Author name: Prasanna

These NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Exercise 12.3

Question 1.
If the cost of 7 m of cloth is ₹ 294, find the cost of 5 m of cloth.
Answer:
Cost of 7 m of cloth = ₹ 294
∴ Cost of 1 m of cloth = \(\frac{294}{7}\) = ₹ 42
∴ Cost of 5 m of cloth = 42 × 5 = ₹ 210
Thus, the cost of 5 m of cloth is ₹ 210.

Question 2.
Ekta earns ₹ 1500 in 10 days. How much will she earn in 30 days?
Answer:
Earning of 10 days = ₹ 1500
∴ Earning of 1 day = \(\frac{1500}{10}\) = ₹ 150
∴ Earning of 30 days = 150 × 30 = ₹ 4500
Thus, the earning of 30 days is ₹ 9000.

Question 3.
If it has rained 276 mm in the last 3 days, how many cm of rain will fall in one full week (7 days)? Assume that the rain continues to fall at the same rate.
Answer:
Rain in 3 days = 276 mm
Rain in 1 day = \(\frac{276}{3}\) = 92 mm
∴ Rain in 7 days = 92 × 7 = 644 mm
Thus, the rain in 7 days is 644 mm.

Question 4.
Cost of 5 kg of wheat is ₹ 30.50.
(a) What will be the cost of 8 kg of wheat?
(b) What quantity of wheat can be purchased in ₹ 183?
Answer:
(a) Cost of 5 kg of wheat = ₹ 30.50
∴ Cost of 1 kg of wheat = \(\frac{30.50}{5}=\frac{3050}{500}\) = ₹ 6.10
∴ Cost of 8 kg of wheat = 6.10 × 8 = ₹ 48.80

(b) From ₹ 30.50, quantity of wheat can be purchased = 5 kg
∴ From ₹ 1, quantity of wheat can be purchased = \(\frac{5}{30.50}\)
From ₹ 61, quantity of wheat can be purchased = \(\frac{5}{30.50}\) × 61
= \(\frac{5}{3050}\) × 6100 = 10 kg

Question 5.
The temperature dropped 15 degree Celsius in the last 30 days. If the rate of temperature drop remains the same, how many degrees will the temperature drop in the next ten days?
Answer:
Degree of temperature dropped in last 30 days = 15 degrees
∴ Degree of temperature dropped in last 30 days = \(\frac{15}{30}=\frac{1}{2}\) degree
∴ Degree of temperature dropped in last 10 days
= \(\frac{1}{2}\) × 10 = 5 degree
Thus, 5 degree Celsius temperature dropped in 10 days.

Question 6.
Shaina pays ₹ 7500 as rent for 3 months. How much does she has to pay for a whole year, if the rent per month remains same?
Answer:
Rent paid for 3 months = ₹ 7500
∴ Rent paid for 1 months = \(\frac{7500}{3}\) = ₹ 2500
∴ Rent paid for 12 months = 2500 × 12 = ₹ 30,000
Thus, the total rent of one year is ₹ 30,000.

Question 7.
Cost of 4 dozens bananas is ₹ 60. How many bananas can be purchased for ₹ 12.50?
Answer:
Cost of 4 dozen bananas = ₹ 60
Cost of 48 bananas = ₹ 60
[4 dozen = 4 × 12 = 48]
∴ From ₹ 60, number of bananas can be purchased = 48
∴ From ₹ 1, number of bananas can be purchased = \(\frac{48}{60}=\frac{4}{5}\)
= \(\frac{4}{5}\) × 12.50 = \(\frac{4}{5} \times \frac{1250}{100}=\frac{250}{25}\)
= 10 bananas
∴ From ₹ 12.50, number of bananas can be purchased
Thus, 10 bananas can be purchased for ₹ 12.50.

Question 8.
The weight of 72 books is 9 kg what is the weight of 40 such books?
Answer:
The weight of 72 books = 9 kg
The weight of 1 book = \(\frac{9}{12}=\frac{1}{8}\)
∴ The weight of 40 books
= \(\frac{1}{8}\) × 40 = 5 kg
Thus, the weight of 40 books is 5 kg.

Question 9.
A truck requires 108 litres of diesel for covering a distance of 594 km. How much diesel will be required by the truck to cover a distance of 1650 km?
Answer:
For covering 594 km, a truck will be required diesel = 108 litres
For covering 1 km, a truck will be required diesel = \(\frac{108}{594}=\frac{2}{11}\)
∴ For covering 1650 km, a truck will be required diesel
= \(\frac{2}{11}\) × 1650 = 300 litres
Thus, 300 litres diesel required by the truck to cover a distance of 1650 km.

Question 10.
Raju purchases 10 pens for ₹ 150 and Manish buys 7 pens for ₹ 84. Can you say who got the pen cheaper?
Answer:
Raju purchase 10 pens for = ₹ 150
∴ Raju purchases 1 pen for = \(\frac{150}{10}\) = ₹ 15
Manish purchases 7 pens for = ₹ 84
∴ Manish purchases 1 pen for = \(\frac{84}{7}\) = ₹ 12
Thus, Manish got the pens cheaper.

Question 11.
Anish made 42 runs in 6 overs and Anup made 63 runs in 7 overs. Who made more runs per over?
Answer:
Anish made in 6 overs = 42 runs
∴ Anish made in 1 overs = \(\frac{42}{6}\) = 7 runs
∴ Anup made in 1 overs = \(\frac{63}{7}\) = 9 runs.
Thus, Anup made more runs per over.

These NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.4

Question 1.
Show that in a right angled triangle, the hypotenuse is the longest side.
Solution:

In right angled triangle ABC
∠A + ∠B + ∠C = 180° (Angle sum property of A)
or, ∠A + ∠C + 90 = 180° (∵ ∠B = 90)
or, ∠A + ∠C = 90°
∴ ∠B > ∠A
Therefore, AC > BC …….(i)
(Side opposite to greater angle is greater)
Again, ∠B > ∠C
∴ AC > AB ……(ii)
(Side opposite to greater angle is greater larger)
From equation (i) and (ii)
AC is longer then both AB and BC.
Therefore AC is longes side of ∆ABC.

Question 2.
In Fig. 7.48 side AB and AC of ∆ABC are extended to point P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.

Solution:
In ∆ABC, we have given,
∠PBC < ∠QCB ……(i)
Now, ∠PBC + ∠ABC = 180° …..(ii) (Linear pair)
Again, ∠QCB + ∠ACB = 180° ……(iii) (Linear pair)
From equation (ii) and (iii)
∠PBC + ∠ABC = ∠QCB + ∠ACB
But ∠PBC < ∠QBC Therefore, ∠ABC > ∠ACB
∴ AC > AB (Side opposite to greater angle is greater)

Question 3.
In Fig. 7.49, ∠B > ∠A and ∠C < ∠D. Show thatn AD < BC.

Solution:
In ∆BAO,
∠B < ∠A
∴ OA < OB ……(i)
(Single opposite to greater angle is greater)
Again, In ∆CDO
∠C < ∠D
∴ OD < OC ……(ii)
(Single opposite to greater angle is greater)
Add equation (i) and (ii)
OA + OD < OB + OC
AD < BC

Question 4.
AB and CD are respectively the smallest and longest sides of quadrilateral ABCD (se Fig. 7.50). Show that ∠A > ∠C and ∠B > ∠D.

Solution:
Given ABCD is a quadrilateral in which AB is smallest and CD is longest side.
To prove that:
(i) ∠A > ∠C
(ii) ∠B > ∠D
Construction: Join A and C.
Proof: In ∆ABC
AB < BC (∵ AB is smallest side)
∴ ∠BCA < ∠BAC …..(i)
(∵ Angle opposite to larger side is greater)
Again, In ∆ACD,
AD < CD (∵ CD is largest side)
∴ ∠ACD < ∠ADC …….(ii)
(Angle opposite to larger side is greater)

Adding equation (i) and (ii)
∠BCA + ∠ACD < ∠BAC + ∠ADC
∠BCD < ∠BAD
or, ∠C < ∠A Similarly by joining B and D we can prove that ∠B > ∠D

Question 5.
In Fig. 7.51, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.

Solution:
In ∆PQR
PR > PQ
∠Q > ∠R
Adding ∠1 both side,
∠Q + ∠1 > ∠R + ∠1
or, ∠Q + ∠1 > ∠R + ∠2 (∵ ∠1 = ∠2)
∠PSR > ∠PSQ
(∵ Exterior angle is equal to the sum of opposite interior angles)

Question 6.
Show that of all line segments drawn from a given point not on it, the perpendicular line segment is shortest.
Solution:
Given: A straight line l and a point P not lying on l. PM ⊥ l and N is any point on l other than M.

To prove that: PM < PN
Proof: In ∆PMN, we have
∠M = 90
∠N < 90
(∵ ∠M = 90
⇒ ∠MPN + ∠PNM = 90
⇒ ∠P + ∠N = 90
⇒ ∠N < 90)
⇒∠N< ∠M
⇒ PM < PN (Side opposite to greater angle is larger)
Hence, PM is shortest of all line segments from P to AB.

These NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 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 BC (see Fig. 7.39). If AD is extended to intersect BC at P, show that
(i) ∆ABD ≅ ∆ACD
(ii) ∆ABP ≅ ∆ACP
(iii) AP bisects ∠A as well as ∠D.
(iv) AP is the perpendicular bisector of BC

Solution:
(i) In ∆ABD and ∆ACD
AB = AC (∵ ABC is an isosceles ∆)
BD = CD (∵ DBC is an isosceles ∆)
AD = AD (Common)
By S-S-S Congruency condition
∆ABD = ∆ACD
∴ ∠BAD = ∠CAD ……(A) (By C.P.CT)

(ii) In ∆ABP and ∆ACP
AB = AC (∵ ABC is an isoscles ∆)
∠BAP = ∠CAP
AP = AP
By S-A-S Congruency Condition
∠ABP = ∠ACP
Therefore, BP = CP …….(i) (By C.P.C.T.)

(iii) We have
∆ABD = ∆ACD (From (A))
∴ ∠BAD = ∠CAD (By C.P.C.T.)
∴ AP bisects ∠A
Again,
In ∆BDP and ∆GDP
BD = CD (∵ ∆BDC is an isoscles ∆)
DP = DP (Common)
BP = CP (From equation (i))
By S-S-S Congruency Condition
∆BDP ≅ ∆CDP ………(B)
∴ ∠BDP = ∠CDP (By C.P.C.T)
Therefore, AP bisects ∠D …….(iii)
From equation (ii) and (iii) we can say AP bisects ∠A as well as ∠D.

(iv) we have
∆BDP = ∆CDP (from (B))
∴ BP = CP ……(iv) (By C.P.C.T.)
Again, ∠BPD = ∠CPD (By C.P.C.T.)
Now, ∠BPD + ∠CPD = 180
or, ∠BPD + ∠BPD = 180
2∠BPD = 180
∴ ∠BPD = 90 ……(v)
Therefore, From equations (iv) and (v) we can say AP is the perpendicular bisector of BC.

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

Solution:
(i) In ∆ABD and ∆ACD
AB = AC (Given)
∠ADB = ∠ADC (each 90°)
AD = AD (Common)
By R.H.S Congruency Condition
∆ABD = ∆ACD ……(i)
∴ BD = CD (By C.P.C.T.)
Therefore, AD bisects BC.

(ii) We have
∆ABD = ∆ACD (from equation (i))
∴ ∠BAD = ∠CAD (By C.P.C.T.)
Therefore, AD bisects ∠A.

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 Fig. 7.40). Show that:
(i) ∆ABM ≅ ∆PQN
(ii) ∆ABC ≅ ∆PQR

Solution:
(i) In ∆ABM and ∆PQN
AB = PQ (Given)
BM = QN (Given)
AM = PN (Given)
By S-S-S Congruency Condition,
∆ABM ≅ ∆PQN …..prove (i)
∴ ∠B = ∠Q (By C.P.C.T)

(ii) Now, In ∆ABC and ∆PQR
AB = PQ (Given)
BM = QN (Given)
∴ 2BM = 2QN
or, BC = QR
(Because M and N are respectively the midpoint of side BC and QR)
∠B = ∠Q (Prove above)
By S-A-S Congruency Condition
∆ABC ≅ ∆PQR …….prove (ii)

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

Solution:
In ∆BFC and ∆CEB
CR = BE (Given)
∠BFC = ∠CEB (Each 90°)
BC = BC (Common)
∴ By R-H-S congruency condition,
∆BFC ≅ ∆CEB
∴ ∠B = ∠C (By C.P.C.T)
Therefore, AB = AC (Sides opposite to equal angles of a ∆)
Hence, ∆ABC is isosceles.

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

Solution:
In ∆ABP and ∆ACP
AB = AC (Given)
∠APB = ∠APC (Each 90°)
AP = AP (Common)
By R-H-S Congruency Condition,
∆ABP ≅ ∆ACP
or, ∠B = ∠C (By C.P.C.T)

These NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Exercise 5.4

Question 1.
What is the measure of
(i) a right angle?
(ii) a straight angle?
Answer:
(i) 90°
(ii) 180°

Question 2.
Say True or False:
(a) The measure of an acute angle < 90°.
(b) The measure of an obtuse angle < 90°. (c) The measure of a reflex angle > 180°.
(d) The measure of on complete revolution = 360°.
(e) If m∠A = 53° and m∠B = 35° then m∠A > m∠B.
Answer:
(a) True
(b) False
(c) True
(d) True
(e) True

Question 3.
Write down the measure of:
(a) some acute angles
(b) some obtuse angles
(give at least two examples of each)
Answer:
(a) 35°, 20°
(b) 110°, 135°

Question 4.
Measure the angles give below, using the protractor and write down the measure:


Answer:
(a) 40°
(b) 130°
(c) 90°
(d) 60°

Question 5.
Which angle has a large measure? First estimate and then measure:
Measure of angle A = ?
Measure of angle B = ?

Answer:
∠B has larger measure.
∠A = 40° and ∠B = 65°

Question 6.
From these two angles which has larger measure? Estimate and then confirm by measuring them:

Answer:
Second angle has larger measure.

Question 7.
Fill in the blanks with acute, obtuse, right or straight:
(a) An angle whose measure is less than that of a right angle is …………………………
(b) An angle whose measure is greater than that of a right angle is …………………………
(c) An angle whose measure is the sum of the measures of two right angles is …………………………
(d) When the sum of the measures of two angles is that of a right angle, then each one of them is …………………………
(e) When the sum of the measures of two angles is that of a straight angle and if one of them is acute then the other should be ……………………….
Answer:
(a) acute angle
(b) obtuse angle
(c) straight angle
(d) acute angle
(e) obtuse angle

Question 8.
Find the measure of the angle shown in each figure. (First estimate with your eyes and then find the actual measure with a protractor).

Answer:
(i) 30°
(ii) 120°
(iii) 60°
(iv) 150°

Question 9.
Find the angle measure between the hands of the clock in each figure:

Answer:
(i) 90° (Right angle)
(ii) 30° (Acute angle)
(iii) 180° (Straight angle)

Question 10.
Investigate:
In the given figure, the angle measures 30°. Look at the same figure through a magnifying glass. Does the angle becomes larger? Does the size of the angle change?

Answer:
No, the measure of angle will be same

Question 11.
Measure and classify each angle:

Answer: