Table of Contents
MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th
Get Chapter Wise MCQ Questions for Class 7 Maths with Answers prepared here according to the latest CBSE syllabus and NCERT curriculum. Students can practice CBSE Class 7 Maths MCQs Multiple Choice Questions with Answers to score good marks in the examination. Students can also visit the most accurate and elaborate NCERT Solutions for Class 7 Maths. Every question of the textbook has been answered here.
Chapter - 10 Practical Geometry
MCQs
Question 1.
In ΔRST, R = 5 cm, and ∠SRT = 45° and ∠RST = 45°. Which criterion can be used to construct ΔRST?
(a) A.S.A. criterion
(b) S.A.S. criterion
(c) S.S.S. criterion
(d) R.H.S. criterion
Answer
Answer: (a) A.S.A. criterion
Hint:
Clearly, from the figure two angles and the included side are given. So, A.S.A. criterion can be used to construct ARST.
Question 2.
Identify the criterion of construction of the equilateral triangle LMN given LM = 6 cm.
(a) S.A.S. criterion
(b) R.H.S. criterion
(c) A.S.A. criterion
(d) S.S.S. criterion
Answer
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.
Question 3.
The idea of equal alternate angles is used to construct which of the following?
(a) A line parallel to a given line
(b) A triangle
(c) A square
(d) Two triangles
Answer
Answer: (a) A line parallel to a given line.
Question 4.
A Given AB = 3 cm, AC = 5 cm,and ∠B = 30°, ΔABC cannot be uniquely constructed, with AC as base, why?
(a) Two sides and included angle are given.
(b) The other two angles are not given.
(c) The vertex B cannot be uniquely located.
(d) The vertex A coincides with the vertex C.
Answer
Answer: (c) The vertex B cannot be uniquely located.
Question 5.
A line panda point X not on it are given. Which of the following is used to draw a line parallel to p through X?
(a) Equal corresponding angles.
(b) Congruent triangles.
(c) Angle sum property of triangles.
(d) Pythagoras’ theorem.
Answer
Answer: (a) Equal corresponding angles.
Hint:
Corresponding angles of parallel lines are equal.
Question 6.
Δ PQR is such that ∠P = ∠Q = ∠R = 60° which of the following is true?
(a) Δ PQR is equilateral.
(b) Δ PQR is acute angled.
(c) Both [a] and [b]
(d) Neither [a] nor [b]
Answer
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.
Question 7.
Which vertex of ΔABC is right angled if AB¯¯¯¯¯¯¯¯ = 8 cm, AC¯¯¯¯¯¯¯¯ = 6 cm,and BC¯¯¯¯¯¯¯¯ = 10 cm,?
(a) ∠C
(b) ∠A
(c) ∠B
(d) A or C
Answer
Answer: (b) ∠A
Hint:
From the given measurements, BC¯¯¯¯¯¯¯¯ is the hypotenuse. The angle opposite to BC¯¯¯¯¯¯¯¯ is ∠A which is a right angle.
Question 8.
An isosceles triangle is constructed as shown in the figure.
Which of the given statements is incorrect?
(a) PR¯¯¯¯¯¯¯¯ is the hypotenuse of ΔPQR.
(b) ΔPQR is an equilateral triangle.
(c) ΔPQR is a right angled triangle.
(d) If right angled ΔPQR has its equal angles measuring 45° each.
Answer
Answer: (b) ΔPQR is an equilateral triangle.
Question 9.
ΔPQR is constructed with all its angles measuring 60° each. Which of the following is correct?
(a) ΔPQR is an equilateral triangle.
(b) ΔPQR is isosceles triangle.
(c) ΔPQR is a scalene triangle.
(d) ΔPQR is a right angled triangle.
Answer
Answer: (a) ΔPQR is an equilateral triangle.
Question 10.
How many perpendicular lines can be drawn to a line from a point not on it?
(a) 1
(b) 2
(c) 0
(d) Infinite
Answer
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.
Question 11.
Identify the false statement.
(a) A triangle with three equal sides is called an equilateral triangle.
(b) A triangle with a right angle is called a right angled triangle.
(c) A triangle with two equal sides is called a scalene triangle.
(d) A right angled triangle has two acute angles and a right angle.
Answer
Answer: (c) A triangle with two equal sides is called a scalene triangle.
Question 12.
ΔPQR is constructed such that PQ = 5 cm, PR = 5 cm and ∠RPQ = 50° Identify the type of triangle constructed.
(a) An isosceles triangle
(b) An acute angled triangle
(c) An obtuse angled triangle
(d) Both [a] and [b]
Answer
Answer: (d) Both [a] and [b]
Question 13.
Which of the following is NOT constructed using a ruler and a set square?
(a) A perpendicular to a line from a point not on it.
(b) A perpendicular bisector of a line segment.
(c) A perpendicular to a line at a point on the line.
(d) A line parallel to a given line through a given point.
Answer
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.
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