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These NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Exercise 12.2

Question 1.
Determine if the following are in proportion.
(a) 15, 45 ; 40, 120
(b) 33, 121; 9, 96
(c) 24, 28, 36, 48
(d) 32,48 :70,210
(e) 4, 6 ; 8, 12
(f) 33, 44 ; 75, 100
Answer:
(a) 15 : 45 = \(\frac{15}{45}=\frac{1}{3}\) = 1 : 3
40 : 120 = \(\frac{40}{120}=\frac{1}{3}\) = 1 : 3
Since, 15 : 45 = 40 : 120
Therefore, 15,45,40,120 are in proportion.

(b) 33 : 121= \(\frac{33}{121}=\frac{3}{11}\) = 3 : 11
9 : 96 = \(\frac{9}{96}=\frac{3}{32}\) = 3 : 32
Since, 33 : 121 ≠ 9 : 96
Therefore, 33,121,9,96 are not in proportion.

(c) 24 : 28 = \(\frac{24}{28}=\frac{6}{7}\) = 6 : 7
36 : 48 = \(\frac{36}{48}=\frac{3}{4}\) = 3 : 4
Since, 24 : 28 ≠ 36 : 48
Therefore, 24, 28,36, 48 are not in proportion.

(d) 32 : 48 = \(\frac{32}{48}=\frac{2}{3}\) = 2 : 3
70 : 210= \(\frac{70}{210}=\frac{1}{3}\) = 1 : 3
Since, 32 : 48 ≠ 70 : 210
Therefore, 32,48, 70, 210 are not in proportion.
(e) 4 : 6 = \(\frac{4}{6}=\frac{2}{3}\) = 2 : 3
8 : 12 = \(\frac{8}{12}=\frac{2}{3}\) = 2 : 3
Since, 4 : 6 = 8 : 12
Therefore, 4,6,8,12 are in proportion.

(f) 33 : 44 = \(\frac{33}{44}=\frac{3}{4}\) = 3 : 4
75 : 100 = \(\frac{75}{100}=\frac{3}{4}\) = 3 : 4
Since, 33 : 44 = 75 : 100
Therefore, 33, 44, 75,100 are in ratio.

Question 2.
Write True (T) or False (F) against each of the following statements:
(a) 16 : 24 :: 20 : 30
(b) 21: 6:: 35 : 10
(c) 12: 18:: 28:12
(d) 8:9:: 24: 27
(e) 5.2 : 3.9 :: 3 :4
(f) 0.9 : 0.36 :: 10 : 4
Answer:
(a) 16 : 24 :: 20 : 30
⇒ \(\frac{16}{24}=\frac{20}{30}\) ⇒ \(\frac{2}{3}=\frac{2}{3}\)
Hence, it is true.

(b) 21 : 6 :: 35 : 10
⇒ \(\frac{21}{6}=\frac{35}{10}\) ⇒ \(\frac{7}{2}=\frac{7}{2}\)
Hence, it is true.

(c) 12 : 18 :: 28 : 12
⇒ \(\frac{12}{18}=\frac{28}{12}\) ⇒ \(\frac{2}{3} \neq \frac{7}{3}\)
Hence, it is false.

(d) 8 : 9 :: 24 : 27
⇒ \(\frac{8}{9}=\frac{24}{27}\) ⇒ \(\frac{8}{9}=\frac{8}{9}\)
Hence, it is true.

(e) 5.2 : 3.9 :: 3 : 4
⇒ \(\frac{5.2}{3.9}=\frac{3}{4}\) ⇒ \(\frac{4}{3} \neq \frac{3}{4}\)
Hence, it is false.

(f) 0.9 : 0.36 :: 10 : 4
⇒ \(\frac{0.9}{0.36}=\frac{10}{4}\) ⇒ \(\frac{5}{2}=\frac{5}{2}\)
Hence, it is true.

Question 3.
Are the following statements true?
(a) 40 persons : 200 persons = ₹ 15: ₹ 75
(b) 7.5 litres : 15 litres = 5 kg : 10 kg
(c) 99 kg : 45 kg = ₹ 44 : ₹ 20
(d) 32 m : 64 m = 6 sec : 12 sec
(e) 45 km : 60 km = 12 hours : 15 hours
Answer:
(a) 40 persons : 200 persons
= \(\frac{40}{200}=\frac{1}{5}\) = 1 : 5
₹ 15 : ₹ 75 = \(\frac{15}{75}=\frac{1}{5}\) = 1 : 5
Since, 40 persons : 200 persons = ₹ 15 : ₹ 75
Hence, the statement is true.

(b) 7.5 litres : 15 litres
= \(\frac{7.5}{15}=\frac{75}{150}=\frac{1}{2}\) = 1 : 2
5 kg : 10 kg = \(\frac{5}{10}=\frac{1}{2}\) = 1 : 2
Since, 7.5 litres : 15 litres = 5 kg: 10 kg
Hence, the statement is true.

(c) 99 kg : 45 kg = \(\frac{99}{45}=\frac{11}{5}\) = 11 : 5
₹ 44 : ₹ 20= \(\frac{44}{20}=\frac{11}{5}\) = 11 : 5
Since, 99 kg: 45 kg = ₹ 44 : ₹ 20 Hence, the statement is true.

(d) 32 m : 64 m = \(\frac{32}{64}=\frac{1}{2}\) = 1 : 2
6 sec : 12 sec = \(\frac{6}{12}=\frac{1}{2}\) = 1 : 2
Since, 32 m : 64 m = 6 sec : 12 sec Hence, the statement is true.

(e) 45 km : 60 km = \(\frac{45}{60}=\frac{3}{4}\) = 3 : 4
12 hours : 15 hours
= \(\frac{12}{15}=\frac{4}{5}\) = 4 : 5
Since, 45 km : 60 km ≠ 12 hours : 15 hours
Hence, the statement is not true.

Question 4.
Determine if the following ratios form a proportion. Also, write the middle terms and extreme terms where the ratios form a proportion.
(a) 25 cm : 1 m and ₹ 40: ₹ 160
(b) 39 litres : 65 litres and 6 bottles : 10 bottles
(c) 2 kg: 80 kg and 25 g : 625 g
(d) 200 mL : 2.5 litre and ₹ 4 : ₹ 50
Answer:
(a) 25 cm : 1 m = 25 cm : (1 × 100) cm
= 25 cm : 100 cm
= \(\frac{25}{100}=\frac{1}{4}\) = 1 : 4
₹ 40 : ₹ 160 = \(\frac{40}{160}=\frac{1}{4}\) = 1 : 4
Since, the ratios are equal, therefore these are in proportion.
Middle terms = 1 m, ₹ 40 and
Extreme terms = 25 cm, ₹ 160

(b) 39 litres : 65 litres
= \(\frac{39}{65}=\frac{3}{5}\) = 3 : 5
6 bottles : 10 bottles
= \(\frac{6}{10}=\frac{3}{5}\) = 3 : 5
Since, the ratios are equal, therefore these are in proportion.
Middle terms = 65 litres, 6 bottles and Extreme terms = 39 litres, 10 bottles

(c) 2 kg : 80 kg = \(\frac{2}{80}=\frac{1}{40}\) = 1 : 40
25 g : 625 g = \(\frac{25}{625}=\frac{1}{25}\) = 1 : 25
Since, the ratios are not equal, therefore these are not in proportion,

(d) 200 ml : 2.5 litres = 200 ml: (25000) litres = 200 ml : 2500 ml = \(\frac{200}{2500}=\frac{2}{25}\) = 2 : 25
₹ 4: ₹ 50 = \(\frac{4}{50}=\frac{2}{25}\) = 2 : 25
Since, the ratios are equal, therefore these are in proportion.
Middle terms = 2.5 litres, ₹ 4 and Extreme terms = 200 ml, ₹ 50

These NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.2

Question 1.
In an issosceies 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
Solution:
(i) In ∆ABC,
We have given that
AB = AC
∴ ∠B = ∠C (Angle opposite to equal sides are equal)
or, \(\frac {1}{2}\) ∠B = \(\frac {1}{2}\) ∠C
∴ ∠1 = ∠2
Now in ∠OBC,
∠1 = ∠2
∴ OB = OC (Side opposite to equal angles are equal)

(ii) In ∆AOB and ∆AOC
AB = AC
OB = OC
∠B = ∠C
∴ \(\frac {1}{2}\) ∠B = \(\frac {1}{2}\) ∠B
or ∠3 = ∠4
Therefore, by S-A-S Congruency Condition,
∆AOB ≅ ∆AOC
∴ ∠BAO = ∠CAO (by C.P.C.T)
∴ AO is the bisector of ∠A.

Question 2.
In ∆ABC, AD is the perpendicular bisector of BC (see Fig. 7.30). Show that ∆ABC is an isosceles triangle in which AB = AC.

Solution:
In ∆ABD and ∆ACD
BD = CD (∵ AD bisects BC)
∠ADB = ∠ADC (Each 90°)
AD = AD (Common)
By S-A-S Congurency Condition,
∆ABD ≅ ∆ACD
Therefore, AB = AC (By C.P.C.T)
∴ ABC is an isosceles triangle.

Question 3.
ABC is an isosceles triangle in which altitudes BE and CF are drawn to side AC and AB respectively (see Fig. 7.31). Show that these altitudes are equal.

Solution:
In ∆ABE and ∆ACF,
∠A = ∠A (Common)
∠AEB = ∠AFC (each 90°)
AB = AC (Given)
By, A-A-S Congruency Condition
∆ABE = ∆ACP
Therefore, BE = CF (By C.P.C.T)

Question 4.
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig. 7.32). Show that
(i) ∆ABE ≅ ∆ACF
(ii) AB = AC, i.e. ∆ABC is an isosceles triangle.

Solution:
(i) ∆ABE and ∆ACF,
BE = CF (Given)
∠A = ∠A (Common)
∠AEB = ∠AFC (Each 90°)
By A-A-S Congruency Condition
∆ABE = ∆ACF

(ii) Since ∆ABE ≅ ∆ACF
So, AB = AC (By C.P.C.T.)
or, ∆ABC is an isosceles triangle.

Question 5.
ABC and DBC are two isosceles triangles on the same base BC (see Fig. 7.33). Show that ∠ABD = ∠ACD.

Solution:
In ∆ABC,
AB = AC
∴ ∠ABC = ∠ACB …….(i)
(Angle opposite to equal sides of a ∆ are equal)
Again, In ∆DBC
DB = DC
∠DBC = ∠DCB ……(ii)
(Angle opposite to equal sides of a ∆ are equal)
Adding equation (i) and (ii)
∠ABC + ∠DBC = ∠ACB + ∠DCB
or ∠ABD = ∠ACD.

Question 6.
∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig. 7.34). Show that ∠BCD is a right angle.

Solution:
In ∆ABC,
AB = AC
∠ABC = ∠ACB ……(i)
(Angles opposite to equal sides of a ∆ABC)
Now, In ∆ACD,
AC = AD
∠ACD = ∠ADC ……(ii)
(Angles opposite to equal sides of ∆ACD)
Now, ∠BAC + ∠CAD = 180° ……(iii) (Linear pair)
Also, ∠CAD = ∠ABC + ∠ACB (Exterior angle of ∆ABC)
∠CAD = 2∠ACB …….(iv)
From equation (i),
∠ABC = ∠ACB
and ∠BAC = ∠ACD + ∠ADC (Exterior angle of ∆ADC)
∠BAC = 2∠ACD …….(v)
(From equation (ii), ∠ACD = ∠ADC)
From equation (iii) we have,
∠BAC + ∠CAD = 180°
⇒ 2∠ACD + 2∠ACB = 180° (From equation (iv) and (v))
⇒ 2(∠ACD + ∠ACB) = 180°
⇒ 2(∠BCD) = 180°
⇒ ∠BCD = 90°

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

Solution:
In ∆ABC,
AB = AC
∴ ∠B = ∠C (Angle opposite to equal sides of ∆ABC)
Now, ∠A + ∠B + ∠C = 180° (Angle sum property)
or, 90° + ∠B + ∠B = 180° (∵ ∠A = 90° and ∠B = ∠C)
or, 2∠B = 90°
∴ ∠B = 45°
Therefore, ∠B = ∠C = 45°.

Question 8.
Show that the angles of equilateral triangles are 60° each.

Solution:
In ∆ABC,
AB = BC = AC (Because ABC is an equilateral triangle)
∴ ∠C = ∠A = ∠B (Angle opposite to equal sides of ∆ABC)
Now, ∠A + ∠B + ∠C = 180° (Angle sum property of ∆)
or, ∠A + ∠A + ∠A = 180° (∵ ∠A = ∠B = ∠C)
or, 3∠A = 180
Therefore, ∠A = 60°
Again, ∠A = ∠B = ∠C
∴ ∠A = ∠B = ∠C = 60°
Therefore, each angle of an equilateral ∆ is 60°.

These NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Exercise 5.1

Question 1.
What is the disadvantage in comparing line segments by mere observation?
Answer:
Comparing the line segments simply by ‘observation’ may not be accurate. For example, the line segments AB and CD (in the following figure) seem to be equal, but actually they are not.

Question 2.
Why is it better to use a divider than a ruler, while measuring the length of a line segment?
Answer:
It is better to use a divider than a ruler, because the thickness of the ruler may cause difficulties in reading off the length. However, divider gives up accurate measurement.

Question 3.
Draw any line segment, say \(\begin{equation}
\overline{\mathbf{A B}}
\end{equation}\). Take any point C lying in between A and B. Measure the lengths of AB, BC and AC. Is AB = AC + CB?
[Note: If A, B, C are any three points on a line, such that AC + CB = AB, then we can be sure that C lies between A and B.]
Answer:
Yes.

\(\begin{equation}
\overline{\mathbf{A C}}
\end{equation}\) = 6.3 cm
\(\begin{equation}
\overline{\mathbf{B C}}
\end{equation}\) = 2.7 cm
\(\begin{equation}
\overline{\mathbf{A B}}
\end{equation}\) – 9.0 cm
∵ AC + BC = 6.3 cm + 2.7 cm = 9.0 cm and AB = 9.0 cm
∴ AC + BC = AB Hence verified.

Question 4.
If A, B, C are three points on a line such that AB = 5 cm, BC = 3 cm and AC = 8 cm, which one of them lies between the other two?
Answer:
∵ AB = 5 cm
∵ BC = 3 cm
∴ AB + BC = 5 cm + 3 cm = 8 cm
But AC = 8 cm
The point B lies between A and C.

Question 5.
Verify whether D is the mid-point of \(\begin{equation}
\overline{\mathbf{A G}}
\end{equation}\).

Answer:
AD = 3 units, DG = 3 units

AD = DG.
Thus, D is the mid-point.

Question 6.
If B is the mid-point of \(\begin{equation} \overline{\mathbf{A C}}\end{equation}\) and C is the mid-point of \(\begin{equation}\overline{\mathbf{B D}}\end{equation}\) where A, B, C, D lie on a straight line, say why AB = CD?
Ans.
B is the mid-point of \(\begin{equation} \overline{\mathbf{A C}}\end{equation}\).

∴ AB = BC … (i)
And C is the mid-point of \(\begin{equation} \overline{\mathbf{BD}}\end{equation}\).
∴ BC = CD … (ii)
From equation (i) and (ii), we get AB = CD

Question 7.
Draw five triangles and measure their sides. Check in each case, of the sum of the lengths of any two sides is always less than the third side.
Answer:
Yes, sum of two sides of a triangle is always greater than the third side.

These NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.1

Question 1.
In quadrilateral ACBD (See Fig. 7.16), AC = AD, and AB bisects ∠A. Show that ΔABC ≅ ΔABD. What can you say about BC and BD?

Solution:
In ΔABC and ΔABD
AC = AD (Given)
∠CAB = ∠DAB (Given)
AB = AB (Common)
Therefore, By SAS congruency condition
ΔABC ≅ ΔABD
So, BC = BD (By C.P.C.T)

Question 2.
ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see Fig. 7.17)
Prove that:
(i) ΔABD = ΔBAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC

Solution:
(i) In ΔABD and ΔBAC
AD = BC
∠DAB = ∠CBA (Given)
and AB = BA (Common)
By SAS Congruency Condition
ΔABD = ΔBAC
(ii) BD= AC (By C.P.C.T)
(iii) ∠ABD = ∠BAC (Again by C.P.C.T)

Question 3.
AD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB.

Solution:
In ΔAOD and ΔBOC,
AD = BC (Given)
∠OAD = ∠OBC (each 90°)
∠AOD = ∠BOC (Vertically opposite angles)
Therefore, by ASA congruency condition.
ΔAOD = ΔBOC
So, OA = OB (by C.P.C.T)
Hence, CD bisects line segment AB.

Question 4.
l and m are two parallel lines intersected by another pair of parallel lines p and q (see Fig. 7.19). Show that ΔABC = ΔCDA.

Solution:
We have given that l || m and p || q
Therefore, In ΔABC and ΔCDA
∠BAC = ∠DCA
(Alternate interior angles as AB || DC)
∠ACB = ∠CAD
(Alternate interior angles as BC || DA)
AC = CA (common)
So, By A-S-A congruency condition,
ΔABC = ΔCDA

Question 5.
Line 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 Fig. 7.20) show that:
(i) ΔAPB ≅ ΔAQB
(ii) BP = BQ or B is equidistance from the arms of ∠A.

Solution:
In ΔABP and ΔABQ,
∠BAP = ∠BAQ (Given)
∠APB = ∠AQB (Each 90°)
AB = AB (Common)
By A-A-S congruency condition.
So, ΔABP ≅ ΔABQ
(ii) BP = BQ (By C.P.C.T)

Question 6.
In Fig. 7.21, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

Solution:
In ΔBAC and ΔDAE
AB = AD (Given)
AC = AE (Given)
∠BAD = ∠EAC ……(i) (Given)
Adding ∠DAC both side in equation (i)
∴ ∠BAD + ∠DAC = ∠EAC + ∠DAC
∠BAC = ∠DAE
Therefore by S-A-S Congruency Condition
ΔBAC = ΔDAE
So, BO = DE (By C.P.C.T)

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 Fig. 7.22) show that
(i) ΔDAP ≅ ΔEBP
(ii) AD = BE

Solution:
(i) In ΔDAP and ΔEBP
∠DAP= ∠EBP (Given)
∠APE = ∠DPB (Given)
∴ ∠APE + ∠EPD = ∠DPB + ∠EPD (Add ∠EPD both side)
∠APD = ∠BPE
AP = BP (Given P is the mid point of AB)
∴ By A-S-A Congruency Condition,
ΔDAP ≅ ΔEBP
(ii) AD = BE (By C.P.C.T.)

Question 8.
In the right triangle ABC right angled at C, M is the midpoint of hypotenuse AB. C is joined to M and produces to a point D such that DM = CM. Point D is joined to point B (see Fig. 7.23). Show that
(i) ΔAMC ≅ ΔBMD
(ii) ∠DBC is right angle.
(iii) ΔDBC ≅ ΔACB
(iv) CM = \(\frac {1}{2}\) AB

Solution:
(i) In ΔAMC and ΔBMD,
AM = BM (Given)
CM = DM (Given)
∠AMC = ∠BMD (Vertically opposite angles)
∴ By S-A-S Congruency Condition.
ΔAMC = ΔBMD

(ii) ∠CAM = ∠DBM (by C.P.C.T)
Also, ∠CAM + ∠MBC = 90° (Since ∠C = 90°)
∴ ∠DBM + ∠MBC = 90° (∵ ∠CAM = ∠DBM)
or, ∠DBC = 90°

(iii) In ΔDBC and ΔACB,
BC = BC (Common)
DB = AC (∵ ΔBMD ≅ ΔAMC by C.PC.T)
and ∠DBC = ∠ACB (each 90° proved above)
Therefore, by S-A-S Congruency Condition,
∴ ΔDBC ≅ ΔACB

(iv) Since, ΔDBC ≅ ΔACB,
DC = AB
∴ \(\frac {1}{2}\) DC = \(\frac {1}{2}\) AB
CM = AM
or CM = \(\frac {1}{2}\) AB