the area of parallelogram using hero’s formula.
Q.2) The perimeter of a triangle
field is 420 m and sides are in ratio 6: 7: 8. Find the area of triangular field
using hero’s formula.
Q.3) The perimeter of a triangle
is 50 cm. one side of a triangle is 4 cm longer than the smaller side and the
third side is 6 cm less than twice the smaller side. Find the area of
triangular field using hero’s formula.
Q.4) A rhombus shaped sheet with
perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate of
Rs.5 per m2. Find the cost of painting. (use heron’s formula to
calculate area)
Q.5) A design is made on a
rectangular tile of dimensions 50 cm X 70 cm as shown in below figure. The design
shows 8 triangles, each of side 26 cm, 17 cm and 25 cm. Find the total area of
design and the remaining area of tile.
Tuesday 17 October 2017
class 9 / maths
CLASS 9 / PRACTICE QUESTIONS
AREA OF PARALLELOGRAM AND TRIANGLE
Q.1) ABCDE is a pentagon. A line through B parallel to AC
meets DC produced at F.
Show ar (∆ ACB) = ar(∆ ACF).
Q.2) In the
given figure, PQR and QST are two triangles such that S is the mid-point of QR. and QT // PR.
Q.3) ABCD is
a Quadrilateral. A line through D, parallel to AC, meets BC produced in P as
shown in figure. Prove that ar(∆ ABP) = ar(Quad ABCD).
Q.4) In a
triangle ABC, D is the mid-point of AB. P is any point of BC. CQ || PD meets AB
in Q. Show that
Q.5) Prove
that area (∆ AFG) = ½ area(BDEF).
Q.6) If D, E
and F are the mid points of sides AB, BC and AC respectively then show that
(i) area (∆ ADE) = area (∆ AFE)
(ii) area (∆ BDE) = area (∆ CEF)
(iii) area (ADEF)
= ½ area (∆ ABC)
Q7) In ∆
ABC, D is the mid-point of BC, E is the mid-point of BD. If „O‟ is the
mid-point of AE, prove that
ar (∆ BOE) = (1/8) ar (∆ ABC).
Q.8) In the
given figure E is the midpoint of BC and D is the midpoint of AE. PEDB and QEDC
are parallelograms then show that area (∆ PBE) + area (∆ QCE) = ½ area (∆ ABC).
Q.9) In the
given figure, if BE||CF and area (ABCE) = area (BDEF) then prove that AD|| BE.
Q.10) The
area of parallelogram PQRS is 152 cm2.
Find the area of rectangle PQXY. If the base PQ = 19 cm, find the height of the
parallelogram.
Monday 16 October 2017
Sunday 15 October 2017
class 10 / spherical mirrors
Spherical Mirrors
A mirror
whose polished, reflecting surface is a part of a hollow sphere of glass or
plastic is called a spherical mirror.
The
spherical mirror is classified as:
- Concave mirror
- Convex mirror
1. concave mirror – a
spherical mirror whose reflecting surface is bent inwards is called concave
mirror.
2. convex mirror - a
spherical mirror whose reflecting surface is bent outwards is called concave
mirror.\
Terms related to spherical mirrors:
- Pole
·
The center of a spherical
mirror is called its pole .
·
It lies on the
reflecting surface of a spherical mirror
·
Is represented by
letter P .
2. Centre of curvature
·
The reflecting surface
of a spherical mirror forms a part of a sphere. This sphere has a centre. This
point is called the centre of curvature of the spherical mirror.
·
It is represented by
the letter C.
·
The centre of
curvature is not a part of the mirror. It lies outside its reflecting surface.
·
The centre of
curvature of a concave mirror lies in front of it. However, it lies behind the
mirror in case of a convex mirror.
3. Radius of curvature
·
The radius of the
sphere of which the reflecting surface of a spherical mirror forms a part, is
called the radius of curvature of the mirror.
·
It is represented by the letter R.
·
The distance PC is
equal to the radius of curvature.
4. Principal axis
·
Straight line passing
through the pole and the centre of curvature of a spherical mirror is called
principle axis of the mirror.
·
The principal axis is
normal to the mirror at its pole.
5. Aperture of the mirror
·
Portion of the mirror
from which reflection of light actually takes place is called the aperture of
the mirror.
·
Aperture of the mirror
actually represents the size of the mirror.
·
Distance MN represents
the aperture.
Principle
focus and focal length of a Spherical Mirrors
- Consider
light rays parallel to the principal axis are falling on a concave mirror.
By observing the reflected rays we conclude that they are all intersecting
at a point F on the principal axis of the mirror. This point is called the
principal focus of the concave mirror.
- In
case of convex mirror rays get reflected at the reflecting surface of the
mirror and these reflected rays appear to come from point F on the
principle axis and this point F is called principle focus of convex
mirror.
- The
distance between the pole and the principal focus of a spherical mirror is
called the focal length. It is represented by the letter f.
- There
is a relationship between the radius of curvature R, and focal length f,
of a spherical mirror and is given by R=2f which means that that the
principal focus of a spherical mirror lies midway between the pole and
centre of curvature.
Characteristics of Concave and a Convex Mirror
|
Convex Mirror
|
Concave Mirror
|
|
Reflecting
surface is curved outwards.
|
Reflecting
surface is curved inwards.
|
|
The focus is
virtual as the rays of light after reflection appear to come from the focus.
|
The focus is
real as the rays of light after reflection converge at the focus.
|
|
The focus lies behind
the mirror
|
The focus is in front
of the mirror
|
|
Diverging mirror
|
Converging mirror
|
|
Image Cann’t be projected
on a screen
|
Image Can be projected on a
screen
|
Note
: A concave mirror is also
known as a converging mirror as the parallel rays of light after getting
reflected from the concave mirror converge at the focus.
A
convex mirror is known as a diverging mirror as the parallel rays of light
after reflection appear to come from a point, i.e., the rays diverge after
reflection.
Wednesday 6 September 2017
Monday 4 September 2017
Saturday 2 September 2017
CLASS 9 / PHYSICS / NUMERICAL REVISION
ARMY PUBLIC SCHOOL,
GOLCONDA
CLASS 9
Q.1)
A scooter starts from rest moves in a straight line with a constant acceleration
and covers a distance of 64 m in 4 seconds. Calculate its acceleration and
final velocity.
Q.2)
In the given speed-time particle is moving in a straight line motion. Thus speed
can be considered as velocity. Observe the given graph and answer the following
question-
a) Calculate the value of acceleration for t= 0 to
20 sec, t= 20 to 50 sec, t= 50 to 100 sec.
b) What information u will conclude from the above
values of acceleration?
c) Show which part of the graph represents uniform acceleration,
zero acceleration and uniform retardation.
d) Calculate the total distance travelled.
Q.3)
Velocity-time graph for body A, B and C is given. Which covers the maximum
distance in given time?
Q.4)
In the given distance – time graph of a body, answer the following questions
a) Calculate distance travelled in 20, 35 and 45
sec.
b) Distance travelled in last 10 sec.
c) Displacement carried in time 0 to 45 seconds.
Q.5)
Cars
of mass 1800 kg moving with a speed of 10 m/s is bought to rest after covering
a distance of 50 m. Calculate the force acting on the car. Name the force and
identify the direction of applied force.
Q.6)
A body of mass 5 kg is moving with a velocity of 10 m/s. A force is applied to
it so that in 25 seconds, it attains a velocity of 35 m/s. Calculate the value
of force.
Q.7)
The velocity-time graph of a truck is shown as below. The truck weighs 1000 kg.
a) What is the accelerating and breaking force
acting on the truck?
b) Find momentum of car during t = 2 to 5 seconds
Q.8)
How
much momentum will a dumb-bell of mass 10 kg transfer to the ground floor if it
falls from a height of 80 cm?
( g = 10 m/s2).
Q.9)
A rifle of mass 3 kg fires a bullet of mass 0.03 kg. The bullet leaves the
barrel of the riffle at a velocity of 100 m/s. Find the recoil velocity of
rifle.
Q.10)
from a rifle of mass 4 kg, a bullet of mass 50 kg is fired with an initial
velocity of 35 m/s. Calculate the initial recoil velocity of rifle.
Thursday 31 August 2017
Subscribe to:
Posts (Atom)
-
CONCEPT MAP POINTS TO REMEMBER 1. If a ray stands on a line, then the sum of the two adjacent angles so formed is 180° and vice- ...
