NCERT Solutions for Class 9 Science Chapter 8 Motion

NCERT Solutions for Class 9 Science Chapter 8 Motion

These Solutions are part of NCERT Solutions for Class 9 Science. Here we have given NCERT Solutions for Class 9 Science Chapter 8 Motion. LearnInsta.com provides you the Free PDF download of NCERT Solutions for Class 9 Science (Biology) Chapter 8 – Motion solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 8 – Motion Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

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NCERT TEXT BOOK QUESTIONS

IN TEXT QUESTIONS

Question 1.
An object has moved through a distance. Can it have zero displacement ? If yes, support your answer with an example. (CBSE 2015)
Answer:
Yes, The displacement of the object can be zero. Let a boy completes one round of a circular track in 5 minutes. The distance travelled by the boy = circumference of the circular track. However, displacement of the boy is zero because his initial and final positions are same.

Question 2.
A farmer moves along the boundary of a square field of side 10 m in 40 s.
What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds ? (CBSE 2010, 2015)
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 1
Answer:
ABCD is a square field of side 10 m.
The farmer moves along the boundary of the field from the corner A via the corners B, C and D.
After every 40 s, the farmer is again at the corner A, so his displacement after every 40 s is zero.
At the end of 2 minutes 20 seconds = (2 x 60 + 20) = 140 s, the farmer will be at the corner C.
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 2

Question 3.
Distinguish between speed and velocity. (CBSE 2010, 2012, 2013, 2015)
Answer:

Speed

Velocity

1. Distance travelled by an object per unit time is known as its speed. The distance travelled by an object in a particular direction (i.e. displacement) per unit time is known as its velocity.
2. Average speed of a moving object cannot be zero. Average velocity of a moving object can be zero.
3. tells how fast an object moves. Velocity tells how fast an object moves and in which direction it moves.
4. Speed is a scalar quantity. Velocity is a vector quantity.
5. Speed of an object is always positive. Velocity of an object can be positive or negative.

Question 4.
Under what conditions) is the magnitude of the average velocity of an object equal to its average speed ? (CBSE 2012, 2013)
Answer:
When an object moves in one direction along a straight line.

Question 5.
What does the odometer of an automobile measure ? (CBSE 2010, 2011, 2014, 2015)
Answer:
Odometer of an automobile measures distance travelled by the automobile.

Question 6.
What does the path of an object look like when it is in uniform motion ? (CBSE 2012)
Answer:
Straight path.

Question 7.
During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station ? The signal travels at the speed of light, that is, 3 x 108 m s-1
Answer:
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 3

Question 8.
When will you say a body is in

  1. uniform acceleration ?
  2. non-uniform acceleration (CBSE 2012, 2013)

Answer:

  1. A body has uniform acceleration if its velocity changes by an equal amount in equal intervals of time. .
  2. A body has non-uniform acceleration if its velocity changes by unequal amount in equal intervals of time.

Question 9.
A bus decreases its speed from 80 kmh-1 to 60 kmh-1 in 5 s. Find the acceleration of the bus.
(CBSE 2010, 2011, 2012, 2015)
Answer:
Here,
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 4
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 5

Question 10.
A train starting from a railway station and moving with a uniform acceleration attains a speed of 40 km h-1 n 10 minutes. Find its acceleration.
Answer:
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 6

Question 11.
What is the nature of the distance-time graphs for uniform and non-uniform motion of an object ?
(CBSE 2012)
Answer:

  1. For uniform motion, distance-time graph is a straight line having constant gradient or slope as shown in figure.
  2. For non-uniform motion, distance-time graph is a curve having increasing gradient or slope or decreasing gradient as shown in figures.
    NCERT Solutions for Class 9 Science Chapter 8 Motion image - 7

Question 12.
What can you say about the motion of an object whose distance-time graph is a straight line parallel to
the time axis ? (CBSE 2010, 2012)
Answer:
When distance-time graph is a straight line parallel to the time axis, it means, the distance of the object is not changing with time. Thus, the object is stationary.

Question 13.
What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time-axis ?
Answer:
In this case, speed of the object is constant. That is, the object travels equal distances in equal intervals of time along a straight line. Hence, motion of the object is uniform motion.

Question 14.
What is the quantity which is measured by the area occupied below the velocity-time graph ?
(CBSE 2010, 2012)
Answer:
Magnitude of the displacement of a body is measured by the area under the velocify-time graph.

Question 15.
A bus starting from rest moves with a uniform acceleration of 0.1 m s-2 for 2 minutes. Find
(a) the speed acquired,
(b) the distance travelled. (CBSE 2012)
Answer:
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 8

Question 16.
A train is travelling at a speed of 90 km h-1. Brakes are applied so as to produce a uniform acceleration of – 0.5 m s-2 . Find how far the train will go before it is brought to rest ?
Answer:
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 9

Question 17.
A trolley, while going down an inclined plane has an acceleration of 2 cm s-2. What will be its velocity 3 s after the start ?
Answer:
Here, u = 0,v = ?, a = 2 cm s-2, t = 3 s
Using, v = u + at, we get
v = 0 + 2 x 3 = 6 cm s-2.

Question 18.
A racing car has a uniform acceleration of 4 m s-2. What distance will it cover in 10 s after start ?
Answer:
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 10

Question 19.
A stone is thrown in a vertically upward direction with a velocity of 5 m s-1 If the acceleration of the stone during its motion is 10 m s-2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there ?
Answer:
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 11

NCERT CHAPTER END EXERCISE

Question 1.
An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes and 20 s ?
Answer:
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 12
(ii) After every 40 s, athlete reaches his starting point, so after 40 s, his displacement is zero. It means, the athlete completes the circular track 3 times in 120 s and in the next 20 s, he is just opposite to his starting point. Therefore, the magnitude of the displacement of the athlete at the end of the 140 s (or 2 minutes 20 s) = Diameter of the circular track = 200 m.

Question 2.
Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 50 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging
(a) from A to B and
(b) from A to C ? (CBSE 2010, Term I)
Answer:
Distance from A to B – 300 m
Displacement from A to B = 300 m
Time taken to go from A to B =2 minutes 50 s = 170 s
T Solutions for Class 9 Science Chapter 8 Motion image - 13

Question 3.
Abdul, while driving to school, computes the average speed for his trip to be 20 km h-1. On his return trip along the same route, there is less traffic and the average speed is 40 km h-1. What is the average speed for Abdul’s trip ? (CBSE 2011)
Answer:
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 14

Question 4.
A motor boat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s-2 for 8.0 s. How far does the boat travel during this time ? (CBSE 2011, 2013)
Answer:
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 15

Question 5.
A driver of a car travelling at 52 km h-1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h-1 in another car applies his breaks slowly and stops in 10 s. On the same graph paper plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied ?
Answer:
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 16

Question 6.
Figure shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions :
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 17
(a) Which of the three is travelling the fastest ?
(b) Are all three ever at the same point on the road ?
(c) How far has C travelled when B passes A ?
(d) How far has B travelled by the time it passes C ?
Answer:
(a) Speed = Slope of distance-time graph.
Since slope of distance-time graph for object B is the greatest, so object B is travelling the fastest.
(b) All the three objects will be at the same point on the road if all the three distance-time graphs intersect each other at a time. Since, all the three distance-time graphs do not intersect each other at a time, so they are never at the same point on the road.
(c) When B passes A, distance travelled by C = 9.6 – 2 = 7.6 km.
(d) Distance travelled by B by the time it passes C = 6 km.

Question 7.
A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s-2, with what velocity will it strike the ground ? After what time will it strike the ground ?
Answer:
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 18

Question 8.
The speed time graph for a car as shown in figure
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 19
(a) Find how far does the car travel in the firdt four seconds.
Shade the area on the graph that represents the distance travelled by the car during the period.
(b) Which part of the graph represents uniform motion of the car?
Answer:
(a)
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 20
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 21
(b) The straight part of the curve parallel to time axis represents the uniform motion of the car.

Question 9.
State which of the following situations are possible and give an example for each of these :
(a) an object with a constant acceleration but zero velocity.
(b) an object moving with an acceleration but with uniform speed.
(c) an object moving in a certain direction with an acceleration in the perpendicular direction.
(CBSE 2012)
Answer:
(a) When an object is thrown vertically upward, its velocity at the highest point is zero but it has constant acceleration = 9.8 m s-2
(b) Uniform circular motion.
(c) An object moving in a circular path with uniform speed, its acceleration is always perpendicular to the direction of the motion of the object.

Question 10.
An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 . hours to revolve around the earth. (CBSE 2011)
Answer:
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 22

NCERT Solutions for Class 9 Science Chapter 8 Motion

Hope given NCERT Solutions for Class 9 Science Chapter 8 are helpful to complete your science homework.

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HOTS Questions for Class 9 Science Chapter 14 Natural Resources

HOTS Questions for Class 9 Science Chapter 14 Natural Resources

These Solutions are part of HOTS Questions for Class 9 Science. Here we have given HOTS Questions for Class 9 Science Chapter 14 Natural Resources

Question 1.
(a) Fill in the blanks : A, B and C.
HOTS Questions for Class 9 Science Chapter 14 Natural Resources image - 1
(b) Identify the biogeochemical cycle.
Answer:
(a) Fill in the blanks :
A- Organic molecules (C6H12O6).
B- Photosynthesis.
C- Respiration.
(b) Identification: Oxygen cycle.

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Question 2.
(a) Fill in the blanks: A to F.
HOTS Questions for Class 9 Science Chapter 14 Natural Resources image - 2
(b) What will happen if the step A does not take place ?
(c) What will happen if the step F does not take place ?
Answer:
(a) Fill in the blanks :
A- Ammonification.
B- Ammonia.
C- Nitrites.
D- Nitrates.
E- Nitrification.
F- Denitrification.
(b) Absence of step A: No breakdown of proteins to release ammonia.
(c) Absence of step F: There will be no replenishment of atmospheric nitrogen.

Hope given HOTS Questions for Class 9 Science Chapter 14 Natural Resources are helpful to complete your science homework.

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RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS

RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS

Other Exercises

Mark the correct alternative in each of the following:
Question 1.
Two parallelograms are on the same base and between the same parallels. The ratio of their areas is
(a) 1 : 2
(b) 2 : 1
(c) 1 : 1
(d) 3 : 1
Solution:
Two parallelograms which are on the same base and between the same parallels are equal in area
∴ Ratio in their areas =1 : 1 (c)

Question 2.
A triangle and a parallelogram are on the same base and between the same parallels. The ratio of the areas of triangle and parallelogram is
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 1 : 3
Solution:
A triangle and a parallelogram which are on the same base and between the same parallels, then area of triangle is half the area of the parallelogram
∴ Their ratio =1:2 (c)

Question 3.
Let ABC be a triangle of area 24 sq. units and PQR be the triangle formed by the mid-points of sides of ∆ABC. Then the area of ∆PQR is
(a) 12 sq. units
(b) 6 sq. units
(c) 4 sq. units
(d) 3 sq. units
Solution:
Area of ∆ABC = 24 sq. units
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q4.1
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q4.2

Question 4.
The median of a triangle divides it into two
(a) congruent triangle
(b) isosceles triangles
(c) right triangles
(d) triangles of equal areas
Solution:
The median of a triangle divides it into two triangles equal in area (d)

Question 5.
In a ∆ABC, D, E, F are the mid-points of sides BC, CA and AB respectively. If
ar(∆ABC) = 16 cm2, then ar(trapezium FBCE) =
(a) 4 cm²
(b) 8 cm²
(c) 12 cm²
(d) 10 cm²
Solution:
In ∆ABC, D, E and F are the mid points of sides BC, CA and AB respectively
ar(∆ABC) = 16 cm²
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q5.1

Question 6.
ABCD is a parallelogram. P is any point on CD. If ar(∆DPA) = 15 cm² and ar(∆APC) = 20 cm², then ar(∆APB) =
(a) 15 cm²
(b) 20 cm²
(c) 35 cm²
(d) 30 cm²
Solution:
In ||gm ABCD, P is any point on CD
AP, AC and PB are joined
ar(∆DPA) =15 cm²
ar(∆APC) = 20 cm²
Adding, ar(∆ADC) = 15 + 20 = 35 cm²
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q6.1
∵ AC divides it into two triangles equal in area
∴ ar(∆ACB) = ar(∆ADC) = 35 cm²
∵ ∆APB and ∆ACB are on the same base
AB and between the same parallels
∴ ar(∆APB) = ar(∆ACB) = 35 cm²(c)

Question 7.
The area of the figure formed by joining the mid-points of the adjacent sides of a rhombus with diagonals 16 cm and 12 cm is
(a) 28 cm²
(b) 48 cm²
(c) 96 cm²
(d) 24 cm²
Solution:
In rhombus ABCD,
P, Q, R and S are the mid points of sides AB, BC, CD and DA respectively and are joined in order to get a quad. PQRS
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q7.1

Question 8.
A, B, C, D are mid points of sides of parallelogram PQRS. If ar(PQRS) = 36 cm²,then ar(ABCD) =
(a) 24 cm²
(b) 18 cm²
(c) 30 cm²
(d) 36 cm²
Solution:
A, B, C and D are the mid points of a ||gm PQRS
Area of PQRS = 36 cm²
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q8.1
The area of ||gm formed by joining AB, BC, CD and DA
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q8.2

Question 9.
The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm is
(a) a rhombus of area 24 cm²
(b) a rectangle of area 24 cm²
(c) a square of area 26 cm²
(d) a trapezium of area 14 cm²
Solution:
Let P, Q, R, S be the mid points of sides of a rectangle ABCD. Whose sides 8 cm and 6 cm
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q9.1
Their PQRS is a rhombus
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q9.2

Question 10.
If AD is median of ∆ABC and P is a point on AC such that ar(∆ADP) : ar(∆ABD) = 2:3, then ar(∆PDC) : ar(∆ABC) is
(a) 1 : 5
(b) 1 : 5
(c) 1 : 6
(d) 3 : 5
Solution:
AD is the median of ∆ABC,
P is a point on AC such that
ar(∆ADP) : ar(∆ABD) = 2:3
Let area of ∆ADP = 2×2
Then area of ∆ABD = 3×2
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q10.1
But area of AABD = \(\frac { 1 }{ 2 }\) area AABC
∴ Area ∆ABC = 2 x area of ∆ABD
= 2 x 3x² = 6x²
and area of ∆PDC = area ∆ADC – (ar∆ADP) = area ∆ABD – ar ∆ADP
= 3x² – 2x² = x²
∴ Ratio = x² : 6x²
= 1 : 6 (c)

Question 11.
Medians of AABC, intersect at G. If ar(∆ABC) = 27 cm2, then ar(∆BGC) =
(a) 6 cm2
(b) 9 cm2
(c) 12 cm2
(d) 18 cm2
Solution:
In ∆ABC, AD, BE and CF are the medians which intersect each other at G
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q11.1

Question 12.
In a ∆ABC if D and E are mid-points of BC and AD respectively such that ar(∆AEC) = 4 cm², then ar(∆BEC) =
(a) 4 cm²
(b) 6 cm²
(c) 8 cm²
(d) 12 cm²
Solution:
In ∆ABC, D and E are the mid points of BC and AD
Join BE and CE ar(∆AEC) = 4 cm²
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q12.1
In ∆ABC,
∵ AD is the median of BC
∴ ar(∆ABD) = ar(∆ACD)
Similarly in ∆EBC,
ED is the median
∴ ar(∆EBD) = ar(∆ECD)
and in ∆ADC, CE is the median
∴ ar(∆FDC) = ar(∆AEC)
= 4 cm
∴ar∆EBC = 2 x ar(∆EDC)
= 2 x 4 = 8 cm (c)

Question 13.
In the figure, ABCD is a parallelogram. If AB = 12 cm, AE = 7.5 cm, CF = 15 cm, then AD =
(a) 3 cm
(b) 6 cm
(c) 8 cm
(d) 10.5 cm
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q13.1
Solution:
In ||gm ABCD, AB = 12 cm AE = 7.5 cm
∴ Area of ||gm ABCD = base x height = AB x AE = 12 x 7.5 cm² = 90 cm²
Now area ||gm ABCD = 90 cm²
and altitude CF = 15 cm
∴ Base AD = \(\frac { Area }{ Altitude }\) = \(\frac { 90 }{ 15 }\) = 6 cm (b)

Question 14.
In the figure, PQRS is a parallelogram. If X and Y are mid-points of PQ and SR respectively and diagonal SQ is joined. The ratio ar(||gm XQRY) : ar(∆QSR) =
(a) 1 : 4
(b) 2 : 1
(c) 1 : 2
(d) 1 : 1
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q14.1
Solution:
In ||gm PQRS, X and Y are the mid points of PQ and SR respectively XY and SQ are joined.
∵ XY bisects PQ and SR
∴ PXYS and XQRY are also ||gms and ar(∆PXYS) = nr(∆XQRY)
∵ ||gm PQRS and AQSR are on the same base and between the same parallel lines
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q14.2

Question 15.
Diagonal AC and BD of trapezium ABCD, in which AB || DC, intersect each other at O. The triangle which is equal in area of ∆AOD is
(a) ∆AOB
(b) ∆BOC
(c) ∆DOC
(d) ∆ADC
Solution:
In trapezium ABCD, diagonals AC and BD intersect each other at O. AB || DC
∆ABC and ∆ABD are on the same base and between the same parallels
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q15.1
∴ ar(∆ABC) = or(∆ABD)
Subtracting ar(∆AOB)
ar(∆ABC) – ar(∆AOB) = ar(∆ADB) – ar(∆AOB)
⇒ ar(∆BOC) = ar(∆AOD)
ar(∆AOD) = ar(∆BOC) (c)

Question 16.
ABCD is a trapezium in which AB || DC. If ar(∆ABD) = 24 cm² and AB = 8 cm, then height of ∆ABC is
(a) 3 cm
(b) 4 cm
(c) 6 cm
(d) 8 cm
Solution:
In trapezium ABCD, AB || DC
AC and BD are joined
ar(∆ABD) = 24 cm2
AB = 8 cm,
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q16.1

Question 17.
ABCD is a trapezium with parallel sides AB = a and DC = b. If E and F are mid-points of non-parallel sides AD and BC respectively, then the ratio of areas of quadrilaterals ABFE and EFCD is
(a) a : b
(b) (a + 3b) : (3a + b)
(c) (3a + b) : (a + 3b)
(d) (2a + b) : (3a + b)
Solution:
In quadrilateral ABCD, E and F are the mid points of AD and BC
AB = a, CD = b
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q17.1
Let h be the height of trapezium ABCD then height of each quadrilateral
ABFE = altitude of quadrilateral EFCD = \(\frac { h }{ 2 }\)
Now area of trap. ABFE = \(\frac { 1 }{ 2 }\) (sum of parallel sides) x altitude
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q17.2

Question 18.
ABCD is a rectangle with O as any point in its interior. If or(∆AOD) = 3 cm2 and ar(∆BOC) = 6 cm2, then area of rectangle ABCD is
(a) 9 cm2
(b) 12 cm2
(c) 15 cm2
(d) 18 cm2
Solution:
In rectangle ABCD, O is any point
ar(∆AOD) = 3 cm2
and ar(∆BOC) = 6 cm2
Join OA, OB, OC and OD
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q18.1
We know that if O is any point in ABCD Then ar(AOB) + ar(COD) = ar(AOB) + ar(BOC)
= 3 + 6 = 9 cm
∴ ar(rect. ABCD) = 2 x 9 = 18 cm (d)

Question 19.
The mid-points of the sides of a triangle ABC along with any of the vertices as the fourth point make a parallelogram of area equal to
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q19.1
Solution:
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q19.2
P,Q and R the mid points of the sides of a ∆ABC then area of any parallelogram formed by the mid points and one vertex of the given triangle has area = \(\frac { 1 }{ 2 }\) area ∆ABC (b)

Question 20.
In the figure, ABCD and FECG are parallelograms equal in area. If ar(∆AQE) = 12 cm2, then ar(||gm FGBQ) =
(a) 12 cm2
(b) 20 cm2
(c) 24 cm2
(d) 36 cm2
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q20.1
Solution:
In the figure, ABCD and EFCG are parallelograms equal in area and ar(∆AQE) = 12 cm2
In ||gm AQED, AE is the diagonal
∴ ar(∆AQE) = \(\frac { 1 }{ 2 }\) ar(||gm AQED)
⇒ 12 cm2 = \(\frac { 1 }{ 2 }\) ar(||gm AQED)
∴ ar(||gm AQED) = 24 cm2
∵ ar ||gm ABCD = ar ||gm FECG
⇒ ar(||gm ∆QED) + ar(|| gm QBCE)
= ar(||gm QBCE) + ar(||gm FGBQ)
⇒ ar(||gm ∆QED) = ar(||gm FGBQ)
= 24 cm2 (c)

Hope given RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

NCERT Solutions for Class 9 Science Chapter 14 Natural Resources

NCERT Solutions for Class 9 Science Chapter 14 Natural Resources

These Solutions are part of NCERT Solutions for Class 9 Science. Here we have given NCERT Solutions for Class 9 Science Chapter 14 Natural Resources. LearnInsta.com provides you the Free PDF download of NCERT Solutions for Class 9 Science (Biology) Chapter 14 – Natural Resources solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 14 – Natural Resources Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

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NCERT TEXT BOOK QUESTIONS

IN TEXT QUESTIONS

Question 1.
How is our atmosphere different from the atmosphere of Venus and Mars ?
Answer:

Atmosphere over Earth Atmosphere over Venus and Mars
1. Nitrogen and Oxygen. It contains nitrogen and oxygen. The two are absent
2. Carbon Dioxide. Carbon dioxide content is little (0-03%). Carbon dioxide is the major component of atmosphere forming 95-97%.
3. Water Vapours. The atmosphere contains water vapours which form a component of water cycle. The atmosphere does not contain water vapours. Living beings, being absent, have no role in
4. living Beings. Composition of atmosphere is maintained by living beings. determining composition of atmosphere.

Question 2.
How does atmosphere act as a blanket ?
Answer:
Atmosphere or air is a bad conductor of heat. It, therefore, functions as a blanket.

  1. It does not allow sudden increase in temperature during the daylight hours when sun shines overhead.
  2. There is no sudden cooling during night. Atmosphere slows down the escape of heat to the outer space from the area of earth under darkness.
  3. Atmosphere maintains the average temperature of the earth fairly steady not only during the day but throughout the year.

Question 3.
What causes winds ?
Answer:
Winds are basically caused by heating of air in certain parts. The hot air rises upwards. This creates an area of low pressure. Cooler air from adjacent higher pressure areas passes into this area. This creates wind. The factors which control movement of winds in different directions in different parts of the earth are

  1. Uneven heating of land in different parts of earth
  2. Differences in heating and cooling of land and water.
  3. Barrier of mountains
  4. Rotation of earth.

Question 4.
How are clouds formed ? (CCE 2011)
Answer:
Clouds are wet air masses that float in the direction of prevailing wind. They develop when water vapours are formed in large number. There is evaporation from the surface of water bodies and wet areas due to their heating during the daytime. Plants also give out water vapours in transpiration while animals do so in exhaled air and perspiration. Air also heats up during daytime. The hot air along with water vapour rises up. At a height, air expands and becomes cool. Cooling causes the water vapours to condense. Suspended particles of dust and other materials function as nuclei around which water vapours condense. When a large wet air mass collects, cloud is formed.

Question 5.
Answer:
List any three human activities that you think would lead to air pollution.

  1. Burning of fossil fuels in industries, vehicles and thermal plants.
  2. Processing industries like textiles, asbestos, flour mills.
  3. Stone crushing.

Question 6.
Why do organisms need water ? (CCE 2011)
Answer:
Organisms need water due to following reasons :

  1. Component of Living Matter: Water is a major component (60-90%) of living matter.
  2. Solvent: Water is a general solvent for chemicals found in the living world.
  3. Reaction Medium: All biochemical reactions occur in the medium of water.
  4. Transport: Substances are transported in the body of a living organisms only in the dissolved state.
  5. Turgidity: Cells, cell organelles, tissues and other structures maintain their shape only when they contain sufficient water to make them turgid.
  6. Temperature Buffer: Water protects the body from sudden changes of temperature.
  7. Wastes: It helps in separation and elimination of metabolic wastes.

Question 7.
What is the major source of fresh water in the city/town/village where you live ?
Answer:
Ground water which is pumped out by tube wells (In some it is local reservoir, canal or river.

Question 8.
Do you know of any activity which may be polluting this water source ?
Answer:
Dumping of industrial wastes from where pollutants seep into soil to reach ground water (sewage and industrial effluents in case of canal or river water).

Question 9.
How is soil formed ? What is the function of humps in soil ? (CCE 2011, 2012)
Answer:
Soil is formed through two processes of weathering and humification.
Weathering
It is pulverisation of rocks or breaking of rocks into fine particles. There are three types of weathering — physical (atmospheric changes and mechanical forces), chemical and biological. Sun, water, wind and living organisms perform them.

  1. Sun: It causes expansion of rocks by heating. Cooling causes their contraction. Different parts of rocks expand and contract at different rates. Uneven expansion and contraction produces cracks leading to fragmentation of rocks.
  2. Water:
    1. Wetting and Drying: Certain rock components can pick up and lose moisture. They undergo swelling and contraction resulting in fragmentation of rocks.
    2. Frost Action: Water seeping in cracks would swell up and exert a great pressure if it freezes due to low temperature. The rock would undergo fragmentation.
    3. Abrasion: Running water carrying rock fragments would break and grind rocks occurring in the pathway. Rain and hail also cause rock breaking.
  3. Wind: Dust and fine sand carried by wind cause abrasion of the rock surface when wind strikes the same.
  4. Living Organisms: Lichens secrete chemicals to dissolve minerals from the rock surface. This produces crevices where dust collects. Mosses grow there. They cause deepening of crevices and development of small cracks. Roots of short lived plants widen these cracks. Roots of larger plants cause fragmentation of rocks by entering the cracks and growing in size.

Humification
Partially decomposed organic matter or humus mixes with weathered rock particles to form soil. Humus helps in formation of soil crumbs which are essential for maintaining proper hydration and aeration of soil.

Question 10.
What are the methods of preventing or reducing soil erosion ?
Answer:
Soil Erosion: It is removal of top soil by agency of wind or water. Wind and water are also the agencies which cause weathering of rocks and carrying the fine particles to other places for the formation of soil. Removal of top soil by water or wind leaves the underneath subsoil and rocky base exposed. Very little plant growth occurs there.
Factors Promoting Soil Erosion

  1. It destroys herbs, grasses and seedlings. The soil is exposed. Trampling by animals causes compaction of soil which reduces its porosity and percolation.
  2. Removal of litter or scraping of forest floor leaves the ground bare for action of agencies causing erosion.
  3. There is decreased absorption of water as the latter does not stay for long on the slope. Run off water passing along the slope gathers speed and develops high cutting and carrying capacity.
  4. Felling of Trees: Felling of trees in excess of regeneration capacity of a forest causes deforestation. It also leaves large area bare for action of wind and water. Deforestation or clearing of forest not only destroys biodiversity but also leads to large scale soil erosion.
  5. Clean Tilling: Clean tilling of crop fields exposes the soil to erosion.
  6. Heavy Rain and Strong Winds. Uncovered soil is eroded quickly by heavy rain and strong wind.

Question 11.
What are two different states in which water is found during the water cycle ?
Answer:
Liquid and vapour, occasionally solid (snow) as well.

Question 12.
Name two biologically important compounds that contain both oxygen and nitrogen. . (CCE 2014)
Answer:
Proteins and nucleic acids.

Question 13.
List any three human activities which would lead to an increase in the carbon dioxide content of air.
Answer:

  1. Increasing combustion of fossil fuels (coal, petroleum, natural gas) in homes, industries, transportation and power projects.
  2. Increasing use of wood for cooking and heating.
  3. Deforestation leading to reduced utilisation of carbon dioxide in photosynthesis.

Question 14.
What is green house effect ? (CCE 2011, 2012, 2013)
Answer:
Green house effect is keeping an area warm by allowing the solar radiations to pass in but preventing long waves to escape due to presence of radiatively active gases and glass panes.

Question 15.
What are the two forms of oxygen found in the atmosphere ? (CCE 2011, 2014)
Answer:

  1. Diatomic oxygen, O2
  2. Triatomic oxygen or ozone, O3.

NCERT CHAPTER END EXERCISES

Question 1.
Why is the atmosphere essential for life ? (CCE 2012)
Answer:

  1. Oxygen: Atmosphere contains oxygen which is essential for combustion and respiration of most organisms.
  2. Carbon Dioxide: Atmosphere provides carbon dioxide for photosynthesis of plants.
  3. Protection: Atmosphere filters out lethal cosmic rays and high energy ultraviolet rays.
  4. Temperature Buffer: Atmosphere does not allow daytime temperature to rise abnormally nor does it allow night time temperature to fall down drastically. This provides favourable temperature for the living organisms.
  5. Other Functions: Air currents help in dispersal of spores and other dissemules. Water cycle operates through atmosphere and produces rain to replenish fresh water over land.

Question 2.
Why is water essential for life ?
Answer:

Organisms need water due to following reasons :

  1. Component of Living Matter: Water is a major component (60-90%) of living matter.
  2. Solvent: Water is a general solvent for chemicals found in the living world.
  3. Reaction Medium: All biochemical reactions occur in the medium of water.
  4. Transport: Substances are transported in the body of a living organisms only in the dissolved state.
  5. Turgidity: Cells, cell organelles, tissues and other structures maintain their shape only when they contain sufficient water to make them turgid.
  6. Temperature Buffer: Water protects the body from sudden changes of temperature.
  7. Wastes: It helps in separation and elimination of metabolic wastes.

Question 3.
How are living organisms dependent on soil ? Are organisms that live in water totally independent of soil as a resource ? ‘
Answer:
All terrestrial organisms depend upon plants for their food and its contained energy. Plants are dependent on soil for anchorage, absorption of water and nutrients. Without them plants cannot manufacture food. So, all living terrestrial organisms depend upon soil.
Aquatic organisms are apparently not connected with soil. However, aquatic autotrophs require inorganic nutrients for manufacture of food. Nutrients reach water bodies only when rain water passes over and inside the soil. Therefore, aquatic organisms are not totally independent of soil as a resource.

Question 4.
You have seen weather reports on television and in newspapers. How do you think we are able to predict the weather ?
Answer:
Weather reports depict areas of low and high pressure, prevailing direction of wind, dryness or wetness of air masses, clouds and their progress, presence and progress of any cyclone, etc. Predictions are then made whether a particular area will or will not receive rain, have calm weather or high speed wind and dust storm.

Question 5.
We know that many human activities lead to increasing levels of pollution of air, water bodies and soil. Do you think that isolating these activities to specific and limited areas would help in reducing pollution ?
Answer:
Restricting pollution related activities to specific and limited areas will not reduce pollution in those areas. The same may rather increase. There are two benefits of such a practice

  1. Joint pollution treatment plants can be installed
  2. The residential and commercial areas away from such pollution generating regions will be comparatively free from pollution.

Question 6.
Write a note on how forests influence the quality of our air, soil and water resources. (CCE 2012)
Answer:
Air Resources:

  1. Oxygen-Carbon Dioxide Balance. Forests maintain the optimum level of oxygen and carbon dioxide. They function as sink for excess carbon dioxide being produced due to excessive combustion. Forests also release a lot of more oxygen to compensate for excess being consumed elsewhere in respiration and combustion.
  2. Control of Air Pollution. Both suspended particles and gaseous pollutants are picked up by forest plants.

Soil Resources: Roots of the forest plants hold the soil firmly. Forest cover protects the soil from direct pounding by rain drops. Forest soil is also sufficiently porous to reduce run off and increase infiltration of rain water. All the three factors prevent soil erosion.
Water Resources:

  1. Rainfall. Forests help in increasing the amount and periodicity of rain fall.
  2. Forest trees retain a lot of water at their bases. Percolation of water into interior of earth produces springs which form rivulets with perennial flow of water.

SELECTION TYPE QUESTIONS

Alternate Response Type Questions :
(True/False, Right/Wrong, Yes/No)

Question 1.
As 75% of earth’s surface is covered with water, the outer crust of earth is called hydrosphere.
Question 2.
Combustion consumes oxygen and liberates carbon dioxide.
Question 3.
Winds develop due to uneven heating of earth.
Question 4.
Carbon monoxide and carbon dioxide produce acid rain.
Question 5.
The amount of rainfall directly influences the abundance and diversity of like forms.
Question 6.
Soil has no role in supplying nutrients to aquatic biota
Question 7.
Pesticides and fertilizers are harmful to soil as they kill the microorganisms involved in recycling of nutrients.
Question 8.
Green house gases are the ones which allow the heat emitted by earth to pass out.

Matching Type Questions :

Question 9.
Match the articles of the column I and column II (Single Matching)

Column I Column II
(a) Chlorofluorocarbons (i) Bacteria
(b) Carbon dioxide (ii) Fossil fuels
(c) Nitrogen fixation (iii) Ozone depletion
(d) Nitrogen and sulphur oxides (iv) Green house gas

Question 10.
Match the contents of columns I, II and III (Double matching)

Column I Column II Column III
(a) Food (i) Sun, water and wind (p) Soil
(b) Abiotic (ii) Resources (q) Living organisms
(c) Carbon dioxide (iii) Air and Water (r) Energy
(d) Paedogenesis (iv) Photosynthesis (s) Shells

Question 11.
Match the pollutants with the type of pollution—air (A), water (W) and soil (S) (Key or check List matching)

Pollutants Pollution

(a) Hydrogen sulphide

(b) SPM

(c) Algal bloom

(d) Raw manure

Question 12.
Match Stimulus with Appropriate Response.

Conservation Practice Soil-A Water-B Air-C

(i) Pollution under control certificate

(ii) Vegetation cover

(iii) Terracing

(iv) Sewage treatment

Fill In the Blanks

Question 13. The earth ……………… and energy from sun are necessary to meet basic requirements of life forms.
Question 14. Surface temperature of moon varies from …………….. to 110°C.
Question 15. …………….. makes soil porous and allows water and air to penetrate deep underground.
Question 16. Carbon occurs in elemental form in ………………. and graphite.
Question 17. CFCs are carbon compounds having both ……………. and chlorine.

Answers:

SOME TYPICAL QUESTIONS

Question 1.
Name the different physical divisions of biosphere.
Answer:
Three – atmosphere (air), hydrosphere (water) and lithosphere (land).

Question 2.
Differentiate between renewable and non-renewable resources.
Answer:

Renewable Resources Non-renewable Resources
1.     Replacement. The resources are replenished within reasonable time.

2.     Use. The resources can be used forever provided the use is limited.

3.     Life. They are both abiotic and biotic.

4.     Availability. It can be increased only by enhancing replenishment.

Replenishment is absent.

The resources will ultimately dwindle, and get exhausted.

They are abiotic.

This is not possible. Increased exploitation will result in quick exhaustion.

Question 3.
Name the parts of India falling under wet zone.
Answer:
Wet zone (rainfall over 200 cm/yr) comprises Western Ghats, Andaman-Nicobar and North East India.

Question 4.
Name two diseases caused by
(a) Infectious agents in polluted water,
(b) Toxic chemicals in polluted water.
Answer:
(a) Infectious Agents. Jaundice, dysentery.
(b) Toxic Agents. Minamata (mercury), itai-itai (cadmium).

Question 5.
What is ozone hole ? Where is it found ? What is its effect ? (CCE 2011)
Answer:
Ozone hole is thinning of ozone in the stratosphere where it is normally present in high concentration as ozone layer. Ozone hole is formed during spring time over antarctica and to a small extent over north pole. Thinning of ozone layer or ozone hole increases the passage of harmful ultravoilet rays to earth. This has increased incidence of skin cancers, defective eye sight, reduced immunity, increased number of mutations and reduced crop yield in southern countries of southern hemisphere.

NCERT Solutions for Class 9 Science Chapter 14 Natural Resources

Hope given NCERT Solutions for Class 9 Science Chapter 14 are helpful to complete your science homework.

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RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.2

RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.2

Other Exercises

Question 1.
In the figure, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD. [NCERT]
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.2 Q1.1
Solution:
In ||gm ABCD,
Base AB = 16 cm
and altitude AE = 8 cm
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.2 Q1.2
∴ Area = Base x Altitude
= AB x AE
= 16 x 8 = 128 cm2
Now area of ||gm ABCD = 128 cm2
Altitude CF = 10 cm
∴ Base AD = \(\frac { Area }{ Altitude }\) = \(\frac { 128 }{ 10 }\) = 12.8cm

Question 2.
In Q. No. 1, if AD = 6 cm, CF = 10 cm, AE = 8 cm, find AB.
Solution:
Area of ||gm ABCD,
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.2 Q2.1
= Base x Altitude
= AD x CF
= 6 x 10 = 60 cm2
Again area of ||gm ABCD = 60 cm2
Altitude AE = 8 cm
∴ Base AB =\(\frac { Area }{ Altitude }\) = \(\frac { 60 }{ 8 }\) = \(\frac { 15 }{ 2 }\) cm = 7.5 cm

Question 3.
Let ABCD be a parallelogram of area 124 cm2. If E and F are the mid-points of sides AB and CD respectively, then find the area of parallelogram AEFD.
Solution:
Area of ||gm ABCD = 124 cm2
E and F are the mid points of sides AB and CD respectively. E, F are joined.
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.2 Q3.1
Draw DL ⊥ AB
Now area of ||gm ABCD = Base x Altitude
= AB x DL = 124 cm2
∵ E and F are mid points of sides AB and CD
∴ AEFD is a ||gm
Now area of ||gm AEFD = AE x DL
= \(\frac { 1 }{ 2 }\)AB x DL [∵ E is mid point of AB]
= \(\frac { 1 }{ 2 }\) x area of ||gm ABCD
= \(\frac { 1 }{ 2 }\) x 124 = 62 cm2

Question 4.
If ABCD is a parallelogram, then prove that ar(∆ABD) = ar(∆BCD) = ar(∆ABC) = ar(∆ACD) = \(\frac { 1 }{ 2 }\)ar( ||gm ABCD).
Solution:
Given : In ||gm ABCD, BD and AC are joined
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.2 Q4.1
To prove : ar(∆ABD) = ar(∆BCD) = ar(∆ABC) = ar(∆ACD) = \(\frac { 1 }{ 2 }\)ar(||gm ABCD)
Proof: ∵ Diagonals of a parallelogram bisect it into two triangles equal in area When BD is the diagonal, then
∴ ar(∆ABD) = ar(∆BCD) = \(\frac { 1 }{ 2 }\)ar(||gm ABCD) …(i)
Similarly, when AC is the diagonal, then
ar(∆ABC) = ar(∆ADC) = \(\frac { 1 }{ 2 }\)ar(||gm ABCD) …(ii)
From (i) and (ii),
ar(∆ABD) = ar(∆BCD) = ar(∆ABC) = ar(∆ACD) = \(\frac { 1 }{ 2 }\) ar(||gm ABCD)

Hope given RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.2 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4

RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4

Other Exercises

Question 1.
In a ∆ABC, D, E and F are respectively, the mid-points of BC, CA and AB. If the lengths of side AB, BC and CA are 7cm, 8cm and 9cm, respectively, find the perimeter of ∆DEF.
Solution:
In ∆ABC, D, E and F are the mid-points of sides,
BC, CA, AB respectively
AB = 7cm, BC = 8cm and CA = 9cm
∵ D and E are the mid points of BC and CA
∴ DE || AB and DE =\(\frac { 1 }{ 2 }\) AB =\(\frac { 1 }{ 2 }\) x 7 = 3.5cm
Similarly,
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 Q1.1

Question 2.
In a triangle ∠ABC, ∠A = 50°, ∠B = 60° and ∠C = 70°. Find the measures of the angles of the triangle formed by joining the mid-points of the sides of this triangle.
Solution:
In ∆ABC,
∠A = 50°, ∠B = 60° and ∠C = 70°
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 Q2.1
D, E and F are the mid points of sides BC, CA and AB respectively
DE, EF and ED are joined
∵ D, E and F are the mid points of sides BC, CA and AB respectively
∴ EF || BC
DE || AB and FD || AC
∴ BDEF and CDEF are parallelogram
∴ ∠B = ∠E = 60° and ∠C = ∠F = 70°
Then ∠A = ∠D = 50°
Hence ∠D = 50°, ∠E = 60° and ∠F = 70°

Question 3.
In a triangle, P, Q and R are the mid-points of sides BC, CA and AB respectively. If AC = 21 cm, BC = 29cm and AB = 30cm, find the perimeter of the quadrilateral ARPQ.
Solution:
P, Q, R are the mid points of sides BC, CA and AB respectively
AC = 21 cm, BC = 29 cm and AB = 30°
∵ P, Q, R and the mid points of sides BC, CA and AB respectively.
∴ PQ || AB and PQ = \(\frac { 1 }{ 2 }\) AB
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 Q3.1
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 Q3.2

Question 4.
In a ∆ABC median AD is produced to X such that AD = DX. Prove that ABXC is a parallelogram.
Solution:
Given : In ∆ABC, AD is median and AD is produced to X such that DX = AD
To prove : ABXC is a parallelogram
Construction : Join BX and CX
Proof : In ∆ABD and ∆CDX
AD = DX (Given)
BD = DC (D is mid points)
∠ADB = ∠CDX (Vertically opposite angles)
∴ ∆ABD ≅ ∆CDX (SAS criterian)
∴ AB = CX (c.p.c.t.)
and ∠ABD = ∠DCX
But these are alternate angles
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 Q4.1
∴ AB || CX and AB = CX
∴ ABXC is a parallelogram.

Question 5.
In a ∆ABC, E and F are the mid-points of AC and AB respectively. The altitude AP to BC intersects FE at Q. Prove that AQ = QP.
Solution:
Given : In ∆ABC, E and F are the mid-points of AC and AB respectively.
EF are joined.
AP ⊥ BC is drawn which intersects EF at Q and meets BC at P.
To prove: AQ = QP
proof : In ∆ABC
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 Q5.1
E and F are the mid points of AC and AB
∴ EF || BC and EF = \(\frac { 1 }{ 2 }\)BC
∴ ∠F = ∠B
In ∆ABP,
F is mid point of AB and Q is the mid point of FE or FQ || BC
∴ Q is mid point of AP,
∴ AQ = QP

Question 6.
In a ∆ABC, BM and CN are perpendiculars from B and C respectively on any line passing through A. If L is the mid-point of BC, prove that ML NL.
Solution:
In ∆ABC,
BM and CN are perpendicular on a line drawn from A. L is the mid point of BC. ML and NL are joined.
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 Q6.1
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 Q6.2
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 Q6.3

Question 7.
In the figure triangle ABC is right-angled at B. Given that AB = 9cm. AC = 15cm and D, E are the mid points of the sides AB and AC respectively, calculate.
(i) The length of BC
(ii) The area of ∆ADC
Solution:
In ∆ABC, ∠B = 90°
AC =15 cm, AB = 9cm
D and E are the mid points of sides AB and AC respectively and D, E are joined.
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 Q7.1
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 Q7.2

Question 8.
In the figure, M, N and P are the mid points of AB, AC and BC respectively. If MN = 3 cm, NP = 3.5cm and MP = 2.5cm, calculate BC, AB and AC.
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 Q8.1
Solution:
In ∆ABC,
M, N and P are the mid points of side, AB, AC and BC respectively.
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 Q8.2
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 Q8.3

Question 9.
In the figure, AB = AC and CP || BA and AP is the bisector of exterior ∠CAD of ∆ABC. Prove that (i) ∠PAC = ∠BCA (ii) ABCP is a parallelogram.
Solution:
Given : In ABC, AB = AC
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 Q9.1
nd CP || BA, AP is the bisector of exterior ∠CAD of ∆ABC
To prove :
(i) ∠PAC = ∠BCA
(ii) ABCP is a ||gm
Proof : (i) In ∆ABC,
∵ AB =AC
∴ ∠C = ∠B (Angles opposite to equal sides) and ext.
∠CAD = ∠B + ∠C
= ∠C + ∠C = 2∠C ….(i)
∵ AP is the bisector of ∠CAD
∴ 2∠PAC = ∠CAD …(ii)
From (i) and (ii)
∠C = 2∠PAC
∠C = ∠CAD or ∠BCA = ∠PAC
Hence ∠PAC = ∠BCA
(ii) But there are alternate angles,
∴ AD || BC
But BA || CP
∴ ABCP is a ||gm.

Question 10.
ABCD is a kite having AB = AD and BC = CD. Prove that the figure formed by joining the mid-points of the sides, in order, is a rectangle.
Solution:
Given : In fne figure, ABCD is a kite in which AB = AD and BC = CD.
P, Q, R and S are the mid points of the sides AB, BC, CD and DA respectively.
To prove : PQRS is a rectangle.
Construction : Join AC and BD.
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 Q10.1
Proof: In ∆ABD,
P and S are mid points of AB and AD
∴ PS || BD and PS = \(\frac { 1 }{ 2 }\) BD …(i)
Similarly in ∆BCD,
Q and R the mid points of BC and CD
∴ QR || BD and
QR = \(\frac { 1 }{ 2 }\) BD …(ii)
∴ Similarly, we can prove that PQ || SR and PQ = SR …(iii)
From (i) and (ii) and (iii)
PQRS is a parallelogram,
∵ AC and BD intersect each other at right angles.
∴ PQRS is a rectangle.

Question 11.
Let ABC be an isosceles triangle in which AB = AC. If D, E, F be the mid-points of the sides BC, CA and AB respectively, show that the segment AD and EF bisect each other at right angles.
Solution:
In ∆ABC, AB = AC
D, E and F are the mid points of the sides BC, CA and AB respectively,
AD and EF are joined intersecting at O
To prove : AD and EF bisect each other at right angles.
Construction : Join DE and DF.
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 Q11.1
Proof : ∵ D, E and F are the mid-points of
the sides BC, CA and AB respectively
∴ AFDE is a ||gm
∴ AF = DE and AE = DF
But AF = AE
(∵ E and F are mid-points of equal sides AB and AC)
∴ AF = DF = DE = AE
∴AFDE is a rhombus
∵ The diagonals of a rhombus bisect each other at right angle.
∴ AO = OD and EO = OF
Hence, AD and EF bisect each other at right angles.

Question 12.
Show that the line segments joining the mid points of the opposite sides of a quadrilateral bisect each other.
Solution:
Given : In quad. ABCD,
P, Q, R and S are the mid points of sides AB, BC, CD and DA respectively.
PR and QS to intersect each other at O
To prove : PO = OR and QO = OS
Construction: Join PQ, QR, RS and SP and also join AC.
Proof: In ∆ABC
P and Q are mid-points of AB and BC
∴ PQ || AC and PQ = \(\frac { 1 }{ 2 }\) AC …(i)
Similarly is ∆ADC,
S and R are the mid-points of AD and CD
∴ SR || AC and SR = \(\frac { 1 }{ 2 }\) AC ..(ii)
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 Q12.1
from (i) and (ii)
PQ = SQ and PQ || SR
PQRS is a ||gm (∵ opposite sides are equal area parallel)
But the diagonals of a ||gm bisect each other.
∴ PR and QS bisect each other.

Question 13.
Fill in the blanks to make the following statements correct :
(i) The triangle formed by joining the mid-points of the sides of an isosceles triangle is …
(ii) The triangle formed by joining the mid-points of the sides of a right triangle is …
(iii) The figure formed by joining the mid-points of consecutive sides of a quadrilateral is …
Solution:
(i) The triangle formed by joining the mid-points of the sides of an isosceles triangle is an isosceles triangle.
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 Q13.1
(ii) The triangle formed by joining the mid-points of the sides of a right triangle is right triangle.
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 Q13.2
(iii) The figure formed by joining the mid-points of consecutive sides of a quadrilateral is a parallelogram.
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 Q13.3

Question 14.
ABC is a triangle and through A, B, C lines are drawn parallel to BC, CA and AB respectively intersecting at P, Q and R. Prove that the perimeter of ∆PQR is double the perimeter of ∆ABC.
Solution:
Given : In ∆ABC,
Through A, B and C, lines are drawn parallel to BC, CA and AB respectively meeting at P, Q and R.
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 Q14.1
To prove : Perimeter of ∆PQR = 2 x perimeter of ∆ABC
Proof : ∵ PQ || BC and QR || AB
∴ ABCQ is a ||gm
∴ BC = AQ
Similarly, BCAP is a ||gm
∴ BC = AP …(i)
∴ AQ = AP = BL
⇒ PQ = 2BC
Similarly, we can prove that
QR = 2AB and PR = 2AC
Now perimeter of ∆PQR.
= PQ + QR + PR = 2AB + 2BC + 2AC
= 2(AB + BC + AC)
= 2 perimeter of ∆ABC.
Hence proved

Question 15.
In the figure, BE ⊥ AC. AD is any line from A to BC intersecting BE in H. P, Q and R are respectively the mid-points of AH, AB and BC. Prove that PQR = 90°.
Solution:
Given: In ∆ABC, BE ⊥ AC
AD is any line from A to BC meeting BC in D and intersecting BE in H. P, Q and R are respectively mid points of AH, AB and BC. PQ and QR are joined B.
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 Q15.1
To prove : ∠PQR = 90°
Proof: In ∆ABC,
Q and R the mid points of AB and BC 1
∴ QR || AC and QR = \(\frac { 1 }{ 2 }\) AC
Similarly, in ∆ABH,
Q and P are the mid points of AB and AH
∴ QP || BH or QP || BE
But AC ⊥ BE
∴ QP ⊥ QR
∴ ∠PQR = 90°

Question 16.
ABC is a triangle. D is a point on AB such that AD = \(\frac { 1 }{ 4 }\) AB and E is a point on AC such that AE = \(\frac { 1 }{ 4 }\) AC. Prove that DE = \(\frac { 1 }{ 4 }\) BC.
Solution:
Given : In ∆ABC,
D is a point on AB such that
AD = \(\frac { 1 }{ 4 }\) AB and E is a point on AC such 1
that AE = \(\frac { 1 }{ 4 }\) AC
DE is joined.
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 Q16.1
To prove : DE = \(\frac { 1 }{ 4 }\) BC
Construction : Take P and Q the mid points of AB and AC and join them
Proof: In ∆ABC,
∵ P and Q are the mid-points of AB and AC
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 Q16.2

Question 17.
In the figure, ABCD is a parallelogram in which P is the mid-point of DC and Q is a point on AC such that CQ = \(\frac { 1 }{ 4 }\) AC. If PQ produced meets BC at R, prove that R is a mid-point of BC.
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 Q17.1
Solution:
Given : In ||gm ABCD,
P is the mid-point of DC and Q is a point on AC such that CQ = \(\frac { 1 }{ 4 }\) AC. PQ is produced meets BC at R.
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 Q17.2
To prove : R is mid point of BC
Construction : Join BD
Proof : ∵ In ||gm ABCD,
∵ Diagonal AC and BD bisect each other at O
∴ AO = OC = \(\frac { 1 }{ 2 }\) AC …(i)
In ∆OCD,
P and Q the mid-points of CD and CO
∴ PQ || OD and PQ = \(\frac { 1 }{ 2 }\) OD
In ∆BCD,
P is mid-poiht of DC and PQ || OD (Proved above)
Or PR || BD
∴ R is mid-point BC.

Question 18.
In the figure, ABCD and PQRC are rectangles and Q is the mid-point of AC.
Prove that (i) DP = PC (ii) PR = \(\frac { 1 }{ 2 }\) AC.
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 Q18.1
Solution:
Given : ABCD are PQRC are rectangles and Q is the mid-point of AC.
To prove : (i) DP = PC (ii) PR = \(\frac { 1 }{ 2 }\) AC
Construction : Join diagonal AC which passes through Q and join PR.
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 Q18.2
Proof : (i) In ∆ACD,
Q is mid-point of AC and QP || AD (Sides of rectangles)
∴ P is mid-point of CD
∴ DP = PC
(ii) ∵PR and QC are the diagonals of rectangle PQRC
∴ PR = QC
But Q is the mid-point of AC
∴ QC = \(\frac { 1 }{ 2 }\) AC
Hence PR = \(\frac { 1 }{ 2 }\) AC

Question 19.
ABCD is a parallelogram, E and F are the mid points AB and CD respectively. GFI is any line intersecting AD, EF and BC at Q P and H respectively. Prove that GP = PH.
Solution:
Given : In ||gm ABCD,
E and F are mid-points of AB and CD
GH is any line intersecting AD, EF and BC at GP and H respectively
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 Q19.1
To prove : GP = PH
Proof: ∵ E and F are the mid-points of AB and CD
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 Q19.2
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 Q19.3

Question 20.
BM and CN are perpendiculars to a line passing, through the vertex A of a triangle ABC. If L is the mid-point of BC, prove that LM = LN.
Solution:
In ∆ABC,
BM and CN are perpendicular on a line drawn from A.
L is the mid point of BC.
ML and NL are joined.
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 Q20.1
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 Q20.2

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NCERT Exemplar Solutions for Class 9 Science Chapter 13 Why Do we Fall Ill

NCERT Exemplar Solutions for Class 9 Science Chapter 13 Why Do we Fall Ill

These Solutions are part of NCERT Exemplar Solutions for Class 9 Science . Here we have given NCERT Exemplar Solutions for Class 9 Science Chapter 13 Why Do we Fall Ill

Question 1.
Give two examples for each of the following :

  1. Acute diseases
  2. Chronic diseases
  3. Infectious diseases
  4. Non-infectious diseases.

Answer:

  1. Acute Diseases. Typhoid, Malaria, Influenza.
  2. Chronic Diseases. TB (tuberculosis), Elephantiasis.
  3. Infectious Diseases. Typhoid, Chicken Pox.
  4. Non-infectious Diseases. Diabetes, Goitre.

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Question 2.
Name two diseases caused by protozoans. What are their causal organisms ?
Answer:

  1. Sleeping Sickness, caused by Trypanosoma gambiense.
  2. Kala-azar, caused by Leishmania donovani.

Question 3.
Which bacterium causes peptic ulcers ? Who discovered the pathogen for the first time ?
Answer:
Bacterium Causing Peptic Ulcers. Helicobacter pylori Discovery. Warren (1984). Marshall and Warren (1985).

Question 4.
What is antibiotic ? Give two examples. (CCE 2012)
Answer:
Antibiotic. It is a biochemical produced by a microbe which kills or blocks growth of other microbes (e.g. bacteria) by blocking their life processes without harming human cells, e.g., Penicillin, Streptomycin.

Question 5.
Fill in the blocks :

  1. Pneumonia is an example of ……………… disease.
  2. Many skin diseases are caused by ………….. .
  3. Antibiotics commonly block biochemical pathways important for the growth of …………….. .
  4. Living organisms carrying the infecting agents from one person to another are called ……………. .

Answer:

  1. Communicable (infectious)
  2. fungi
  3. bacteria
  4. vectors.

Question 6.
Name the target organs of the following diseases :

  1. Hepatitis targets …………… .
  2. Fit or unconsciousness targets ……………… .
  3. Pneumonia targets ……………. .
  4. Fungal disease targets ……………. .

Answer:

  1. Liver
  2. Brain
  3. Lungs
  4. Skin.

Question 7.
(a) Who discovered Vaccine” for the first time.
(b) Name two diseases which can be prevented by using vaccines.
Answer:
(a) Edward Jenner
(b) Pertussis, diphtheria, tuberculosis, polio.

Question 8.
Fill in the blanks :

  1. ……………… disease continues for many days and causes ………………….. on body.
  2.  …………….. disease continues for a few days and causes no long term effect on body.
  3. …………….. is defined as physical, mental and social well being and comfort.
  4. Common cold is ……………… disease.
  5. Many skin diseases are caused by …………………. .

Answer:

  1. Chronic, long-term effect
  2. Acute
  3. Health
  4. Infectious (communicable)
  5. Fungi.

Question 9.
Classify the following diseases as infectious and non-infectious:

  1. AIDS
  2. Tuberculosis
  3. Cholera
  4. High blood pressure
  5. Heart disease
  6. Pneumonia
  7. Cancer.

Answer:

  1. AIDS—infectious,
  2. Tuberculosis—infectious,
  3. Cholera—infectious,
  4. High Blood Pressure—non-infectious.
  5. Heart Disease—non-infectious.
  6. Pneumonia—infectious,
  7. Cancer—non-infectious.

Question 10.
Name any two groups of microorganisms from which antibiotics could be extracted.
Answer:
Bacteria, Fungi.

Question 11.
Name any three diseases transmitted through vectors.
Malaria (vector Anopheles), Dengue (vector Aedes), Kala-azar (vector Sandfly).

Question 12.
Explain giving reasons :

  1. Balanced diet is necessary for maintaining healthy body.
  2. Health of an organism depends upon the surrounding environmental conditions.
  3. Our surrounding area should be free of stagnant water. (CCE 2012)
  4. Social harmony and good economic conditions are necessary for good health.

Answer:

  1. Balanced diet provides all the nutrients for metabolic activities of the organism.
  2. Environment contributes the immediate, second level and third level of causes.
  3. Stagnant water becomes the breeding site for mosquitoes.
  4. Good health is not only being disease free but also physical, mental and social well being.

Question 13.
What is disease ? How many types of diseases have you studied ? Give examples. (CCE 2011)
Answer:
Disease. It is a condition of derangement or disturbed functioning of the body or its part.
Types of Diseases :

  1. On the Basis of Duration. Acute and chronic.
  2. On the Basis of Period of Occurrence. Congenital and acquired.
  3. On the Basis of Causal Agent. Infectious and non-infectious.

Infectious or communicable diseases can be contagious or non-contagious. Non-infectious disease may be deficiency disease, metabolic disease, degenerative disease, allergy, cancer and injury.
Examples. Influenza, tuberculosis, pneumonia (infectious), cancer (non-infectious).

Question 14.
What do you mean by disease symptoms ? Explain giving two examples.
Answer:
Symptoms: They are manifestations or evidences of the presence of diseases. Symptoms are in the form of structural and functional changes in the body or body parts. They indicate that there is something wrong in the body,
e.g., wound with pus, cough, cold, loose motions, pain in abdomen, headache, fever. Symptoms do not give any exact cause of the disease. For instance, headache is due to some dozen different diseases. There may be problem of eye sight, blood pressure, examination and other stress, meningitis, etc.

Question 15.
Why is immune system essential for our health ?
Answer:
Immune system is body defence system against various types of pathogens. It has many components—phagocytic cells, natural killer cells, T-lymphocytes and B-lymphocytes. B-lymphocytes produce antibodies against pathogens and their toxins. Immune system keeps the body healthy by killing infecting microbes.

Question 16.
What precautions would you take to justify “prevention is better than cure”. (CCE 2011, 2012)
Answer:
Prevention is always better than cure as a disease always causes some damage to the body, loss of working days, besides expenditure on medication. The important precautions for preventing diseases are

  1. Hygienic environment
  2. Personal hygiene
  3. Proper nutrition
  4. Clean food
  5. Clean water
  6. Regular exercise and
  7. Relaxation: Every body should also be aware of diseases and their spread. A regular medical check up is also reQuired.
  8. Immunisation programme should be followed.

Question 17.
Why do some children fall ill more frequently than others living in the same locality ?
Answer:
Children fall ill more frequendy due to

  1. Poor personal hygiene
  2. Poor domestic hygiene
  3. Playing over floor
  4. Eating with unclean hands.
  5. Lack of proper nutrition and balanced diet. All these make the immune system weak.

Question 18.
Why are antibiotics not effective for viral diseases ? (CCE 2011, 2012)
Answer:
Antibiotics are effective against bacteria and some other non-viral pathogens as they block some of their biosynthetic pathways without affecting human beings. However, viruses do not have their own metabolic machinery. There are very few biochemical processes that can block viral multiplication. Antibiotics are not effective against them. They can be over-powered only by specific anti-viral drugs.

Question 19.
Becoming exposed to or infected with a microbe does not necessarily mean developing noticeable disease. Explain.
Answer:
An infectious microbe is able to cause a disease only if the immune system of the person is unable to put proper defence against it. Many persons have strong immune system or have acquired immunity against the pathogen or the pathogen attack is less than the infective dose. In such cases, despite exposure to infective microbe, the person will not catch the disease.

Question 20.
Give any four factors necessary for a healthy person. (CCE 2012)
Answer:

  1. Environment,
    1. A clean physical environment with the help of public health services,
    2. A congenial social environment.
  2. Personal Hygiene. Personal cleanliness prevents catching up of infectious diseases.
  3. Nourishment. A proper balanced diet keeps the immune system strong.
  4. Vaccination. Timely vaccination against major diseases protects oneself from catching those diseases.
  5. Avoiding overcrowded areas
  6. Regular exercise and relaxation.

Question 21.
Why is AIDS considered to be a “syndrome” and not a disease ? (CCE 2012)
Answer:
Syndrome is a group of symptoms, signs, physical and physiological disturbances that are due to a common cause. AIDS is also a complex of diseases and symptoms which develop due to failure of the body to fight off even minor infections. HIV that causes AIDS damages immune system of the patient by destroying T4 lymphocytes. As a result, even small cold leads to development of pneumonia, a minor gut infection leads to severe diarrhoea and blood loss, while skin rashes develop into ulcers.

Hope given NCERT Exemplar Solutions for Class 9 Science Chapter 13 Why Do we Fall Ill are helpful to complete your science homework.

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HOTS Questions for Class 9 Science Chapter 13 Why Do we Fall Ill

HOTS Questions for Class 9 Science Chapter 13 Why Do we Fall Ill

These Solutions are part of HOTS Questions for Class 9 Science. Here we have given HOTS Questions for Class 9 Science Chapter 13 Why Do we Fall Ill

Question 1.
(a) Identify the figure.
HOTS Questions for Class 9 Science Chapter 13 Why Do we Fall Ill image - 1
(b) What do A and B depict.
Answer:
(a) Viral particles (as of SARS) coming out of host cells.
(b) A — host cell.
B — Virus.

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Question 2.
Name a disease in which
(a) Antiboitic has no role.
(b) Kissing does not spread the disease while sexual contact transfers the same.
(c) Mass scale immunisation is going on.
(d) Virus, bacterium and protozoan can be causal agent.
Answer:
(a) Malaria
(b) AIDS
(c) Polio
(d) Diarrhoea.

Question 3.
What is correct ? Give reason.
(a) A person strolling in the lawn of his house is relaxing or doing exercise.
(b) Wearing socks and full sleeves at night will prevent the attack from dengue.
(c) Regular use of ORS cures diarrhoea.
Answer:
(a) Strolling is no exercise as a person moves slowly. It is a way of relaxation where stress and strain can be relieved.
(b) No. Dengue is caused by bite of Aedes mosquito which is active during day time only.
(c) Yes, ORS prevents dehydration. Diarrhoea is generally cured automatically after 1-2 days because it is mostly viral infection.

Hope given HOTS Questions for Class 9 Science Chapter 13 Why Do we Fall Ill are helpful to complete your science homework.

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RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS

RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS

Other Exercises

Question 1.
In a parallelogram ABCD, write the sum of angles A and B.
Solution:
In ||gm ABCD,
∠A + ∠B = 180°
(Sum of consecutive angles of a ||gm)
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS Q1.1

Question 2.
In a parallelogram ABCD, if ∠D = 115°, then write the measure of ∠A.
Solution:
In ||gm ABCD,
∠D = 115°
But ∠A + ∠D = 180°
(Sum of consecutive angles of a ||gm)
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS Q2.1
⇒ ∠A + 115°= 180° ∠A = 180°- 115°
∴ ∠A = 65°

Question 3.
PQRS is a square such that PR and SQ intersect at O. State the measure of ∠POQ.
Solution:
In a square PQRS,
Diagonals PR and QS intersects each other at O.
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS Q3.1
∵ The diagonals of a square bisect each other at right angles.
∴ ∠POQ = 90°

Question 4.
If PQRS is a square then write the measure of ∠SRP.
Solution:
In square PQRS,
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS Q4.1
Join PR,
∵Diagonals of a square bisect are opposite angles
∴∠SRP = \(\frac { 1 }{ 2 }\)x ∠SRQ
= \(\frac { 1 }{ 2 }\) x 90° = 45°

Question 5.
If ABCD is a rhombus with ∠ABC = 56°, find the measure of ∠ACD.
Solution:
In rhombus ABCD,
Diagonals bisect each other at 0 at right angles.
∠ABC = 56°
But ∠ABC + ∠BCD = 180° (Sum of consecutive angles)
⇒ 56° + ∠BCD = 180°
⇒ ∠BCD = 180° – 56° = 124°
∵ Diagonals of a rhombus bisect the opposite angle
∴ ∠ACD = \(\frac { 1 }{ 2 }\) ∠BCD = \(\frac { 1 }{ 2 }\) x 124°
= 62°
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS Q5.1

Question 6.
The perimeter of a parallelogram is 22 cm. If the longer side measures 6.5 cm, what is the measure of the shorter side.
Solution:
Perimeter of a ||gm ABCD = 22cm
∴ Sum of two consecutive sides = \(\frac { 22 }{ 2 }\)
= 11cm
i.e. AB + BC = 11 cm
AB = 6.5 cm and let BC = x cm
∴ 6.5 + x = 11 cm
x = 11 – 6.5 = 4.5
∴ Shorter side = 4.5 cm
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS Q6.1

Question 7.
If the angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. Then find the measure of the smallest angle.
Solution:
Ratio in the angles of a quadrilateral = 3 : 5 : 9 : 13
Let first angle = 3x
Second angle = 5x
Third angle = 9x
and fourth angle = 13x
∵ The sum of angles of a quadrilateral = 360°
∴ 3x + 5x + 9x + 13x = 360°
⇒ 30x = 360° ⇒ x = \(\frac { { 360 }^{ \circ } }{ 30 }\)  = 12
∴ Smallest angle = 3x = 3 x 12° = 36°

Question 8.
In parallelogram ABCD if ∠A = (3x – 20°), ∠B = (y + 15)°, ∠C = (x + 40°), then find the value of x and y.
Solution:
In a ||gm ABCD,
∠A = (3x – 20°), ∠B = y + 15°,
∠C = x + 40°
Now, ∠A = ∠C (Opposite angles of a ||gm)
⇒ 3x – 20 = x + 40°
⇒ 3x – x = 40° + 20° ⇒ 2x = 60°
⇒ x = \(\frac { { 60 }^{ \circ } }{ 2 }\)  = 30°
and ∠A + ∠B = 180° (Sum of the consecutive angles)
⇒ 3x-20° + y + 15° = 180°
⇒ 3x + y – 5° = 180°
⇒ 3 x 30° +y- 5° = 180°
⇒ 90° – 5° + y = 180
y = 180° – 90° + 5 = 95°
∴ x = 30°, y = 95°

Question 9.
If measures opposite angles of a parallelogram are (60 – x)° and (3x – 4)°, then find the measures of angles of the parallelogram.
Solution:
Opposite angles of a ||gm ABCD are (60 – x)° and (3x – 4°)
But opposite angles of a ||gm are equal, the
60° – x° = 3x – 4° ⇒ 60° + 4° = 3x + x
⇒ 4x = 64° ⇒ x = \(\frac { { 64 }^{ \circ } }{{ 4 }^{ \circ } }\)  = 16°
∴ ∠A = 60° – x = 60° – 16° = 44°
But ∠A + ∠B = 180° (sum of consecutive angle)
⇒ 44° + ∠B = 180°
⇒ ∠B = 180° – 44°
⇒ ∠B = 136°
But ∠A = ∠C and ∠B = ∠D (Opposite angles)
∴ Angles are 44°, 136°, 44°, 136°

Question 10.
In a parallelogram ABCD, the bisectors of ∠A also bisect BC at x, find AB : AD.
Solution:
In ||gm ABCD,
Bisectors of ∠A meets BC at X and BX = XC
Draw XY ||gm AB meeting AD at Y
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS Q10.1
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS Q10.2

Question 11.
In the figure, PQRS in an isosceles trapezium find x and y.
Solution:
∵ PQRS is an isosceles trapezium in which
SP = RQ and SR || PQ
∴ ∠P + ∠S = 180° (Sum of co-interior angles)
3x + 2x = 180° ⇒ 5x = 180°
⇒ x = \(\frac { { 180 }^{ \circ } }{ 5 }\)  = 36°
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS Q11.1
But ∠P = ∠Qm (Base angles of isosceles trapezium)
y = 2x = 2 x 36° = 12°
∴ y = 12°
Hence x = 36°, y = 12°

Question 12.
In the figure ABCD is a trapezium. Find the values of x and y.
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS Q12.1
Solution:
In trapezium ABCD,
AB || CD
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS Q12.2
∴ ∠A + ∠D = 180° (Sum of cointerior angles)
x + 20° + 2x + 10° = 180°
3x + 30° = 180°
⇒ 3x= 180° – 30°
3x = 150°
x = \(\frac { { 150 }^{ \circ } }{ 3 }\)  = 50°
Similarly, ∠B + ∠C = 180°
⇒ y + 92° = 180°
⇒ y = 180° – 92° = 88°
∴ x = 50°, y = 88°

Question 13.
In the figure, ABCD and AEFG are two parallelograms. If ∠C = 58°, find ∠F.
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS Q13.1
Solution:
In the figure, ABCD and AEFG are two parallelograms ∠C = 58°
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS Q13.2
∵ DC || GF and CB || FE (Sides of ||gms)
∴ ∠C = ∠F
But ∠C = 58°
∴ ∠F = 58°

Question 14.
Complete each of the following statements by means of one of those given in brackets against each:
(i) If one pair of opposite sides are equal and parallel, then the figure is ……… (parallelogram, rectangle, trapezium)
(ii) If in a quadrilateral only one pair of opposite sides are parallel, the quadrilateral is …….. (square, rectangle, trapezium)
(iii) A line drawn from the mid-point of one side of a triangle ………. another side intersects the third side at its mid-point, (perpendicular to, parallel to, to meet)
(iv) If one angle of a parallelogram is a right angle, then it is necessarily a …….. (rectangle, square, rhombus)
(v) Consecutive angle of a parallelogram are ……… (supplementary, complementary)
(vi) If both pairs of opposite sides of a quadrilateral are equal, then it is necessarily a ……… (rectangle, parallelogram, rhombus)
(vii) If opposite angles of a quadrilateral are equal, then it is necessarily a ………. (parallelogram, rhombus, rectangle)
(viii)If consecutive sides of a parallelogram are equal, then it is necessarily a …….. (kite, rhombus, square)
Solution:
(i) If one pair of opposite sides are equal and parallel, then the figure is parallelogram.
(ii) If in a quadrilateral only one pair of opposite sides are parallel, the quadrilateral is trapezium.
(iii) A line drawn from the mid-point of one side of a triangle parallel to another side intersects the third side at its mid-point,
(iv) If one angle of a parallelogram is a right angle, then it is necessarily a rectangle.
(v) Consecutive angle of a parallelogram are supplementary.
(vi) If both pairs of opposite sides of a quadrilateral are equal, then it is necessarily a parallelogram.
(vii) If opposite angles of a quadrilateral are equal, then it is necessarily a parallelogram.
(viii) If consecutive sides of a parallelogram are equal, then it is necessarily a rhombus.

Question 15.
In a quadrilateral ABCD, bisectors of A and B intersect at O such that ∠AOB = 75°, then write the value of ∠C + ∠D.
Solution:
In quadrilateral ABCD,
Bisectors of ∠A and ∠B meet at O and ∠AOB = 75°
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS Q15.1
In AOB, ∠AOB = 75°
∴ ∠OAB + ∠OBA = 180° – 75° = 105°
But OA and OB are the bisectors of ∠A and ∠B.
∴ ∠A + ∠B = 2 x 105° = 210°
But ∠A + ∠B + ∠C + ∠D = 360° (Sum of angles of a quad.)
∴ 210° + ∠C + ∠D = 360°
⇒ ∠C + ∠D = 360° – 210° = 150°
Hence ∠C + ∠D = 150°

Question 16.
The diagonals of a rectangle ABCD meet at O. If ∠BOC = 44° find ∠OAD.
Solution:
In rectangle ABCD,
Diagonals AC and BD intersect each other at O and ∠BOC = 44°
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS Q16.1
But ∠AOD = ∠BOC (Vertically opposite angles)
∴ ∠AOD = 44°
In ∆AOD,
∠AOD + ∠OAD + ∠ODA = 180° (Sum of angles of a triangle)
⇒ 44° + ∠OAD + ∠OAD = 180° [∵ OA = OD, ∠OAD = ∠ODA]
⇒ 2∠OAD = 180° – 44° = 136°
∴ ∠OAD = \(\frac { { 136 }^{ \circ } }{ 2 }\)  = 68°

Question 17.
If ABCD is a rectangle with ∠BAC = 32°, find the measure if ∠DBC.
Solution:
In rectangle ABCD,
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS Q17.1
Diagonals bisect each other at O
∠BAC = 32°
∵ OA = OB
∴ ∠OBA Or ∠DBA = ∠BAC = 32°
But ∠ABC = 90° (Angle of a rectangles)
∴ ∠DBC = ∠ABC – ∠DBA
= 90° – 32° = 58°

Question 18.
If the bisectors of two adjacent angles A and B of a quadrilateral ABCD intersect at a point O. Such that ∠C + ∠D = k(∠AOB), then find the value of k.
Solution:
In quadrilateral ABCD,
Bisectors of ∠A and ∠B meet at O
Such that ∠C + ∠D = k (∠AOB)
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS Q18.1
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS Q18.2

Question 19.
In the figure, PQRS is a rhombus in which the diagonal PR is produced to T. If ∠SRT = 152°, find x, y and z.
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS Q19.1
Solution:
In rhombus PQRS,
Diagonal PR and SQ bisect each other at right angles and PR is produced to T such that ∠SRT = 152°
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS Q19.2
But ∠SRT + ∠SRP = 180° (Linear pair)
⇒ 152° +∠SRP = 180°
⇒ ∠SRP =180°- 152° = 28°
But ∠SPR = ∠SRP (∵ PR bisects ∠P and ∠R)
⇒ z = 28°
y = 90° (∵ Diagonals bisect each other at right angles)
∠RPQ = z = 28°
∴ In ∆POQ,
z + x = 90° ⇒ 28° + x = 90°
⇒ x = 90° – 28° = 62°
∴ x = 62°, y = 90°, z = 28°

Question 20.
In the figure, ABCD is a rectangle in which diagonal AC is produced to E. If ∠ECD = 146°, find ∠AOB.
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS Q20.1
Solution:
In rectangle ABCD,
Diagonals AC and BD bisect each other at O
AC is produced to E and ∠DCE = 146°
RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS Q20.2
∠DCE + ∠DCA = 180° (Linear pair)
⇒ 146°+ ∠DCA= 180°
⇒ ∠DCA = 180°- 146°
⇒ ∠DCA = 34°
∴ ∠CAB = ∠DCA (Alternate angles)
= 34°
Now in ∆AOB,
∠AOB = 180° – (∠DAB + ∠OBA)
= 180° – (34° + 34°)
= 1803 – 68° = 112°

Hope given RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS are helpful to complete your math homework.

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Value Based Questions in Science for Class 9 Chapter 7 Diversity in Living Organisms

Value Based Questions in Science for Class 9 Chapter 7 Diversity in Living Organisms

These Solutions are part of Value Based Questions in Science for Class 9. Here we have given Value Based Questions in Science for Class 9 Chapter 7 Diversity in Living Organisms

Question 1.
“Raman lives in a coastal village. He is son of a fisherman. Whenever any unwanted animal comes in the net, instead of killing it, he puts back the same in the sea.” Answer the following questions based on above information

  1. What would have happened, had he killed those animals ?
  2. Give one reason to justify that Raman’s action is environment friendly.
  3. How can you contribute in the preservation of flora and fauna around you ? Mention any two steps.
    (Sample Paper 2012—13)

Answer:

  1. Killing of unwanted animals would have contributed to disturbing ecological balance.
  2. Raman is helping in conserving biodiversity.
    1. Spreading awareness about importance of biodiversity amongst classmates, family members and community members.
    2. Nonuse of products derived from wild animals.
    3. By becoming member of PETA (People for Ethical Treatment of Animals) and developing empathy and love for all living organisms.

More Resources

Question 2.
Seeing a bat flying over the roof of her house, Saira asked her papa

  1. What is this night flying bird ?
  2. How does it see during night ?
  3. What does it eat and how does it obtain the same ? What would be reply of her papa ?

Answer:

  1. Bat is not a bird. It is a mammal that has patagia in the fore limbs to function as wings and help in flight.
  2. Bat does not require sharp vision for its flight. It flies through écholocation or sending echo waves that are interpreted to know the obstacles. Bat, therefore, “sees through its ears.”
  3. Bat feeds on small flying insects. The insects are located through sound waves produced by them. Feeding on insects functions as biocontrol method on the population of night flying insects. It is rule of nature and keeps ecological balance.

Question 3.
On a rainy day, Raghav found small brownish worm like animals crawling slowly over the ground of his school. On close examination he found that the animal has faintly segmented body.

  1. What is the possible identity of the animal ?
  2. Why is it seen only in the rainy season ?
  3. What is its ecological importance ?

Answer:

  1. The identity of the crawling animal is earthworm.
  2. It lives in burrows inside the soil. In rainy season, the burrows get filled up with rain water. So the earthworms come out of them.
  3. Earthworm feeds on decaying fallen leaves and other organic remains. It pulverises the same. The worm castings are good source of soil nutrients. Earthworm is called farmer’s friend as it ploughs the soil by its burrowing habit and converts organic remains into manure.

Question 4.
While on a visit to hill station, Shaurya found extensive mat-like growth of very small erect green leafy plants over the wet rocks. The plants possess knobbed stalks over their tips.

  1. What are the plants seen by Shaurya ?
  2. What is the name of knobbed stalks.
  3. What is the reason of abundant growth of the plants.
  4. Write down about the impact of the plants on the rocks.

Answer:

  1. Shaurya saw moss plants growing on wet rocks.
  2. The knobbed stalks present over tips of moss plants are sporophytes. The knobbed structures are capsules that bear spores.
  3. Moss plant has a high reproductive potential through spores and fragmentation of filamentous protenema that develops from them.
  4. By their extensive growth moss plants trap soil particles, corrode rock surface and develop cracks in them. This builds up soil that results in growth of first herbaceous plants and then larger plants like shrubs and trees. The phenomenon is called ecological succession.

Hope given Value Based Questions in Science for Class 9 Chapter 7 Diversity in Living Organisms are helpful to complete your science homework.

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RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.1

RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.1

Other Exercises

Question 1.
Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and two parallels: [NCERT]
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.1 Q1.1
Solution:
(i) ΔPCD and trapezium ABCD are on the same base CD and between the same parallels AB and DC.
(ii) Parallelograms ABCD and APQD are on the same base AD and between the same parallels AD and BQ.
(iii) Parallelogram ABCD and ΔPQR are between the same parallels AD and BC but they are not on the same base.
(iv) ΔQRT and parallelogram PQRS are on the same base QR and between the same parallels QR and PS.
(v) Parallelogram PQRS and trapezium SMNR on tire same base SR but they are not between the same parallels.
(vi) Parallelograms PQRS, AQRD, BCQR are between the same parallels. Also, parallelograms PQRS, BPSC, and APSD are between the same parallels.

 

Hope given RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.1 are helpful to complete your math homework.

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