Bhagya

Students can download Maths Chapter 5 Statistics Ex 5.3 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 5 Statistics Ex 5.3

Question 1.
Read the given Bar Graph which shows the percentage of marks obtained by Brinda in different subjects in an assessment test.

Observe the Bar Graph and answer the following questions.
(i) 1 Unit = ………… % of marks on vertical line.
(ii) Brinda has scored maximum marks in …….. subject.
(iii) Brinda has scored minimum marks in …….. subject.
(iv) The percentage of marks scored by Brinda in Science is ……..
(v) Brinda scored 60 % marks in the subject ……..
(vi) Brinda scored 20% more in …….. subject than ……… subject.
Solution:
(i) 10
(ii) Mathematics
(iii) Language
(iv) 65%
(v) English
(vi) Mathematics, Science

Question 2.
Chitra has to buv Laddus in order to distribute to her friends as follows:

Draw a Bar Graph for this data.
Solution:
Distribution of Laddus by Chitra to her friends
Scale : 1 unit = 10 Laddus

Question 3.
The fruits liked by the students of a class are as follows:

Solution:
Draw a bar graph for data this data
Scale 1 unit = 2 fruits

Question 4.
The pictograph below gives the number of absentees on different days of the week in class six. Draw the Bar graph for the same.

Solution:
No of absentees on different days of the week in a class
Scale 1 unit = 2 students

Objective Type Questions

Question 5.
A bar graph can be drawn using
(a) Horizontal bars only
(b) Vertical bars only
(c) Both horizontal bars and Vertical bars
(d) Either horizontal bars or vertical bars
Solution:
(d) Either horizontal bars or vertical bars

Question 6.
The spaces between any two bars in a bar graph
(a) can be different
(b) are the same
(c) are not the same
(d) all of these
Solution:
(b) are the same

Students can download Maths Chapter 5 Statistics Ex 5.2 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 5 Statistics Ex 5.2

Question 1.
Fill in the blanks.

(iii) Representation of data by using pictures is known as ………
Solution:
(i) 150
(ii)
(iii) Pictograph

Question 2.
Draw a pictograph for the given data.

(Choose your own suitable scale)
Solution:
Number of computers sold
Scale : 1 Unit = 100 computers

Question 3.
The following table shows the number of tourists who visited the places in the month of May. Draw a pictograph

(Choose your own suitable scale)
Solution:
Number of Tourists who visited the places in the month of May.

Question 4.
The following pictograph shows the number of students playing different games in a school.

Answer the following questions.
(i) Which is the most popular game among the students?
(ii) Find the number of students playing Kabaddi.
(iii) Which two games are played by an equal number of students?
(iv) What is the difference between the number of students playing Kho-Kho and Hockey?
(v) Which is the least popular game among the students?
Solution:
(i) Kabaddi is the most popular game among students.
(ii) There are 11 × 10 = 110 students playing kabaddi.
(iii) Kho-Kho and Hockey are played by an equal number of students.
(iv) Difference is 90 – 90 = 0.
(v) Basketball is the least popular game among students.

Objective Type Questions

Question 5.
The pictorial representation for a phrase is a ____
(a) Picto
(b) Tally mark
(c) Frequency
(d) Data
Solution:
(d) Data

Question 6.
A pictograph is also known as
a) Pictoword
(b) Pictogram
(c) Pictophrase
(d) Pictograph
Solution:
(b) Pictogram

Students can download Maths Chapter 5 Statistics Ex 5.1 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 5 Statistics Ex 5.1

Question 1.
Fill in the blanks.

1. The collected information is called ……..
2. An example of Primary data is ………
3. An example of Secondary Data is ……….
4. The tally marks for number 8 in standard form is ………

Solution:

1. Data
2. List of absentees in a class
3. Cricket scores gathered from a website
4. 

Question 2.
Viji threw a die 30 times and noted down the result each time as follows. Prepare a table on the numbers shown using Tally Marks.
1, 4, 3, 5, 5, 6, 6, 4, 3, 5, 4, 5, 6, 5, 2, 4, 2, 6, 5, 5, 6, 6, 4, 5, 6, 6, 5, 4, 1, 1
Solution:

Question 3.
The following list tells colours liked by 25 students. Prepare a table using Tally Marks.

Solution:

Question 4.
The following are the marks obtained by 30 students in a class test out of 20 in Mathematics subject.

Prepare a table using Tally Marks.

Question 5.
The table shows the number of calls recorded by a Fire Service Station in one year.

Complete the table and answer the following questions.
(i) Which type of call was recorded the most?
(ii) Which type of call was recorded the least?
(iii) How many calls were recorded in all?
(iv How many calls were recorded as False Alarms?
Solution:

(i) The call for “Other Fires” was recorded the most
(ii) The call for “Rescues” was recorded the least
(iii) The total of 35 calls was recorded
(iv) There are 7 calls were recorded as False alarm.

Objective Type Questions

Question 6.
The tally marks for the number 7 in standard form is ………

Solution:
(b)

Question 7.

(a) 5
(b) 8
(c) 9
(d) 10
Solution:
(c) 9

Question 8.
The plural form of ‘datum’ is ____
(a) datum
(b) datums
(c) data
(d) dates
Solution:
(c) data

Students can download Maths Chapter 4 Geometry Additional Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Additional Questions

I. Multiple choice questions

Question 1.
If a straight line intersects the sides AB and AC of a ∆ABC at D and E respectively and is parallel to BC, then \(\frac { AE }{ AC } \) = …………
(1) \(\frac { AD }{ DB } \)
(2) \(\frac { AD }{ DB } \)
(3) \(\frac { DE }{ BC } \)
(4) \(\frac { AD }{ EC } \)
Answer:
(2) \(\frac { AD }{ DB } \)

Hint:
By BPT theorem,
\(\frac { AE }{ AC } \) = \(\frac { AD }{ AB } \)

Question 2.
In ∆ABC, DE is || to BC, meeting AB and AC at D and E. If AD = 3 cm, DB = 2 cm and AE = 2.7 cm, then AC is equal to ……..
(1) 6.5 cm
(2) 4.5 cm
(3) 3.5 cm
(4) 3.5 cm
Answer:
(2) 4.5 cm
Hint:
By BPT theorem,

\(\frac { AD }{ DB } \) = \(\frac { AE }{ EC } \)
\(\frac { 3 }{ 2 } \) = \(\frac { 2.7 }{ EC } \)
∴ EC = \(\frac{2.7 \times 2}{3}\) = 1.8 CM
AC = AE + EC
= 2.7 + 1.8 = 4.5 cm

Question 3.
In ∆PQR, RS is the bisector of ∠R. If PQ = 6 cm, QR = 8 cm, RP = 4 cm then PS is equal to …………..
(1) 2 cm
(2) 4 cm
(3) 3 cm
(4) 6 cm
Answer:
(2) 2 cm
Hint:

By ABT theorem,
\(\frac { PS }{ SQ } \) = \(\frac { PR }{ QR } \) ⇒ \(\frac { x }{ 6-x } \) = \(\frac { 4 }{ 8 } \)
24 – 4x = 8x ⇒ 24 = 12x
x = 2
PS = 2 cm

Question 4.
In figure, if \(\frac { AB }{ AC } \) = \(\frac { BD }{ DC } \), ∠B = 40° and ∠C = 60°, then ∠BAD = ……………
(1) 30°
(2) 50°
(3) 80°
(4) 40°
Answer:
(4) 40°
Hint:
\(\frac { AB }{ AC } \) = \(\frac { BD }{ DC } \)

AD is the internal bisector of ∠BAC
∠A + ∠B + ∠C = 180°
∠A + 40° + 60° = 180° ⇒ ∠A = 80°
∴ ∠BAC = \(\frac { 80 }{ 2 } \) = 40°

Question 5.
In the figure, the value x is equal to …………
(1) 4.2
(2) 3.2
(3) 0.8
(4) 0.4

Answer:
(2) 3.2
Hint:
By Thales theorem
(DE || BC)
\(\frac { AD }{ BD } \) = \(\frac { AE }{ EC } \)
\(\frac { x }{ 8 } \) = \(\frac { 4 }{ 10 } \) ⇒ x = \(\frac{8 \times 4}{10}\) = 3.2

Question 6.
In triangles ABC and DEF, ∠B = ∠E, ∠C = ∠F, then ………….
(1) \(\frac { AB }{ DE } \) = \(\frac { CA }{ EF } \)
(2) \(\frac { BC }{ EF } \) = \(\frac { AB }{ FD } \)
(3) \(\frac { AB }{ DE } \) = \(\frac { BC }{ EF } \)
(4) \(\frac { CA }{ FD } \) = \(\frac { AB }{ EF } \)
Answer:
(3) \(\frac { AB }{ DE } \) = \(\frac { BC }{ EF } \)
Hint:

Question 7.
From the given figure, identify the wrong statement.

(1) ∆ADB ~ ∆ABC
(2) ∆ABD ~ ∆ABC
(3) ∆BDC ~ ∆ABC
(4) ∆ADB ~ ∆BDC
Answer:
(2) ∆ ABD ~ ∆ ABC

Question 8.
If a vertical stick 12 m long casts a shadow 8 m long on the ground and at the same time a tower casts a shadow 40 m long on the ground, then the height of the tower is …………..
(1) 40 m
(2) 50 m
(3) 75 m
(4) 60 m
Answer:
(4) 60 m
Hint:

\(\frac { AB }{ DE } \) = \(\frac { BC }{ EF } \)
\(\frac { 12 }{ h } \) = \(\frac { 8 }{ 40 } \); h = \(\frac{40 \times 12}{8}\) = 60

Question 9.
The sides of two similar triangles are in the ratio 2:3, then their areas are in the ratio ………….
(1) 9 : 4
(2) 4 : 9
(3) 2 : 3
(4) 3 : 2
Answer:
(2) 4 : 9
Hint:
Ratio of the Area of two similar triangle
= 2² : 3² = 4 : 9
[squares of their corresponding sides]

Question 10.
Triangles ABC and DEF are similar. If their areas are 100 cm² and 49 cm² respectively and BC is 8.2 cm then EF = …………….
(1) 5.47 cm
(2) 5.74 cm
(3) 6.47 cm
(4) 6.74 cm
Answer:
(2) 5.74 cm
Hint:

Question 11.
The perimeters of two similar triangles are 24 cm and 18 cm respectively.
If one side of the first triangle is 8 cm, then the corresponding side of the other triangle is ………
(1) 4 cm
(2) 3 cm
(3) 9 cm
(4) 6 cm
Answer:
(4) 6 cm
Hint:

Question 12.
A point P is 26 cm away from the centre O of a circle and PT is the tangent drawn from P to the circle 10 cm, then OT is equal to …………
(1) 36 cm
(2) 20 cm
(3) 18 cm
(4) 24 cm
Answer:
(4) 24 cm
Hint:

OT² = OP² – PT²
= 26² – 10²
= (26 + 10) (26 – 10)
OT² = 36 × 16
OT = 6 × 4 = 24 cm

Question 13.
In the figure, if ∠PAB = 120° then ∠BPT = ……….

(1) 120°
(2) 30°
(3) 40°
(4) 60°
Answer:
(4) 60°
Hint:
∠BCP + ∠PAB = 180°
(sum of the opposite angles of a cyclic quadrilateral)
∠BCP = 180° – 120° = 60°
∠BPT = 60°
(By tangent chord theorem)

Question 14.
If the tangents PA and PB from an external point P to circle with centre O are inclined to each other at an angle of 40°, then ∠POA = ………………
(1) 70°
(2) 80°
(3) 50°
(4) 60°
Answer:
(1) 70°
Hint:
∠OPA = \(\frac { 40 }{ 2 } \) = 20°
In ∆OAP.
∠POA + ∠OAP + ∠APO = 180°
(sum of the angles of a triangle)

∠POA + 90° + 20° = 180°
∠POA = 180° – 110° = 70°

Question 15.
In the figure, PA and PB are tangents to the circle drawn from an external point P. Also CD is a tangent to the circle at Q. If PA = 8 cm and CQ = 3 cm, then PC is equal to ……….

(1) 11 cm
(2) 5 cm
(3) 24 cm
(4) 38 cm
Answer:
(2) 5 cm
Hint:
PA = PB (tangent of a circle)
PB = 8 cm
PC + BC = 8
PA + QC = (BC = QC tangent)
PC + 3 = 8
∴ PC = 8 cm – 3 cm = 5 cm

Question 16.
∆ABC is a right angled triangle where ∠B = 90° and BD ⊥ AC. If BD = 8 cm, AD = 4 cm, then CD is …………
(1) 24 cm
(2) 16 cm
(3) 32 cm
(4) 8 cm
Answer:
(2) 16 cm
Hint:
∆DCB ~ ∆DBA
\(\frac { DC }{ DB } \) = \(\frac { DB }{ DA } \)

DB² = DC × DA
8² = DC × 4
DC = \(\frac { 64 }{ 4 } \) = 16 cm

Question 17.
The areas of two similar triangles are 16 cm2 and 36 cm2 respectively. If the altitude of the first triangle is 3 cm, then the corresponding altitude of the other triangle is
(1) 6.5 cm
(2) 6 cm
(3) 4 cm
(4) 4.5 cm
Answer:
(4) 4.5 cm
Hint:

Question 18.
The perimeter of two similar triangles ∆ABC and ∆DEF are 36 cm and 24 cm respectively. If DE =10 cm, then AB is …………
(1) 12 cm
(2) 20 cm
(3) 15 cm
(4) 18 cm
Answer:
(3) 15 cm
Hint:
Perimeter of

Question 19.
In the given diagram θ is ………….
(1) 15°
(2) 30°
(3) 45°
(4) 60°

Answer:
(2) 30°
Hint:
∠BAD = 180° – 150° = 30°
= 180° – 150° = 30°
∠DAC = θ = 30°

Question 20.
If AD is the bisector of ∠A then AC is …………

(1) 12
(2) 16
(3) 18
(4) 20
Answer:
(4) 20
Hint. In ∆ABC, AD is the internal bisector of ∠A
\(\frac { BD }{ DC } \) = \(\frac { AB }{ AC } \)
\(\frac { 4 }{ 10 } \) = \(\frac { 8 }{ x } \)
4x = 10 × 8
x = \(\frac{10 \times 8}{4}\) = 20 cm

Question 21.
In ∆ABC and ∆DEF, ∠A = ∠E and ∠B = ∠F. Then AB : AC is ………….
(1) DE : DF
(2) DE : EF
(3) EF : ED
(4) DF : EF
Answer:
(3) EF : ED
Hint:

\(\frac { AB }{ EF } \) = \(\frac { BC }{ FD } \) = \(\frac { AC }{ ED } \)

Question 22.
Two circles of radius 8.2 cm and 3.6 cm touch each other externally, the distance between their centres is ………….
(1) 1.8 cm
(2) 4.1 cm
(3) 4.6 cm
(4) 11.8 cm
Answer:
(4) 11.8 cm
Hint:
Distance between the two centres

= r₁ + r₂
= 8.2 + 3.6
= 11.8 cm

Question 23.
In the given diagram PA and PB are tangents drawn from P to a circle with centre O. ∠OPA = 35° then a and b is …………

(1) a = 30°, b = 60°
(2) a = 35°, b = 55°
(3) a = 40°, b = 50°
(4) a = 45°, b = 45°
Answer:
(2) a = 35°, b = 55°
Hint:
∠OAP = 90° (tangent of the circle)
∠AOP + ∠OPA + ∠PAO = 180°
b + 35° + 90° = 180°
b = 180° – 125°
= 55°
OP is the angle bisector of ∠P
∴ a = 35°

II. Answer the following questions

Question 1.
The image of a man of height 1.8 m, is of length 1.5 cm on the film of a camera. If the film is 3 cm from the lens of the camera, how far is the man from the camera?
Solution:
Let AB be the height of the man, CD be the height of the image of the man of height 1.8 m (180 cm). LM be the distance between man and lens. LN be the distance between Lens and the film.

Given, AB = 1.8 m (180 cm)
CD = 1.5 cm and LN = 3 cm
Consider ∆ LAB and ∆ LCD
∠ALB = ∠DLC (vertically opposite angles)
∠LAB = ∠LDC (alternate angles) (AB || CD)
∴ ∆ LAB ~ ∆ LDC (AA similarity)
∴ \(\frac { AB }{ CD } \) = \(\frac { LM }{ LN } \) ⇒ \(\frac { 180 }{ 1.5 } \) = \(\frac { x }{ 3 } \) ⇒ 1.5x = 180 × 3
x = \(\frac{180 \times 3}{1.5}\) = \(\frac{180 \times 3 \times 10}{15}\)
x = 360 cm (or) 3.6 m
∴ The distance between the man and the camera = 3.6 cm

Question 2.
A girl of height 120 cm is walking away from the base of a lamp-post at a speed of 0. 6 m/sec. If the lamp is 3.6 m above the ground level, then find the length of her shadow after 4 seconds.
Solution:
Let AB be the height of the lamp-post above the ground level.
AB = 3.6 m = 360 cm
Let CD be the height of the girl.
CD = 1.2 m = 120 cm
The distance travelled by the girl in 4 seconds (AC)
= 4 × 0.6 = 2.4m = 240 cm
Consider ∆ECD and ∆EAB
Given (CD || AB)
∠EAB = ∠ECD = 90°
∠E is common

∴ ∆ EAB = ∆ ECD = 90°
\(\frac { AB }{ CD } \) = \(\frac { AE }{ CE } \)
= \(\frac { x+240 }{ x } \) ⇒ 3 = \(\frac { x+240 }{ x } \)
3x = x + 240
2x = 240 ⇒ x = \(\frac { 240 }{ 2 } \) = 120 cm
∴ Lenght of girls shadow after 4 seconds = 120 cm

Question 3.
In ∆ ABC, AB = AC and BC = 6 cm. D is a point on the side AC such that AD = 5 cm and CD = 4 cm. Show that ∆BCD ~ ∆ACB and hence find BD.
Solution:
Given, In ∆ ABC, AB = AC = 9 cm and BC = 6 cm, AD = 5 cm and CD = 4 cm
\(\frac { BC }{ AC } \) = \(\frac { 6 }{ 9 } \) = \(\frac { 2 }{ 3 } \) …..(1)
\(\frac { CD }{ CB } \) = \(\frac { 4 }{ 6 } \) = \(\frac { 2 }{ 3 } \) ……..(2)

From (1) and (2) we get,
\(\frac { BC }{ AC } \) = \(\frac { CD }{ CB } \)
In ∆ BCD and ∆ ACB
∠C = ∠C (common angle)
\(\frac { BC }{ AC } \) = \(\frac { CD }{ CB } \)
∴ ∆ BCD ~ ∆ ACB
\(\frac { BD }{ AB } \) = \(\frac { BC }{ AC } \) ⇒ \(\frac { BD }{ 9 } \) = \(\frac { 6 }{ 9 } \) ⇒ ∴ BD = 6 cm

Question 4.
The lengths of three sides of a triangle ABC are 6 cm, 4 cm and 9 cm. ∆PQR ~ ∆ABC. One of the lengths of sides of APQR is 35 cm. What is the greatest perimeter possible for ∆PQR?
Solution:
Given, ∆ PQR ~ ∆ ABC

Perimeter of ∆ ABC = 6 + 4 + 9 = 19 cm
When the perimeter of ∆ PQR is the greatest only the corresponding side QR must be equal to 35 cm

Question 5.
A man sees the top of a tower in a mirror which is at a distance of 87.6 m from the tower. The mirror is on the ground, facing upward. The man is 0.4 m away from the mirror, and the distance of his eye level from the ground is 1.5 m. How tall is the tower? (The foot of man, the mirror and the foot of the tower lie along a straight line).
Solution:
Let AB and ED be the heights of the man and the tower respectively. Let C be the point of incidence of the tower in the mirror.
In ∆ ABC and ∆ EDC, we have

∠ABC = ∠EDC = 90°
∠BCA = ∠DCE
(angular elevation is same at the same instant, i.e., the angle of incidence and the angle of reflection are same.)
∴ ∆ ABC ~ ∆ EDC (AA similarity criterion)
Thus,
\(\frac { ED }{ AB } \) = \(\frac { DC }{ BC } \) (corresponding sides are proportional)
ED = \(\frac { DC }{ BC } \) × AB = \(\frac { 87.6 }{ 0.4 } \) × 1.5 = 328.5
Hence, the height of the tower is 328.5 m.

Question 6.
In ∆ PQR, given that S is a point on PQ such that ST || QR and \(\frac { PS }{ SQ } \) = \(\frac { 3 }{ 5 } \). If PR = 5.6 cm, then find PT.
Solution:
In ∆ PQR, we have ST || QR and by Thales theorem,
\(\frac { PS }{ SQ } \) = \(\frac { PT }{ TR } \) …….(1)
Let PT = x
Thus, TR = PR – PT = 5.6 – x

From (1), we get PT = TR (\(\frac { PS }{ SQ } \))
x = (5.6 – x) (\(\frac { 3 }{ 5 } \))
5x = 16.8 – 3x
8x = 16.8
x = \(\frac { 16.8 }{ 8 } \) = 2.1
That is PT = 2.1 cm

Question 7.
In a ∆ ABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 4x – 3, BD = 3x – 1, AE = 8x – 7 and EC = 5x – 3, then find the value of x.
Solution:
In ∆ ABC, DE || BC. By BPT theorem. (Thales theorem)
We get \(\frac { AD }{ DB } \) = \(\frac { AE }{ EC } \) ⇒ \(\frac { 4x-3 }{ 3x-1 } \) = \(\frac { 8x-7 }{ 5x-3 } \)

(8x – 7) (3x – 1) = (4x – 3) (5x – 3)
24x² – 8x – 21x + 7 = 20x² – 12x – 15x + 9
24x² – 8x – 21x + 7 = 20x² – 27x + 9
24x² – 29x + 7 – 20x² + 27x – 9 = 0
∴ 4x² – 2x – 2 = 0
÷ 2 ⇒ 2x² – x – 1 = 0
(x – 1) (2x + 1) = 0

x – 1 = 0 (or) 2x + 1 = 0
x = 1 (or) 2x = – 1
x = – \(\frac { 1 }{ 2 } \) ⇒ since, x ≠ – \(\frac { 1 }{ 2 } \)
∴ The value of x = 1

Question 8.
In the figure AC || BD and CE || DF. if OA = 12 cm, AB = 9cm, OC = 8 cm and EF = 4.5 cm, then find FO.
Solution:

In OBD, AC || BD
∴ \(\frac { OA }{ AB } \) = \(\frac { OC }{ CD } \) (By Thales theorem)
\(\frac { 12 }{ 9 } \) = \(\frac { 8 }{ CD } \)
∴ CD = \(\frac{9 \times 8}{12}\) = 6 CM
In ODF, CE || DF
\(\frac { OC }{ CD } \) = \(\frac { OE }{ EF } \) (By Thales theorem)
\(\frac { 8 }{ 6 } \) = \(\frac { OE }{ 4.5 } \) ⇒ OE = \(=\frac{8 \times 4.5}{6}\) = 6 cm
FO = FE + EO = 4.5 + 6 = 10.5 cm
∴ The value of FO = 10.5 cm

Question 9.
Check whether AD is the bisector of ∠A of ∆ABC in each of the following.
(i) AB = 4 cm, AC = 6 cm, BD 1.6 cm, and CD = 2.4 cm.
Solution:
\(\frac { BD }{ DC } \) = \(\frac { 1.6 }{ 2.4 } \) = \(\frac { 16 }{ 24 } \) = \(\frac { 2 }{ 3 } \) …..(1)
\(\frac { AB }{ AC } \) = \(\frac { 4 }{ 6 } \) = \(\frac { 2 }{ 3 } \) ……..(2)
From (1) and (2) we get,

\(\frac { BD }{ DC } \) = \(\frac { AB }{ AC } \)
By the converse of angle bisector theorem we have,
∴ AD is the internal bisector of ∠A

(ii) AB = 6 cm, AC = 8 cm, BD = 1.5 cm and CD = 3 cm
Solution:
\(\frac { BD }{ DC } \) = \(\frac { 1.5 }{ 3 } \) = 0.5 …….(1)
\(\frac { AB }{ AC } \) = \(\frac { 6 }{ 8 } \) = \(\frac { 3 }{ 4 } \) …….(2)

From (1) and (2) we get,
\(\frac { BD }{ DC } \) ≠ \(\frac { AB }{ AC } \)
Hence AD is not the bisector of ∠A.

Question 10.
In a ∆ABC, AD is the internal bisector of ∠A, meeting BC at D. If AB 5.6 cm, AC = 6 cm and DC = 3 cm, find BC.
Solution:
Given, AB = 5.6 cm, AC = 6 cm, DC = 3 cm
Let BD be x
In ∆ ABC, AD is the internal bisector of ∠A.
By Angle bisector theorem, we have,

\(\frac { BD }{ DC } \) = \(\frac { AB }{ AC } \) ⇒ \(\frac { x }{ 3 } \) = \(\frac { 5.6 }{ 6 } \)
x = \(\frac{3 \times 5.6}{6}\) = 2.8 cm
∴ BC = BD + DC = 2.8 + 3 = 5.8 cm
Length of BC = 5.8 cm

Question 11.
In the figure, tangents PA and PB are drawn to a circle with centre O from an external point P. If CD is a tangent to the circle at E and AP = 15 cm, find the perimeter of ∆PCD.
Solution:
We know that the lengths of the two tangents from an exterior point to a circle are equal.
∴ CA = CE, DB = DE and PA = PB
Now, the perimeter of ∆PCD

= PC + CD + DP
= PC + CE + ED + DP
= PC + CA + DB + DP
= PA + PB = 2 PA (PB = PA)
Thus, the perimeter of APCD = 2 × 15 = 30 cm

Question 12.
ABCD is a quadrilateral such that all of its sides touch a circle. If AB = 6 cm, BC = 6.5 cm and CD = 7 cm, then find the length of AD.
Solution:
Let P, Q, R and S be the points where the circle touches the quadrilateral.
We know that the lengths of the two tangents drawn from an exterior point to a circle are equal. Thus, we have,

AP = AS, BP = BQ, CR = CQ and DR = DS
Hence, AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ AB + CD = AD + BC
⇒ AD = AB + CD – BC
= 6 + 7 – 6.5 = 6.5
Thus, AD = 6.5 cm

Question 13.
A man goes 10 m due east and then 24 m due north. Find the distance from the starting point.
Solution:
Let the initial position of the man be “O” and the final position be B.
In the ∆AOB,

OB² = OA² + AB²
OB² = 10² + 24²
= 100 + 576 = 676
OB = \(\sqrt { 676 }\) = 26 m
The man is at a distance of 26 m from the starting point.

Question 14.
Suppose AB, AC and BC have lengths 13, 16 and 20 respectively. If \(\frac { AF }{ FB } \) = \(\frac { 4 }{ 5 } \) and \(\frac { CE }{ EA } \) = \(\frac { 5 }{ 12 } \) Find BD and DC.
Solution:
Given that AB = 13, AC = 16 and BC = 20. Let BD = JC and DC = y.
Using Ceva’s theorem we have

\(\frac { BD }{ DC } \) × \(\frac { CE }{ EA } \) × \(\frac { AF }{ FB } \) = 1 ………(1)
Substitute the given values
\(\frac { x }{ y } \) × \(\frac { 5 }{ 12 } \) × \(\frac { 4 }{ 5 } \) = 1 ⇒ \(\frac { x }{ y } \) × \(\frac { 1 }{ 3 } \) = 1
\(\frac { x }{ y } \) = 3 ⇒ x = 3y
x – 3y = 0 ……(2)
Given BC = 20
x + y = 20 …..(3)
Subtract (2) and (3)

Question 15.
ABC is a right – angled triangle at B. Let D and E be any two points on AB and BC respectively. prove that AE² + CD² = AC² + DE².
Solution:
In the right ∆ ABE right- angled at B.
AE² = AB² + BE² …..(1)
In the right ∆ DBC,CD² = BD² + BC² ………(2)
Adding (1) and (2) we get
AE² + CD² = AB² + BE² + BD² + BC²
= (AB² + BC²) + (BC² + BD²)
AE² + CD² = AC² + DE²
Hence it is proved

[AC² = AB² + BC²]
[DE² = BE² + BD²]

III. Answer the following questions

Question 1.
A point O in the interior of a rectangle ABCD is joined to each of the vertices A, B, C and D. Prove that OA² + OC² = OB² + OD².
Solution:
Given: O is any point inside the rectangle ABCD.
To prove: OA² + OC² = OB² + OD²
Construction: Through “O” draw EF || AB.
Proof: Using Pythagoras theorem

In the right ∆OEA,
∴ OA²= OE² + AE² …(1)
(By Pythagoras theorem)
In the right ∆OFC,
OC² = OF² + FC² …(2)
(By Pythagoras theorem)
OB² = OF² + FB² … (3)
(By Pythagoras theorem)
In the right ∆OED,
OD² = OE² + ED² … (4)
(By Pythagoras theorem)
By adding (3) and (4) we get
OB² + OD² = OF² + FB² + OE² + ED²
= (OE² + FB²) + (OF² + ED²)
= (OE² + EA²) + (OF² + FC²)
[FB = EA and ED = FC]
OB² + OD² = OA² + OC² using (1) and (2)

Question 2.
A lotus is 20 cm above the water surface in a pond and its stem is partly below the water surface. As the wind blew, the stem is pushed aside so that the lotus touched the water 40 cm away from the original position of the stem. How much of the stem was below the water surface originally?
Solution:
Let O be the bottom of the stem immersed in water.
Let B be the lotus, AB be the length of the stem above the water surface
AB = 20 cm
Let OA be the length of the stem below the water surface
Let OA = x cm
Let C be the point where the lotus touches the water surface when the wind blow.
OC = OA + AB
OC = x + 20 cm
In ∆ AOC, OC² = OA² + AC²
(x + 20)² = x² + 40²
x² + 400 + 40x = x² + 1600

40x = 1600 – 400
40x = 1200
x = \(\frac { 1200 }{ 40 } \) = 30 cm
The stem is 30 cm below the water surface.

Question 3.
In the figure, DE || BC and \(\frac { AD }{ BD } \) = \(\frac { 3 }{ 5 } \), calculate the value of

Solution:
(i) Given, DE || BC and \(\frac { AD }{ BD } \) = \(\frac { 3 }{ 5 } \)
Let AD = 3k and BD = 5k; AB = 3k + 5k = 8k
In ∆^le ABC and ∆ADE
∠A = ∠A (common angle)
∠ABC = ∠ADE (corresponding angle)
Since DE || BC
∴ ∆ ABC ~ ∆ ADE

(ii) Let Area of ∆ ADE be 9k and Area of ∆ ABC be 64 k
Area of ∆ BCED = Area of ∆ ABC – Area of ∆ ADE
= 64 k – 9 k
= 55 k

Question 4.
A boy is designing a diamond shaped kite, as shown in the figure where AE = 16 cm, EC = 81 cm. He wants to use a straight cross bar BD. How long should it be?
Solution:
Let in AB be “x”, BE be “z” and BC be “y”
In the right ∆ AEB,
AB² = AE² + BE²
x² = 16² + Z² …….(1)
In the right ∆ BEC,
BC² = EC² + BE²
y² = 81² + z² ……(2)

In the right ∆ ACD,
AC² = AD² + DC²
97² = x² + y² ….(3)
Add (1) and (2) ⇒ x² + y² = 162 + z² + 81² + z²
x² + y² = 2z² + 16² + 81²
97² = 2z² + 16² + 81² (from 3)
9409 = 2z² + 256 + 6561
= 2z² + 6817
2z² = 9409 – 6817 = 2592
z² = \(\frac { 2592 }{ 2 } \) = 1296
z = \(\sqrt { 1296 }\) = 36
∴ Length of cross bar BD = 2 × BE = 2 × 36 = 72 cm

Question 5.
Find the unknown values in each of the following figures. All lengths are given in centimetres (All measures are not in scale)

Solution:
(i) In ∆ ABC and ∆ ADE,
∠A = ∠A (common angle)
∠ABC = ∠ADE (corresponding angle) [BC || DE]
∆ ABC ~ ∆ ADE (By AAA similarity)

In ∆ EAG and ∆ ECF,
∠E || ∠E (common angle)
∠ECF = ∠EAG (corresponding angle)
Given CF || AG
∆ EAG ~ ∆ ECF
\(\frac { EC }{ EA } \) = \(\frac { CF }{ AG } \) ⇒ \(\frac { 8 }{ x+8 } \) = \(\frac { 6 }{ y } \) ⇒ \(\frac { 8 }{ 12 } \) = \(\frac { 6 }{ y } \) (x = 4)
8y = 6 × 12
y = \(\frac{12 \times 6}{8}\) = 9 cm
∴ The value of x = 4 cm and y = 9 cm.

(ii) In ∆ HBC and ∆ HFG,
∠H = ∠H (common angle)
∠HFG = ∠HBC (corresponding angle)
Given FG || BC
\(\frac { HF }{ HB } \) = \(\frac { FG }{ BC } \) ⇒ \(\frac { 4 }{ 10 } \) = \(\frac { x }{ 9 } \) ⇒ 10x = 36
x = 3.6 cm
In ∆ FBD and ∆ FHD,
∠BFD = ∠HFG (vertically opposite angle)
∠FBD = ∠FHG (Alternate angles)
By AA similarity
∆ FBD ~ ∆ FHG
\(\frac { FG }{ FD } \) = \(\frac { FH }{ FB } \) ⇒ \(\frac { x }{ 3+y } \) = \(\frac { 4 }{ 6 } \)
4 (3 + y) = 3.6 × 6
3 + y = \(\frac{3.6 \times 6}{4}\) = 5.4 ⇒ y = 5.4 – 3 = 2.4 cm
In ∆ AEG and ∆ ABC,
∠A = ∠A (common angle)
∠AEG = ∠ABC (corresponding angles)
Given EG || BC
\(\frac { AE }{ AB } \) = \(\frac { EG }{ BC } \)
\(\frac { z }{ z+5 } \) = \(\frac { x+y }{ 9 } \) ⇒ \(\frac { z }{ z+5 } \) = \(\frac { 3.6+2.4 }{ 9 } \)
\(\frac { z }{ z+5 } \) = \(\frac { 6 }{ 9 } \) ⇒ \(\frac { z }{ z+5 } \) = \(\frac { 2 }{ 3 } \)
3z = 2z + 10 ⇒ 3z – 2z = 10 ⇒ z = 10
∴ The values of x = 3.6 cm, y = 2.4 cm and z = 10 cm.

Question 6.
The internal bisector of ∠A of AABC meets BC at D and the external bisector of ∠A meets BC produced at E. Prove that \(\frac { BD }{ BE } \) = \(\frac { CD}{ CE } \)
Solution:
Given: In ∆ ABC, AD is the internal bisector of ∠A meets BC at D. AE is the external bisector of ∠A meets BC produced to E.
To Proof: \(\frac { BD }{ BE } \) = \(\frac { CD }{ CE } \)

Proof: In ∆ ABC, AD is the internal bisector of ∠A.
By ABT theorem we have,
∴ \(\frac { BE }{ CD } \) = \(\frac { AB }{ AC } \) ……….(1)
In ∆ABC, AE is the external bisector of ∠A.
By ABT theorem we have,
∴ \(\frac { BE }{ CE } \) = \(\frac { AB }{ AC } \) ……….(2)
From (1) and (2) we get
\(\frac { BD }{ CD } \) = \(\frac { BE }{ CE } \) ⇒ ∴ \(\frac { BD }{ BE } \) = \(\frac { CD }{ CE } \)

Question 7.
In a quadrilateral ABCD, the bisectors of ∠B and ∠D intersect on AC at E. Prove that \(\frac { AB }{ BC } \) = \(\frac { AD }{ DC } \)
Solution:
Given: ABCD is a quadrilateral. BE is the bisector of ∠B intersecting AC at E, DE is the bisector of ∠D intersecting AC at E.
To proove: \(\frac { AB }{ BC } \) = \(\frac { AD }{ DC } \)
Proof: In ∆ABC, BE is the internal bisector of ∠D.
By Angle bisector theorem we have,
\(\frac { AE }{ EC } \) = \(\frac { AB }{ BC } \) ……..(1)

In ∆ ACD, DE is the internal bisector of ∠C.
By Angle bisector theorem we have,
∴ \(\frac { AE }{ EC } \) = \(\frac { AD }{ DC } \) ……….(2)
From (1) and (2) we get,
\(\frac { AB }{ BC } \) = \(\frac { AD }{ DC } \)

Question 8.
ABCD is a quadrilateral with AB parallel to DC. A line drawn parallel to AB meets AD at P and BC at Q. Prove that \(\frac { AP }{ PD } \) = \(\frac { BQ }{ QC } \)
Solution:
Given: ABCD is a quadrilateral. AB || DC.
The line PQ intersect AD at P and BC at Q

To prove: \(\frac { AP }{ PD } \) = \(\frac { BQ }{ QC } \)
Proof: In the ∆ABC, OQ || AB
By BPT, theorem we have, \(\frac { AO }{ OC } \) = \(\frac { BQ}{ QC } \) ……………(1)
In the ∆ACD, PO || DC
By BPT, theorem we have, \(\frac { AO }{ OC } \) = \(\frac { AP }{ PD } \) ……..(2)
From (1) and (2) we get, \(\frac { AP }{ PD } \) = \(\frac { BQ }{ QC } \)

Question 9.
D is the midpoint of the side BC of AABC. If P and Q are points on AB and on AC such that DP bisects ∠BDA and DQ bisects ∠ADC, then prove that PQ || BC.
Solution:
In ∆ABD,DP is the angle bisector of ∠BDA.
∴ \(\frac { AP }{ PB } \) = \(\frac { AD }{ BD } \) (angle bisector theorem) ……..(1)
In ∆ADC, DQ is the bisector of ∠ADC
∴ \(\frac { AQ }{ QC } \) = \(\frac { AD }{ DC } \) (angle bisector theorem) ……….(2)

But, BD = DC (D is the midpoint of BC)
Now (2) ⇒ ∴ \(\frac { AQ }{ QC } \) = \(\frac { AD }{ BD } \)
From (1) and (3) We get,
∴ \(\frac { AP }{ PB } \) = \(\frac { AQ }{ QC } \)
Thus PQ || BC (converse of thales theorem)

Question 10.
ABCD is a trapezium with AB || DC. The diagonal AC and BD intersect at E. If ∆AED ~ ∆BEC. Prove that AD = BC.
Solution:
By given data ABCD is a trapezium with AB || DC.
In ∆ ECD and ∆ ABE

∠EDC = ∠EBA
∠ECD = ∠EAB
∴ ∆ DEC ~ ∆ BEA by AA – similarity
\(\frac { DE }{ BE } \) = \(\frac { EC }{ EA } \) = \(\frac { DC }{ BA } \)
\(\frac { DE }{ BE } \) = \(\frac { EC }{ EA } \)
\(\frac { DE }{ EC } \) = \(\frac { BE }{ EA } \) ……….(1)

Also given ∆ DEA ~ ∆ CEB
\(\frac { DE }{ CE } \) = \(\frac { EA }{ EB } \) = \(\frac { DA }{ CB } \)
\(\frac { DE }{ CE } \) = \(\frac { EA }{ EB } \) …………(2)
From (1) and (2) we get
\(\frac { BE }{ EA } \) = \(\frac { EA }{ EB } \) ⇒ EB² = EA²
∴ EB = EA
Substitute in (2) we get
\(\frac { EA }{ EA } \) = \(\frac { DA }{ CB } \)
1 = \(\frac { DA }{ CB } \) ⇒ ∴ AD = DC Hence it is proved

Students can download Maths Chapter 3 Ratio and Proportion Ex 3.5 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 3 Ratio and Proportion Ex 3.5

Miscellaneous Practice Problems

Question 1.
The maximum speed of some of the animals are given below:
the Elephant = 20 km/h; the
Lion = 80 km/h;
the Cheetah =100 km/h
Find the following ratios of their speeds in simplified form and find which ratio is the least?
(i) the Elephant and the Lion
(ii) the Lion and the Cheetah
(iii) the Elephant and the Cheetah
Solution:
(i) The Elephant: the Lion
= 20 : 80 = \(\frac{20}{80}\) = \(\frac{1}{4}\) = 1 : 4

(ii) the Lion : the Cheetah
= 80 : 100 = \(\frac{80}{100}\) = \(\frac{4}{5}\) = 4 : 5

(iii) the Elephant: the Cheetah
= 20 : 100 = \(\frac{20}{100}\) = \(\frac{1}{5}\) = 1 : 5
The ratio of Elephant to Cheetah is the least.

Question 2.
A particular high school has 1500 students 50 teachers and 5 administrators. If the school grows to 1800 students and the ratios are maintained, then find the number of teachers and administrators.
Solution:
Administrators : teachers : students = 5 : 50 : 1500 = 1 : 10 : 300
If the school grows to 1800 students then 10 parts = teachers
1 part = administrators
300 parts = 1800
1 part = \(\frac{1800}{300}\) = 6
10 parts = 6 × 10 = 60
So, if the school grows to 1800 students the new ratio is administrators : teachers: students
6 : 60 : 1800

Question 3.
I have a box which has 3 green, 9 blue, 4 yellow, 8 orange coloured cubes in it.
(a) What is the ratio of orange to yellow cubes?
(b) What is the ratio of green to blue cubes?
(c) How many different ratios can be formed, when you compare each colour to any one of the other colours?
Solution:
Number of green cubes = 3
Number of blue cubes = 9
Number of yellow cubes = 4
Number of orange cubes = 8
(a) Ratio of orange to yellow cubes \(\frac{\text { Number of orange cubes }}{\text { Number of yellow cubes }}=\frac{8}{4}=\frac{2}{1}=2: 1\)
Ratio of orange to yellow cubes = 2 : 1
(b) \(\frac{\text { Number of green cubes }}{\text { Number of blue cubes }}=\frac{3}{9}=\frac{1}{3}\)
Ratio of green to blue cubes = 1 : 3
(c) The ratios can be Orange : Yellow, Orange: blue, Orange : green, Yellow : Orange, yellow : blue, yellow : green, blue : green, blue : orange, blue : yellow, green : orange, green : yellow, green : blue. Thus 12 ratios can be formed.

Question 4.
A gets double of what B gets and B gets double of what C gets. Find A : B and B : C and verify whether the result is in proportion or not.
Solution:
Let x be the part owned by C then A : B : C = 2(2x) : 2x : x = 4x : 2x : x
A : B = 4x : 2x = 2 : 1
B : C = 2x : x = 2 : 1
A : B : : B : C. i.e, They are in proportion.

Question 5.
The ingredients required for the preparation of Ragi Kali, a healthy dish of Tamilnadu is given below.

(a) If one cup of ragi flour is used then, what would be the amount of raw rice required?
(b) If 16 cups of water are used, then how much ragi flour should be used?
(c) Which of these ingredients cannot be expressed as a ratio? Why?
Solution:
(i) \(\frac{1}{4}\) cup
(ii) 8 cups
(iii) Ragi flour, Raw rice, and water are in one unit. Sesame oil and salt are in different units. These different units cannot be compared and cannot be expressed as a ratio because the two quantities of a ratio should be in the same unit.

Question 6.
Antony brushes his teeth in the morning and night on all days of the week. Shabeen brushes her teeth only in the morning. What is the ratio of the number of times they brush their teeth in a week?
Solution:
Number of times Antony brushes a day = 2
Number of times Antony brushes a week = 2 × 7 = 14
Number of times Shabeen brushes a day = 1
Number of times Shabeen brushes a week = 1 × 7 = 7
Number of times Antony brushes : Number of times Antony brushes = 14 : 7 = 2 : 1
The required ratio = 2 : 1

Question 7.
Thirumagal’s mother wears a bracelet made of 35 red beads and 30 blue beads. Thirumagal wants to make smaller bracelets using the same two coloured beads in the same ratio. In how many different ways can she make the bracelets?

Solution:
Red : blue = 35 : 30 = 7 : 6
Different ways (i) 7 : 6
(ii) 14 : 12;
(iii) 21 : 18;
(iv) 28 : 24

Question 8.
Team A wins 26 matches out of 52 matches. Team B wins three fourth of 52 matches played. Which team has a better winning record?
Solution:
Team A = \(\frac{26}{52}\) = \(\frac{1}{2}\)
Team B = \(\frac{3}{4}\) × 52 = 39
Team B has a better winning record.

Question 9.
In a school excursion, 6 teachers and 12 students from 6th standard and 9 teachers and 27 students from 7th standard, 4 teachers and 16 students from 8th standard took part. Which class has the least teacher to student ratio?
Solution:
Std VI – teachers: students = 6 : 12 = 1 : 2
Std VII – teachers : students = 9 : 27 = 1 : 3
Std VIII – teachers : students = 4 : 16 = 1 : 4
Std VIII has the least ratio.

Question 10.
Fill the boxes using any set of suitable numbers 6 : ___ : : ___ : 15
Solution:
6 : ……. = …….. : 15
Product of the extremes = 6 × 15 = 90
Set of suitable numbers
1 and 90, 2 and 45, 3 and 30, 5 and 18, 6 and 15

Question 11.
From your school diary, write the ratio of the number of holidays to the number of working days in the current academic year.
Solution:
Number of holidays = 145
Number of working days = 220
Holidays : working days = 145 : 220
= \(\frac{145}{220}\)
= \(\frac{29}{44}\)
= 29 : 44

Question 12.
If the ratio of Green, Yellow and Black balls in a bag is 4 : 3 : 5, then
(a) Which is the most likely ball that you can choose from the bag?
(b) How many balls in total are there in the bag if you have 40 black balls in it?
(c) Find the number of green and yellow balls in the bag.
Solution:
Green : Yellow : Black = 4 : 3 : 5
(i) Blackballs;
(ii) 96 balls (32 + 24 + 40);
(iii) green balls = 32
yellow balls = 24

Students can download Maths Chapter 3 Ratio and Proportion Ex 3.4 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 3 Ratio and Proportion Ex 3.4

Question 1.
Fill in the blanks.
(i) If the cost of 3 pens is Rs 18, then the cost of 5 pens is ……..
(ii) If Karkuzhali earns Rs 1800 in 15 days, then she earns Rs 3000 in …….. days
Solution:
(i) ₹ 30
Hint: \(5 \times \frac{18}{3}\) = 5 × 6 = ₹ 30
(ii) 25 Days
Hint:
\(\frac{1800}{3000}=\frac{15}{x}\)
⇒ x = \(\frac{15 \times 3000}{1800}\) = 25 days

Question 2.
Say True or False.
(i) If the weight of 40 books is 8 kg, then the weight of 15 books is 3 kg.
(ii) A car travels 90 km in 3 hours with constant speed. It will travel 140 km in 5 hours at the same speed.
Solution:
(i) True
Hint: Weight of 1 book = \(\frac{8}{40}=\frac{1}{5} \mathrm{kg}\)
Flence Weight of 15 books = \(\frac{1}{5} \times 15=3 \mathrm{kg}\)
(ii) False
1 hour the car travels = \(\frac{90}{3}\) = 30 km
In 5 hours the car travels = 30 × 5 = 150 km

Question 3.
If a person reads 20 pages of a book in 2 hours, how many pages will he read in 8 hours at the same speed?
Solution:
In 2 hours, pages read = 20
In 1 hour, pages read = \(\frac{20}{2}\) = 10
In 8 hours, pages read = 10 × 8
= 80 pages

Question 4.
The cost of 15 chairs is ₹ 7500. Find the number of such chairs that can be purchased for ₹ 12,000?
Solution:
Cost of 15 chairs = Rs 7500
Cost of 1 chair = Rs \(\frac{7500}{15}\) = Rs 500
Number of chairs that can be purchased for Rs 12000 = \(\frac{12000}{500}\) = 24 chairs

Question 5.
A car covers a distance of 125 km in 5 kg of LP Gas. How much distance will it cover in 3 kg of LP Gas?
Solution:
In 5 kg of LPG gas, distance covered = 125 km
In 1 kg of LPG gas, distance covered = \(\frac{125}{5}\) = 25 km
In 3 kg of LPG gas, distance covered = 3 × 25 km = 75 km

Question 6.
Cholan walks 6 km in 1 hour at a constant speed. Find the distance covered by him in 20 minutes at the same speed.
Solution:
In 1 hour (60 minutes), distance covered = 6 km
In 1 minute, distance covered = \(\frac{6 km}{60 min}\) = \(\frac{6000 m}{60}\) = 100 m
In 20 minutes, distance covered = 20 × 100 m = 2000 m = 2 km

Question 7.
The number of correct answers given by Kaarmugilan and Kavitha in a quiz competition are in the ratio 10 : 11. If they had scored a total of 84 points in the competition, then how many points did Kavitha get?
Solution:
Total points scored = 84 Ratio = 10 : 11
Sum of the ratio = 10 + 11 = 21
21 parts = 84 points
1 part = \(\frac{84}{21}\) = 4 points
Kavitha = 11 parts
Kaarmugilan = 10 parts
Foints scored by Kavitha = 11 parts = 11 × 4 points = 44 points

Question 8.
Karmegam made 54 runs in 9 overs and Asif made 77 runs in 11 overs. Whose run rate is better? (run rate = ratio of runs to overs)
Solution:
Karmegam Runs made in 9 overs = 54
Runs made in 1 over = \(\frac{54}{9}\) = 6 runs
Asif Runs made in 11 overs = 77
Runs made in 1 over = \(\frac{77}{11}\) = 7 runs
∴ Asif’s run rate is better than Karmegam.

Question 9.
You purchase 6 apples for Rs 90 and your friend purchases 5 apples for Rs 70. Whose purchase is better?
Solution:
Myself
Cost of 6 apples = Rs 90
Cost of 1 apple = \(\frac{Rs 90}{6}\) = Rs 15
Friend’s purchase
Cost of 5 apples = Rs 70
Cost of 1 apple = \(\frac{70}{5}\) = Rs 14
∴ Friend’s purchase is better than mine.

Objective Type Questions

Question 10.
If a Barbie doll costs ₹ 90, then the cost of 3 such dolls is ₹ _____
(a) 260
(b) 270
(c) 30
(d) 93
Solution:
(b) 270
Hint:
Cost of 3 dolls = 90 × 3 = ₹ 270

Question 11.
If 8 oranges cost Rs 56, then the cost of 5 oranges is Rs …….
(a) 42
(b) 48
(c) 35
(d) 24
Solution:
(c) 35

Question 12.
If a man walks 2 km in 15 minutes, then he will walk _____ km in 45 minutes.
(a) 10
(b) 8
(c) 6
(d) 12
Solution:
(c) 6
Hint:
1 min he walks = \(\frac{2}{15}\) km
45 min he walks = \(\frac{2}{15}\) × 45 = 6 km.

Students can download Maths Chapter 3 Ratio and Proportion Ex 3.2 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 3 Ratio and Proportion Ex 3.2

Question 1.
Fill in the blanks of the given equivalent ratios.
(i) 3 : 5 = 9 : ……
(ii) 4 : 5 = …… : 10
(iii) 6 : …… = 1 : 2
Solution:
(i) 15
Hint: \(\frac{3}{5}=\frac{3 \times 3}{5 \times 3}=\frac{9}{15}\)
(ii) 8
Hint: \(\frac{4}{5}=\frac{4 \times 2}{5 \times 2}=\frac{8}{10}\)
(iii) 12
Hint: \(\frac{1}{2}=\frac{1 \times 6}{2 \times 6}=\frac{6}{12}\)

Question 2.
Complete the table.

Solution:
(i) 1 feet = 12 inches
3 feet = 3 × 12 inches = 36 inches
72 inches = 6 × 12 inches = 6 feet

(ii) 1 week = 7 days
2 weeks = 2 × 7 days = 14 days
63 days = 9 × 7 days = 9 weeks

Question 3.
Say True or False.
(i) 5 : 7 is equivalent to 21 : 15
(ii) If 40 is divided in the ratio 3 : 2, then the larger part is 24
Solution:
(i) False
Hint: \(\frac{21}{15}=\frac{7}{5}=7: 5\)
(ii) True
Hint: \(\frac{3}{5} \times 40=24\)

Question 4.
Give two equivalent ratios for each of the following.
(i) 3 : 2
(ii) 1 : 6
(iii) 5 : 4
Solution:
(i) 3 : 2

3 : 2 = 6 : 4 = 9 : 6

(ii) 1 : 6

1 : 6 = 2 : 12 = 3 : 18

(iii) 5 : 4

5 : 4 = 10 : 8 = 15 : 12

Question 5.
Which of the two ratios is larger?
(i) 4 : 5 or 8 : 15
(ii) 3 : 4 or 7 : 8
(iii) 1 : 2 or 2 : 1
Solution:
(i) 4 : 5 (or) 8 : 15

4 : 5 > 8 : 15

(ii) 3 : 4 (or) 7 : 8

7 : 8 > 3 : 4

(iii) 1 : 2 (or) 2 : 1
1 : 2 = \(\frac{1}{2}\)
2 : 1 = \(\frac{2}{1}\)
= 2
\(\frac{2}{1}\) > \(\frac{1}{2}\)
2 : 1 > 1 : 2

Question 6.
Divide the numbers given below in the required ratio.
(i) 20 in the ratio 3 : 2
(ii) 27 in the ratio 4 : 5
(iii) 40 in the ratio 6 : 14.
Solution:
(i) Ratio = 3 : 2
Sum of the ratio = 3 + 2 = 5
5 parts = 20
1 part = \(\frac{20}{5}\) = 4
3 parts = 3 × 4 = 12
2 parts = 2 × 4 = 8
20 can be divided in the form as 12, 8.

(ii) Ratio = 4 : 5
Sum of the ratio = 4 + 5 = 9
9 parts = 27
1 part = \(\frac{27}{9}\) = 3
4 parts = 4 × 3 = 12
5 parts = 5 × 3 =15
27 can be divided in the form as 12, 15.

(iii) 40 in the ratio 6 : 14
Ratio = 6 : 14
Sum of the ratio = 6 + 14 = 20
20 parts = 40
1 part = \(\frac{40}{20}\) = 2
6 parts = 2 × 6 = 12
14 parts = 2 × 14 = 28
40 can be divided in the form as 12, 28.

Question 7.
In a family, the amount spent in a month for buying Provisions and Vegetables are in the ratio 3 : 2. If the allotted amount is Rs 4000, then what will be the amount spent for
(i) Provisions and
(ii) Vegetables?
Solution:
Allotted amount = Rs 4000
Ratio = 3 : 2
Sum of the ratio = 3 + 2 = 5
5 parts = Rs 4000
1 part = Rs \(\frac{4000}{5}\) = Rs 800
Provisions : Vegetables = 3 : 2
3 parts = 3 × Rs 800 = Rs 2400
2 parts = 2 × Rs 800 = Rs 1600
Amount spent for provisions = Rs 2400
Amount spent for vegetables = Rs 1600

Question 8.
A line segment 63 cm long is to be divided into two parts in the ratio 3 : 4. Find the length of each part.
Solution:
Total length = 63 cm Ratio = 3 : 4
Sum of the ratio = 3 + 4 = 7
7 parts = 63 cm
1 part = \(\frac{63}{7}\) = 9 cm
3 parts = 3 × 9 cm = 27 cm
4 parts = 4 × 9 cm = 36 cm
∴ 63 cm can be divided into the parts as 27 cm and 36 cm.

Objective Type Questions

Question 9.
If 2 : 3 and 4 : …… are equivalent ratios, then the missing term is
(a) 6
(b) 2
(c) 4
(d) 3
Solution:
(a) 6

Question 10.
An equivalent ratio of 4 : 7 is
(a) 1 : 3
(b) 8 : 15
(c) 14 : 8
(d) 12 : 21
Solution:
(d) 12 : 21

Question 11.
Which is not an equivalent ratio of \(\frac{16}{24}\)?
(a) \(\frac{6}{9}\)
(b) \(\frac{12}{18}\)
(c) \(\frac{10}{15}\)
(d) \(\frac{20}{28}\)
Solution:
(d) \(\frac{20}{28}\)

Question 12.
If Rs 1600 is divided
(a) Rs 480
(b) Rs 800
(c) Rs 1000
(d) Rs 200
Solution:
(c) Rs 1000

Students can download Maths Chapter 4 Geometry Ex 4.1 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Ex 4.1

Question 1.
Check whether the which triangles are similar and find the value of x.


Solution:
(i) In ∆ABC and ∆AED
\(\frac { AB }{ AD } \) = \(\frac { AC }{ AE } \)
\(\frac { 8 }{ 3 } \) = \(\frac{11}{\frac{2}{2}}\)
\(\frac { 8 }{ 3 } \) = \(\frac { 11 }{ 4 } \) ⇒ 32 ≠ 33
∴ The two triangles are not similar.

(ii) In ∆ABC and ∆PQC
∠PQC = 70°
∠ABC = ∠PQC = 70°
∠ACB = ∠PCQ (common)
∆ABC ~ ∆PQC
\(\frac { 5 }{ X } \) = \(\frac { 6 }{ 3 } \)
6x = 15
x = \(\frac { 15 }{ 6 } \) = \(\frac { 5 }{ 2 } \)
∴ x = 2.5
∆ ABC and ∆PQC are similar. The value of x = 2.5

Question 2.
A girl looks the reflection of the top of the lamp post on the mirror which is 66 m away from the foot of the lamppost. The girl whose height is 12.5 m is standing 2.5 m away from the mirror. Assuming the mirror is placed on the ground facing the sky and the girl, mirror and the lamppost are in a same line, find the height of the lamp post.
Solution:
Let the height of the tower ED be “x” m. In ∆ABC and ∆EDC.
∠ABC = ∠CED = 90° (vertical Pole)
∠ACB = ∠ECD (Laws of reflection)
∆ ABC ~ ∆DEC
\(\frac { AB }{ DE } \) = \(\frac { BC }{ EC } \)
\(\frac { 1.5 }{ x } \) = \(\frac { 0.4 }{ 87.6 } \)

x = \(\frac{1.5 \times 87.6}{0.4}\) = \(\frac{1.5 \times 876}{4}\)
= 1.5 × 219 = 328.5
The height of the Lamp Post = 328.5 m

Question 3.
A vertical stick of length 6 m casts a shadow 400 cm long on the ground and at the same time a tower casts a shadow 28 m long. Using similarity, find the height of the tower.
Solution:
In ∆ABC and ∆PQR,
∠ABC = ∠PQR = 90° (Vertical Stick)
∠ACB = ∠PRQ (Same time casts shadow)
∆BCA ~ ∆QRP

\(\frac { AB }{ PQ } \) = \(\frac { BC }{ QR } \)
\(\frac { 6 }{ x } \) = \(\frac { 4 }{ 28 } \)
4x = 6 × 28 ⇒ x = \(\frac{6 \times 28}{4}\) = 42
Length of the lamp post = 42m

Question 4.
Two triangles QPR and QSR, right angled at P and S respectively are drawn on the same base QR and on the same side of QR. If PR and SQ intersect at T, prove that PT × TR = ST × TQ.
Solution:
In ∆PQT and ∆STR we have
∠P = ∠S = 90° (Given)
∠PTQ = ∠STR (Vertically opposite angle)

By AA similarity
∆PTQ ~ ∆STR we get
\(\frac { PT }{ ST } \) = \(\frac { TQ }{ TR } \)
PT × TR = ST × TQ
Hence it is proved.

Question 5.
In the adjacent figure, ∆ABC is right angled at C and DE ⊥ AB. Prove that ∆ABC ~ ∆ADE and hence find the lengths of AE and DE.

Solution:
In ∆ABC and ∆ADE
∠ACB = ∠AED = 90°
∠A = ∠A (common)
∴ ∆ABC ~ ∆ADE (By AA similarity)
\(\frac { BC }{ DE } \) = \(\frac { AB }{ AD } \) = \(\frac { AC }{ AE } \)
\(\frac { 12 }{ DE } \) = \(\frac { 13 }{ 3 } \) = \(\frac { 5 }{ AE } \)
In ∆ABC, AB² = BC² + AC²
= 122 + 52 = 144 + 25 = 169
AB = \(\sqrt { 169 }\) = 13
Consider, \(\frac { 13 }{ 3 } \) = \(\frac { 5 }{ AE } \)
∴ AE = \(\frac{5 \times 3}{13}\) = \(\frac { 15 }{ 13 } \)
AE = \(\frac { 15 }{ 13 } \) and DE = \(\frac { 36 }{ 13 } \)
Consider, \(\frac { 12 }{ DE } \) = \(\frac { 13 }{ 3 } \)
DE = \(\frac{12 \times 3}{13}\) = \(\frac { 36 }{ 13 } \)

Question 6.
In the adjacent figure, ∆ACB ~ ∆APQ. If BC = 8 cm, PQ = 4 cm, BA = 6.5 cm and AP = 2.8 cm, find CA and AQ.

Solution:
Given ∆ACB ~ ∆APQ
\(\frac { AC }{ AP } \) = \(\frac { BC }{ PQ } \) = \(\frac { AB }{ AQ } \)
\(\frac { AC }{ 2.8 } \) = \(\frac { 8 }{ 4 } \) = \(\frac { 6.5 }{ AQ } \)
Consider \(\frac { AC }{ 2.8 } \) = \(\frac { 8 }{ 4 } \)
4 AC = 8 × 2.8
AC = \(\frac{8 \times 2.8}{4}\) = 5.6 cm
Consider \(\frac { 8 }{ 4 } \) = \(\frac { 6.5 }{ AQ } \)
8 AQ = 4 × 6.5
AQ = \(\frac{4 \times 6.5}{8}\) = 3.25 cm
Length of AC = 5.6 cm; Length of AQ = 3.25 cm

Question 7.
If figure OPRQ is a square and ∠MLN = 90°. Prove that

(i) ∆LOP ~ ∆QMO
(ii) ∆LOP ~ ∆RPN
(iii) ∆QMO ~ ∆RPN
(iv) QR2 = MQ × RN.
Solution:
(i) In ∆LOP and ∆QMO
∠OLP = ∠OQM = 90°
∠LOP = ∠OMQ (Since OQRP is a square OP || MN)
∴ ∆LOP~ ∆QMO (By AA similarity)

(ii) In ∆LOP and ∆RPN
∠OLP = ∠PRN = 90°
∠LPO = ∠PNR (OP || MN) .
∴ ∆LOP ~ ∆RPN (By AA similarity)

(iii) In ∆QMO and ∆RPN
∠MQO = ∠NRP = 90°
∠RPN = ∠QOM (OP || MN)
∴ ∆QMO ~ ∆RPN (By AA similarity)

(iv) We have ∆QMO ~ ∆RPN
\(\frac { MQ }{ PR } \) = \(\frac { QO }{ RN } \)
\(\frac { MQ }{ QR } \) = \(\frac { QR }{ RN } \)
QR² = MQ × RN
Hence it is proved.

Question 8.
If ∆ABC ~ ∆DEF such that area of ∆ABC is 9cm² and the area of ∆DEF is 16 cm² and BC = 2.1 cm. Find the length of EF.
Solution:
Given ∆ABC ~ ∆DEF

\(\frac { 9 }{ 16 } \) = \(\frac{(2.1)^{2}}{\mathrm{E} \mathrm{F}^{2}}\)
(\(\frac { 3 }{ 4 } \))² = (\(\frac { 2.1 }{ EF } \))²
\(\frac { 3 }{ 4 } \) = \(\frac { 2.1 }{ EF } \)
EF = \(\frac{4 \times 2.1}{3}\) = 2.8 cm
Legth of EF = 2.8 cm

Question 9.
Two vertical poles of heights 6 m and 3 m are erected above a horizontal ground AC. Find the value of y.

Solution:
In the ∆PAC and ∆BQC
∠PAC = ∠QBC = 90°
∠C is common
∆PAC ~ QBC
\(\frac { AP }{ BQ } \) = \(\frac { AC }{ BC } \)
\(\frac { 6 }{ y } \) = \(\frac { AC }{ BC } \)
∴ \(\frac { BC }{ AC } \) = \(\frac { y }{ 6 } \) …..(1)
In the ∆ACR and ∆QBC
∠ACR = ∠QBC = 90°
∠A is common
∆ACR ~ ABQ
\(\frac { RC }{ QB } \) = \(\frac { AC }{ AB } \)
\(\frac { 3 }{ y } \) = \(\frac { AC }{ AB } \)
\(\frac { AB }{ AC } \) = \(\frac { y }{ 3 } \) ……..(2)
By adding (1) and (2)
\(\frac { BC }{ AC } \) + \(\frac { AB }{ AC } \) = \(\frac { y }{ 6 } \) + \(\frac { y }{ 3 } \)
1 = \(\frac{3 y+6 y}{18}\)
9y = 18 ⇒ y = \(\frac { 18 }{ 9 } \) = 2
The Value of y = 2m

Question 10.
Construct a triangle similar to a given triangle PQR with its sides equal to \(\frac { 2 }{ 3 } \) of the corresponding sides of the triangle PQR (scale factor \(\frac { 2 }{ 3 } \) ).
Solution:
Given ∆PQR, we are required to construct another triangle whose sides are \(\frac { 2 }{ 3 } \) of the corresponding sides of the ∆PQR

Steps of construction:
(i) Construct a ∆PQR with any measurement.
(ii) Draw a ray QX making an acute angle with QR on the side opposite to the vertex P.
(iii) Locate 3 points Q₁, Q₂ and Q₃ on QX.
So that QQ₁ = Q₁Q₂ = Q₂Q₃
(iv) Join Q₃ R and draw a line through Q₂ parallel to Q₃ R to intersect QR at R’.
(v) Draw a line through R’ parallel to the line RP to intersect QP at P’. Then ∆ P’QR’ is the required triangle.

Question 11.
Construct a triangle similar to a given triangle LMN with its sides equal to \(\frac { 4 }{ 5 } \) of the corresponding sides of the triangle LMN (scale factor \(\frac { 4 }{ 5 } \) ).
Solution:
Given a triangle LMN, we are required to construct another ∆ whose sides are \(\frac { 4 }{ 5 } \) of the corresponding sides of the ∆LMN.

Steps of Construction:

1. Construct a ∆LMN with any measurement.
2. Draw a ray MX making an acute angle with MN on the side opposite to the vertex L.
3. Locate 5 Points Q₁, Q₂, Q₃, Q₄, Q₅ on MX.
   So that MQ₁ = Q₁Q₂ = Q₂Q₃ = Q₃Q₄ = Q₄Q₅
4. Join Q₅ N and draw a line through Q₄. Parallel to Q₅N to intersect MN at N’.
5. Draw a line through N’ parallel to the line LN to intersect ML at L’.
   ∴ ∆L’ MN’ is the required triangle.

Question 12.
Construct a triangle similar to a given triangle ABC with its sides equal to \(\frac { 6 }{ 5 } \) of the corresponding sides of the triangle ABC (scale factor \(\frac { 6 }{ 4 } \)).
Solution:
Given triangle ∆ABC, we are required to construct another triangle whose sides are \(\frac { 6 }{ 5 } \) of the corresponding sides of the ∆ABC.
Steps of construction
(i) Construct an ∆ABC with any measurement.
(ii) Draw a ray BX making an acute angle with BC.
(iii) Locate 6 points Q₁, Q₂, Q₃, Q₄, Q₅, Q₆ on BX such that
BQ₁ = Q₁Q₂ = Q₂Q₃ = Q₃Q₄ = Q₅Q₆

(iv) Join Q₅ to C and draw a line through Q₆ parallel to Q₅ C intersecting the extended line BC at C’.
(v) Draw a line through C’ parallel to AC intersecting the extended line segment AB at A’.
∆A’BC’ is the required triangle.

Question 13.
Construct a triangle similar to a given triangle PQR with its sides equal to \(\frac { 7 }{ 3 } \) of the corresponding sides of the triangle PQR (scale factor \(\frac { 7 }{ 3 } \)).
Solution:
Given triangle ABC, we are required to construct another triangle whose sides are \(\frac { 7 }{ 3 } \) of the corresponding sides of the ∆ABC.
Steps of construction
(i) Construct a ∆PQR with any measurement.
(ii) Draw a ray QX making an acute angle with QR on the side opposite to the vertex P.
(iii) Locate 7 points Q₁, Q₂, Q₃, Q₄, Q₅, Q₆, Q₇ on QX.
So that
QQ₁ = Q₁Q₂ = Q₂Q₃ = Q₃Q₄ = Q₅Q₆ = Q₆Q₇

(iv) Join Q₃ to R and draw a line through Q₇ parallel to Q₃R intersecting the extended line segment QR at R’.
(v) Draw a line through parallel to RP.
Intersecting the extended line segment QP at P’.
∴ ∆P’QR’ is the required triangle.

Students can download Maths Chapter 4 Geometry Ex 4.3 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Ex 4.3

Question 1.
A man goes 18 m due east and then 24 m due north. Find the distance of his current position from the starting point?
Solution:
Let the initial position of the man be “O” and his final
position be “B”.
By Pythagoras theorem
In the right ∆ OAB,

OB² = OA² + AB²
= 18² + 24²
= 324 + 576 = 900
OB = \(\sqrt { 900 }\) = 30
The distance of his current position is 30 m

Question 2.
There are two paths that one can choose to go from Sarah’s house to James house. One way is to take C street, and the other way requires to take A street and then B street. How much shorter is the direct path along C street? (Using figure).

Solution:
Distance between Sarah House and James House using “C street”.
AC² = AB² + BC²
= 2² + 1.5²
= 4 + 2.25 = 6.25
AC = \(\sqrt { 6.25 }\)
AC = 2.5 miles

Distance covered by using “A Street” and “B Street”
= (2 + 1.5) miles = 3.5 miles
Difference in distance = 3.5 miles – 2.5 miles = 1 mile

Question 3.
To get from point A to point B you must avoid walking through a pond. You must walk 34 m south and 41 m east. To the nearest meter, how many meters would be saved if it were possible to make a way through the pond?
Solution:
In the right ∆ABC,
By Pythagoras theorem
AC²= AB² + BC² = 34² + 41²
= 1156 + 1681 = 2837
AC = \(\sqrt { 2837 }\)
= 53.26 m

Through A one must walk (34m + 41m) 75 m to reach C.
The difference in Distance = 75 – 53.26
= 21.74 m

Question 4.
In the rectangle WXYZ, XY + YZ = 17 cm, and XZ + YW = 26 cm.
Calculate the length and breadth of the rectangle?

Solution:
Let the length of the rectangle be “a” and the breadth of the rectangle be “b”.
XY + YZ = 17 cm
b + a = 17 …….. (1)
In the right ∆ WXZ,
XZ² = WX² + WZ²
(XZ)² = a² + b²

XZ = \(\sqrt{a^{2}+b^{2}}\)
Similarly WY = \(\sqrt{a^{2}+b^{2}}\) ⇒ XZ + WY = 26
2 \(\sqrt{a^{2}+b^{2}}\) = 26 ⇒ \(\sqrt{a^{2}+b^{2}}\) = 13
Squaring on both sides
a² + b² = 169
(a + b)² – 2ab = 169
17² – 2ab = 169 ⇒ 289 – 169 = 2 ab
120 = 2 ab ⇒ ∴ ab = 60
a = \(\frac { 60 }{ b } \) ….. (2)
Substituting the value of a = \(\frac { 60 }{ b } \) in (1)
\(\frac { 60 }{ b } \) + b = 17
b² – 17b + 60 = 0
(b – 2) (b – 5) = 0
b = 12 or b = 5

If b = 12 ⇒ a = 5
If b = 6 ⇒ a = 12
Lenght = 12 m and breadth = 5 m

Question 5.
The hypotenuse of a right triangle is 6 m more than twice of the shortest side. If the third side is 2 m less than the hypotenuse, find the sides of the triangle.
Solution:
Let the shortest side of the right ∆ be x.
∴ Hypotenuse = 6 + 2x
Third side = 2x + 6 – 2
= 2x + 4

In the right triangle ABC,
AC² = AB² + BC²
(2x + 6)² = x² + (2x + 4)²
4x² + 36 + 24x = x² + 4x² + 16 + 16x
0 = x² – 24x + 16x – 36 + 16
∴ x² – 8x – 20 = 0
(x – 10) (x + 2) = 0
x – 10 = 0 or x + 2 = 0

x = 10 or x = -2 (Negative value will be omitted)
The side AB = 10 m
The side BC = 2 (10) + 4 = 24 m
Hypotenuse AC = 2(10) + 6 = 26 m

Question 6.
5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4m high. If the foot of the ladder is moved 1.6 m towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall.
Solution:
“C” is the position of the foot of the ladder “A” is the position of the top of the ladder.
In the right ∆ABC,
BC² = AC² – AB² = 5² – 4²
= 25 – 16 = 9
BC = \(\sqrt { 9 }\) = 3m.
When the foot of the ladder moved 1.6 m toward the wall.
The distance between the foot of the ladder to the ground is
BE = 3 – 1.6 m
= 1.4 m

Let the distance moved upward on the wall be “h” m
The ladder touch the wall at (4 + h) M
In the right triangle BED,
ED² = AB² + BE²
5² = (4 + h)² + (1.4)²
25 – 1.96= (4 + h)²
∴ 4 + h = \(\sqrt { 23.04 }\)
4 + h = 4. 8 m
h = 4.8 – 4
= 0.8 m
Distance moved upward on the wall = 0.8 m

Question 7.
The perpendicular PS on the base QR of a ∆PQR intersects QR at S, such that QS = 3 SR. Prove that 2PQ² = 2PR² + QR².
Solution:
Given QS = 3SR
QR = QS + SR
= 3SR + SR = 4SR
SR = \(\frac { 1 }{ 4 } \) QR …..(1)
QS = 3SR
SR = \(\frac { QS }{ 3 } \) ……..(2)

From (1) and (2) we get
\(\frac { 1 }{ 4 } \) QR = \(\frac { QS }{ 3 } \)
∴ QS = \(\frac { 3 }{ 4 } \) QR ………(3)
In the right ∆ PQS,
PQ² = PS² + QS² ……….(4)
Similarly in ∆ PSR
PR² = PS² + SR² ………..(5)
Subtract (4) and (5)
PQ² – PR² = PS² + QS² – PS² – SR²
= QS² – SR²

PQ² – PR² = \(\frac { 1 }{ 2 } \) QR²
2PQ² – 2PR² = QR²
2PQ² = 2PR² + QR²
Hence the proved.

Question 8.
In the adjacent figure, ABC is a right angled triangle with right angle at B and points D, E trisect BC. Prove that 8AE² = 3AC² + 5AD².
Solution:
Since the Points D, E trisect BC.
BD = DE = CE
Let BD = DE = CE = x
BE = 2x and BC = 3x

In the right ∆ABD,
AD² = AB² + BD²
AD² = AB² + x² ……….(1)
In the right ∆ABE,
AE² = AB² + 2BE²
AE² = AB² + 4X² ………..(2) (BE = 2x)
In the right ∆ABC
AC² = AB² + BC²
AC² = AB² + 9x² …………… (3) (BC = 3x)
R.H.S = 3AC² + 5AD²
= 3[AB² + 9x²] + 5 [AB² + x²] [From (1) and (3)]
= 3AB² + 27x² + 5AB² + 5x²
= 8AB² + 32x²
= 8 (AB² + 4 x²)
= 8AE² [From (2)]
= R.H.S.
∴ 8AE² = 3AC² + 5AD²

Students can download Maths Chapter 7 Mensuration Additional Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 7 Mensuration Additional Questions

I. Choose the Correct Answer

Question 1.
If the sides of a triangles are 5 cm, 6 cm and 7 cm then the area is ……..
(a) 18 cm²
(b) 6 √2 cm²
(c) 6 √6 cm²
(d) 6 √3 cm²
Solution:
(c) 6 √6 cm²

Question 2.
The perimeter of an equilateral triangle is 60 cm then the area is ………
(a) 60 √3 cm²
(b) 20 √3 cm²
(c) 50 √3 cm²
(d) 100 √3 cm²
Solution:
(d) 100 √3 cm²

Question 3.
The total surface area of the cuboid with dimension 20 cm × 30 cm × 15 cm is ………
(a) 2700 cm²
(b) 1500 cm²
(c) 2500 cm²
(d) 3000 cm²
Solution:
(a) 2700 cm²

Question 4.
The number of bricks each measuring 70 cm × 80 cm × 40 cm that will be required to build a wall whose dimensions are 7 m × 8 m × 4 m is ……..
(a) 4000
(b) 3000
(c) 2000
(d) 1000
Solution:
(d) 1000

Question 5.
The volume of a cube is 4913 m² then the length of its side is ……..
(a) 13 m
(b) 17 m
(c) 34 m
(d) 27 m
Solution:
(b) 17 m

II. Answer the Following Questions

Question 6.
A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non parallel sides are 14 m and 13 m. Find the area of the field.
Solution:
The non parallel sides are 13 m and 14 m. Draw BE || AD. Such that BE = 13 m
∴ ABED is a parallelogram

To find Area of a ΔBCE
a = 13 m, b = 15 m and c = 14 m
s = \(\frac{a+b+c}{2}\)
= \(\frac{13+15+14}{2}\)
= \(\frac{42}{2}\)
= 21 m
s – a = 21 – 13 = 8 m
s – b = 21 – 15 = 6 m
s – c = 21 – 14 = 7 m
Area of a ΔBCE

= 2² × 3 × 7
= 84 m²
Let the height of the triangle BF be x
Area of the ΔBEC = 84 m²
= \(\frac{1}{2}\) × b × h = 84
= \(\frac{1}{2}\) × 15 × h = 84
x = \(\frac{84×2}{15}\)
= \(\frac{56}{5}\) m
= 11.2 m
Area of parallelogram ABED = base × height sq. units
= 10 × 11.2 m²
= 112 m²
∴ Area of the field = Area of ΔBCE + Area of parallelogram ABED
= 84 m² + 112 m²
= 196 m²
(OR)
Area of the field = Area of the trapezium ABCD
= \(\frac{1}{2}\) h (a + b)
= \(\frac{1}{2}\) × 11.2 (25 + 10)
= \(\frac{1}{2}\) × 11.2 (35)
= 196 m²

Question 7.
Find the area of a quadrilateral ABCD in which AB = 8 cm, BC = 6 cm, CD = 8 cm, DA = 10 cm and AC = 10 cm and ⌊B = 90°.
Solution:

In ΔABC, ⌊B = 90°
∴ ABC is a right angle triangle
Area of the right ΔABC = \(\frac{1}{2}\) × AB × BC sq.units
= \(\frac{1}{2}\) × 8 × 6 cm²
= 24 cm²
In ΔACD a = 10 cm, b = 8 cm and c = 10 cm
s = \(\frac{a+b+c}{2}\)
= \(\frac{10+8+10}{2}\)
= \(\frac{28}{2}\)
= 14 cm
s – a = 14 – 10 = 4 cm
s – b = 14 – 8 = 6 cm
s – c = 14 – 10 = 4 cm
Area of ΔACD = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{14×4×6×4}\)
= \(\sqrt{2×7×4×2×3×4}\)
= 4 × 2 \(\sqrt{21}\) cm²
= 8\(\sqrt{21}\) cm²
= 8 × 4.58
= 36.64 cm²
Area of the quadrilateral ABCD
= Area of ΔABC + Area of ΔACD
= 24 cm² + 36.64 cm²
= 60.64 cm²
Area of the quadrilateral = 60.64 cm²

Question 8.
The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs 7.50 per m².
Solution:
Length of the room (l) = 5 m
Breadth of the room (b) = 4 m
Height of the room (h) = 3 m
Area for white washing = Lateral surface area of four walls + Area of the ceiling
= 2(l + b) × h + (l × b)
= 2(5 + 4) × 3 + (5 × 4) m²
= (2 × 9 × 3 + 20) m²
= (54 + 20) m²
= 74 m²
Cost of white washing for one m² = Rs 7.50
Cost of white washing for 74 m² = Rs 74 × 7.50
= Rs 555
The required cost = Rs 555

Question 9.
How many hollow blocks of size 30 cm × 15 cm × 20 cm are needed to construct a wall 60 m in length 0.3 m in breadth and 2 m in height.
Solution:
Length of a wall = 60 m = 6000 cm
Breadth of a wall = 0.3 m = 30 cm
Height of a wall = 2 m = 200 cm
Volume of the wall = l × b × h sq. unit
= 6000 × 30 × 200 cm³
For hollow block
l = 30 cm, b = 15 cm, h = 20 cm
Volume of one hollow block = l × b × h
= 30 × 15 × 20 cm²
Number of hollow blocks required

= 4000
∴ Number of bricks = 4000

Question 10.
Find the number of cubes of side 3 cm that can be cut from a cuboid of dimensions 10 cm × 9 cm × 6 cm.
Solution:
Side of a cube = 3 cm
Volume of a cube = a³ cm
= 3 × 3 × 3 cm³
Length of the cuboid (l) = 10 cm
Breadth of the cuboid (b) = 9 cm
Height of the cuboid (h) = 6 cm
Volume of the cuboid = l × b × h cu. unit
= 10 × 9 × 6 cm
Number of cubes

∴ Number of cubes = 20