Bhagya

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 5 Fractions Ex 5.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 5 Fractions Ex 5.2

1. Classify the following as proper and improper fractions:

Question (i)

Solution:

2. Express each of the following as mixed fractions, Also represent with diagrams:

Question (i)
\(\frac {27}{5}\)
Solution:

∴ Mixed fraction
= Quotient \(\frac {Remainder}{Divisor}\) = \(5\frac {2}{5}\)

Question (ii)
\(\frac {13}{4}\)
Solution:

∴ Mixed fraction
= Quotient \(\frac {Remainder}{Divisor}\) = \(3\frac {1}{4}\)

Question (iii)
\(\frac {43}{8}\)
Solution:

∴ Mixed fraction
= Quotient \(\frac {Remainder}{Divisor}\) = \(5\frac {3}{8}\)

Question (iv)
\(\frac {51}{7}\)
Solution:

∴ Mixed fraction
= Quotient \(\frac {Remainder}{Divisor}\) = \(7\frac {2}{7}\)

Question (v)
\(\frac {20}{3}\)
Solution:

∴ Mixed fraction
= Quotient \(\frac {Remainder}{Divisor}\) = \(6\frac {2}{3}\)

3. Express each of the following mixed fractions as improper fractions:

Question (i)
(i) \(\)2 \frac{1}{3}\(\)
(ii) \(\)5 \frac{2}{7}\(\)
(iii) \(\)4 \frac{3}{5}\(\)
(iv) \(\)3 \frac{3}{4}\(\)
(v) \(\)9 \frac{5}{8}\(\)
Solution:

4. Express the shaded portion as Improper Fraction and Mixed fraction:

Question (i)

Solution:

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 5 Fractions Ex 5.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 5 Fractions Ex 5.1

1. Write the fraction representing the shaded portion:

Solution:

2. Colour the part according to the given fraction:

Solution:

3. Write the fraction for each of the following:

Question (i)
(i) Three-Fourth
(ii) Seven-Tenth
(iii) A Quarter
(iv) Five-Eighth
(v) Three-Twelvth.
Solution:
(i) \(\frac {3}{4}\)
(ii) \(\frac {7}{10}\)
(iii) \(\frac {1}{4}\)
(iv) \(\frac {5}{8}\)
(v) \(\frac {3}{12}\)

4. Write the fraction for the followings:

Question (i)
numerator = 5
denominator = 9
Answer:
\(\frac {5}{9}\)

Question (ii)
numerator = 2
denominator = 11
Answer:
\(\frac {2}{11}\)

Question (iii)
numerator = 6
denominator = 7
Answer:
\(\frac {6}{7}\)

5. Write the numerator and the denominator for the followings:

Question (i)
\(\frac {2}{3}\)
Solution:
Given fraction is \(\frac {2}{3}\) = \(\frac {Numerator}{Denominator}\)
∴ Numerator = 2
and Denominator = 3

Question (ii)
\(\frac {1}{4}\)
Solution:
Given fraction is \(\frac {1}{4}\) = \(\frac {Numerator}{Denominator}\)
∴ Numerator = 1
and Denominator = 4

Question (iii)
\(\frac {5}{11}\)
Solution:
Given fraction is \(\frac {5}{11}\) = \(\frac {Numerator}{Denominator}\)
∴ Numerator = 5
and Denominator = 11

Question (iv)
\(\frac {9}{13}\)
Solution:
Given fraction is \(\frac {9}{13}\) = \(\frac {Numerator}{Denominator}\)
∴ Numerator = 9
and Denominator = 13

Question (iv)
\(\frac {17}{16}\) = \(\frac {Numerator}{Denominator}\)
∴ Numerator = 17 and
Denominator = 16

6. Express:

Question (i)
1 day as a fraction of 1 week.
Solution:
We know 1 week = 7 days
∴ Required fraction= \(\frac {1}{7}\)

Question (ii)
40 seconds as a fraction of 1 minute.
Solution:
We know 1 minute = 60 seconds
∴ Required fraction = \(\frac {40}{60}\) or \(\frac {2}{3}\)
(Dividing both terms by 20)

Question (iii)
15 hours as fraction of 1 day.
Solution:
We know 1 day = 24 hours
∴ Required fraction = \(\frac {15}{24}\) or \(\frac {5}{8}\)
(Dividing both terms by 3)

Question (iv)
2 months as a fraction of 1 year.
Solution:
We know 1 year = 12 months
∴ Required fraction = \(\frac {2}{12}\) or \(\frac {1}{6}\)
(Dividing both terms by 2)

Question (v)
45 cm as a fraction of 1 metre.
Solution:
We know 1 metre = 100 cm
∴ Required fraction = \(\frac {45}{100}\) or \(\frac {9}{20}\)
(Dividing both terms by 5)

7. Write the numbers from 1 to 25.

Question (i)
What fraction of them are even numbers?
Solution:
Numbers from 1 to 25 are:
1,2, 3, 4,5,6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25 i.e. 25 in number.
Even numbers out of these numbers are:
2, 4, 6, 8, 10, 12, 14, 16, 18; 20, 22, 24 i.e. 12 in number
∴ Required fraction = \(\frac {12}{25}\)

Question (ii)
What fraction of them are prime numbers?
Solution:
Prime number out of these numbers are:
2, 3, 5, 7, 11, 13, 17, 19, 23 i.e. 9 in number
∴ Required fraction = \(\frac {9}{25}\)

Question (iii)
What fraction of them are multiples of 3?
Solution:
Multiples of 3 out of these numbers are:
3, 6, 9, 12, 15, 18, 21, 24 i.e. 8 in number
∴ Required fraction = \(\frac {8}{25}\)

8. In class 6th, there are 24 boys and 18 girls. What fraction of total students represent boys and girls?
Solution:
Boys = 24
Girls = 18
Total students = 24 + 18 = 42
Fraction which represents boys
= \(\frac {24}{42}\) or \(\frac {4}{7}\)
(Dividing both terms by 6)
Fraction which represents girls
= \(\frac {18}{42}\) or \(\frac {3}{7}\)
(Dividing both terms by 6)

9. A bag contains 6 red balls and 7 blue balls. What fraction of balls represent red and blue colour?
Solution:
Red balls = 6
Blue balls = 7
Total number of red and blue balls = 6 + 7 = 13
Fraction which represents red balls = \(\frac {6}{13}\)
Fraction which represents blue balls = \(\frac {7}{13}\)

10. Sidharth has a cake. He cuts it into 10 equal parts. He gave 2 parts to Naman, 3 parts to Nidhi, 1 parts to Seema and the remaining four parts he kept for himself. Find:

Question (i)
What fraction of cake, he gave to Naman?
Solution:
Total parts = 10
Fraction of cake, he gave to Naman = \(\frac {2}{10}\) or \(\frac {1}{5}\)

Question (ii)
What fraction of cake, he gave to Nidhi?
Solution:
Fraction of cake, he gave to Nidhi = \(\frac {3}{10}\)

Question (iii)
What fraction of cake, he kept for himself?
Solution:
Fraction of cake, he kept for himself = \(\frac {4}{10}\) or \(\frac {2}{5}\)

Question (iv)
Who has more cake than others?
Solution:
Sidharth has more cake than others

11. In a box, there are 12 apples, 7 oranges and 5 guavas. What fraction of fruits in box represents each?
Solution:
Apples = 12, Oranges = 7
and Gauvas = 5
Total fruits = 12 + 7 + 5 = 24.
Fraction which represents apples
= \(\frac {12}{24}\) or \(\frac {1}{2}\)
Fraction which represents oranges
= \(\frac {7}{24}\)
Fraction which represents Gauvas
= \(\frac {5}{24}\)

12. Dishmeet has 20 pens. He gives one-fourth to Balkirat. How many pens Dishmeet and Balkirat have?
Solution:
Total pens = 20
Pens Balkirat has = One-fourth of 20
= \(\frac {22}{7}\) × 20 = 5
Pens Dishmeet has = 20 – 5 = 15

13. Represent the following fraction on the number line?

Question (i)
\(\frac {2}{5}\)
Solution:
In order to represent \(\frac {2}{5}\) on number line, we divide the gap between 0 and 1 into 5 equal parts, which are, (as shown)

Question 2.
\(\frac {5}{7}\)
Solution:
In order to represent \(\frac {5}{7}\) on number line, we divide the gap between 0 and 1 into 7 equal parts, which are, (as shown)

Question 3.
\(\frac {3}{10}\), \(\frac {5}{10}\), \(\frac {1}{10}\)
Solution:
In order to represent \(\frac {3}{10}\), \(\frac {5}{10}\), \(\frac {1}{10}\) on number line, we divide the gap between 0 and 1 into 10 equal parts, which are (as shown)

Question 4.
\(\frac {3}{8}\), \(\frac {5}{8}\), \(\frac {7}{8}\)
Solution:
In order to represents \(\frac {3}{8}\), \(\frac {5}{8}\), \(\frac {7}{8}\) on number line, we divide the gap between 0 and 1 into 8 equal parts, which are (as shown)

14. Find:

(i) \(\frac {3}{5}\) of 20 books
(ii) \(\frac {5}{8}\) of 32 pens
(iii) \(\frac {1}{6}\) of 36 copies 6
(iv) \(\frac {4}{7}\) of 21 apples
(v) \(\frac {3}{4}\) of 28 pencils.
Solution:

15. Balkirat had a box of 36 erasers. He gave \(\frac {1}{2}\) of them to Rani, \(\frac {2}{9}\) of them to Yuvraj and keeps the rest.

Question (i)
(i) How many erasers does Rani get?
(ii) How many erasers does Yuvraj get?
(iii) How many erasers does Harnik keep?
Solution:
Total erasers = 36

(iii) Erasers Harnik gets = 36 – (18 + 8)
= 36 – 26
= 10

16. State True/False

Question (i)

Solution:
(i) False
(ii) True
(iii) True
(iv) True

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 7 Triangles Ex 7.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.5

Question 1.
ABC is a triangle. Locate a point in the interior of ∆ ABC which is equidistant from all the vertices of ∆ ABC.
Answer:

In ∆ ABC, draw l, the perpendicular bisector of side AB and m, the perpendicular bisector of side BC. Name the point of intersection of l and m as P.
P is a point on the perpendicular bisector of AB. Hence, P is equidistant from A and B.
∴ PA = PB
P is a point on the perpendicular bisector of BC. Hence, P is equidistant from B and C.
∴ PB = PC
Thus, PA = PB = PC
Hence, P is the required point which is equidistant from all the vertices of ∆ ABC.
Note: Since ∆ ABC given here is an acute angled triangle, point P lies in the interior of ∆ ABC. If ∆ ABC is a right angled Mangle, point P lies on the hypotenuse. Actually, in that case, point P will be the midpoint of the hypotenuse. Lastly, if ∆ ABC is an obtuse angled triangle, point P lies in the exterior of ∆ ABC. This point P is called the circumcentre of ∆ ABC.

Question 2.
In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.

Answer:
In ∆ ABC, draw the bisectors of ∠B and ∠C to intersect each other at point I.
I is a point on the bisector of ∠ B. Hence, I is equidistant from sides BA and BC. Similarly, I is a point on the bisector of ZC. Hence, I is equidistant from sides BC and CA.
Thus, point I is the required point which is equidistant from all the three sides AB, BC and CA of ∆ ABC.
This point I is called the incentre of ∆ ABC. It always lies in the interior of ∆ ABC irrespective of the type of ∆ ABC.

Question 3.
In a huge park, people are concentrated at three points (see the given figure):

A: where there are different slides and swings for children.
B: near which a man-made lake is situated.
C: which is near to a large parking and exit.
Where should an ice cream parlour be set up so that maximum number of persons can approach it ?
(Hint: The parlour should be equidistant from A, B and C.)
Answer:
First of all, construct ∆ ABC with the given points A, B and C as vertices. Then, as shown in example 1, draw the perpendicular s bisectors of any two sides of ∆ ABC and name their point of intersection as P.

Now, the ice cream parlour should be set up at the location given by point P as it is equidistant from all the three places (points) A, B and C.

Question 4.
Complete the hexagonal and star shaped < Rangolies [see figure (1) and (2)] by filling them with as many equilateral triangles of > side 1 cm as you can. Count the number < of triangles in each case. Which has more triangles?

Answer:
In figure (1), if we join the opposite vertices, we get three longest diagonals of hexagon ABCDEE By the intersection of these diagonals we get point O and six equilateral triangles – ∆ OAB, ∆ OBC, ∆ OCD, ∆ ODE, ∆ OEF and ∆ OFA. Each side of all these equilateral triangles will measure 5 cm. In each of these six triangles, we can fill 25 (1 + 3 + 5 + 7 + 9) equilateral triangles with side 1 cm each. Hence in the hexagonal Rangoli ABCDEF, we can fill 25 × 6 = 150 triangles.

Similarly, in figure (2), we can fill 150 triangles in the inner hexagon and 150 triangles in the six triangles on the boundary of the hexagon. Thus, in figure (2), 150 + 150 = 300 triangles can be filled.
Hence, more triangles can be filled in figure (2).

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 7 Triangles Ex 7.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4

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

In ∆ ABC, ∠ B is a right angle.
∴ ∠ B = 90° and AC is the hypotenuse.
In ∆ ABC,
∠A + ∠B + ∠C = 180°
∴ ∠ A + 90° + ∠ C = 180°
∴ ∠ A + ∠ C = 90°
Now, ∠ A and ∠ C are both positive (in degrees) and their sum is 90°.
∴ ∠ A < 90° and ∠ C < 90°
∴ ∠ A < Z B and ∠ C < ∠ B
∴ BC < AC and AB < AC (Theorem 7.7)
Hence, hypotenuse AC is greater than each of the other two sides BC and AB. Thus, the hypotenuse is the longest side in a right angled triangle.

Question 2.
In the given figure, sides AB and AC of ∆ ABC are extended to points P and Q respectively. Also, ∠PBC < ∠ QCB. Show that AC > AB.

Answer:
∠ PBC < ∠ QCB (Given)
∴ – ∠ PBC > – ∠ QCB (Multiplying an inequality by (-1), it reverses)
∴ 180° – ∠ PBC > 180° – ∠ QCB (Adding 180° on both the sides) …………. (1)
Now, ∠ ABC and ∠ PBC as well as ∠ ACB and ∠ QCB from a linear pair.
∴ ∠ ABC = 180° – ∠ PBC and
∠ ACB = 180° – ∠ QCB
Substituting these values in (1), we get
∠ ABC > ∠ ACB
Now, in ∆ ABC, ∠ ABC > ∠ ACB A
∴ AC > AB (Theorem 7.7)

Question 3.
In the given figure, ∠ B < ∠ A and ∠ C < ∠ D. Show that AD < BC.

Answer:
In ∆ OAB, ∠ B < ∠ A
∴ OA < OB (Theorem 7.7) …………….. (1)
In ∆ OCD, ∠ C < ∠ D
∴ OD < OC (Theorem 7.7) ……………… (2)
Adding (1) and (2),
OA + OD < OB + OC
∴ AD < BC (As O is the point of intersection of AD and BC, it lies on both the line segments.)

Question 4.
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD ,(see the given figure). Show that ∠A > ∠C and ∠B > ∠D.

Answer:

Construction: In quadrilateral ABCD, draw diagonal AC.
AB is the smallest side of ABCD and CD is the longest side of ABCD.
∴ AB < BC and AD < CD.
In ∆ ABC, AB < BC
∴ ∠ ACB < ∠ BAC …… (1)
In ∆ CDA, AD < CD
∴ ∠ DCA < ∠ DAC ……… (2)
Adding (1) and (2),
∠ ACB + ∠ DCA < ∠ BAC + ∠ DAC
∴ ∠ BCD < ∠ BAD
∴ ∠ BAD > ∠ BCD
Thus, in quadrilateral ABCD, ∠ A > ∠ C. Similarly, after constructing diagonal BD and using the inequalities in A ABD and A CBD, it can be proved that ∠ B > ∠ D.

Question 5.
In the given figure, PR > PQ and PS bisects ∠ QPR. Prove that ∠ PSR > ∠ PSQ.

Answer:
PS is the bisector of ∠QPR.
∴ ∠QPS = ∠RPS = \(\frac{1}{2}\) ∠QPR ……………. (1)
∠ PSR is an exterior angle of A PQS and ∠ PSQ is an exterior angle of A PRS.
∴ PSR = ∠ Q + ∠QPS and
∠PSQ = ∠R + ∠RPS …………….. (2)
Now, in A PQR, PR > PQ
∴ ∠ Q > ∠ R
∴ ∠Q + \(\frac{1}{2}\) ∠ QPR > ∠R + \(\frac{1}{2}\) ∠ QPR
∴ ∠Q + ∠QPS > ∠R + ∠RPS [from (1)]
∴ ∠PSR > ∠PSQ

Question 6.
Show that of all line segments drawn from a given point not on a given line, the perpendicular line segment is the shortest.
Answer:

AB is a line and P is a point not on AB.
PM is the perpendicular line segment drawn from P to line AB.
N is any point on AB, other than M.
In ∆ PMN, ∠ M = 90°
∴ ∠ N < 90°
Thus, in ∆ PMN, ZN < ZM.
∴ PM < PN
This is true for any location of point N.
Hence, of all the line segments drawn from a point not on a given line, the, perpendicular line segment is the shortest.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 7 Triangles MCQ Questions with Answers.

PSEB 9th Class Maths Chapter 7 Triangles MCQ Questions

Multiple Choice Questions and Answer

Answer each question by selecting the proper alternative from those given below each question to make the statement true:

Question 1.
In ∆ ABC, ∠A = ∠C, AC = 5 and BC = 4. Then, the perimeter of ∆ ABC is ……………. .
A. 9
B. 14
C. 13
D. 15
Answer:
C. 13

Question 2.
In ∆ PQR, PQ = PR, QR is extended to S and ∠PRS = 110°. Then, ∠PQR = ……………… .
A. 30°
B. 50°
C. 80°
D. 70°
Answer:
D. 70°

Question 3.
In ∆ ABC and ∆ DEF, AB = DE, BC = EF and ∠B = ∠E. If the perimeter of ∆ ABC is 20, then the perimeter of ∆ DEF is ……………. .
A. 10
B. 20
C. 15
D. 40
Answer:
B. 20

Question 4.
In ∆ ABC and ∆ PQR, AB = PQ, ∠A = ∠P and ∠B = ∠Q. If ∠A + ∠C = 130°, then ∠Q = …………….. .
A. 65°
B. 130°
C. 50°
D. 100°
Answer:
C. 50°

Question 5.
In ∆ PQR, ∠P = ∠Q = ∠R. If PQ = 6, then the perimeter of ∆ PQR is ………………. .
A. 12
B. 9
C. 18
D. 24
Answer:
C. 18

Question 6.
In ∆ ABC, AB < AC. Then …………… holds good.
A. ∠A < ∠B
B. ∠B < ∠C
C. ∠C < ∠A
D. ∠C < ∠B
Answer:
D. ∠C < ∠B

Question 7.
In ∆ PQR, ∠R > ∠Q. Then, ………………. holds good.
A. PQ > PR
B. QR > PQ
C. PR > PQ
D. PQ > QR
Answer:
A. PQ > PR

Question 8.
In ∆ ABC, AB > BC and BC > AC. Then, the smallest angle of ∆ ABC is …………………. .
A. ∠A
B. ∠C
C. ∠B
D. ∠A or ∠C
Answer:
C. ∠B

Question 9
………………….. cannot be the measures of the sides of a triangle.
A. 10, 12, 14
B. 2, 3, 4
C. 8, 9, 10
D. 2, 4, 10
Answer:
D. 2, 4, 10

Question 10.
In ∆ PQR, PQ = 4, QR = 6 and PR = 5. Then, …………….. is the angle with greatest measure in ∆ PQR.
A. ∠P
B. ∠Q
C. ∠R
D. ∠QRP
Answer:
A. ∠P

Question 11.
In ∆ XYZ, ∠X = 45° and ∠Z = 60°. Then, …………….. is the longest side of ∆ XYZ.
A. XY
B. YZ
C. XZ
D. XY or YZ
Answer:
C. XZ

Question 12.
In ∆ ABC, ∠B = 30° and BC is extended to D. If ∠ACD = 110°, then the longest side of ∆ ABC is ………………. .
A. AB
B. BC
C. CA
D. AB or AC
Answer:
B. BC

Question 13.
In ∆ ABC, AB = 4 and BC = 7, Then, ……………… holds good.
A. AC < 7
B. AC > 4
C. 4 < AC < 7
D. 3 < AC < 11
Answer:
D. 3 < AC < 11

Question 14.
In ∆ PQR, PQ = 3 and QR = 7. Then, …………….. holds good.
A. PR = 4
B. PR = 10
C. 10 > PR > 4
D. 7 > PR > 3
Answer:
C. 10 > PR > 4

Question 15.
In ∆ ABC, the bisectors of ∠B and ∠C intersect at I. If ∠A = 70°, then ∠BIC = ………………… .
A. 35°
B. 75°
C. 100°
D. 125°
Answer:
D. 125°

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 11 Perimeter and Area Ex 11.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.1

1. Find the perimeter and the area of a rectangle having :
(i) Length = 28 cm, Breadth = 15 cm
(ii) Length = 9.4 cm Breadth = 2.5 cm
Solution:
(i) Given length of rectangle = 28 cm
Breadth of rectangle = 15 cm
Perimeter of rectangle = 2 [length + Breadth]
= 2 [28 + 15]
= 2 × 43
= 86 cm

Area of rectangle = length × Breadth
= 28 × 15
= 420 cm²

(ii) Perimeter of rectangle = 2 [9.4 + 2.5]
= 2 × 11.9
= 23.8 cm
Area of rectangle = 9.4 × 2.5
= 23.5 cm²

2. Find the perimeter and the area of a square whose side measures
(i) 29 cm
(ii) 8.3 cm
Solution:
(i) Given side of square = 29 cm
Perimeter of square = 4 × side
= 4 × 29
= 116 cm
Area of square = (side)²
= (29)²
= 841 cm²

(ii) Perimeter of square = 4 × 8.3
= 33.2 cm
Area of square = 8.3 × 8.3
= 68.89 cm²

3. The perimeter of a square park is 148 m. Find its area.
Solution:
Given the perimeter of square park = 148 m
Side of the square park = \(\frac{perimeter}{4}\)
= \(\frac {148}{4}\)
Area of the square park = (side)²
= (37)²
= 1369 m²

4. The area of a rectangle is 580 cm². Its length is 29 cm. Find its breadth and also, the perimeter.
Solution:
Given area of rectangle = 580 cm²
Length of the rectangle = 29 cm
Let breadth of the rectangle = b cm
Area of the rectangle = length × breadth
580 = 29 × b
\(\frac {580}{29}\) = b
b = 20 cm
Perimeter of rectangle = 2 [length + breadth]
= 2 [29 + 20]
= 2 × 49
= 98 cm

5. A wire is in the shape of a rectangle. Its length is 48 cm and breadth is 32 cm. If the same wire is rebent into the shape of a square, what will be the measure of each side. Also, find which shape encloses more area and by how much ?
Solution:
Given length of the rectangle = 48 cm
Breadth of the rectangle = 32 cm
Perimeter of the rectangle = 2 [length + breadth]
= 2 [48 + 32]
= 2 × 80
= 160 cm
Let side of square = a cm
Perimeter of the square = 4 × a
Since wire is rebent into the shape of a square
Perimeter of square = Perimeter of rectangle
4 a = 160
Therefore, a = \(\frac {160}{4}\)
= 40 cm
Area of square = (side)²
= 40 × 40
= 1600 cm²
Area of rectangle = length × breadth
= 48 × 32
= 1536 cm²
∴ Square encloses more area by 64 cm²

6. The area of a square park is the same as that of a rectangular park. If the side of the square park is 75 m and the length of the rectangular park is 125 m, find the breadth of the rectangular park. Also, find the perimeter of rectangular park.
Solution:
Given side of square park = 75 m
Area of square park = (75)²
= 75 × 75
= 5625 m²
Length of rectangular park = 125 m
Let breadth of rectangular park = b m
Area of rectangular park = length × breadth
= 125 × b m²
Given that
Area of rectangular park = Area of square park
125 × b = 5625
= 45 m
Perimeter of rectangular park = 2 [length + breadth]
= 2 [125 + 45]
= 2 × 170
= 340 m

7. A door of length 2.5 m and breadth 1.5 m is fitted in a wall. The length of wall, is 9 m and breadth is 6 m. Find the cost of painting the wall, if the rate of painting the wall is ₹ 30 per m².

Solution:
Length of door = 2.5 m
Breadth of door = 1.5 m
Area of door = length × breadth
= 2.5 × 1.5
= 3.75 m²
Area of wall = 9 × 6
= 54 m²
Area of wall painting = Area of wall including door – Area of door
= 54 – 3.75
= 50.25 m²
Cost of painting 1 m² of wall = ₹ 30
Cost of painting 50.25 m² of wall = ₹ 50.25 × 30
= ₹ 1507.50

8. A door of dimensions 3 m × 2 m and a window of dimensions 2.5 m × 1.5 m is fitted in a wall. The length of the wall is 7.8 m and breadth is 3.9 m. Find the cost of painting the wall, if the rate of painting the wall is ₹ 25 per m².
Solution:
Area of door = 3 × 2 = 6 m²
Area of window = 2.5 m × 1.5 m
= 3.75 m²
Area of wall = 7.8 m × 3.9 m
= 30.42 m²
Area of painting the wall = Area of wall – Area of door – Area of window
= 30.42 – 6 – 3.75
= 20.67 m²
Cost of painting the wall = ₹ 25 × 20.67
= ₹ 516.75

9. Find the area and the perimeter of the following figures.

Solution:
(i) Perimeter of the given figure
= AB + BC + CD + DE + EF + FG + GH + HA
= 2 + 3.5 + 3 + 2 + 5 + 3.5 + 10 + 9
= 38 cm²
Area of the figure = Area of rectangle ABCJ + Area of rectangle JDEI + Area of rectangle IFGH
= 2 × 3.5 + 5 × 2 + 10 × 3.5
= 7 + 10 + 35
= 52 cm²

(ii) Perimeter of the given figure
= 8cm + 5 cm + 1.5 cm + 2.5 cm + 2.5 cm + 1.5 cm + 1.5 cm + 1.5 cm + 2.5 cm + 1.5 cm
= 29 cm
Area of the given figure = Area of rectangle I + Area of rectangle II + Area of rectangle III
= 8 cm × 1.5 cm + 3.5 cm × 1.5 cm + 1.5 cm × 1.5 cm
= 12 cm² + 5.25 cm² + 2.25 cm
= 19.5 cm²

10. Multiple Choice Questions :

Question (i).
What is the area of a rectangle of dimensions 12 cm × 10 cm ?
(a) 44 cm²
(b) 120 cm²
(c) 1200 cm²
(d) 1440 cm²
Answer:
(b) 120 cm²

Question (ii).
Find the breadth of a rectangle whose length is 12 cm and perimeter is 36 cm.
(a) 6 cm
(b) 3 cm
(c) 9 cm
(d) 12 cm
Answer:
(a) 6 cm

Question (iii).
If each side of a square is 1 m then its area is ?
(a) 10 cm²
(b) 100 cm²
(c) 1000 cm²
(d) 10000 cm²
Answer:
(d) 10000 cm²

Question (iv).
Find the area of a square whose perimeter is 96 cm.
(a) 576 cm²
(b) 626 cm²
(c) 726 cm²
(d) 748 cm².
Answer:
(a) 576 cm²

Question (v).
The area of a rectangular sheet is 500 cm². If the length of the sheet is 25 cm, what is its breadth ?
(a) 30 cm
(b) 40 cm
(c) 20 cm
(d) 25 cm.
Answer:
(c) 20 cm

Question (vi).
What happens to the area of a square, if its side is doubled ?
(a) The area becomes 4 times, the area of original square.
(b) The area becomes \(\frac {1}{4}\) times, the area of original square.
(c) The area becomes 16 times, the area of original square.
(d) The area becomes \(\frac {1}{6}\) times, 6 the area of original square.
Answer:
(a) The area becomes 4 times, the area of original square.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 4 Integers Ex 4.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 4 Integers Ex 4.2

1. Using number line write the integer which is:

Question (a)
5 less than -1
Solution:
5 less than -1 = ?
We need to find the integer which is 5 less than -1.
So, we shall start with ‘-1 ’ and proceed 5 steps to the left of -1 as shown below:

Therefore 5 less than -1 is -6

Question (b)
5 more than -5
Solution:
5 more than -5 = ?
We need to find the integer which is 5 more than -5.
So, we shall start with -5 and proceed 5 steps to the right of -5 as shown below:

Therefore 5 more than -5 is zero.

Question (c)
2 less than 5
Solution:
2 less than 5 = ?
We need to find the integer which is 2 less than 5.
So, we shall start with 5 and proceed
2 steps to the left of 5 as shown below:

Therefore 2 less than 5 is 3.

Question (d)
3 less than -2.
Solution:
3 less than – 2 = ?
We need to find the integer which is 3 less than -2.
So, we shall start with -2 and proceed 3 steps to the left of -2 as shown below:

Therefore 3 less than -2 is -5.

2. Using number line, add the following integers:

Question (a)
9 + (-3)
Solution:
On number line we shall start from 0 and move 9 steps to the right of zero. Then we shall move three steps to the left of ‘+9’. We finally reach at +6. Thus, +6 is the answer.

Hence, 9 + (-3) = +6

Question (b)
5 + (-11)
Solution:
On number line we shall start from 0 and move 5 steps to the right of zero. Then we shall move eleven steps to the left of ‘+5’. We finally reach at ‘-6’. Thus, -6 is the answer.

Hence, 5 + (-11) = -6

Question (c)
(-1) + (-4)
Solution:
On number line we first move 1 step to the left of zero. Then we moved ahead 4 steps to the left of ‘-1\ We finally reaches at -5. Thus -5 is the answer.

Hence, (-1) + (-4) = -5

Question (d)
(-5) + 12
Solution:
On number line we shall move 5 steps to the left of zero. Then we shall move 12 steps to the right of ‘-5’ and finally reach at ‘+7’. Thus +7 is the answer.

Hence, (-5) + 12 = +7

Question (e)
(-1) + (-2) + (-4)
Solution:
Step I: On number line we shall move one step to the left of zero suggested by minus sign of ‘-1’.
Step II: Then we shall move 2 steps to the left of -1 suggested by minus sign of -2 and we reach at ‘-3’.
Step III: Then again we shall move 4 steps to the left of ‘-3’ suggested by minus sign of -4 and finally we reach at-7.

Hence, (-1) + (-2) + (-4) = -7

Question (f)
(-2) + 4 + (-5)
Solution:
Step I: On number line we shall move 2 steps to the left of zero suggested by minus sign of ‘-2’.
Step II: Then we shall move 4 steps to the right of ‘-2’ suggested by plus sign of ‘44’ and we reach at +2.
Step III: Then we shall move 5 steps to the left of ‘+2’ as suggested by minus sign of ‘-5’ and finally we reach at -3. Thus answer is ‘-3’.

Hence, (-2) + 4 + (-5) = -3

Question (g)
(-3) + (5) + (-4).
Solution:
Step I: On number line we shall move 3 steps to the left of zero as suggested by minus sign of ‘-3’.
Step II: Then we shall move 5 steps to the right of ‘-3’ as suggested by plus sign of ‘+5’ and we reach at ‘+2’.
Step III: Then we shall move 4 steps to the left of ‘+2’ as suggested by minus sign of ‘4’ and finally we reach at ‘-2’. Thus ‘-2’ is the answer.

Hence, (-3) + (5) + (-4) = -2

3. Add without using number line:

Question (a)
18 + 13
Solution:
18 + 13 = 31

Question (b)
18 + (- 13)
Solution:
18 + (- 13) = + (18 – 13) = +5

Question (c)
(-18) + 13
Solution:
(-18) + 13 = – (18 – 13) = -5

Question (d)
(-18) + (-13)
Solution:
(-18) + (-13) = -(18 + 13) = -31

Question (e)
180 + (-200)
Solution:
180 + (-200) = -(200 – 180) = -20

Question (f)
111 + (-67)
Solution:
111 + (-67) = (777 – 67) = 710

Question (g)
1262 + (-366) + (-962)
Solution:
1262 + (-366) + (-962)
= 1262 – (366 + 962)
= 1262 – 1328
= -(1328 – 1262) = -66

Question (h)
30 + (-27) + 21 + (-19) + (-3) + 11 + (-9)
Solution:
30 + (-27) + 21 + (-19) + (-3) + 11 + (-9)
= 30 + 21 + 11 + (-27) + (-19) + (-3) + (-9)
= 62 + (-58) = 62 – 58 = 4

Question (i)
(-7) + (-9) + 4 + 16
Solution:
(-7) + (-9) + 4+16 = (-16) + 20 = 4

Question (j)
37 + (-2) + (-65) + (-8).
Solution:
37 + (-2) + (-65) + (-8)
= 37 – (2 + 65 + 8)
= 37 – 75 = -38

4. Write the successor and predecessor of the following:

Question (a)
-15
Solution:
Successor of -15 = -15 + 1 = -14
Predecessor of -15 = -15 – 1 = -16

Question (b)
27
Solution:
Successor of 27 = 27 + 1 = 28
Predecessor of 27 = 27 – 1 = 26

Question (c)
-79
Solution:
Successor of -79 = -79 + 1 = -78
Predecessor of -79 = -79- 1 = -80

Question (d)
0
Solution:
Successor of 0 = 0+ 1 = 1
Predecessor of 0 = 0 – 1 = -1

Question (e)
29
Solution:
Successor of 29 = 29 + 1 = 30
Predecessor of 29 = 29 – 1 = 28

Question (f)
-18
Solution:
Successor of -18 = -18 + 1 = -17
Predecessor of -18 = -18 – 1 = -19

Question (g)
-21
Solution:
Successor of -21 = -21 + 1 = -200
Predecessor of -21 = -21 – 1 = -22

Question (h)
99
Solution:
Successor of 99 = 99 + 1 = 100
Predecessor of 99 = 99 – 1 = 98

Question (i)
-1
Solution:
Successor of-1 = -1 + 1 = 0
Predecessor of -1 = -1 – 1 = -2

Question (j)
-13.
Solution:
Successor of -13 = -13 + 1 = -12
Predecessor of -13 = -13 – 1 = -14

5. Complete the following addition table:

+	-3	-4	-2	+1	+2	+3
-2
-3
0
+1
+2

Solution:

+	-3	-4	-2	+ 1	+2	+3
-2	-5	-6	-4	-1	0	+ 1
-3	-6	-7	-5	-2	-1	0
0	-3	-4	-2	+ 1	+2	+3
+ 1	-2	-3	-1	+2	+3	+4
+2	-1	-2	0	+3	+4	+5

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

PSEB Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3

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

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

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

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

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

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

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

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

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

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

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

PSEB Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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