The DAV Maths Class 8 Solutions and DAV Class 8 Maths Chapter 12 Brain Teasers Solutions of Construction of Quadrilaterals offer comprehensive answers to textbook questions.
DAV Class 8 Maths Ch 12 Brain Teasers Solutions
Question 1.
Tick (✓) the correct option.
(i) The number of parts required to construct a quadrilateral is
(a) 5
(b) 4
(c) 3
(d) 2
Solution:
To construct a quadrilateral, we shall need a measure of five parts.
Thus, option (a) is the correct answer.
(ii) To construct a parallelogram, the minimum number of dimensions needed is
(a) 2
(b) 3
(c) 4
(d) 1
Solution:
To construct a parallelogram, we shall need a minimum three number of dimensions.
Thus, option (b) is the correct answer.
(iii) The minimum number of dimensions needed to construct a square is
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
We shall need only the side of the square to construct it.
Thus option (a) is the correct answer.
(iv) To construct a rectangle, the minimum number of dimensions needed is
(a) 1
(b) 4
(c) 3
(d) 2
Solution:
To construct a rectangle, we shall need only two dimensions.
Thus, option (d) is the correct answer.
(v) Which of the fifth dimensions is needed to construct the given quadrilateral?
(a) BD
(b) ∠C
(c) DC
(d) cannot be constructed
Solution:
To construct the given quadrilateral, we need DC as the fifth dimension.
Thus, option (c) is the correct answer.
Question 2.
Construct the following quadrilaterals if possible. Justify your answer if construction is not possible.
(a) Construct a quadrilateral ABCD in which \(\overline{\mathrm{AB}}\) = \(\overline{\mathrm{CD}}\) = \(\overline{\mathrm{BC}}\) = \(\overline{\mathrm{AD}}\) = 5 cm and diagonal \(\overline{\mathrm{AC}}\) = 6.5 cm.
Solution:
Steps of Construction:
1. Draw \(\overline{\mathrm{AB}}\) = 5 cm.
2. Draw arcs from A and B with radii 6.5 cm and 5 cm respectively to meet at C.
3. Join \(\overline{\mathrm{AC}}\) and \(\overline{\mathrm{BC}}\).
4. Draw arcs from A and C with a radius of 5 cm each to meet at D.
5. Join \(\overline{\mathrm{CD}}\) and \(\overline{\mathrm{DA}}\).
6. ABCD is the required quadrilateral.
(b) Construct a quadrilateral PQRS in which \(\overline{\mathrm{PQ}}\) = \(\overline{\mathrm{QR}}\) = \(\overline{\mathrm{RS}}\) = 5 cm, diagonals \(\overline{\mathrm{PR}}\) = \(\overline{\mathrm{QS}}\) = 7.5 cm.
Solution:
Steps of Construction:
1. Draw \(\overline{\mathrm{PQ}}\) = 5 cm.
2. Draw arcs from P and Q with radii 7.5 cm and 5 cm respectively to meet at R.
3. Join \(\overline{\mathrm{PR}}\) and \(\overline{\mathrm{QR}}\).
4. Draw arcs from Q and R with radii 7.5 cm and 5 cm respectively to meet at S.
5. Join \(\overline{\mathrm{RS}}\), \(\overline{\mathrm{QS}}\) and \(\overline{\mathrm{QP}}\).
6. PQRS is the required quadrilateral.
(c) Construct a quadrilateral ABCD in which \(\overline{\mathrm{AB}}\) = \(\overline{\mathrm{CD}}\) = 5 cm, \(\overline{\mathrm{BC}}\) = 4 cm, diagonals \(\overline{\mathrm{AC}}\) = 9 cm, BD = 8 cm.
Solution:
Steps of Construction:
\(\overline{\mathrm{AB}}\) = 5 cm, \(\overline{\mathrm{BC}}\) = 4 cm and \(\overline{\mathrm{AC}}\) = 9 cm.
∴ \(\overline{\mathrm{AB}}+\overline{\mathrm{BC}}=\overline{\mathrm{AC}}\)
5 + 4 = 9
9 cm = 9 cm
Point C will lie on AB produced.
Hence quadrilateral ABCD is not possible to form.
(d) Construct a quadrilateral ABCD in which \(\overline{\mathrm{AB}}\) = \(\overline{\mathrm{BC}}\) = \(\overline{\mathrm{CD}}\) = \(\overline{\mathrm{DA}}\) = 5 cm and ∠A = 120°. What type of quadrilateral is it?
Solution:
Steps of Construction:
1. Draw \(\overline{\mathrm{AB}}\) = 5 cm.
2. Draw ∠A = 120° and cut \(\overline{\mathrm{AD}}\) = 5 cm.
3. Draw two arcs at B and D with a radius of 5 cm each to meet at C.
4. Join \(\overline{\mathrm{BC}}\) and \(\overline{\mathrm{DC}}\).
5. ABCD is the required quadrilateral.
6. The constructed quadrilateral is a rhombus.
(e) Construct a quadrilateral PQRS in which \(\overline{\mathrm{PQ}}\) = \(\overline{\mathrm{RQ}}\) = 4 cm, and ∠P = ∠Q = ∠R = ∠S = 90°. What type of quadrilateral is it?
Solution:
Steps of Construction:
1. Draw \(\overline{\mathrm{PQ}}\) = 4 cm.
2. Draw ∠Q = 90° and cut \(\overline{\mathrm{QR}}\) = 4 cm.
3. Draw arcs from P and R with a radius of 4 cm each to meet at S.
4. Join \(\overline{\mathrm{RS}}\) and \(\overline{\mathrm{PS}}\).
5. PQRS is the required quadrilateral.
6. PQRS is a square.
(f) Construct a quadrilateral ABCD in which \(\overline{\mathrm{AB}}\) = \(\overline{\mathrm{BC}}\) = \(\overline{\mathrm{CD}}\) = \(\overline{\mathrm{AD}}\) = 5 cm, diagonal \(\overline{\mathrm{AC}}\) = 7 cm and diagonals bisect each other at right angles. What type of quadrilateral is it?
Solution:
Steps of Construction:
1. Draw \(\overline{\mathrm{AB}}\) = 5 cm.
2. Draw arcs from A and B with radii 7 cm and 5 cm respectively to meet at C.
3. Draw the right bisector of \(\overline{\mathrm{AC}}\) at O.
4. Make \(\overline{\mathrm{CD}}\) = 5 cm.
5. Join D to A.
6. ABCD is the required quadrilateral.
7. The construct quadrilateral ABCD is a rhombus.
(g) Construct a quadrilateral ABCD in which \(\overline{\mathrm{AB}}\) = \(\overline{\mathrm{CD}}\) = 3 cm, \(\overline{\mathrm{BC}}\) = \(\overline{\mathrm{AD}}\) = 4 cm and diagonals are 5 cm each. What type of quadrilateral is it?
Solution:
Steps of Construction:
1. Draw \(\overline{\mathrm{AB}}\) = 3 cm.
2. Draw arcs from A and B with radii 5 cm and 4 cm respectively to meet at C.
3. Join \(\overline{\mathrm{AC}}\) and \(\overline{\mathrm{BC}}\).
4. Draw arcs at A and C with radius 4 cm and 3 cm respectively to meet at D.
5. Join \(\overline{\mathrm{AD}}\) and \(\overline{\mathrm{CD}}\).
6. ABCD is the required quadrilateral.
7. The construct quadrilateral ABCD is a rectangle.
(h) Construct a quadrilateral ABCD in which all sides are 5 cm each and ∠A = ∠B = 90°.
Solution:
Steps of Construction:
1. Draw \(\overline{\mathrm{AB}}\) = 5 cm.
2. Draw ∠A = ∠B = 90°.
3. Make \(\overline{\mathrm{BC}}\) = \(\overline{\mathrm{AD}}\) = 5 cm.
4. Join \(\overline{\mathrm{CD}}\).
5. ABCD is the required quadrilateral.
DAV Class 8 Maths Chapter 12 HOTS
Question 1.
Construct a trapezium ABCD with AB || CD where AB = 6 cm, CD = 4 cm, ∠A = 70°, ∠ABD = 30°.
Solution:
Steps of Construction:
1. Draw \(\overline{\mathrm{AB}}\) = 6 cm.
2. At A, construct ∠BAX = 70°.
3. At B, construct ∠ABY = 30°. Arm \(\overrightarrow{\mathrm{BY}}\) of ∠ABY interesect \(\overrightarrow{\mathrm{AX}}\). Name the point of intersection as D.
4. Since interior angles of the same side of transversal (AD) of parallel lines (AB || CD) are supplementary. So, ∠ADZ = 180° – 70°= 110°. Draw ∠ADZ = 110° at D.
5. Cut off \(\overline{\mathrm{DC}}\) = 4 cm on ray DZ.
6. Join B to C.
7. ABCD is the required trapezium.
Question 2.
Construct a rectangle whose length is twice its width and whose perimeter is equal to the perimeter of a rhombus of side 6 cm.
Solution:
To construct the rectangle, we need its length and width.
Let the width of the rectangle be x cm.
Then, the length of the rectangle = 2x
According to the question,
Perimeter of the rectangle = Perimeter of rhombus
⇒ 2(length + width) = 4 × side
⇒ 2(2x + x) = 4 × 6
⇒ 2 × 3x = 24
⇒ 6x = 24
⇒ x = 4
∴ Length of the rectangle = 8 cm and width = 4 cm.
Steps of Construction:
1. Draw the line segment \(\overline{\mathrm{AB}}\) = 8 cm
2. At A and B, construct ∠BAX = 90° and ∠ABY = 90° respectively.
3. Cut off \(\overline{\mathrm{AD}}\) = 4 cm on ray AX and \(\overline{\mathrm{BC}}\) = 4 cm on ray BY.
4. Join C to D.
5. ABCD is the required rectangle.
DAV Class 8 Maths Chapter 12 Enrichment Questions
Question 1.
Draw an equilateral triangle ABD of side 4.5 cm. Taking one of its sides as a diagonal, complete the rhombus ABCD.
Solution:
Steps of Construction:
1. Draw line segment \(\overline{\mathrm{AB}}\) = 4.5 cm.
2. Taking A and B as centres and a radius of 4.5 cm each time, draw two arcs that intersect each other at D.
3. Join A to D and B to D. ΔABD is an equilateral triangle.
Note: Two sides of rhombus ABCD i.e., AB and AD, and one diagonal BD are given here. We know that diagonals of rhombus bisect each other at right angles.
4. Draw a perpendicular bisector (AOX) of diagonal BD which meets AB at A.
5. Cut off \(\overline{\mathrm{OC}}\) = \(\overline{\mathrm{OA}}\).
6. Join D to C and B to C.
7. ABCD is the required rhombus.
Question 2.
Construct a rhombus of perimeter 20 cm and one base angle the supplement of 130°.
Solution:
The perimeter of the rhombus = 20 cm (given)
⇒ 4 × side = 20 cm
⇒ side = 5 cm
Let one base angle of the rhombus be x.
Supplement of x = 130° (given)
⇒ 180° -x = 130 (∵ x + 130° = 180°)
⇒ x = 50°
We have one angle of rhombus = 50° and one side = 5 cm.
Steps of Construction:
1. Draw a line segment AB =5 cm
2. Draw ∠BAX = 50° at A.
3. Cut off AD = 5 cm.
4. Taking radius = 5 cm, draw arcs from D and B which cut each other at C.
5. Join D to C and B to C.
6. ABCD is the required rhombus.
Additional Questions:
Question 1.
Construct a quadrilateral PQRS in which \(\overline{\mathrm{PQ}}\) = 4 cm, \(\overline{\mathrm{QR}}\) = 5.5 cm, \(\overline{\mathrm{RS}}\) = 5 cm, ∠Q = 60° and ∠R = 75°.
Solution:
Steps of Construction:
1. Draw \(\overline{\mathrm{PQ}}\) = 4 cm.
2. Draw ∠Q = 60°.
3. Make \(\overline{\mathrm{QR}}\) = 5.5 cm.
4. Draw ∠R = 75°.
5. Cut \(\overline{\mathrm{RS}}\) = 5 cm.
6. Join S to P.
7. PQRS is the required quadrilateral.
Question 2.
Construct the following quadrilateral if possible:
\(\overline{\mathrm{AB}}\) = 4 cm, \(\overline{\mathrm{BC}}\) = 4.5 cm, \(\overline{\mathrm{CD}}\) = 4 cm and the diagonal \(\overline{\mathrm{AC}}\) = 9 cm.
Solution:
Steps of Construction:
1. Draw \(\overline{\mathrm{AB}}\) = 4 cm.
2. Draw arcs from A and B with radii of 9 cm and 4.5 cm respectively.
3. The two arcs do not cut each other anywhere
Reason: For a triangle ABC
\(\overline{\mathrm{AB}}+\overline{\mathrm{BC}}>\overline{\mathrm{AC}}\)
But 4 cm + 4.5 cm < 9 cm
⇒ 8.5 cm < 9 cm
So, the construction of quadrilateral ABCD is not possible.
Question 3.
Construct a quadrilateral PQRS where \(\overline{\mathrm{PQ}}\) = 4 cm, \(\overline{\mathrm{QR}}\) = 6 cm, \(\overline{\mathrm{RS}}\) = 5 cm, \(\overline{\mathrm{PS}}\) = 5.5 cm and \(\overline{\mathrm{PR}}\) = 7 cm.
Solution:
Steps of Construction:
1. Draw \(\overline{\mathrm{PQ}}\) = 4 cm.
2. Draw two arcs from P and Q with radii 7 cm and 6 cm respectively to meet at R.
3. Join \(\overline{\mathrm{PR}}\) and \(\overline{\mathrm{QR}}\).
4. Draw two arcs from P and R with radii 5.5 cm and 5 cm respectively to meet at S.
5. Join \(\overline{\mathrm{PS}}\) and \(\overline{\mathrm{RS}}\).
7. PQRS is the required quadrilateral.
Question 4.
Construct a MIST where MI = 3.5 cm, IS = 6.5 cm, ∠M = 75°, ∠I = 105° and ∠S = 120°.
Solution:
Steps of Construction:
1. Draw \(\overline{\mathrm{MI}}\) = 3.5 cm.
2. Draw ∠M = 75° and ∠I = 105°.
3. Cut \(\overline{\mathrm{IS}}\) = 6.5 cm.
4. Draw ∠S = 120°.
5. The ray of ∠S and the ray of ∠M intersect each other at T.
6. MIST is the required quadrilateral.
Question 5.
Construct a quadrilateral ABCD where \(\overline{\mathrm{AB}}\) = 4 cm, \(\overline{\mathrm{BC}}\) = 5 cm, \(\overline{\mathrm{CD}}\) = 6.5 cm and ∠B = 105° and ∠C = 80°.
Solution:
Steps of Construction:
1. Draw \(\overline{\mathrm{AB}}\) = 4 cm.
2. Draw ∠B = 105° and make \(\overline{\mathrm{BC}}\) = 5 cm.
3. Draw ∠C = 80° and make \(\overline{\mathrm{CD}}\) = 6.5 cm.
4. Join D to A.
5. ABCD is the required quadrilateral.
Question 6.
Construct a quadrilateral TRUE in which TR = 3.5 cm, RU = 3 cm, UE = 4 cm, ∠R = 75° and ∠U = 120°.
Solution:
Steps of Construction:
1. Draw TR 3.5 cm.
2. Draw ∠R = 75° and cut \(\overline{\mathrm{RU}}\) = 4 cm.
3. Draw ∠U = 120° and cut \(\overline{\mathrm{UE}}\) = 4 cm.
4. Join E to T.
5. TRUE is the required quadrilateral.
