Extra Questions for Class 7 Maths PDF are provided here. Students can download the pdf of these solutions from the given links. Practical Geometry Class 7 Extra Questions Maths Chapter 10 provided in accordance with the latest syllabus of CBSE which, in turn, help the students to build a strong foundation and secure excellent marks in their board exams. https://meritbatch.com/practical-geometry-class-7-extra-questions/
Practical Geometry Class 7 Extra Questions Maths Chapter 10
Extra Questions for Class 7 Maths Chapter 10 Practical Geometry
Practical Geometry Class 7 Extra Questions Very Short Answer Type
Practical Geometry Class 7 Extra Questions Question 1.
State whether the triangle is possible to construct if
(a) In ΔABC, m∠A = 80°, m∠B = 60°, AB = 5.5 cm
(b) In ΔPQR, PQ = 5 cm, QR = 3 cm, PR = 8.8 cm
Solution:
(a) m∠A = 80°, m∠B = 60°
m∠A + m∠B = 80° + 60° = 140° < 180°
So, ΔABC can be possible to construct.
(b) PQ = 5 cm, QR = 3 cm, PR = 8.8 cm
PQ + QR = 5 cm + 3 cm = 8 cm < 8.8 cm
or PQ + QR < PR
So, the ΔPQR can not be constructed.
Practical Geometry Class 7 Worksheets With Answers Question 2.
Draw an equilateral triangle whose each side is 4.5 cm.
Solution:
Steps of construction:
(i) Draw AB = 4.5 cm.
(ii) Draw two arcs with centres A and B and same radius of 4.5 cm to meet each other at C.
(iii) Join CA and CB.
(iv) ΔCAB is the required triangle.
Practical Geometry Class 7 Worksheet Question 3.
Draw a ΔPQR, in which QR = 3.5 cm, m∠Q = 40°, m∠R = 60°.
Solution:
Steps of construction:
(i) Draw QR = 3.5 cm.
(ii) Draw ∠Q = 40°, ∠R = 60° which meet each other at P.
(iii) ΔPQR is the required triangle.
Extra Questions On Practical Geometry For Class 7 Question 4.
There are four options, out of which one is correct. Choose the correct one:
(i) A triangle can be constructed with the given measurement.
(a) 1.5 cm, 3.5 cm, 4.5 cm
(b) 6.5 cm, 7.5 cm, 15 cm
(c) 3.2 cm, 2.3 cm, 5.5 cm
(d) 2 cm, 3 cm, 6 cm
(ii) (a) m∠P = 40°, m∠Q = 60°, AQ = 4 cm
(b) m∠B = 90°, m∠C = 120° , AC = 6.5 cm
(c) m∠L = 150°, m∠N = 70°, MN = 3.5 cm
(d) m∠P = 105°, m∠Q = 80°, PQ = 3 cm
Solution:
(i) Option (a) is possible to construct.
1.5 cm + 3.5 cm > 4.5 cm
(ii) Option (a) is correct.
m∠P + m∠Q = 40° + 60° = 100° < 180°
Practical Geometry Class 7 Worksheets With Answers Pdf Question 5.
What will be the other angles of a right-angled isosceles triangle?
Solution:
In right angled isosceles triangle ABC, ∠B = 90°
∠A + ∠C = 180° – 90° = 90°
But ∠A = ∠B
∠A = ∠C = \(\frac { 90 }{ 2 }\) = 45°
Hence the required angles are ∠A = ∠C = 45°
Practical Geometry Class 7 Questions Question 6.
What is the measure of an exterior angle of an equilateral triangle?
Solution:
We know that the measure of each interior angle = 60°
Exterior angle = 180° – 60° = 120°
Class 7 Maths Practical Geometry Extra Questions Question 7.
In ΔABC, ∠A = ∠B = 50°. Name the pair of sides which are equal.
Solution:
∠A = ∠B = 50°
AC = BC [∵ Sides opposite to equal angles are equal]
Hence, the required sides are AC and BC.
Practical Geometry Extra Questions Class 7 Question 8.
If one of the other angles of a right-angled triangle is obtuse, whether the triangle is possible to construct.
Solution:
We know that the angles other than right angle of a right-angled triangle are acute angles.
So, such a triangle is not possible to construct.
Extra Questions Of Practical Geometry Class 7 Question 9.
State whether the given pair of triangles are congruent.
Solution:
Here, AB = PQ = 3.5 cm
AC = PR = 5.2 cm
∠BAC = ∠QPR = 70°
ΔABC = ΔPQR [By SAS rule]
Practical Geometry Class 7 Extra Questions Short Answer Type
Ncert Class 7 Maths Chapter 10 Extra Questions Question 10.
Draw a ΔABC in which BC = 5 cm, AB = 4 cm and m∠B = 50°.
Solution:
Steps of construction:
(i) Draw BC = 5 cm.
(ii) Draw ∠B = 50° and cut AB = 4 cm.
(iii) Join AC.
(iv) ΔABC is the required triangle.
Practical Geometry Questions For Class 7 Question 11.
Draw ΔPQR in which QR = 5.4 cm, ∠Q = 40° and PR = 6.2 cm.
Solution:
Steps of construction:
(i) Draw QR = 5.4 cm.
(ii) Draw ∠Q = 40°.
(iii) Take R as the centre and with radius 6.2 cm, draw an arc to meet the former angle line at P.
(iv) Join PR.
(v) ΔPQR is the required triangle.
Questions On Practical Geometry For Class 7 Question 12.
Construct a ΔPQR in which m∠P = 60° and m∠Q = 30°, QR = 4.8 cm.
Solution:
m∠Q = 30°, m∠P = 60°
m∠Q + m∠P + m∠R = 180° (Angle sum property of triangle)
30° + 60° + m∠R = 180°
90° + m∠R = 180°
m∠R = 180° – 90°
m∠R = 90°
Steps of construction:
(i) Draw QR = 4.8 cm.
(ii) Draw ∠Q = 30°.
(iii) Draw ∠R = 90° which meets the former angle line at P.
(iv) ∠P = 180° – (30° + 90°) = 60°
(v) ΔPQR is the required triangle.
Practical Geometry Class 7 Extra Questions Higher Order Thinking Skills [HOTS] Type
Question 13.
Draw an isosceles right-angled triangle whose hypotenuse is 5.8 cm.
Solution:
Right angled triangle is an isosceles triangle
Each of its acute angles = \(\frac { 90 }{ 2 }\) = 45°
Steps of construction:
(i) Draw AB = 5.8 cm.
(ii) Construct ∠A = 45° and ∠B = 45° to meet each other at C.
(iii) ∠C = 180° – (45° + 45°) = 90°
(iv) ΔACB is the required isosceles right angle triangle.
Question 14.
Construct a ΔABC such that AB = 6.5 cm, AC = 5 cm and the altitude AP to BC is 4 cm.
Solution:
Steps of construction:
(i) Draw a line l and take any point P on it.
(ii) Construct a perpendicular to l at P.
(iii) Cut AP = 4 cm.
(iv) Draw two arcs with centre A and radii 6.5 cm and 5 cm to cut the line l at B and C respectively.
(v) Join AB and AC.
(vi) ΔABC is the required triangle.
Question 15.
Construct an equilateral triangle whose altitude is 4.5 cm.
Solution:
Steps of construction:
(i) Draw any line l and take a point D on it.
(ii) Construct a perpendicular to l at D and cut AD = 4.5 cm.
(iii) Draw the angle of 30° at on either side of AD to meet the line l at B and C.
(iv) ΔABC is the required equilateral triangle.