CHAPTER – 10 Practical Geometry | CLASS 7TH |NCERT MATHS IMPORTANT QUESTIONS & MCQS | EDUGROWN

Answer: (a) A.S.A. criterion
Hint:

Answer: (d) S.S.S. criterion
Hint:
Since ALMN is equilateral the measurement of one side is used for the other two sides of the triangle. Hence ALMN can be constructed by S.S.S. criterion.

Answer: (a) A line parallel to a given line.

Answer: (c) The vertex B cannot be uniquely located.

Answer: (a) Equal corresponding angles.
Hint:
Corresponding angles of parallel lines are equal.

Answer: (c) Both [a] and [b]
Hint:
In ΔPQR since all the angles are acute, it is acute angled. Also since all the angles are equal, it is equilateral.

Answer: (b) ∠A
Hint:
From the given measurements, BC¯¯¯¯¯¯¯¯ is the hypotenuse. The angle opposite to BC¯¯¯¯¯¯¯¯ is ∠A which is a right angle.

Answer: (b) ΔPQR is an equilateral triangle.

Answer: (a) ΔPQR is an equilateral triangle.

Answer: (a) 1
Hint:

As can be seen from the given figure, one and only one perpendicular line can be drawn to a given line from a point not on it.

Answer: (c) A triangle with two equal sides is called a scalene triangle.

Answer: (d) Both [a] and [b]

Answer: (b) A perpendicular bisector of a line segment.

Important 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.

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.

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.

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°

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 = 902 = 45°
Hence the required angles are ∠A = ∠C = 45°

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°

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.

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.

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

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.

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.

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 = 902 = 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.