Study MaterialsImportant QuestionsPractical Geometry Class 7 Extra Questions Maths Chapter 10

Practical Geometry Class 7 Extra Questions Maths Chapter 10

Practical Geometry Class 7 Extra Questions Maths Chapter 10

Extra Questions for Class 7 Maths Chapter 10 Practical Geometry

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    Practical Geometry Class 7 Extra Questions Very Short Answer Type

    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.
    Practical Geometry Class 7 Extra Questions Maths Chapter 10 Q2
    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°.
    Practical Geometry Class 7 Extra Questions Maths Chapter 10 Q3
    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?
    Practical Geometry Class 7 Extra Questions Maths Chapter 10 Q5
    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°

    Question 6.
    What is the measure of an exterior angle of an equilateral triangle?
    Practical Geometry Class 7 Extra Questions Maths Chapter 10 Q6
    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.
    Practical Geometry Class 7 Extra Questions Maths Chapter 10 Q7
    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.
    Practical Geometry Class 7 Extra Questions Maths Chapter 10 Q9
    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°.
    Practical Geometry Class 7 Extra Questions Maths Chapter 10 Q10
    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.
    Practical Geometry Class 7 Extra Questions Maths Chapter 10 Q11
    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.
    Practical Geometry Class 7 Extra Questions Maths Chapter 10 Q12
    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°
    Practical Geometry Class 7 Extra Questions Maths Chapter 10 Q13
    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.
    Practical Geometry Class 7 Extra Questions Maths Chapter 10 Q14
    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.
    Practical Geometry Class 7 Extra Questions Maths Chapter 10 Q15
    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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