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Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 7 Triangles Ex 7.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3

Question 1.
∆ ABC and ∆ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see the given figure). If AD is extended to intersect BC at D show that,

(i) ∆ ABD ≅ ∆ ACD
(ii) ∆ ABP ≅ ∆ ACP
(iii) AP bisects ∠ A as well as ∠ D.
(iv) AP is the perpendicular bisector of BC.
Answer:
∆ ABC and ∆ DBC are isosceles triangles on the same base BC.
∴ In ∆ ABC, AB = AC and in ∆ DBC, DB = DC.
In ∆ ABD and ∆ ACD,
AB = AC
DB = DC
and AD = AD (Common)
∴ ∆ ABD s ∆ ACD (SSS rule) [Result (i)]
∴ ∠ BAD = ∠ CAD (CPCT)
In ∆ ABP and ∆ ACP
AB = AC
∠ BAP = ∠ CAP (∵ ∠ BAD = ∠ CAD)
and AP = AP (Common)
∴ ∆ ABP ≅ ∆ ACP (SAS rule) [Result (ii)]
∴ BP = CP (CPCT)
In ∆ DBP and ∆ DCR
DB = DC
BP = CP
and DP = DP (Common)
∴ ∆ DBP ≅ ∆ DCP (SSS rule)
From ∆ ABP ≅ ∆ ACR ∠ BAP = ∠ CAP (CPCT)
∴ AP bisects ∠A.
From ∆ DBP ≅ ∆ DCR ∠BDP = ∠ CDP (CPCT)
∴ DP bisects ∠D.
Thus, AP bisects ∠A as well as ∠D. [Result (iii)]
∆ A ABP ≅ ∆ ACP
∴ BP = CP and ∠ APB = ∠ APC (CPCT)
But, ∠ APB + ∠ APC = 180° (Linear pair)
∴ ∠ APB = ∠ APC = \(\) = 90°
Thus, BP = CP and AP ⊥ BC.
∴ AP is the perpendicular bisector of BC. [Result (iv)]

Question 2.
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that:
(i) AD bisects BC
(ii) AD bisects ∠ A

Answer:
AD is an altitude of A ABC.
∴ ∠ ADB = ∠ ADC = 90°
In ∆ ADB and ∆ ADC,
hypotenuse AB = hypotenuse AC (Given)
∠ ADB = ∠ ADC (Right angles)
AD = AD (Common)
∴ ∆ ADB ≅ ∆ ADC (RHS rule)
∴ BD = CD and ∠ BAD = ∠ CAD (CPCT)
Now, BD = CD means D is the midpoint of BC.
Hence, AD bisects BC. [Result (i)]
Moreover, ∠ BAD = ∠ CAD and
∠ BAD + ∠ CAD = ∠ BAC.
Hence, AD bisects ∠A. [Result (ii)]

Question 3.
Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ∆ PQR (see the given figure). Show that:
(i) ∆ ABM ≅ ∆ PQN
(ii) ∆ ABC ≅ ∆ PQR

Answer:
In ∆ ABC, AM is a median.
∴ BM = CM = \(\frac{1}{2}\) BC
In ∆ PQR, PN is a median.
∴ QN = RN = \(\frac{1}{2}\) QR
Now, BC = QR (Given)
∴ \(\frac{1}{2}\) BC = \(\frac{1}{2}\) QR
∴ BM = QN
In ∆ ABM and ∆ PQN,
AB = PQ (Given)
AM = PN (Given)
BM = QN (Proved)
∴ ∆ ABM ≅ ∆ PQN (SSS rule) [Result (i)]
∴ ∠ ABM = ∠ PQN (CPCT)
∴ ∠ ABC = ∠ PQR
Now, in ∆ ABC and ∆ PQR,
AB = PQ
∠ ABC = ∠ PQR
BC = QR .
∴ ∆ ABC ≅ ∆ PQR (SAS rule) [Result (ii)]

Question 4.
BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

Answer:
In ∆ FBC and ∆ ECB,
CF = BE (Given)
∠ CFB = ∠ BEC = 90° (Given)
BC = CB (Common)
∴ A FBC ≅ A ECB (RHS rule)
∴ ∠ FBC = ∠ ECB (CPCT)
∴ ∠ ABC = ∠ ACB
Now, in ∆ ABC, ∠ ABC = ∠ ACB
∴ AC = AB (Theorem 7.3)
Hence, ∆ ABC is an isosceles triangle.

Question 5.
ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠ B = ∠ C.
Answer:

In ∆ ABC, AP is an altitude.
∴ ∠ APB = ∠ APC = 90°
In ∆ APB and ∆ APC,
∠ APB = ∠ APC = 90°
AB = AC (Given)
AP = AP (Common)
∴ ∆ APB ≅ ∆ APC (RHS rule)
∴ ∠ ABP = ∠ AGP (CPCT)
∴ ∠ ABC = ∠ ACB
Thus, in ∆ ABC, ∠ B = ∠ C.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 7 Triangles Ex 7.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2

Question 1.
In an isosceles triangle ABC, with AB = AC, the bisectors of ∠ B and ∠ C intersect each other at O. Join A to O. Show that:
(i) OB = OC
(ii) AO bisects ∠ A
Answer:

In ∆ ABC, AB = AC
∴ ∠ ABC = ∠ ACB (Theorem 7.2)
∴ \(\frac{1}{2}\) ∠ ABC = \(\frac{1}{2}\) ∠ ACB
∴ ∠ OBC = ∠ OCB (BO bisects ∠ ABC and CO bisects ∠ ACB)
Now, in ∆ OBC, ∠ OBC = ∠ OCB
∴ OB = OC (Theorem 7.3)
Similarly, ∠ ABC = ∠ ACB gives
∴ \(\frac{1}{2}\) ∠ ABC = \(\frac{1}{2}\) ∠ ACB
∴ ∠ ABO = ∠ ACO
Now, in ∆ ABO and ∆ ACO,
AB = AC (Given)
∠ ABO = ∠ ACO
and OB = OC
∴ ∆ ABO ≅ ∆ ACO (SAS rule)
∴ ∠ BAO = ∠ CAO (CPCT)
But, ∠ BAO + ∠ CAO = ∠ BAC (Adjacent angles)
∴ ∠ BAO = ∠ CAO = \(\frac{1}{2}\) ∠ BAC
Thus, AO bisects ∠ A.

Question 2.
In ∆ ABC, AD is the perpendicular bisector of BC (see the given figure). Show that ∆ ABC is an isosceles triangle in which AB = AC.

Answer:
In ∆ ABC, AD is the perpendicular bisector of BC.
∴ BD = CD and ∠ ADB = ∠ ADC = 90°
In ∆ ADB and ∆ ADC,
AD = AD (Common)
∠ ADB = ∠ ADC (Right angles)
and BD = CD
∴ ∆ ADB ≅ ∆ ADC (SAS rule)
∴ AB = AC (CPCT)
Now, in ∆ ABC, AB = AC.
Hence, ∆ ABC is an isosceles triangle in which AB = AC.

Question 3.
ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see the given figure). Show that these altitudes are equal.

Answer:
In ∆ ABC, AC = AB
∴ ∠ ABC = ∠ ACB
∴ ∠ FBC = ∠ ECB
Now, in ∆ FBC and ∆ ECB,
∠ FBC = ∠ ECB
∠ BFC = ∠ CEB (Right angles)
BC = CB (Common)
∴ ∆ FBC ≅ ∆ ECB (AAS rule)
∴ CF = BE (CPCT)
Thus, the altitudes CF and BE on equal sides AB and AC respectively of ∆ ABC are equal.

Question 4.
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see the given figure). Show that
(i) ∆ ABE ≅ ∆ ACF
(ii) AB = AC, i.e., ABC is an isosceles triangle.

Answer:
In ∆ ABE and ∆ ACF,
∠ AEB = ∠ AFC (Right angles)
∠ A = ∠ A (Common)
BE = CF (Given)
∴ ∆ ABE ≅ ∆ ACF (AAS rule)
∴ AB = AC (CPCT)
Thus, ∆ ABC is an isosceles triangle.

Question 5.
ABC and DBC are two isosceles triangles on the same base BC (see the given figure). Show that ∠ ABD = ∠ ACD.

Answer:
∠ ABC and ∠ DBC are adjacent angles.
∴ ∠ ABC + ∠ DBC = ∠ ABD ………… (1)
∠ ACB and ∠ DCB are adjacent angles.
∴ ∠ ACB + ∠ DCB = ∠ ACD ………….. (2)
In ∆ ABC, AB = AC.
∴ ∠ ABC = ∠ ACB (Theorem 7.2)
In ∆ DBC, DB = DC.
∴ ∠ DBC = ∠ DCB (Theorem 7.2)
∴ ∠ ABC + ∠ DBC = ∠ ACB + ∠ DCB
∴ ∠ ABD = ∠ ACD [From (1) and (2))

Question 6.
∆ ABC is an isosceles triangle in which AB = AC. side BA is produced to D such that AD = AB (see the given figure). Show that ∠ BCD is a right angle.

Answer:
AB = AC and AD = AB
∴ AC = AD
In ∆ ABC, AB = AC
∴ ∠ ACB = ∠ ABC (Theorem 7.2) ……………… (1)
In A ADC, AC = AD
∴ ∠ ACD = ∠ ADC (Theorem 7.2) ……………… (2)
Adding (1) and (2),
∠ ACB + ∠ ACD = ∠ ABC + ∠ ADC
∴ ∠ BCD = ∠ DBC + ∠ BDC
(Adjacent angles and A lies on BD)
In ∆ BCD,
∠ DBC + ∠ BDC + ∠ BCD = 180°
∴ ∠ BCD + ∠ BCD = 180° (from (3)]
∴ 2 ∠ BCD = 180°
∴ ∠ BCD = 90°
Thus, ∠ BCD is a right angle.

Question 7.
ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠ B and ∠ C.
Answer:

In ∆ ABC, AB = AC
∴ Z C = Z B (Theorem 7.2)
In ∆ ABC,
∠ A + ∠ B + ∠ C = 180°
∴ 90° + ∠ B + ∠ B = 180° (Given and ∠ C = ∠ B)
∴ 2 ∠ B = 90°
∴ ∠ B = 45°
∴ ∠ C = 45°

Question 8.
Show that the angles of an equilateral triangle are 60° each.
Answer:

∆ ABC is an equilateral triangle.
∴ AB = BC = AC
In ∆ ABC, AB = BC
∴ ∠ C = ∠ A (Theorem 7.2)
In ∆ ABC, AB = AC
∴ ∠ C = ∠ B (Theorem 7.2)
Hence, ∠ A = ∠B = ∠ C.
Now, in ∆ ABC, ∠ A + ∠ B + Z C = 180°
∴ ∠ A = ∠ B = ∠ C = \(\frac{180^{\circ}}{3}\) = 60°
Thus, the angles of ah equilateral triangle are 60° each.

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4

1. Construct ΔABC, given AB = 6 cm, ∠A = 30° and ∠B = 75°.
Solution:
Given. One side of ΔABC as AB = 6 cm, m∠A = 30° and m∠B = 75°.
To construct: A triangle with one side and these two angles.
Steps of Construction :
Step 1. We first draw a rough sketch of ΔABC and indicate the measures of side and two angles.

Step 2. Draw a ray AB of length 6 cm.

Step 3. At A; draw a ray AX making an angle 30° with AB.

Step 4. With the help of compass. At B; draw a ray BY making an angle of 75° with AB.

Step 5. Both rays AX and BY intersect, at a point. So the point of intersection of the two rays is C Then. ΔABC is now obtained.

2. Construct an isosceles ΔABC such that AB = 5.3 cm and each base angle = 45°.
Solution:
Given : Isosceles ΔABC with AB = 5.3 cm each base angle = 45°.
To Construct: A triangle with one side and two base angles.
Steps of construction :
Step 1. Draw a rough sketch of ΔABC with given measures

Step 2. Draw a line segment AB = 5.3 cm.

Step 3. Taking A as centre with the help of compass. Draw a ray AX making an angle 45° with AB.

Step 4. With the help of compass and taking B as a centre. Draw a ray BY making an angle 45° with the line segment AB.

Step 5. Rays AX and BY intersect, at a point say C, then ABC is the required triangle.

3. Construct ΔXYZ if XY = 4 cm, ∠X = 45° and ∠Z = 60°.
[Hint : ∠Y = 180° – 45° – 60° = 75°]
Solution:
One side of ΔXYZ as
XY = 4 cm,
∠X = 45°
and ∠Z = 60°.
As we know that by angle-sum property of a triangle; sum of all three angles of a triangle is equal to 180°.
∴ ∠X + ∠Y + ∠Z = 180°
⇒ 45° + ∠Y + 60° = 180°
⇒ 105° + ∠Y = 180°
⇒ ∠Y = 75°.
Now it will be easy to construct triangle with side
XY = 4 cm,
∠X = 45°
and ∠Y = 75°.
Steps of Construction :
Step 1. We first draw a rough sketch of ΔXYZ and indicate the measure of side and two angles.

Step 2. Draw a ray XY of length 4 cm.

Step 3. At X draw a ray XA making an angle of 45° with XY.

Step 4. At Y; draw a ray YB making an angle of 75° with XY.

Step 5. Z has to lie on both rays XA and YB. So, the point of intersection of two rays is Z.
ΔXYZ is now obtained.

4. Examine whether you can construct ΔPQR such that ∠P = 100°, ∠Q = 90° and PQ = 4.3 cm If not possible given reason.
Solution:
No, we cannot construct given ΔPQR.
Reason :
As we know that by angle sum property of a triangle; sum of all three angles a triangle is equal to 180°. But in given question sum of two angles;
m∠P + m∠Q
= 100° + 90°
= 190°
The sum of these two angles should be less than 180°. So triangle with given measures cannot be constructed as it violates the angle sum property of a triangle.

5. Question (i).
In which of the following cases a unique triangle can be drawn ?
(a) BC = 5 cm, ∠B = 90° and ∠C = 100°
(b) AB = 4 cm, BC = 7 cm and CA = 2 cm
(c) XY = 5 cm, ∠X = 45°, ∠Y = 60°
(d) An isosceles triangle with length of each equal side equal to 5 cm.
Answer:
(c) XY = 5 cm, ∠X = 45°, ∠Y = 60°

Question (ii).
A triangle can be constructed by taking two of its angles as.
(a) 110°, 40°
(b) 70°, 115°
(c) 135°, 45°
(d) 90°, 90°
Answer:
(a) 110°, 40°

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 4 Integers Ex 4.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 4 Integers Ex 4.1

1. Write two examples from day-to-day life in which we can use positive and negative integers.
Solution:
1. If positive represents above sea level, then negative represents below sea level.
2. If positive represents a deposit, negative represents a withdrawal.

2. Write the opposite of the following:

Question (a)
A profit of ₹ 500
Solution:
A loss of ₹ 500

Question (b)
A withdrawal of ₹ 70 from bank account
Solution:
Deposit of ₹ 70 in bank account

Question (c)
A deposit of ₹ 1000
Solution:
Withdrawal of ₹ 1000

Question (d)
326 B.C
Solution:
326 AD

Question (e)
500 m below sea level
Solution:
500 m above sea level

Question (f)
25° above 0°C.
Solution:
25° below 0°C.

3. Represent the situations mentioned in integers.
Solution:
(a) + 500
(b) – 70
(c) + 1000
(d) – 326
(e) – 500 m
(f) + 25.

4. Represent the following situations in integers.

Question (a)
A deposit of ₹ 500.
Solution:
+ 500

Question (b)
An Aeroplane is flying at a height two thousand metre above the sea level.
Solution:
+ 2000

Question (c)
A withdrawal of ₹ 700 from Bank Account.
Solution:
– 700

Question (d)
A diver dives to a depth of 6 feet below ground level.
Solution:
– 6.

5. Represent the following numbers on number line.

Question (i)
(a) – 5
(b) + 6
(c) o
(d) + 1
(e) – 9
(f) – 4
(g) + 8
(h) + 3.
Solution:

6. Integers are represented on a horizontal number line as shown where A represents – 2. With reference to the number line, answer the following questions:

(a) Which point represent – 3?
(b) Locate the point which represents the opposite of B and name it P.
(c) Write integers for the points C and E.
(d) Which point marked on the number line has the least value?
Solution:

(a) Point B represents – 3.
(b) Point P represents + 3.
(c) Point C represents -7 and Point E represents + 4.
(d) Point C has the least value – 7.

7. In each of the following pairs, which number is to the right of other on the number line?

Question (i)
(a) 2 9
(b) -3, -8
(c) 0, -5
(d) -11, 10
(e) -9, 9
(f) 2, – 200.
Solution:
(a) 9
(b) – 3
(c) 0
(d) 10
(e) 9
(f) 2

8. Write all the integers between the given pairs (write them in increasing order)

Question (a)
0 and -6
Solution:
-5, -4, -3, -2, -1

Question (b)
-6 and +6
Solution:
-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5

Question (c)
-9 and -17
Solution:
-16, -15, -14, -13, -12, -11, -10

Question (d)
-19 and -5.
Solution:
-18, -17, -16, -15, -14, -13, -12, -11, -10, -9, -8, -7, -6.

9.

Question (a)
Write five negative integers greater than ‘-15’.
Solution:
Five negative integers greater than ‘-15’ are:
-14, -13, -12, -11, -10

Question (b)
Write five integers smaller than ‘-20’.
Solution:
Five integers smaller than ‘-20’ are:
-21, -22, -23, -24, -25

Question (c)
Write five integers greater than 0.
Solution:
Five integers greater than 0 are:
1,2, 3, 4, 5

Question (d)
Write five integers smaller than 0.
Solution:
Five integers smaller than 0 are:
-1, -2, -3, -4, -5.

10. Encircle the greater integer in each given pair.

(a) -5, -7
(b) 0,-3
(e) 5, 7
(d) -9, 0
(e) -9, -11
(f) -4, 4
(g) -10, -100
(h) 10, 100.
Solution:
(a) -5
(b) 0
(c) 7
(d) 0
(e) -9
(f) 4
(g) -10
(h) 100.

11. Arrange the following integers in ascending order:

Question (a)
0, -7, -9, 5, -3, 2, -4
Solution:
Ascending order of given integers is:
-9, -7, -4, -3, 0, 2, 5

Question (b)
8, -3, 7, 0, -9, -6.
Solution:
Ascending order of given integers is:
-9, -6, -3, 0, 7, 8.

12. Arrange the following integers in descending order:

Question (a)
-9, 3, 4, -6, 8, -3
Solution:
8, 4, 3, -3 -6, -9

Question (b)
4, 8,-3,-2, 5, 0.
Solution:
8, 5, 4, 0, -2, -3.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 7 Triangles Ex 7.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1

Question 1.
In quadrilateral ACBD. AC = AD and AB bisects ∠ A (see the given figure). Show that ∆ ABC ≅ ∆ ABD. What can you say about BC and BD?

Answer:
In ∆ ABC and ∆ ABD,
AC = AD (Given)
∠ BAC = ∠ BAD (AB bisects ∠ A)
AB = AB (Common)
∴ ∆ ABC ≅ ∆ ABD (SAS rule)
∴ BC = BD (CPCT)
Thus, BC and BD are equal.

Question 2.
ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see the given figure). Prove that (i) ∆ ABD ≅ ∆ BAC, (ii) BD = AC and (iii) ∠ ABD = ∠ BAC

Answer:
In ∆ ABD and ∆ BAC,
AD = BC (Given)
∠ DAB = ∠ CBA (Given)
AB = BA (Common)
∴ ∆ ABD ≅ ∆ BAC (SAS rule)
∴ BD = AC (CPCT)
∴ ∠ ABD = ∠ BAC (CPCT)

Question 3.
AD and BC are equal perpendiculars to a line segment AB (see the given figure). Show that CD bisects AB.

Answer:
AD and BC are equal perpendiculars to line segment AB.
∴ AD = BC and ∠ OAD = ∠ OBC = 90°.
Now, in ∆ ADO and ∆ BCO,
AD = BC
∠ OAD = ∠ OBC
∠ AOD = ∠ BOC (Vertically opposite angles)
∴ ∆ ADO ≅ ∆ BCO (AAS rule)
∴ OA = OB (CPCT)
CD intersects AB at O and OA = OB.
Hence, CD bisects AB.

Question 4.
l and m are two parallel lines intersected by another pair of parallel lines p and q (see the given figure). Show that:
∆ ABC ≅ ∆ CDA.

Answer:
l || m and AC is their transversal.
∴ ∠ BCA = ∠ DAC (Alternate angles)
p l| q and AC is their transversal.
∴ ∠ BAC = ∠ DCA (Alternate angles)
Now, in ∆ ABC and ∆ CDA,
∠ BCA = ∠ DAC
∠ BAC = ∠ DCA
AC = CA (Common)
∴ ∆ ABC ≅ ∆ CDA (ASA rule)

Question 5.
Ray l is the bisector of an angle ∠ A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠ A (see the given figure). Show that:
(i) ∆ APB ≅ ∆ AQB
(ii) BP = BQ or B is equidistant from the arms of ∠ A.

Answer:
l is the bisector of ∠ PAQ and B is any point on l.
∴ ∠ PAB = ∠ QAB
BP and BQ are perpendiculars from B to AP and AQ.
∴ ∠ BPA = ∠ BQA = 90°.
Now, in ∆ APB and ∆ AQB,
∠ PAB = ∠ QAB
∠ BPA = ∠ BQA
AB = AB (Common)
∴ ∆ APB ≅ ∆ AQB (AAS rule)
∴ BP = BQ (CPCT)
BP and BQ are perpendiculars from B to arms AP and AQ of ∠ A.
∴ BP is the distance of B from AP and BQ is the distance of B from AQ.
Thus, B is equidistant from the arms of ∠ A.

Question 6.
In the given figure, AC = AE, AB = AD and ∠ BAD = ∠ EAC. Show that BC = DE

Answer:
∠ BAD = ∠ EAC
∴ ∠ BAD + ∠ DAC = ∠ EAC + ∠ DAC
∴ ∠ BAC = ∠ DAE (Adjacent angles)
Now, in ∆ BAC and ∆ DAE,
AC = AE (Given)
AB = AD (Given)
∠ BAC = ∠ DAE
∴ ∆ BAC ≅ ∆ DAE (SAS rule)
∴ BC = DE (CPCT)

Question 7.
AB is a line segment and P is its midpoint. D and E are points on the same side of AB such that ∠ BAD = ∠ ABE and ∠ EPA = ∠ DPB (see the given figure). Show that:
(i) ∆ DAP ≅ ∆ EBP
(ii) AD = BE

Answer:
∠ BAD = ∠ ABE
∴ ∠ PAD = ∠PBE (∵ P lies on AB.)
∠ EPA = ∠ DPB
∴ ∠ EPA + ∠ EPD = ∠ DPB + ∠ EPD
∴ ∠ APD = ∠ BPE (Adjacent angles)
P is the midpoint of AB.
∴ AP = BP
Now, in ∆ DAP and ∆ EBP
∠ PAD = ∠ PBE
∠ APD = ∠ BPE
AP = BP
∴ ∆ DAP ≅ ∆ EBP (ASA rule)
∴ AD = BE (CPCT)

Question 8.
In right triangle ABC, right angled at C, M is the midpoint of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see the given figure). Show that:
(i) ∆ AMC ≅ ∆ BMD
(ii) ∠ DBC is a right angle
(iii) ∆ DBC ≅ ∆ ACB
(iv) CM = \(\frac{1}{2}\) AB

Answer:
In ∆ AMC and ∆ BMD,
AM = BM (∵ M is the midpoint of AB.)
CM = DM (Given)
∠ AMC = ∠ BMD (Vertically opposite angles)
∴ By SAS rule, ∆ AMC ≅ ∆ BMD [Result (i)]
∴ ∠ MCA = ∠ MDB (CPCT)
∠ MCA and ∠ MDB are alternate angles formed by transversal CD of lines AC and BD and they are equal.
∴ AC || BD
Now, ∠ DBC and ∠ ACB are interior angles on the same side of transversal BC of AC || BD.
∴ ∠ DBC + ∠ ACB = 180°
∴ ∠ DBC + 90° = 180° (Given : ∠ C = 90°)
∴ ∠ DBC = 90°
Thus, ∠ DBC is a right angle. [Result (ii)]
Now, ∆ AMC ≅ ∆ BMD
∴ AC = BD
In ∆ DBC and ∆ ACB,
BD = CA
∠ DBC = ∠ ACB (Right angles)
BC = CB (Common)
∴ ∆ DBC ≅ ∆ ACB [Result (iii)]
∴ DC = AB (CPCT)
DM = CM and M lies on line’ segment CD.
∴ DC = 2 CM
∴ AB = 2CM
∴ \(\frac{1}{2}\)AB = CM
∴ CM = \(\frac{1}{2}\)AB

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3

1. Construct ΔABC such that AB = 4 cm, ∠B = 30°, BC = 4 cm. Also name the type of triangle on the basis of sides.
Solution:
Given : Two sides of ΔABC as AB = 4 cm, BC = 4 cm and ∠B = 30°.
To construct: A triangle with these two sides and included angle.
Step of Construction :
Step 1. We first draw a rough sketch of the ΔABC and indicate the measure of these two sides and included angle.

Step 2. Draw a line segment BC of length 4 cm.

Step 3. At B draw BX making an angle of 30° with BC (The point A must be somewhere on this ray of the angle).

Step 4. (To fix A, the distance AB has been given) With B as centre, draw an arc of radius 3 cm. It cuts BX at the point A.

Step 5. Join AC.
ΔDEF is now obtained.

Since two sides of triangle are equal.
Therefore ΔABC is an isosceles triangle.

2. Construct ΔABC with AB = 7.5 cm, BC = 5 cm and ∠B = 30°.
Solution:
Given. Two sides of ΔABC as AB = 7.5 cm,
BC = 5 cm
and ∠B = 30°
To construct A triangle with these two sides and included angle.
Steps of Construction.
Step 1. We first draw a rough sketch of the ΔABC and indicate the measures of these two sides and included angle.

Step 2. Draw a line segment BC of length 5 cm.

Step 3. At B draw BX making an angle of 30° with BC. (The point A must be somewhere on this ray of the angle)

Step 4. (To fix A; the distance BC has been given) With B as centre draw an arc of radius 7.5 cm. It cuts CX at the point A.

Step 5 : Join AC.
ΔABC is now obtained.

3. Construct a triangle XYZ, such that XY = 6 cm, YZ = 6 cm and ∠Y = 60°. Also name the type of this triangle.
Solution:
Step 1. Draw a rough sketch of XYZ with given measures.

Step 2. Draw a line segment XY of length 6 cm.

Step 3. With the help of compass, at Y, draw a ray YA making an angle 60°

Step 4. With Y as centre and radius 6 cm. draw an arc intersecting the ray YX at point Z.

Step 5. Join XZ.ΔXYZ is required triangle, Measure the third side. We see that ZX = 6 cm
∴ In Δ XYZ
XY = YZ = ZX = 6 cm
Therefore XYZ is an equilateral triangle.

4. Which of the following triangle can be constructed using SAS criterion.
(a) AB = 5 cm, BC = 5 cm, CA = 6 cm
(b) AB = 5 cm, BC = -5 cm, ∠B = 40°
(c) ∠A = 60°, ∠B = 60°, ∠C = 60°
(d) BC = 5 cm, ∠B = ∠C = 45°
Answer:
(b) AB = 5 cm, BC = -5 cm, ∠B = 40°

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.2

1. Construct ΔABC in which AB = 3.5 cm, BC = 5 cm and CA = 7 cm.
Solution:
Given : Three sides of a triangle as AB = 3.5 cm, BC = 5 cm and CA = 7 cm.
To construct : A triangle with these three sides.
Step 1. We first draw a rough sketch of the triangle ABC and indicate the lengths of the three sides.

Step 2. Draw a line segment BC = 5 cm.

Step 3 : From B; point A is at a distance 3.5 cm. So, with B as centre, draw an arc of radius
3.5 cm (Now point A will be some where on this arc. Our job is to find where exactly A is.)

Step 4. From C, point A is at a distance of 7 cm. So, with C as centre; draw an arc of radius 7 cm. (A will be some where on this arc. We have to fix it).

Step 5. A has to be on both the arcs drawn. So it is the point of intersection of arcs.
Mark the point of intersection of arcs as A. Join AB and AC.
Thus we obtain ΔABC.

2. Construct a triangle ABC in which AB = BC = 6.5 cm and CA = 4 cm. Also name the kind of triangle drawn.
Solution:
Given : Three sides of triangle as AB = BC = 6.5 cm. and CA = 4 cm.
To construct : A triangle with these three sides.
Steps of Construction :
Step 1. We first draw a rough sketch of the triangle ABC and indicate the lengths of the three sides.

Step 2. Draw a line segment AC = 4 cm.

Step 3. From A; point B is at a distance of 6.5 cm. So, with A as centre, draw an arc of radius 6.5 cm. (Now point B will be somewhere on this arc. Our job is to find where exactly A is.)

Step 4. From C; point B is at a distance of 6.5 cm. So; with C as centre; draw an arc of radius 6.5 cm. (B will be some where on this arc. We have to fix it.)

Step 5. B has to be on both the arcs drawn. So it is the point of intersection of arcs. Mark the point of intersection of arcs as B. Join AB and BC.

We observe that AB = BC = 6.5 cm.
Since two sides are of equal length. Thus we obtain an isosceles ΔABC.

3. Construct a triangle XYZ such that length of each side is 5 cm. Also name the kind of triangle drawn.
Solution:
Given : A triangle XYZ in which XY = YZ = ZX = 5 cm..
To Construct. A triangle XYZ with each side 5 cm.
Steps of Construction :
Step 1. We first draw a rough sketch of the triangle XYZ and indicate the lengths of the three sides.

Step 2. Draw a line segment YZ = 5 cm.

Step 3. From Y; point X is at a distance of 5 cm. So, with Y as centre, draw an arc of radius 5 cm. (Now point X will be somewhere on this arc. Our job is to find where exactly X is.)

Step 4. From Z, point X is at a distance of 5 cm. So, with Z as centre, draw an arc of radius 5 cm. (X will be somewhere oh this arc. We have to fix it.)

Step 5. Point X has to be on both the arcs drawn. So, it is the point of intersection of arcs.
Mark the point of intersection of arcs as X. Join XY and XZ.

Thus we obtain an equilateral ΔXYZ each of whose side is 5 cm.

4. Construct a triangle PQR such that PQ = 2.5 cm, QR = 6 cm and RP = 6.5 cm. Measure ∠PQR and also name the kind of triangle drawn.
Solution:
Given. Three sides of triangle as PQ = 2.5 cm, QR = 6 cm and RP = 6.5 cm.
To construct. A triangle with these three sides.
Steps of Construction :
Step 1. We first draw a rough sketch of the triangle PQR and indicate the lengths of the three sides.

Step 2. Draw a line segment QR of length 6 cm.

Step 3. From Q; point P is at a distance of 2.5 cm. So, with Q as centre, draw an arc of radius 2.5 cm. (Now point P will be some where on this arc. Our job is to find where exactly P is.)

Step 4. From R; point P is at a distance of 6.5 cm. So; with R as centre; draw an arc of radius 6.5 cm. (P will be somewhere on this arc. We have to fix it.)

Step 5. Point P has to be on both the arcs drawn. So, it is the point of intersection of arcs. Mark a point of intersection of arcs as P. Join PQ and PR.

Thus we obtain ΔPQR; on measuring ∠PQR, we observe that ∠PQR = 90° so it is a right angled triangle.

5. Construct a triangle ABC, in which AB = 6 cm, BC = 2 cm, CA = 3 cm. (If possible). If not possible give the reason.
Solution:
Since AB = 6 cm, BC = 2 cm, CA = 3 cm
Here BC + CA = 2 cm + 3 cm
= 5 cm < 6 cm < AB
Which is not possible because the sum of two sides of a triangle is always greater than third side of the triangle.

6. Question (i).
Which of the following can be used to construct a triangle ?
(a) The lengths of the three sides
(b) The perimeter of the triangle
(c) The measures of three angles
(d) The name of three vertices
Answer:
(a) The lengths of the three sides

Question (ii).
A triangle can be constructed by taking its sides as :
(a) 1.8 cm, 2.6 cm, 4.4 cm
(b) 3 cm, 4 cm, 8 cm
(c) 4 cm, 7 cm, 2 cm
(d) 5 cm, 4 cm, 4 cm.
Answer:
(d) 5 cm, 4 cm, 4 cm.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 3 Playing with Numbers Ex 3.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.5

1. Find LCM of following numbers by prime factorization method:

Question (i)
45, 60
Solution:

∴ 45 = 3 × 3 × 5
60 = 2 × 2 × 3 × 5
We find that in these prime factorizations, 2 occurs maximum two times, 3 occurs maximum two times and 5 occurs maximum once
∴ L.C.M. of 45 and 60
= 2 × 2 × 3 × 3 × 5 = 180

Question (ii)
52, 56
Solution:

We find that in these prime fatorisation, 2 occurs maximum 3 times, 13 and 7 occurs maximum once.
∴ L.C.M. of 52 and 56
= 2 × 2 × 2 × 13 × 7 = 728

Question (iii)
96, 360
Solution:

∴ 96 = 2 × 2 × 2 × 2 × 2 × 3
360 = 2 × 2 × 2 × 3 × 3 × 5
We find that in these prime factorisation, 2 occurs maximum 5 times, 3 occurs maximum 2 times and 5 occurs maximum once.
∴ L.C.M. of 96 and 360
= 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 = 1440

Question (iv)
36, 96, 180
Solution:

∴ 36 = 2 × 2 × 3 × 3
96 = 2 × 2 × 2 × 2 × 2 × 3
and 180 = 2 × 2 × 3 × 3 × 5
We find that in these factorisation, 2 occurs maximum 5 times, 3 occurs maximum 2 times and 5 occurs maximum once.
∴ L.C.M. of 36, 96 and 182
= 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 = 1440

Question (v)
18, 42, 72.
Solution:

∴ 18 = 2 × 3 × 3
42 = 2 × 3 × 7
72 = 2 × 2 × 2 × 3 × 3
We find that in these factorization 2 occurs maximum 3 times, 3 occurs maximum 2 times and 7 occurs maximum once.
∴ L.C.M. of 18, 42 and 72
= 2 × 2 × 2 × 3 × 3 × 7 = 504

2. Find LCM of the following by common division method:

Question (i)
24, 64
Solution:

∴ L.C.M. of 24, 64
= 2 × 2 × 2 × 3 × 8 = 192

Question (ii)
42, 63
Solution:

∴ L.C.M. of 42 and 63
= 3 × 7 × 2 × 3 = 126

Question (iii)
108, 135, 162
Solution:

∴ L.C.M. of 108, 135 and 162
= 2 × 3 × 3 × 3 × 2 × 5 × 3 = 1620

Question (iv)
16, 18, 48
Solution:

∴ L.C.M. of 16, 18 and 48
= 2 × 2 × 2 × 2 × 3 × 3 = 144

Question (v)
48, 72, 108
Solution:

∴ L.C.M. of 48, 72 and 108
= 2 × 2 × 2 × 3 × 3 × 2 × 3 = 144

3. Find the smallest number which is divisible by 6, 8 and 10.
Solution:
We know that the smallest number divisible by 6, 8 and 10 is their L.C.M.
So, we calculate L.C.M. 6, 8 and 10

∴ L.C.M. = 2 × 3 × 4 × 5 = 120
Hence, required number =120

4. Find the least number when divided by 10,12 and 15 leaves remainder 7 in each case.
Solution:
We know that the least number divisible by 10, 12 and 15 is their L.C.M.

So, the required number will be 7 more than their L.C.M.
We calculate their L.C.M.
L.C.M of 10, 12 and 15 = 2 × 3 × 5 × 2 = 60
Hence, Required number = 60 + 7 = 67

5. Find the greatest 4-digit number exactly divisible by 12, 18 and 30.
Solution:
First find the L.C.M. of 12, 18, 30

∴ L.C.M. of 12, 18, 30
= 2 × 3 × 2 × 3 × 5 = 180
Now the greatest 4 digit number = 9999
We find that when 9999 is divided by 180, the remainder is 99.
Hence, the greatest number of 4 digits which is exactly divisible by 12, 18, 30
= 9999 – 99 = 9900

6. Find the sandiest 4-digit number exactly divisible by 15, 24 and 36.
Solution:
First find L.C.M. of 15, 24, 36

L.C.M. of 15, 24, 36
= 2 × 2 × 3 × 5 × 2 × 3 = 360 Now, 4 digit smallest number is 1000 We find that when 1000 is divided by 360, the remainder is 280.
∴ Smallest 4 digits number, which is exactly divisible by 15, 24 and 36
= 1000 + (360 – 280) = 1000 + 80 = 1080.
Hence, required number = 1080

7. Four bells toll at intervals of 4, 7, 12 and 14 seconds. The bells toll together at 5 a.m. When will they again toll together?
Solution:
The bells will toll together at a time which is multiple of four intervals 4, 7, 12 and 14 seconds
So, first we find L.C.M. of 4, 7, 12 and 14

∴ L.C.M. = 2 × 2 × 7 × 3 = 84
Thus the bells will toll together after 84 seconds or 1 minute 24 seconds.
First they toll together at 5 a.m., then they will toll together after 1 minutes 24 seconds i.e. 5 : 01 : 24 a.m.

8. Three boys step off together from the same spot their steps measures 56 cm, 70 cm and 63 cm respectively. At what distance from the starting point will they again step together?
Solution:
The distance covered by each one of them has to be same as well as minimum walk So, the minimum distance each should their steps will be L.C.M. of the distances L.C.M. of the measure of their steps.

∴ L.C.M. = 2 × 7 × 4 × 5 × 9 = 2520cm
Hence, the will again step to gether after a distance of 2520 cm.

9. Can two numbers have 15 as their HCF and 65 as their LCM. Give reasons in support of your answer.
Solution:
We know that H.C.F. of given numbers is a factor of their L.C.M.
But 15 is not a factor of 65
So, there can not be two numbers with H.C.F. 15 and L.C.M. 65.

10. Can two numbers have 12 as their HCF and 72 as their LCM. Give reasons in support of your answer.
Solution:
We know that H.C.F. of given numbers is a factor of their L.C.M.
Here, 12 divides 72 exactly. So 12 is a factor of 72
Hence, there can be two numbers with H.C.F. 12 and L.C.M 72.

11. The HCF and LCM of two numbers are 13 and 182 respectively. If one of the numbers is 26. Find other numbers.
Solution:
H.C.F. = 13 and L.C.M. = 182,
1st number = 25
Now, 1st number × 2nd number = H.C.F. × L.C.M.
26 × 2nd number = 13 × 182
∴ 2nd number = \(\frac {13×182}{26}\)
= 91

12. The LCM of two co-prime numbers is 195. If one number is 15 then find the other number.
Solution:
L.C.M. of two co-prime numbers = 195
H.C.F. of two co-prime numbers = 1
One number = 15
1st number × 2nd number = H.C.F. × L.C.M.
15 × 2nd number= 1 × 195
∴ 2nd number = \(\frac {1×195}{15}\)
= 13

13. The H.C.F. of two numbers is 6 and product of two numbers is 216. Find their L.C.M.
Solution:
H.C.F. of two numbers = 6
Product of two numbers = 216
We know that
H.C.F. × L.C.M. = Product of two numbers
∴ 6 × L.C.M. = 216
∴ L.C.M. = \(\frac {216}{6}\) = 36

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 6 Triangles Ex 6.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 6 Triangles Ex 6.1

1. In ΔABC, P is midpoint of BC, then
(i) BP = ……………..
(ii) AP is a …………….. of ΔABC
(iii) ∠ADC = ……………..
(iv) BD = BC (True/False)
(v) AD is an …………….. of ΔABC

Solution:
(i) PC
(ii) Median
(iii) 90°
(iv) False
(v) Altitude

2. (a) Draw AD, BE, CF three medians in a ΔABC.
(b) Draw an equilateral triangle and its medians. Also compare the lengths of the medians.
(c) Draw an isosceles triangle ABC in which AB = BC. Also draw its altitudes.
Solutions:
(a) We are given ΔABC D, E and F are mid points of the sides BC, CA and AB respectively. Join AD, BE and CF.
The AD, BE and CF are the required medians.

(b) Draw an equilateral triangle ABC, D,E and F are the mid points of sides BC, CA and AB respectively. On joining AD, BE and CF, we get the required medians AD, BE and CF. Measure the lengths of AD, BE and CF we observe that the three medians AD, BE and CF are equal in length.
∴ AD = BE = CF.

(c) Draw an isosceles ΔABC in which AB = BC Altitude can be drawn as below :
AD is the altitude from A to D.


3. Find the value of the unknown exterior angles.

Question (i).

Answer:
In the given triangle,
By exterior angle property of a triangle
Exterior angle = sum of interior opposed angles
x = 100° + 40°
∴ x = 140°

Question (ii).

Answer:
By exterior angle property of a triangle
Exterior angle = Sum of interior opp. angles
x = 20° + 30°
∴ x = 50°

Question (iii).

Answer:
By exterior angle property of a triangle
Exterior angle = Sum of interior opp. angles
x = 60° + 60°
∴ x = 120°

Question (iv).

Answer:
By exterior angle property of a triangle
Exterior angle = Sum of interior opp. angles
x = 90° + 30°
∴ x = 120°

4. Find the value of x in the following figures.

Question (i).

Answer:
By exterior angle property of a triangle
Sum of interior opp. angles = Exterior angle
40° + x = 120°
x = 120° – 40°
x = 80°.

Question (ii).

Answer:
By exterior angle property of a triangle
Sum of interior opp. angles = Exterior angle
x + 90° = 135°
x = 135° – 90°
x = 45°

Question (iii).

Answer:
By exterior angle property of a triangle
Sum of interior opp. angles = Exterior angle
x + 80° = 130°
x = 130° – 80°
x = 50°

Question (iv).

Answer:
By exterior angle property of a triangle
Sum of interior opp. angles = Exterior angle
x + 25° = 155°
x = 155° – 25°
x = 130°

5. Find the value of y in following figures.

Question (i).

Answer:
By exterior angle property of a triangle
Sum of interior opp. angles = Exterior angle
y + y = 140°
2y = 140°
y = \(\frac{140^{\circ}}{2}\)
y = 70°

Question (ii).

Answer:
By exterior angle property of a triangle
Sum of interior opp. angles = Exterior angle
y + 90° = 160°
y = 160° – 90°
y = 70°

Question (iii).

Answer:
By exterior angle property of a triangle
exterior angle = Sum of interior opp. angles
5y = y + 80°
5y – y = 80°
4y = 80°
y = \(\frac{80^{\circ}}{4}\)
y = 20°.

Punjab State Board PSEB 9th Class Science Book Solutions Chapter 12 Sound Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Science Chapter 12 Sound

PSEB 9th Class Science Guide Sound Textbook Questions and Answers

Question 1.
What is sound and how is it produced?
Answer:
Sound: Sound is a form of energy which produces in our ears the sensation of hearing. It is produced due to vibration of a body about its mean position.

How to produce sound? We can produce sound in different bodies by plucking, by rubbing, by blowing or by giving jolt. In other words, by producing vibration in bodies sound can be produced. By vibration we mean moving a body rapidly to and fro about its mean position.

Question 2.
Describe with the help of diagram, how compressions and rarefactions are produced in air near a source of sound?
Answer:
Sound in air gets propagated in the form of longitudinal wave motion consisting of regions of compressions and rarefactions. Consider, for example, a tuning, fork in a state of vibrations. [Fig.(a)] As prong moves towards right, it compresses the layer of air in contact with it. As air has elasticity, the compressed air tends to relieve itself of its strain and moves forward to right to compress the next layer and so on.

Thus, a wave of compression moves towards the right. At the point of compression, there is an increase of pressure and is shown in form of crest C. At the point of rarefaction of concentration of particles is least and has been shown as trough R.

When the prong moves towards left, a region of reduced pressure or rarefaction is produced towards right [Fig. (b)].

Thus, a wave of rarefaction starts moving towards right. This way a series of compressions arid rarefactions move in forward direction.

Question 3.
Cite an experiment to show that sound needs a material medium for its propagation.
Or
Describe an experiment to show that sound needs a material medium for its propagation.
Or
Describe an activity to show that sound is a mechanical wave and needs a material medium for its propagation.
Answer:
Sound needs material medium for propagation: Sound is a mechanical wave which needs a material medium to travel, (propagate) It can travel through air, water, steel, etc but cannot travel through vacuum. This can be demonstrated by the following experiment.

Experiment: Take an electric bell and a glass bell-jar. Suspend the electric bell in a bell jar with the help of a cork fitted in the mouth of the jar. Connect the bell jar to a vacuum pump as shown in Fig. Press the switch of electric bell when sound is heard.

Now work the exhaust pump and remove air from the jar slowly. As air is removed the sound becomes fainter and fainter. After sometime when most of the air is removed, a feeble sound will be heard. If the whole of the air from the jar is removed no sound of electric bell will be heard. This proves that material medium is needed for the propagation of sound.

Question 4.
Why is sound wave called longitudinal wave?
Answer:
Sound waves when travel through a medium, the particles of the medium move to and fro in the same direction in which the disturbance (wave) travels. That is why, the sound waves are called longitudinal waves.

Question 5.
Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room.
Answer:
On the basis of quality or timbre of sound, we can identify our friend’s voice.

Question 6.
Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen. Why?
Answer:
The speed of sound (344 m s^-1) is much smaller than the speed of light (3 × 10⁸ m s^-1). So thunder is heard a few seconds after the flash is seen although these are produced at the same time.

Question 7.
A person has a hearing range from 20 Hz to 20 KHz. What is the typical wavelength of sound waves in air corresponding to these frequencies? Take the speed of sound in air as 344 ms^-1.
Solution:
Given speed of sound (υ) = 344 m s^-1
Lower limit of frequency (ν₁) = 20 Hz
Upper limit of audible frequency (ν₂) = 20 KHz
= 20 × 1000 Hz.
= 20000 Hz

Question 8.
Two children are at opposite er Is of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in air and in aluminium to reach the second child.
Solution:

Question 9.
The frequency of source of sound is 100 Hz. How many times does it vibrate in a minute?
Solution:
Frequency of source sound = 100 Hz
i.e, Number of vibrations produced in 1 second = 100
Number of vibrations produced in 1 min = 60 s = 100 × 60 = 6000

Question 10.
Does sound follow the same laws of reflection as light does? Explain.
Answer:

Yes, sound follows the same laws of reflection as light does. Like light sound is reflected from solid or liquid surface.
These laws are:

• First Law: The directions of incident sound and reflected sound make equal angles with the normal to the surface at the point of incidence.
  i.e. \(\angle i=\angle r\)
• Second Law: The incident sound wave, the reflected sound wave and normal to the reflecting surface at the point of incidence all lie in the same plane.

Question 11.
When a sound is reflected from a distant object, an echo is produced. Let the distance between the reflecting surface and the source of sound production remains the same. Do you hear echo sound on a hotter day?
Answer:
On a hotter day the speed of sound increases with the increase of temperature. So on that day reflected sound returns tq source earlier than 0.1 s. Hence a clear echo sound can not be heard.

Question 12.
Give two practical applications of reflection of sound waves.
Answer:

Stethoscope is doctor’s device which is used to hear the sound produced inside heart or lungs. The speed of sound of a patient’s heart beat is guided along the tube to the doctor’s ears by multiple reflection of sound.

2. The front part of musical instruments like megaphone or loudspeaker, horn, shehnai is made open and conical so that the sound waves produced may be reflected repeatedly and may be reflected repeatedly and may be sent forward towards the listeners.

Question 13.
A stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower. When is the splash heard at the top? Given, g = 10 m s^-2 and speed of sound = 340 m s^-2.
Solution:
Here, initial velocity of sound (u) = 0
Height of the lower (i.e. distance covered) (S) = 500 m
Acceleration due to gravity (g) = 10 ms^-2

Question 14.
A sound wave travels at a speed of 339 m s^-1. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?
Solution:
Speed of sound (υ) = 339 ms^-1
Wavelength of sound (λ) = 1.5 cm
= \(\frac {1.5}{100}\)
= 0.015 m
Frequency of wave (ν) =?
We know, frequency (ν) = \(\frac{υ}{\lambda}\)
= \(\frac {339}{0.015}\)
= 22600 Hz
Yes, sound waves are inaudible because these have frequency 22600 Hz which is not within the audible range 20 Hz to 20,000 Hz.

Question 15.
What is reverberation? How can it be reduced?
Answer:
Reverberation: The persistance of sound due to repeated reflection of sound is called reverberation. If sound after its production is allowed to suffer repeated reflection from walls and ceiling of big halls of concert so that it persists is called reverberation. It is unwanted sound because of which sound is not distinctly heard. To reduce reverberation effect of sound, walls and ceiling should be covered with sound absorbing materials like compressed fiber or heavy curtains having folds etc.

Question 16.
What is loudness of sound? What factors does it depend on?
Answer:
Loudness of sound is the measure of sensitivity of human ears. Like intensity it is not sound energy passing through unit area in 1 second. Two sounds can have same frequencies but still these may be heard having different loudness.

Question 17.
Explain how bats use ultrasounds to catch a prey?
Answer:

In darkness bats while flying in search of their prey emit ultrasound waves and then detect these waves after reflection. Very high-frequency ultrasonic squeaks of bat are reflected from prey and returned to bat’s ear. Amount and time delay of reflected wave helps bat in estimating the position and distance of prey.

Question 18.
How is ultrasound used for cleaning?
Answer:
Ultrasounds are used to clean parts located in hard-to-reach places e.g., complicated electronic components, watches, spiral or odd shaped parts. Appliances to be cleaned are placed in cleaning solutions and ultrasonic waves are sent through cleaning solution. Due to high frequency of ultrasounds, the dust, oil, grease and dirt get detached.

Question 19.
Explain the working and applications of SONAR.
Or
Write the full name of SONAR. How will you determine the depth of a sea using echo ranging?
Or
Write full form of SONAR. List any two purposes for which, it is used and explain its working for any one such purpose.
Answer:
SONAR: The acronym Sonar stands for Sound Navigation and Ranging. Sonar is a device that uses ultrasonic waves to measure the distance, direction and speed of underwater objects.
Principle: It uses the phenomenon of echoes in determining the sea-depth and locating the presence of underwater objects.

Working: Sonar consists of a transmitter T, and a detector D, installed below a ship as shown in Fig. The wave produced by transmitter travel through water and are reflected by sea-bed of obstacle. Reflected waves are sensed by the detector. Detector converts ultrasonic waves into electric signal. These signals are interpreted by detector.

If time interval between transmission and reception is t and speed of sound in sea-water is v, then 2d = υ × t or d = υt/2, where d is the depth of the sea. This method is also called echo sounding.
Practical Applications: It is used to locate underwater submarines, icebergs, sunken ships and underwater hills etc.

Question 20.
A sonar device on a submarine sends out a signal and receives an echo 5 s later. Calculate the speed of sound in water if the distance of the object from submarine is 3,625 m.
Solution:
Time between emission of sound and its collection (t) = 5 s
Depth of sea (2d) = 2 × 3625 m = 7250 m
We know, 2d = Speed of sound × Time
7250 = Speed of sound × 5
∴ Speed of sound (υ) = \(\frac {7250}{5}\)
= 1450 m s^-1

Question 21.
Explain how defects in a metal block can be detected using ultrasound?
Answer:
In industries metallic components are used in the construction of big structures like buildings, bridges, machines, scientific equipments, etc. ultrasounds (ultrasonic waves) are used to detect the internal defects or cracks in big metallic blocks which are not visible from outside.

Ultrasonic waves are allowed to pass through the metal block and detectors are used to detect the transmitted waves. If there is even a small defect, the ultrasound gets reflected back and does not reach the detector, as shown in figure. This indicates the presence of a defect.

Question 22.
Explain how the human ear works.
Answer:
Ear is very sensitive device used to hear sound. It converts compressions and rarefactions of frequency range 20 Hz to 20,000 Hz into electric signals that travel to brain via auditory nerve.

The ear consists of three sections:

1. the outer ear,
2. the middle ear and
3. the inner ear.

The outer ear consists of Pinna and Auditory Canal. Pinna is a cup-shaped fleshy part of the outer ear. Pinna collects and amplifies sound waves which then pass on the auditory canal. At the end of auditory canal, there is a thin membrane called tympanic membrane or eardrum. When compression reaches the eardrum, the pressure on membrane increases and the ear drum is forced inwards. When rarefaction reaches the eardrum, it moves outwards.

The vibrations are amplified by lever action of three bones called hammer, anvil and stirrup in the middle ear. In turn, the middle ear transmits the amplified pressure variations to the inner ear. The amplified pressure variations are converted into electric signals by cochlea in the inner ear. The electric signals generated are conveyed to the brain

via the auditory nerve. The brain interprets them as sound. In fact we do not hear with ear. We hear with brain through ears.

Science Guide for Class 9 PSEB Sound InText Questions and Answers

Question 1.
How does the sound produced by the vibrating object in a medium reach your ear?
Answer:
When the vibrating object (such as tuning fork or school bell) moves forward then it compresses the air particles lying just ahead of it which results in production of high pressure region. This region is called compression. This pressure moves forward in the direction in which the object is vibrating, when this vibrat ing object moves backward then a region of low pressure is produced which is known as rarefaction.

When the vibrating object rapidly moves to and fro then a series of compression and rarefaction pulses are formed i.e. sound wave is produced.
In this way the transmission of sound is caused in the form of transmission of change in density which reaches our ear and forces the tympanic membrane to vibrate. This produces the sensation of hearing in us.

Question 2.
Explain how sound is produced by your school bell?
Answer:
When the school bell is hit with a hammer, it begins to vibrate which produces sound waves. If we gently touch the bell, we feel vibrations. Wave is a disturbance which produces motion in the neighbouring particles of the medium. These particles handover the disturbance in the next particles lying close to the vibrating particles so that sound waves reach us. The particles of the medium do not move from one place to another, it is only the disturbance that travels forward.

Question 3.
Why are sound waves called mechanical waves?
Answer:
Sound is a kind of energy which cannot be produced by itself. To produce it some mechanical enery is required which may be by clapping or by striking bell with a hammer. This sound energy is transmitted in the form of waves by producing disturbance of the particles of the medium. Therefore, sound waves are called mechanical waves.

Question 4.
Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?
Answer:
For propagation of sound, air or some other material medium is required. On the surface of moon there is no such medium present as a result of which sound can not be propagated from one place to another place. So you can neither talk to your friend nor the sound produced by your friend can be heard by you.

Question 5.
Which wave property determines
(a) loudness
(b) pitch?
Answer:
(a) Loudness: The loudness of a sound wave is determined by its amplitude. The amplitude of sound wave depends upon magnitude of force. The more the force, the loud is sound produced. Loud sound traverses more distance because it has more energy in it. The more the sound is away from the source, the less is its loudness. Hence, loudness depends upon square of the amplitude.

(b) Pitch: The frequency of sound produced is called pitch. Frequency determines the aitch of a sound. The more is the vibration of the source, the higher will be its pitch.

So more the frequency, higher the pitch of sound.
In sound of high pitch the number of compressions passing through a fixed point in a unit time will be more.

Question 6.
Guess, which sound has a higher pitch: guitar or a car horn?
Answer:
Though sound of car horn is louder than that of guitar but guitar has higher pitch than car horn.

Question 7.
What are wavelength, frequency, time period and amplitude of a sound wave?
Answer:
1. Wavelength of wave: The distance travelled by the wave during the time, the particle of the medium completes 1 vibration is called wavelength.
Or
The distance between two consecutive compressions or rarefactions in a longitudinal wave or the distance between the consecutive crests or two consecutive troughs is called wavelength. It is denoted by a greek letter ‘λ’ (lambda). SI unit of wavelength is meter (m).

2. Frequency: In any medium when wave propagates the number of vibrations made by a particle of the medium is called frequency. It is denoted by ‘ν’. S.I. unit of frequency is Hertz (Hz). It is determined by the number of compressions or rarefactions passing through a fixed point.

3. Time Period: It is the time taken by a particle to complete one vibration during the propagation of wave. It is denoted by “T”. S.I. unit of time period is second.
Or
Time taken by two nearest compressions or rarefactions of sound waves to cross a point is called time period.

4. Amplitude: The maximum displacement of a particle of the medium on either side of mean position is called amplitude, of wave. It is denoted by letter ‘A’. For sound wave its unit is same as that of pressure or density. The loudness of sound depends on its amplitude.

Question 8.
How are the wavelength and frequency of a sound wave related to its speed?
Answer:
Velocity of sound wave (υ) = Wavelength (λ) × Frequency (ν).

Question 9.
Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m s^-1 in a given medium.
Solution:
Velocity of sound wave (υ) = 440 ms^-1
Frequency of sound (ν) = 220 Hz
Wavelength of sound wave (λ) = ?
We know, υ = ν × λ
440 = 220 × λ
or λ = \(\frac {440}{220}\)
or wavelength (λ) = 2 m

Question 10.
A person is listening to a tone of 500 Hz sitting at a distance of 450 m from the source of sound. What is the time interval between successive compressions from the source?
Solution:
Given frequency of sound (ν) = 500 Hz
Time taken between two successive compressions (T) = ?
We know, time period (T) = \(\frac{1}{\text { Frequency }(ν)}\)
= \(\frac {1}{500}\)
= 0.002 s

Question 11.
Distinguish between loudness and intensity of sound.
Answer:
Difference between Loudness and Intensity:

Loudness	Intensity
1. The loudness of sound is the measure of senstivity of ears.	It is the sound energy passing through a unit area in 1 second.
2. The loudness of sound can not be measured.	The intensity of sound can be measured.
3. For different observers the loudness of sound is different.	The intensity of sound is same for different persons.
4. The loudness of ultrasonic and infrasonic waves is zero because they are inaudible.	There is a possibility of intensity in ultrasonic and infrasonic sound in ultrasonic and infrasonic sound waves.

Question 12.
In which of the three media, air, water or iron sound travels the fastest at a particular temperature?
Answer:
Sound travels fastest in iron as compared to air and water. The velocity of sound in iron is 5950 m s^-1, followed by water [1500 m s^-1], air [350 m s^-1],

Question 13.
An echo returned in 3 s. What is the distance of reflecting surface from the source? Given that the speed of sound is 342 m s^-1.
Solution:
Velocity of sound (v) = 342 ms^-1
Time taken for echo to be heard (t) = 3s
Distance travelled by sound (S) = υ × t
= 342 × 3
= 1026 m
i.e. sound takes 3 s to travel from source to reflecting surface and then back to source and during this time the distance travelled is 1026 m.
Distance between source and reflecting surface = \(\frac {1026}{2}\) m
= 513 m

Question 14.
Why are the ceilings of concert halls curved?
Answer:

Ceilings of concert halls are made curved as is shown in Fig so that sound after reflection from all surfaces of hall may spread evenly to all parts and heard equally clear.

Question 15.
What is audible range of average human ear?
Answer:
For average human ear the audible range of sound is 20 Hz to 20,000 Hz.

Question 16.
What is the range of frequencies associated with
(a) infrasound
(b) ultrasound?
Answer:
(a) For infrasound the frequency range is less than 20 Hz.
(b) For ultrasound the frequency range is more than 20 KHz (i.e. 20,000 Hz)

Question 17.
A submarine emits a sonar pulse, which returns from an underwater cliff in 1.02 s. If the speed of sound in salt water is 1531 m s^-1, how far away is the cliff?
Solution:
Time is taken by the sound to travel from submarine to cliff and back to the submarine
= 1.02 s
Speed of sound in saltwater = 1531 ms^-1
Distance travelled by sound (2d) = Speed of sound × Time taken
= 1531 × 1.02 [∵ d is the distance between submarine and cliff]
= 1561.62 m
or d = \(\frac {1561.62}{2}\) m
i.e. Distance between submarine and cliff (d) = 780.81 m