Bhagya

Students can download Maths Chapter 4 Geometry Ex 4.3 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 2 Chapter 4 Geometry Ex 4.3

Miscellaneous Practice Problems

Question 1.
What are the angles of an isosceles right-angled triangle?
Solution:
Since it is a right-angled triangle
One of the angles is 90°
The other two angles are equal because it is an isosceles triangle.
The other two angles must be 45° and 45°
Angles are 90°, 45°, 45°.

Question 2.
Which of the following correctly describes the given triangle?

(a) It is a right isosceles triangle
(b) It is an acute isosceles triangle
(c) It is an obtuse isosceles triangle
(d) It is an obtuse scalene triangle
Solution:
(c) It is an obtuse isosceles triangle

Question 3.
Which of the following is not possible?
(a) An obtuse isosceles triangle.
(b) An acute isosceles triangle.
(c) An obtuse equilateral triangle.
(d) An acute equilateral triangle.
Solution:
(c) An obtuse equilateral triangle.

Question 4.
If one angle of an isosceles triangle is 124°, then find the other angles
Solution:
In an isosceles triangle, any two sides are equal. Also, the two angles are equal.
Sum of three angles of a triangle = 180°
Given one angle = 124°
Sum of other two angles = 180° – 124° = 56°
Other angles are = \(\frac{56}{2}\) = 28°
28° and 28°.

Question 5.
The diagram shows a square ABCD. If the line segment joins A and C, then mention the type of triangle so formed.
Solution:

Isosceles right-angled triangles

Question 6.
Draw a line segment AB of length 6 cm. At each end of this line segment AB, draw a line perpendicular to the line segment AB. Are these lines parallel?
Solution:

yes, they are parallel

Challenge Problems

Question 7.
Is a triangle possible with the angles 90°, 90°, and 0°, Why?
Solution:
No, a triangle cannot have more than one right angle.

Question 8.
Which of the following statements is true? Why?
(a) Every equilateral triangle is an isosceles triangle.
(b) Every isosceles triangle is an equilateral triangle
Solution:
“(a)” is true, because an isosceles triangle need not have three equal sides

Question 9.
If one angle of an isosceles triangle is 70°, then find the possibilities for the other two angles.
Solution:
(i) Given one angle = 70°
Also, it is an isosceles triangle.
Another one angle also can be 70°.
Sum of these two angles = 70° + 70° = 140°
We know that the sum of three angles in a triangle = 180°.
Third angle = 180° – 140° = 40°
One possibility is 70°, 70°, and 40°
(ii) Also if one angle is 70°
Sum of other two angles = 180° – 70° = 110°
Both are equal. They are \(\frac{110}{2}\) = 55°.
Another possibility is 70°, 55° and 55°.

Question 10.
Which of the following can be the sides of an isosceles triangle?
(a) 6 cm, 3 cm, 3 cm
(b) 5 cm, 2 cm, 2 cm
(c) 6 cm, 6 cm, 7 cm
(d) 4 cm, 4 cm, 8 cm
Solution:
(c) 6 cm, 6 cm, 7 cm

Question 11.
Study the given figure and identify the following triangles,

(a) equilateral triangle
(b) isosceles triangle
(c) scalene triangles
(d) acute triangle
(e) obtuse triangle
(f) right triangle
Solution:(a) BC = 1 + 1 + 1 + 1 = 4 cm
AB = AC = 4 cm
∆ABC is an equilateral triangle.
(b) ∆ABC and ∆AEF are isosceles triangles.
Since AB = AC = 4 cm Also AE = AF.
(c) In a scalene triangle, no two sides are equal.
∆AEB, ∆AED, ∆ADF, ∆AFC, ∆ABD, ∆ADC, ∆ABF, and ∆AEC are scalene triangles.
(d) In an acute-angled triangle all the three angles are less than 90°.
∆ABC, ∆AEF, ∆ABF, and ∆AEC are acute-angled triangles.
(e) In an obtuse-angled triangle any one of the angles is greater than 90°.
∆AEB and ∆AFC are obtuse angled triangles.
(f) In a right triangle, one of the angles is 90°.
∆ADB, ∆ADC, ∆ADE, and ∆ADF are right-angled triangles.

Question 12.
Two sides of the triangle are given in the table. Find the third side of the triangle.

Solution:
(i) between 3 and 11
(ii) between 0 and 16
(iii) between 4 and 11
(iv) between 4 and 24

Question 13.
Complete the following table:

Solution:
(i) Always acute angles
(ii) Acute angle
(iii) Obtuse angle

Students can download Maths Chapter 4 Geometry Ex 4.2 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 2 Chapter 4 Geometry Ex 4.2

Question 1.
Draw a line segment AB = 7 cm and mark a point P on it. Draw a line perpendicular to the given line segment at P.
Solution:

Step 1 : Draw a line AB = 7 cm and take a point P anywhere on the line.
Step 2 : Place the set square on the line in such a way that the vertex which forms right angle coincides with P and one arm of the right angle coincides with the line AB.
Step 3 : Draw a line PQ through P along the other arm of the right angle of the set square.
Step 4 : The line PQ is perpendicular to the line AB at P. That is, PQ ⊥ AB
∠APQ = ∠BPQ = 90°

Question 2.
Draw a line segment LM = 6.5 cm and mark a point X not lying on it. Using a set square construct a line perpendicular to LM through X.
Solution:

Step 1 : Draw a line LM = 6.5 cm and take a point X anywhere above the line LM.
Step 2 : Place one of the arms of the right angle of a set square along the line LM and the other arm of its right angle touches the point X.
Step 3 : Draw a line through the point X meeting LM at Y.
Step 4 : The line XY is perpendicular to the line LM at Y. That is, LM ⊥ XY.

Question 3.
Find the distance between the given lines using a set square at two different points on each of the pairs of lines and check whether they are parallel.

Solution:
They are parallel

Question 4.
Draw a line segment measuring 7.8 cm. Mark a point B above it at a distance of 5 cm. Through B draw a line parallel to the given segment.
Solution:

Step 1 : Draw a line. Mark two points M and N on the line such that MN = 7.8 cm. Mark a point B any where above the line.
Step 2 : Place the set square below B in such a way that one of the edges that form a right angle lies along MN Place the scale along the other edge of the set square.
Step 3 : Holding the scale firmly, Slide the set square along the edge of the scale until the other edge of the set square reaches the point B. Through B draw a line.
Step 4 : The line MN is parallel to AB. That is, MN || AB.

Question 5.
Draw a line and mark a point R above it at a distance of 5.4 cm Through R draw a line parallel to the given line.
Solution:

Step 1 : Using a scale draw a line AB and mark a point Q on the line.
Step 2 : Place the set square in such a way that the vertex of the right angle coincides with Q and one of the edges of right angle lies along AB. Mark the point R such that QR = 5.4 cm
Step 3 : Place the scale and the set square as shown in the figure.
Step 4 : Hold the scale firmly and slide the set square along the edge of the scale until the other edge touches the point R. Draw a line RS through R.
Step 5 : The line RS is parallel to AB. That is, RS || AB.

Students can download Maths Chapter 4 Geometry Ex 4.4 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 4 Geometry Ex 4.4

Miscellaneous Practise Problems

Question 1.
Find the type of lines marked in thick lines (Parallel, intersecting or perpendicular)

Solution:
(i) Parallel lines
(ii) Parallel lines
(iii) Parallel and Perpendicular lines
(iv) Intersecting lines

Question 2.
Find the parallel and intersecting line segments in the picture given below.

Solution:
(a) Parallel line segments

• \(\overline{\mathrm{YZ}} \text { and } \overline{\mathrm{DE}}\)
• \(\overline{\mathrm{EA}} \text { and } \overline{\mathrm{ZV}}\)
• \(\overline{\mathrm{VW}} \text { and } \overline{\mathrm{AB}}\)
• \(\overline{\mathrm{WX}} \text { and } \overline{\mathrm{BC}}\)
• \(\overline{\mathrm{YX}} \text { and } \overline{\mathrm{DC}}\)
• \(\overline{\mathrm{YD}} \text { and } \overline{\mathrm{XC}}\)
• \(\overline{\mathrm{XC}} \text { and } \overline{\mathrm{WB}}\)
• \(\overline{\mathrm{WB}} \text { and } \overline{\mathrm{VA}}\)
• \(\overline{\mathrm{VA}} \text { and } \overline{\mathrm{ZE}}\)
• \(\overline{\mathrm{ZE}} \text { and } \overline{\mathrm{YD}}\)

(b) Intersecting line segments

• DE and ZV
• WX and DC

Question 3.
Name the following angles as shown in the figure.

(i) ∠1 =
(ii) ∠2 =
(iii) ∠3 =
(iv) ∠1 + ∠2 =
(v) ∠2 + ∠3 =
(vi) ∠1 + ∠2 + ∠3 =
Solution:
(i) ∠1 = ∠DBC or ∠CBD
(ii) ∠2 = ∠DBE or ∠EBD
(iii) ∠3 = ∠ABE or ∠EBA
(iv) ∠1 + ∠2 = ∠EBC or ∠CBE
(v) ∠2 + ∠3 = ∠ABD or ∠DHA
(vi) ∠1 + ∠2 + ∠3 = ∠ABC or ∠B or ∠CBA

Question 4.
Measure the angles of the given figures using a protractor and identify the type of angle as acute, obtuse, right or straight.

Solution:
(i) right angle
(ii) acute angle
(iii) straight angle
(iv) obtuse angle

Question 5.
Draw the following angles using the protractor.
(i) 45°
(ii) 120°
(iii) 65°
(iv) 135°
(v) 0°
(vi) 180°
(vii) 38°
(viii) 90°
Solution:

Question 6.
From the figures given below, classify the following pairs of angles into complementary and non-complementary.

Solution:
We know that the two angles are complementary if they add up to 90°.
Therefore (a) (i) is complementary.
In (v) ∠ABC and ∠CBD are complementary
(b) (ii), (iii), (iv) and (v) are non-complementary

Question 7.
From the figures given below, classify the following pairs of angles into supplementary and non supplementary.

Solution:
Ans: (ii) and (iv) are supplementary angles.
(i), and (iii) non-supplementary angles.

Question 8.
From the figure

(i) name a pair of complementary angles.
(ii) name a pair of supplementary angles.
Solution:
(i) ∠FAE; ∠EAD
(ii) ∠FAD; ∠DAC
∠BAC; ∠CAE
∠FAB; ∠BAC
∠FAB; ∠FAE

Question 9.
Find the complementary angle of
(i) 30°
(ii) 26°
(iii) 85°
(iv) 0°
(v) 90°
Solution:

Question 10.
Find the supplementary angle of
(i) 70°
(ii) 35°
(iii) 165°
(iv) 90°
(v) 0°
(vi) 180°
(vii) 95°
Solution:

Challenging Problems

Question 11.
Think and write an object having
(i) Parallel lines (1) ……….. (2) ………. (3) ………..
(ii) Perpendicular lines (1) ……… (2) ……… (3) ………..
(iii) Intersecting lines (1) ……….. (2) ………. (3) ……….
Solution:
(i) Legs of the table, railway tracks, edges of the scale
(ii) Adjacent sides of a Board, Crossbars of windows, Adjacent sides of the textbook
(iii) Crossbars of windows, Ladder, blades of a scissor.

Question 12.
Which angle is equal to twice its complement.
Solution:
Let the angle be x
According to the problem, x = 2 × (90 – x)
x = 180 – 2x
x + 2x = 180
3x = 180
x = \(\frac{180}{3}\)
x = 60
∴ The angle is 60°

Question 13.
Which angle is equal to two-thirds of its supplement.
Solution:
Let the angle be x
According to the problem,
x = \(\frac{2}{3}\) × (180° – x)
3x = 2(180 – x)
3x = 360 – 2x
3x + 2x = 360°
5x = 360°
x = \(\frac{360°}{5}\)
x = 72°
∴ The angle is 72°

Question 14.
Given two angles are supplementary and one angle is 20° more than the other. Find the two angles.
Solution:
Let the angles be x and x + 20°
According to the problem,
x + x + 20 = 180°
2x + 20° = 180°
2x = 180° – 20°
2x = 160°
x = \(\frac{160°}{2}\)
x = 80°
x + 20 = 80° + 20°
= 100°
∴ The two angles are 80° and 100°

Question 15.
Two complementary angles are in ratio 7 : 2. Find the angles.
Solution:
Let the angles be 7x, 2x
According to the problem,
7x + 2x = 90
9x = 90
x = \(\frac{90}{9}\)
x = 10
7x = 7 × 10
= 70
2x = 2 × 10
= 20
∴ Two angles are 70° and 20°

Question 16.
Two supplementary angles are in ratio 5 : 4. Find the angles.
Solution:
Total of two supplementary angles = 180°
Given they are in the ratio 5 : 4
Dividing total angles to 5 + 4 = 9 equal parts.
One angle \(=\frac{5}{9} \times 180=100^{\circ}\)
Another angle \(=\frac{4}{9} \times 180=80^{\circ}\)
Two angles are 100° and 80°.

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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.2

Question 1.
What is the slope of a line whose inclination with positive direction of x -axis is
(i) 90°
(ii) 0°
Solution:
Here θ = 90°
Slope (m) = tan θ
Slope = tan 90°
= undefined.

(ii) Here θ = 0°
Slope (m) = tan θ
Slope = tan 0°
= 0

Question 2.
What is the inclination of a line whose slope is
(i) 0
(ii) 1
Solution:
(i) m = 0
tan θ = 0 ⇒ θ = 0°
(ii) m = 1 ⇒ tan θ = tan 45° ⇒ 0 = 45°

Question 3.
Find the slope of a line joining the points
(i) (5,\(\sqrt { 5 }\)) with the origin
(ii) (sin θ, -cos θ) and (-sin θ, cos θ)
Solution:
(i) The given points is (5,\(\sqrt { 5 }\)) and (0, 0)
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\) = \(\frac{0-\sqrt{5}}{0-5}\)
= \(\frac{\sqrt{5}}{5}=\frac{1}{\sqrt{5}}\)

(ii) The given points is (sin θ, -cos θ) and (-sin θ, cos θ)
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{\cos \theta+\cos \theta}{-\sin \theta-\sin \theta}\)
= \(\frac{2 \cos \theta}{-2 \sin \theta}\) = – cot θ

Question 4.
What is the slope of a line perpendicular to the line joining A(5,1) and P where P is the mid-point of the segment joining (4,2) and (-6,4).
Solution:
Mid point of XY = \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\) = (\(\frac { 4-6 }{ 2 } \),\(\frac { 2+4 }{ 2 } \))
= (\(\frac { -2 }{ 2 } \),\(\frac { 6 }{ 2 } \)) = (-1, 3)

Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\) = (\(\frac { 3-1 }{ -1-5 } \))
= \(\frac { 2 }{ -6 } \) = – \(\frac { 1 }{ 3 } \)

Question 5.
Show that the given points are collinear: (-3, -4), (7,2) and (12, 5)
Solution:
The vertices are A(-3, -4), B(7, 2) and C(12, 5)
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of AB = \(\frac { 2+4 }{ 7+3 } \) = \(\frac { 6 }{ 10 } \) = \(\frac { 3 }{ 5 } \)
Slope of BC = \(\frac { 5-2 }{ 12-7 } \) = \(\frac { 3 }{ 5 } \)
Slope of AB = Slope of BC = \(\frac { 3 }{ 5 } \)
∴ The three points A,B,C are collinear.

Question 6.
If the three points (3, -1), (a, 3) and (1, -3) are collinear, find the value of a.
Solution:
The vertices are A(3, -1), B(a, 3) and C(1, -3)
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of AB = \(\frac { 3+1 }{ a-3 } \) = \(\frac { 4 }{ a-3 } \)
Slope of BC = \(\frac { 3+3 }{ a-1 } \) = \(\frac { 6 }{ a-1 } \)
Since the three points are collinear.
Slope of AB = Slope BC
\(\frac { 4 }{ a-3 } \) = \(\frac { 6 }{ a-1 } \)
6 (a – 3) = 4 (a – 1)
6a – 18 = 4a – 4
6a – 4a = -4 + 18
2a = 14 ⇒ a = \(\frac { 14 }{ 2 } \) = 7
The value of a = 7

Question 7.
The line through the points (-2, a) and (9,3) has slope –\(\frac { 1 }{ 2 } \) Find the value of a.
Solution:
The given points are (-2, a) and (9, 3)
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
– \(\frac { 1 }{ 2 } \) = \(\frac { 3-a }{ 9+2 } \) ⇒ – \(\frac { 1 }{ 2 } \) = \(\frac { 3-a }{ 11 } \)
2(3 – a) = -11 ⇒ 6 – 2a = -11
-2a = -11 – 6 ⇒ -2a = -17 ⇒ a = – \(\frac { 17 }{ 2 } \)
∴ The value of a = \(\frac { 17 }{ 2 } \)

Question 8.
The line through the points (-2, 6) and (4, 8) is perpendicular to the line through the points (8,12) and (x, 24). Find the value of x.
Solution:
Find the slope of the line joining the point (-2, 6) and (4, 8)
Slope of line (m1) = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac { 8-6 }{ 4+2 } \) = \(\frac { 2 }{ 6 } \) = \(\frac { 1 }{ 3 } \)
Find the slope of the line joining the points (8, 12) and (x, 24)
Slope of a line (m2) = \(\frac { 24-12 }{ x-8 } \) = \(\frac { 12 }{ x-8 } \)
Since the two lines are perpendicular.
m₁ × m₂ = -1
\(\frac { 1 }{ 3 } \) × \(\frac { 12 }{ x-8 } \) = -1 ⇒ \(\frac{12}{3(x-8)}=-1\)
-1 × 3 (x – 8) = 12
-3x + 24 = 12 ⇒ – 3x = 12 -24
-3x = -12 ⇒ x = \(\frac { 12 }{ 3 } \) = 4
∴ The value of x = 4

Question 9.
Show that the given points form a right angled triangle and check whether they satisfies Pythagoras theorem.
(i) A(1, -4) , B(2, -3) and C(4, -7)
(ii) L(0, 5), M(9,12) and N(3,14)
Solution:
(i) The vertices are A(1, -4), B(2, -3) and C(4, -7)
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of AB = \(\frac { -3+4 }{ 2-1 } \) = \(\frac { 1 }{ 1 } \) = 1
Slope of BC = \(\frac { -7+3 }{ 4-2 } \) = \(\frac { -4 }{ 2 } \) = -2
Slope of AC = \(\frac { -7+4 }{ 4-1 } \) = – \(\frac { 3 }{ 3 } \) = -1
Slope of AB × Slope of AC = 1 × -1 = -1

∴ AB is ⊥^r to AC
∠A = 90°
∴ ABC is a right angle triangle
Verification:

20 = 2 + 18
20 = 20 ⇒ Pythagoras theorem verified

(ii) The vertices are L(0, 5), M(9, 12) and N(3, 14)
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of LM = \(\frac { 12-5 }{ 9-0 } \) = \(\frac { 7 }{ 9 } \)

Slope of MN = \(\frac { 14-12 }{ 3-9 } \) = \(\frac { 2 }{ -6 } \) = – \(\frac { 1 }{ 3 } \)
Slope of LN = \(\frac { 14-5 }{ 3-0 } \) = \(\frac { 9 }{ 3 } \) = 3
Slope of MN × Slope of LN = – \(\frac { 1 }{ 3 } \) × 3 = -1
∴ MN ⊥ LN
∠N = 90°
∴ LMN is a right angle triangle
Verification:


130 = 90 + 40
130 = 130 ⇒ Pythagoras theorem is verified

Question 10.
Show that the given points form a parallelogram:
A (2.5,3.5), B(10, -4), C(2.5, -2.5) and D(-5, 5).
Solution:
Let A(2.5, 3.5), B(10, -4), C(2.5, -2.5) and D(-5, 5) are the vertices of a parallelogram.


Slope of AB = Slope of CD = -1
∴ AB is Parallel to CD ……(1)

Slope of BC = Slope of AD
∴ BC is parallel to AD
From (1) and (2) we get ABCD is a parallelogram.

Question 11.
If the points A(2, 2), B(-2, -3), C(1, -3) and D(x, y) form a parallelogram then find the value of x and y.
Solution:
Let A(2, 2), B(-2, -3), C(1, -3) and D(x, y) are the vertices of a parallelogram.

Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of AB = \(\frac { -3-2 }{ -2-2 } \) = \(\frac { -5 }{ -4 } \) = \(\frac { 5 }{ 4 } \)
Slope of BC = \(\frac { -3+3 }{ -2-1 } \) = \(\frac { 0 }{ -3 } \) = 0
Slope of CD = \(\frac { y+3 }{ x-1 } \)
Slope of AD = \(\frac { y-2 }{ x-2 } \)
Since ABCD is a parallelogram
Slope of AB = Slope of CD
\(\frac { 5 }{ 4 } \) = \(\frac { y+3 }{ x-1 } \)
5(x – 1) = 4 (y + 3)
5x – 5 = 4y + 12
5x – 4y = 12 + 5
5x – 4y = 17 ……(1)
Slope of BC = Slope of AD
0 = \(\frac { y-2 }{ x-2 } \)
y – 2 = 0
y = 2
Substitute the value of y = 2 in (1)
5x – 4(2) = 17
5x -8 = 17 ⇒ 5x = 17 + 18
5x = 25 ⇒ x = \(\frac { 25 }{ 5 } \) = 5
The value of x = 5 and y = 2.

Question 12.
Let A(3, -4), B(9, -4) , C(5, -7) and D(7, -7). Show that ABCD is a trapezium.
Solution:
Let A(3, -4), B(9, -4), C(5, -7) and D(7, -7) are the vertices of a quadrilateral.

Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of AB = \(\frac { -4+4 }{ 9-3 } \) = \(\frac { 0 }{ 6 } \) = 0
Slope of BC = \(\frac { -7+4 }{ 5-9 } \) = \(\frac { -3 }{ -4 } \) = \(\frac { 3 }{ 4 } \)
Slope of CD = \(\frac { -7+7 }{ 7-5 } \) = \(\frac { 0 }{ 2 } \) = 0
Slope of AD = \(\frac { -7+4 }{ 7-3 } \) = \(\frac { -3 }{ 4 } \) = – \(\frac { 3 }{ 4 } \)
The slope of AB and CD are equal.
∴ AB is parallel to CD. Similarly the slope of AD and BC are not equal.
∴ AD and BC are not parallel.
∴ The Quadrilateral ABCD is a trapezium.

Question 13.
A quadrilateral has vertices at A(-4, -2), B(5, -1) , C(6, 5) and D(-7, 6). Show that the mid-points of its sides form a parallelogram.
Solution:
Let A(-4, -2), B(5, -1), C(6, 5) and D(-7, 6) are the vertices of a quadrilateral.





Slope of EF = Slope of GH = \(\frac { 7 }{ 10 } \)
∴ EF || GH …….(1)
Slope of FG= Slope of EH = – \(\frac { 7 }{ 12 } \)
∴ FG || EH ……(2)
From (1) and (2) we get EFGH is a parallelogram.
The mid point of the sides of the Quadrilateral ABCD is a Parallelogram.

Question 14.
PQRS is a rhombus. Its diagonals PR and QS intersect at the point M and satisfy QS = 2PR. If the coordinates of S and M are (1, 1) and (2, -1) respectively, find the coordinates of P.
Solution:
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of SM = \(\frac { 1+1 }{ 1-2 } \) = \(\frac { 2 }{ -1 } \) = -2
Slope of PM = \(\frac { 1 }{ 2 } \) (Since SM and PM are ⊥^r)
Let the point p be (a,b)
Slope of PM = \(\frac { 1 }{ 2 } \)
\(\frac { b+1 }{ a-2 } \) = \(\frac { 1 }{ 2 } \) ⇒ a – 2 = 2b + 2

a – 2b = 4
a = 4 + 2b ……(1)
Given QS = 2PR
\(\frac { QS }{ 2 } \) = PR
∴ SM = PR
SM = 2PM (PR = 2PM)

Squaring on both sides

∴ (b + 1)² = \(\frac { 1 }{ 4 } \) ⇒ b + 1 = ± \(\frac { 1 }{ 2 } \)
b = \(\frac { 1 }{ 2 } \) – 1 (or) b = – \(\frac { 1 }{ 2 } \) – 1
= – \(\frac { 1 }{ 2 } \) – 1 (or) b = –\(\frac { 1 }{ 2 } \) – 1
= – \(\frac { 1 }{ 2 } \) (or) – \(\frac { 3 }{ 2 } \)
a = 4 + 2b
a = 4 + 2 (\(\frac { -1 }{ 2 } \))
a = 3
a = 4 + 2 (\(\frac { -3 }{ 2 } \))
a = 4 – 3
a = 1
The point of p is (3,\(\frac { -1 }{ 2 } \)) (or) (1,\(\frac { -3 }{ 2 } \))

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Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 4 Geometry Ex 4.3

Question 1.
Observe the diagram and fill in the blanks.

(i) ‘A’, ‘O’ and ‘B’ are ……… points.
(ii) ‘A’, ‘O’ and ‘C’ are ……….. points.
(iii) ‘A’,‘B’ and ‘C’are ……… points.
(iv) ……… is the point of concurrency.
Solution:
(i) collinear points
Hint: Points on a line.
(ii) non-collinear points
Hint: Points not on a line
(iii) endpoints/non-collinear points
(iv) O is the point of concurrency.
Hint: A points where lines meet

Question 2.
Draw any line and mark any 3 points that are collinear.
Solution:

Question 3.
Draw any line and mark any 4 points that are not collinear.
Solution:

Question 4.
Draw any 3 lines to have a point of concurrency.
Solution:

Question 5.
Draw any 3 lines that are not concurrent. Find the number of points of intersection.
Solution:

Number of points of intersection = 3

Objective Type Questions

Observe the Diagram and give answers

Question 6.
A set of collinear points in the figure are
(a) A, B, C
(b) A, F, C
(c) B, C, D
(d) A, C, D
Solution:
(b) A, F, C
Hint: Collinear points are points on a line.

Question 7.
A set of non-collinear points in the figure are ……….
(a) A, F, C
(b) B, F, D
(c) E, F, G
(d)A,D,C
Solution:
(d) A, D, C

Question 8.
A point of concurrency in the figure is ______
(a) E
(b) F
(c) G
(d) H
Solution:
(b) F

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Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 4 Geometry Ex 4.2

Question 1.
Use any number of the given dots to make different angles.
(i) An Accute angle

(ii) An Obtuse Angle

(iii) A Right Angle

(iv) A Straight Angle

Solution:
(i) An Accute angle

(ii) An Obtuse Angle

(iii) A Right Angle

(iv) A Straight Angle

Question 2.
Name the vertex and sides that form each angle.

Solution:
(i) D, DE and DF
(ii) D, DE and DC
(iii) P, PQ and PR
(iv) S, SV and ST

Question 3.
Pick out the Right angles from the given figures.

Solution:
(i), (iii), (v)

Question 4.
Pick out the Accute angles from the given figures.

Solution:
(i), (iii), (iv) are the Acute Angles.

Question 5.
Pick out the Obtuse angles from the given figures.

Solution:
(i) and (ii) are the Obtuse Angles.

Question 6.
Name the angle in each figure given below in all the possible ways.

Solution:
(i) ∠M or ∠LMN or ∠NML
(ii) ∠Q or ∠PQR or ∠RQP
(iii) ∠N or ∠MNO or ∠ONM
(iv) ∠A or ∠TAS or ∠SAT
(v) ∠Y or ∠XYZ or ∠ZYX
(vi) There are 3 angles in (vi)

• ∠ADC or ∠CDA
• ∠ CDB or ∠BDC
• ∠D or ∠ADB or ∠BDA

Question 7.
Say True or False.
(i) 20° and 70° are complementary.
(ii) 88° and 12° are complementary.
(iii) 80° and 180° are supplementary.
(iv) 0° and 180° are supplementary.
Solution:
(i) True
Hint: 20°+ 70° = 90°
(ii) False
Hint: 88° + 180° = 260° ≠ 1
(iii) False
Hint: 80° + 180° = 260° ≠ 1
(iv) True
Hint: 0° + 180° = 180°

Question 8.
Draw and label each of the angles.
(i) ∠NAS = 90°n
(ii) ∠BIG = 35°
(iii) ∠SMC = 145°
Solution:

Question 9.
Identify the types of angles shown by the hands of the given clock.

Solution:
(i) Obtuse angle
(ii) Zero angle
(iii) Straight angle
(iv) Acute angle
(v) Right angle

Question 10.
Find the supplementary/complementary angles in each case.

Solution:

Objective Type Questions

Question 11.
In this figure, which is not the correct way of naming an angle?

(a) ∠Y
(b) ∠ZXY
(c) ∠ZYX
(d) ∠XYZ
Solution:
(b) ∠ZXY

Question 12.
In this figure, ∠AYZ = 45. If point ‘A’ is shifted to point ‘B’ along the ray, then the measure of ∠BYZ is

(a) more than 45°
(b) 45°
(c) less than 45°
(d) 90°
Solution:
(b) 45°

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Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 4 Geometry Ex 4.1

Question 1.
Fill in the blanks.
(i) A line through two endpoints ‘A’ and ‘B’ is denoted by ______
(ii) Aline segment from point ‘B’ to point ‘A’ is denoted by ______
(iii) A ray has ______ endpoint(s).
Solution:
(i) \(\overleftrightarrow { AB }\)
(ii) \(\bar { BA } \)
(iii) one

Question 2.
How many line segments are there in the given line? Name them.

Solution:
10, \(\overline { PQ } \), \(\overline { PA } \), \(\overline { PB } \), \(\overline { PC } \), \(\overline { AB } \), \(\overline { BC } \), \(\overline {CQ} \), \(\overline { AQ } \), \(\overline { BQ } \), \(\overline { AC } \).

Question 3.
Measure the following line segments.

Solution:
\(\overline { XY } \) = 2.4 cm, \(\overline { AB } \) = 3.4 cm, \(\overline { EF } \) = 4 cm, \(\overline { PQ } \) = 3 cm.

Question 4.
Construct a line segment using a ruler and compass.
(1) \(\overline { AB } \) = 7.5 cm
(2) \(\overline { CD } \) = 3.6 cm
(3) \(\overline { QR } \) = 10 cm
Solution:
(1)
(i) Draw line 1 and mark a point A on it.
(ii) Measure 7.5 cm using a compass, placing the pointer at ‘O’ and the pencil pointer at 7.5 cm.
(iii) Place the pointer of the compass at A then draw a small arc on the line 1 with the pencil pointer. It cuts line 1 at a point and name that point as B.
(iv) Now \(\overline { AB } \) is the required line segment of length 7.5 cm.

(2)
(i) Draw a line 1 and mark a point C on it.
(ii) Measure 3.6 cm using a compass, placing the pointer at O and the pencil pointer at 3.6 cm.
(iii) Place the pointer of the compass at C then draw a small arc on the line 1 with the pencil pointer. It cuts line 1 at a point and names the point as D.
(iv) Now \(\overline { CD } \) is the required line segment of length 3.6 cm.

(3)
(i) Draw a line 1 and mark a point Q on it.
(ii) Measure 10 cm using compass placing the pointer at O and the pencil pointer at 10 cm.
(iii) Place the pointer of the compass at Q then draw a small arc on the line 1 with the pencil pointer. It cuts the line 1 at a point and name that point as R.
(iv) Now \(\overline { QR }\) is the required line segment of length 10 cm.

Question 5.
From the given figure
(i) identify the parallel lines
(ii) identify the intersecting lines
(iii) name the points of intersection.

Solution:
(i) Parallel lines:
(a) \(\overrightarrow{\mathrm{CD}} \text { and } \overrightarrow{\mathrm{AB}}\)
(b) \(\overrightarrow{\mathrm{EF}} \text { and } \overrightarrow{\mathrm{GH}}\)
(ii) Intersecting lines:
(a) \(\overrightarrow{\mathrm{CD}} \text { and } \overrightarrow{\mathrm{EF}}\)
(b) \(\overrightarrow{\mathrm{CD}} \text { and } \overrightarrow{\mathrm{GH}}\)
(c) \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{EF}}\)
(d) \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{GH}}\)
(iii) Point of Intersection:
P, Q, R and S are the points of Intersection.
(iii) P, Q, R and S

Question 6.
From the given figure, name the
(i) parallel lines
(ii) intersecting lines
(iii) points of intersection.

Solution:
(i) \(\overleftrightarrow { CD } \) and \(\overleftrightarrow { EF } \), \(\overleftrightarrow { CD } \) and \(\overleftrightarrow { IJ } \), \(\overleftrightarrow { EF } \) and \(\overleftrightarrow { IJ } \)
(ii) \(\overleftrightarrow { AB } \) and \(\overleftrightarrow { CD } \), \(\overleftrightarrow { AB } \) and \(\overleftrightarrow { EF } \), \(\overleftrightarrow { AB } \) and \(\overleftrightarrow { IJ } \), \(\overleftrightarrow { GH } \) and \(\overleftrightarrow { IJ } \), \(\overleftrightarrow { AB } \) and \(\overleftrightarrow { GH } \)
(iii) P, Q and R

Question 7.
From the given figure, name
(i) all pairs of parallel lines.
(ii) all pairs of intersecting lines.
(iii) pair of lines whose point of intersection is ‘V’.
(iv) point of intersection of the lines ‘l₂‘ and ‘l₃‘.
(v) point of intersection of the lines ‘l₁‘, and ‘l₅

Solution:
(i) Pairs of parallel lines:

• l₃ and l₄
• l₄ and l₅
• l₃ and l₅

(ii) Pairs of intersecting lines:

• l₁ and l₂
• l₁ and l₃
• l₁ and l₄
• l₁ and l₅
• l₂ and l₃
• l₂ and l₄
• l₂ and l₅

(iii) l₁ and l₂ intersect at ‘V’
(iv) point of intersection of the lines ‘l₂‘ and; l₅‘ is ‘Q’
(v) point of intersection of the lines ‘l₁‘ and ‘l₅‘ is ‘U’

Objective Type Questions

Question 8.
The number of line segments in
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(c) 3

Question 9.
A line is denoted as __________
(a) AB
(b) \(\overrightarrow{AB}\)
(c) \(\overleftrightarrow {AB} \)
(d) \(\overline { AB }\)
Solution:
(c) \(\overleftrightarrow {AB} \)

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Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 5 Statistics Ex 5.4

Miscellaneous Practice Problems

Question 1.
The heights (in centimeters) of 40 children are.

Prepare a tally marks table.
Solution:

Question 2.
There are 1000 students in a school. Data regarding the mode of transport of the students is given below. Draw a pictograph to represent the data.

Solution:
Mode of transport of the students
Scale: 1 Unit=100 Students

Question 3.
The following pictograph shows the total savings of a group of friends in a year. Each .picture represents a saving of Rs.100. Answer the following questions.

(i) What is the Ratio of Ruby’s saving to that of Thasnim’s?
(ii) What is the ratio of Kuzhali’s savings to that of others?
(iii) How much is Iniya’s savings?
(iv) Find the total amount of savings of all the friends?
(v) Ruby and Kuzhali save the same amount. Say True or False.
Solution:
(i) Ratio of Ruby’s saving to that of Thasnim’s
\(=\frac{\text { Ruby’s saving }}{\text { Thasnim’s saving }}=\frac{5 \times 100}{4 \times 100}=\frac{5}{4}=5: 4\)
Ratio of Ruby’s saving to that of Thasnims = 5 : 4
(ii) Ratio of Kuzhali’s savings to that of others
\(=\frac{\text { kuzhali’s saving }}{\text { others saving }}=\frac{5 \times 100}{19 \times 100}=\frac{5}{19}=5: 19\)
Ratio of Kuzhali’s saving to that of others = 5 : 19
(iii) Iniya’s saving = 3 × 100 = ₹ 300
(iv) Saving of all the friends = (5 + 7 + 4 + 5 + 3) × 100 = 24 × 100 = ₹ 2400.
Total savings = ₹ 2400
(v) True.

Challenging Problems

Question 4.
The table shows the numbers of moons that orbit each of the planets in oar solar system.

Make a Bar graph for the above data.
Solution:
Number of moons that orbit each of the planets in our solar system
Scale: 1 Unit = 2 moons

Question 5.
The prediction of the weather in the month of September is given below.

(i) Make a frequency table of the types of weather by reading the calendar.
(ii) How many days are either cloudy or partly cloudy?
(iii) How many days do not have rain? Give two ways to find the answer?
(iv) Find the ratio of the number of Sunny days to Rainy days.
Solution:
(i)
(ii) 14 days
(iii) 24 days (30 – 6 = 24 days)
(iv) 10 : 6

Question 6.
26 students were interviewed to find out what they want they to become in future. Their responses are given in the following table.

Represent this data using pictograph.
Solution:
Students responses in an interview about their future profession
Scale: 1 Unit = 1 student

Question 7.
Yasmin of class VI was given a task to count the number of books which are biographies, in her school library. The information collected by her is represented as follows.

Observe the pictograph and answer the following questions.
(i) Which title has the maximum number of biographies?
(ii) Which title has the minimum number of biographies?
(iii) Which title has exactly half the number of biographies as Novelists?
(iv) How many biographies are there on the title of sportspersons?
(v) What is the total number of biographies in the library?
Solution:
(i) ‘The title Novelists’ have the maximum number of biographies
(ii) ‘The title Scientists’ have a minimum number of biographies.
(iii) ‘Sportspersons’ title has exactly half the number of biographies as Novelist.
(iv) \((1 \times 20)+\frac{20}{4}=20+5=25\) biographies are there in the title sportsperson.
(v) 8 × 20 = 160 biographies are there in the library.

Question 8.
The bar graph illustrates the results of a survey conducted on vehicles crossing over a Toll Plaza in one hour.

Observe the bar graph carefully and fill up the following table.

Solution:
Vans = 50; Buses = 40; Cars = 65; Others = 15
Total Vehicles = 245

Question 9.
The lengths (in the nearest centimeter) of 30 drumsticks are given as follows.

Draw the bar graph showing the same information.
Solution:
The lengths (in nearest cm) of drumsticks
Scale : 1 Unit = 1 drumstick

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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.1

Question 1.
Find the area of the triangle formed by the points
(i) (1,-1), (-4, 6) and (-3, -5)
(ii) (-10, -4), (-8, -1) and (-3, -5)
Solution:
(i) Let the vertices A (1, -1), B (-4, 6) and C (-3, -5)


= \(\frac { 1 }{ 2 } \) [(6 + 20 + 3) – (4 – 18 – 5)] = \(\frac { 1 }{ 2 } \) [29 – (-19)] = \(\frac { 1 }{ 2 } \) [29 + 19]
= \(\frac { 1 }{ 2 } \) × 48 = 24 sq. units.
Area of ∆ABC = 24 sq. units

(ii) Let the vertices be A(-10, -4), B(-8 -1) and C(-3, -5)

Area of ∆ABC = \(\frac { 1 }{ 2 } \)[(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)]
= \(\frac { 1 }{ 2 } \) [(50 + 3 + 32) – (12 + 40 + 10)]

= \(\frac { 1 }{ 2 } \) [85 – (62)] = \(\frac { 1 }{ 2 } \) [23] = 11.5
Area of ∆ACB = 11.5 sq.units

Question 2.
Determine whether the sets of points are collinear?
(i) (-\(\frac { 1 }{ 2 } \),3)
(ii) (a,b + c), (b,c + a) and (c,a + b)
Solution:
(i) Let the points be A (-\(\frac { 1 }{ 2 } \),3), B (-5, 6) and C(-8, 8)
Area of ∆ABC = \(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₃ + x₁y₃)]
= \(\frac { 1 }{ 2 } \) [(- 3 – 40 – 24) – (-15 – 48 – 4)]

= \(\frac { 1 }{ 2 } \) [-67 + 67] = \(\frac { 1 }{ 2 } \) × 0 = 0
Area of a ∆ is 0.
∴ The three points are collinear.

(ii) Let the points be A (a, b + c), B (b, c + a) and C (c, a + b)
Area of the triangle = \(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)]

Since the area of a triangle is 0.
∴ The given points are collinear.

Question 3.
Vertices of given triangles are taken in order and their areas are provided aside. In each case, find the value of ‘p’

Solution:
(i) Let the vertices be A (0,0) B (p, 8), c (6, 2)
Area of a triangle = 20 sq. units
\(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)] = 20

\(\frac { 1 }{ 2 } \) [(0 + 2p + 0) – (0 + 48 + 0)] = 20
\(\frac { 1 }{ 2 } \) [2p – 48] = 20
2p – 48 = 40 ⇒ 2p = 40 + 48
p = \(\frac { 88 }{ 2 } \) = 44
The value of p = 44

(ii) Let the vertices be A (p, p), B (5, 6) and C (5, -2)
Area of a triangle = 32 sq. units
\(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)] = 32

\(\frac { 1 }{ 2 } \) [6p – 10 + 5p) – (5p + 30 – 2p)] = 32
\(\frac { 1 }{ 2 } \) [11 p – 10 – 3p – 30] = 32
11p – 10 – 3p – 30 = 64
8p – 40 = 64
8p = 64 + 40 ⇒ 8p = 104
p = \(\frac { 104 }{ 8 } \) = 13
The value of p = 13

Question 4.
In each of the following, find the value of ‘a’ for which the given points are collinear.
(i) (2,3), (4, a) and (6, -3)
(ii) (a, 2 – 2a), (-a + 1, 2a) and (-4 -a, 6 – 2a).
Solution:
(i) Let the points be A (2, 3), B(4, a) and C(6, -3).
Since the given points are collinear.
Area of a triangle = 0
\(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)] = 0)

\(\frac { 1 }{ 2 } \) [(2a – 12 + 18) – (12 + 6a – 6)] = 0
2a + 6 – (6 + 6a) = 0
2a + 6 – 6 – 6a = 0
-4a = 0 ⇒ a = \(\frac { 0 }{ 4 } \) = 0
The value of a = 0

(ii) Let the points be A (a, 2 – 2a), B (-a + 1, 2a) C (-4 -a, 6 – 2a).
Since the given points are collinear.
Area of a ∆ = 0

6a² – 2a – 2 – (-2a² – 6a + 2) = 0
6a² – 2a – 2 + 2a² + 6a – 2 = 0
8a² + 4a – 4 = 0 (Divided by 4)
2a² + a – 1 = 0
2a² + 2a – a – 1 = 0
2a (a + 1) – 1 (a + 1) = 0

(a + 1) (2a – 1) = 0
a + 1 = 0 (or) 2a – 1 = 0
a = -1 (or) 2a = 1 ⇒ a = \(\frac { 1 }{ 2 } \)
The value of a = -1 (or) \(\frac { 1 }{ 2 } \)

Question 5.
Find the area of the quadrilateral whose vertices are at
(i) (-9, -2), (-8, -4), (2, 2) and (1, -3)
(ii) (-9, 0), (-8,6), (-1, -2) and (-6, -3)
Solution:
(i) Let the vertices A (-9, -2), B(-8, -4), C(2, 2) and D(1, -3).
Plot the vertices in a graph.

[Note: Consider the points in counter clock wise order]

Area of the Quadrilateral ABDC = \(\frac { 1 }{ 2 } \) [36 + 24 + 2 – 4 – (16 – 4 – 6 – 18)]
= \(\frac { 1 }{ 2 } \) [58 – (-12)] – \(\frac { 1 }{ 2 } \)[58 + 12]
= \(\frac { 1 }{ 2 } \) × 70 = 35 sq. units 2
Area of the Quadrilateral = 35 sq. units

(ii) Let the vertices A(-9, 0), B(-8, 6), C(-1, -2) and D(-6, -3)
Plot the vertices in a graph and take them in counter – clock wise order.
Area of the Quadrilateral DCB

= \(\frac { 1 }{ 2 } \) [33 + 35] = \(\frac { 1 }{ 2 } \) × 68 = 34 sq. units
Area of the Quadrilateral = 34 sq. units

Question 6.
Find the value of k, if the area of a quadrilateral is 28 sq. units, whose vertices are (-4, -2), (-3, k), (3, -2) and (2, 3)
Solution:
Let the vertices A (-A, -2), B (-3, k), C (3, -2) and D (2, 3)
Area of the Quadrilateral = 28 sq. units
\(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₄ + x₄y₁) – (x₂y₁ + x₃y₂ + x₄y₃ + x₁y₄)] = 28

-7k + 21 = 56
-7k = 56 – 21
-7k = 35 ⇒ 7k = – 35
k = – \(\frac { 35 }{ 7 } \) = -5
The value of k = -5

Question 7.
If the points A(-3, 9), B(a, b) and C(4, -5) are collinear and if a + b = 1, then find a and b.
Solution:
Since the three points are collinear
Area of a ∆ = 0

-3b – 5a + 36 – 9a – 4b – 15 = 0
-7b – 14a + 21 = 0
(÷ by 7) – b – 2a + 3 = 0
2a + b – 3 = 0

Substitute the value of a = 2 in (2) ⇒ 2 + b = 1
b = 1 – 2 = -1
The value of a = 2 and b = -1

Question 8.
Let P(11, 7), Q(13.5, 4) and R(9.5, 4) be the midpoints of the sides AB, BC and AC respectively of ∆ABC . Find the coordinates of the vertices A, B and C. Hence find the area of ∆ABC and compare this with area of ∆PQR.
Solution:
Let the vertices of the ∆ABC be A(x₁,y₁), B(x₂,y₂), C(x₃,y₃)

Question 9.
In the figure, the quadrilateral swimming pool shown is surrounded by concrete patio. Find the area of the patio.

Solution:

= \(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₄ + x₄y₁) – (x₂y₁ + x₃y₂ + x₄y₃ + x₁y₄)]
= \(\frac { 1 }{ 2 } \) [(16 + 80 + 36 + 80) – (-64 – 24 – 100 – 24)]
= \(\frac { 1 }{ 2 } \) [212 – (-212)]

= \(\frac { 1 }{ 2 } \) [212 + 212] = \(\frac { 1 }{ 2 } \) [424] = 212 sq. units

= \(\frac { 1 }{ 2 } \) [90 – (-90)]

= \(\frac { 1 }{ 2 } \) [90 + 90]
= \(\frac { 1 }{ 2 } \) × 180 = 90 sq. units
Area of the patio = Area of the Quadrilateral ABCD – Area of the Quadrilateral EFGH
= (212 – 90) sq. units
Area of the patio = 122 sq. units

Question 10.
A triangular shaped glass with vertices at A(-5, -4), B(l, 6) and C(7, -4) has to be painted. If one bucket of paint covers 6 square feet, how many buckets of paint will be required to paint the whole glass, if only one coat of paint is applied.
Solution:
Given the vertices of the triangular glass is A (-5, -4), B (1, 6), and C (7, -4)


= \(\frac { 1 }{ 2 } \) [(20 + 42 – 4) – (-28 – 4 – 30)]
= \(\frac { 1 }{ 2 } \) [58 – (-62)]
= \(\frac { 1 }{ 2 } \) [58 + 62]
= \(\frac { 1 }{ 2 } \) × 120 = 60 sq. feet
Number of cans to paint 6 square feet = 1
∴ Number of cans = \(\frac { 60 }{ 6 } \) = 10 ⇒ Number of cans = 10

Question 11.
In the figure, find the area of
(i) triangle AGF
(ii) triangle FED
(iii) quadrilateral BCEG.

Solution:
Area of a triangle = \(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)]
(i) Area of ∆AGF = \(\frac { 1 }{ 2 } \) [(-2.5 – 13.5 – 6) – (-13.5 – 1 – 15)]
= \(\frac { 1 }{ 2 } \) [-22 – (-29.5)]

= \(\frac { 1 }{ 2 } \) [-22 + 29.5]
= \(\frac { 1 }{ 2 } \) × 7.5 = 3.75 sq.units

(ii) Area of ∆FED = \(\frac { 1 }{ 2 } \) [(-2 + 4.5 + 3) – (4.5 + 1 – 6)]
= \(\frac { 1 }{ 2 } \) [5.5 – (-0.5)]

= \(\frac { 1 }{ 2 } \) [5.5 + 0.5] = \(\frac { 1 }{ 2 } \) × 6 = 3 sq.units

(iii)

= \(\frac { 1 }{ 2 } \) [(4 + 2 + 0.75 + 9) – (-4 -1.5 – 4.5 -2)]
= \(\frac { 1 }{ 2 } \) [15.75 + 12]
= \(\frac { 1 }{ 2 } \) [27.75] = 13.875
= 13.88 sq. units

Students can download Maths Chapter 6 Information Processing Ex 6.3 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 6 Information Processing Ex 6.3

Question 1.
How many Triangles are there in each of the following figures?

Solution:
(i) 12 triangles
(ii) 16 triangles
(iii) 32 triangles
(iv) 35 triangles

Question 2.
Find the number of dots in the tenth figure of the following patterns.

Solution:
(i) 55
(ii) 100

Question 3.

(i) Draw the next pattern.
(ii) Prepare a table for the number of dots used for each pattern.
(iii) Explain the pattern.
(iv) Find the number of dots in the 25th pattern.
Solution:

(iv) 350

Question 4.

Solution:
(i) 20 squares
(ii) 10 squares

Question 5.
How many circles are there in the following figure?

Solution:
7 circles

Question 6.
Find the minimum number of straight lines used in forming the following figures.

Solution:
(i) 10
(ii) 12