PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.3

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3

1. Construct ΔABC such that AB = 4 cm, ∠B = 30°, BC = 4 cm. Also name the type of triangle on the basis of sides.
Solution:
Given : Two sides of ΔABC as AB = 4 cm, BC = 4 cm and ∠B = 30°.
To construct: A triangle with these two sides and included angle.
Step of Construction :
Step 1. We first draw a rough sketch of the ΔABC and indicate the measure of these two sides and included angle.
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.3 1
Step 2. Draw a line segment BC of length 4 cm.
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.3 2
Step 3. At B draw BX making an angle of 30° with BC (The point A must be somewhere on this ray of the angle).
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.3 3
Step 4. (To fix A, the distance AB has been given) With B as centre, draw an arc of radius 3 cm. It cuts BX at the point A.
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.3 4
Step 5. Join AC.
ΔDEF is now obtained.
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.3 5
Since two sides of triangle are equal.
Therefore ΔABC is an isosceles triangle.

PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.3

2. Construct ΔABC with AB = 7.5 cm, BC = 5 cm and ∠B = 30°.
Solution:
Given. Two sides of ΔABC as AB = 7.5 cm,
BC = 5 cm
and ∠B = 30°
To construct A triangle with these two sides and included angle.
Steps of Construction.
Step 1. We first draw a rough sketch of the ΔABC and indicate the measures of these two sides and included angle.
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.3 6
Step 2. Draw a line segment BC of length 5 cm.
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.3 7
Step 3. At B draw BX making an angle of 30° with BC. (The point A must be somewhere on this ray of the angle)
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.3 8
Step 4. (To fix A; the distance BC has been given) With B as centre draw an arc of radius 7.5 cm. It cuts CX at the point A.
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.3 9
Step 5 : Join AC.
ΔABC is now obtained.
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.3 10

PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.3

3. Construct a triangle XYZ, such that XY = 6 cm, YZ = 6 cm and ∠Y = 60°. Also name the type of this triangle.
Solution:
Step 1. Draw a rough sketch of XYZ with given measures.
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.3 11
Step 2. Draw a line segment XY of length 6 cm.
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.3 12
Step 3. With the help of compass, at Y, draw a ray YA making an angle 60°
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.3 13
Step 4. With Y as centre and radius 6 cm. draw an arc intersecting the ray YX at point Z.
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.3 14
Step 5. Join XZ.ΔXYZ is required triangle, Measure the third side. We see that ZX = 6 cm
∴ In Δ XYZ
XY = YZ = ZX = 6 cm
Therefore XYZ is an equilateral triangle.
PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.3 15

PSEB 7th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.3

4. Which of the following triangle can be constructed using SAS criterion.
(a) AB = 5 cm, BC = 5 cm, CA = 6 cm
(b) AB = 5 cm, BC = -5 cm, ∠B = 40°
(c) ∠A = 60°, ∠B = 60°, ∠C = 60°
(d) BC = 5 cm, ∠B = ∠C = 45°
Answer:
(b) AB = 5 cm, BC = -5 cm, ∠B = 40°

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