ML Aggarwal Solutions for Class 8 Maths Chapter 14 Constructions of Quadrilaterals mainly help students understand the steps to be followed in the construction of quadrilaterals. The steps of construction are explained in an elaborate manner to make it easy for the students to revise before the exams. To make it easier for the students, ML Aggarwal Solutions for Class 8 Maths Chapter 14 Constructions of Quadrilaterals PDF can be downloaded from the links available below.
Chapter 14 has problems on constructing quadrilaterals based on the given dimensions. Students are advised to solve the textbook problems and refer to the solutions to cross-check their answers. It will help them improve their problem-solving and analytical thinking abilities, which are important from the exam perspective.
ML Aggarwal Solutions for Class 8 Maths Chapter 14: Constructions of Quadrilaterals
Access ML Aggarwal Solutions for Class 8 Maths Chapter 14: Constructions of Quadrilaterals
Exercise 14.1
1. Construct a quadrilateral PQRS where PQ = 4.5 cm, QR = 6 cm, RS = 5.5 cm, PS = 5 cm and PR = 6.5 cm.
Solution:
Steps of Construction:
(i) Construct a line segment PR = 6.5 cm.
(ii) Taking P as the centre and 4.5 cm radius and R as the centre and 6 cm radius, construct arcs which intersect each other at Q.
(iii) Now, join PQ and QR.
(iv) Taking P as the centre and 5 cm radius and R as the centre and 5.5 cm radius, construct arcs which intersect each other at S.
(v) Join PS and SR.
Hence, PQRS is the required quadrilateral.
2. Construct a quadrilateral ABCD in which AB = 3.5 cm, BC = 5 cm, CD = 5.6 cm, DA = 4 cm, BD = 5.4 cm.
Solution:
Steps of Construction:
(i) Construct a line segment AB = 3.5 cm.
(ii) Taking A as the centre and 4 cm radius, construct an arc and with B as the centre and 5.4 cm, construct an arc which meets the previous arc at point D.
Now join AD and BD.
(iii) Taking B as the centre and 5 cm radius, construct an arc
Taking D as the centre and 5.6 cm radius, construct an arc which meets the previous arc at point C.
(iv) Join BC and CD.
Hence, ABCD is the required quadrilateral.
3. Construct a quadrilateral PQRS in which PQ = 3 cm, QR = 2.5 cm, PS = 3.5 cm, PR = 4 cm and QS = 5 cm.
Solution:
Steps of Construction:
(i) Construct PQ = 3 cm.
(ii) Taking P as the centre and 4 cm radius, construct an arc
Taking Q as the centre and 2.5 cm radius, construct an arc which meets the previous arc at R
Now join PQ and QR
(iii) Taking P as the centre and 3.5 cm radius, construct an arc
Taking Q as the centre and 5 cm radius, construct an arc which meets the previous arc at S.
(iv) Join PS, QS and SR.
Therefore, PQRS is the required quadrilateral.
4. Construct a quadrilateral ABCD such that BC = 5 cm, AD = 5.5 cm, CD = 4.5 cm, AC = 7 cm and BC = 5.5 cm.
Solution:
Steps of Construction:
(i) Construct a line segment CD = 4.5 cm.
(ii) Taking C as the centre and 5 cm radius and taking D as the centre and 7 cm radius, construct arcs which intersect each other at B.
(iii) Join BC and BD.
(iv) Taking C as the centre and 5.5 cm radius and taking D as the centre and 5.5 cm radius, construct arcs which intersect each other at A.
(v) Now join AC and AD.
(vi) Join AB.
Therefore, ABCD is the required quadrilateral.
5. Construct a quadrilateral ABCD given that BC = 6 cm, CD = 4 cm, ∠B = 450, ∠C = 900 and ∠D = 1200.
Solution:
Steps of Construction:
(i) Construct BC = 6 cm.
(ii) At point B, draw ∠CBP = 450.
(iii) At point C, draw ∠BCQ = 900.
(iv) Cut off CD = 4 cm from CQ.
(v) At point D, draw ∠CDR = 1200.
(vi) Let BP and DR meet at point A.
Therefore, ABCD is the required quadrilateral.
6. Construct a quadrilateral PQRS where PQ = 4 cm, QR = 6 cm, ∠P = 600, ∠Q = 900 and ∠R = 1200.
Solution:
Steps of Construction:
(i) Construct a line segment QR = 6 cm.
(ii) At point Q, construct a ray QX making an angle of 900 and cut off QP = 4 cm.
(iii) At point P, construct a ray making an angle of 600 and at R, a ray making an angle 1200, which meets each other at point S.
Therefore, PQRS is the required quadrilateral.
7. Construct a quadrilateral ABCD such that AB = 5 cm, BC = 4.2 cm, AD = 3.5 cm, ∠A = 900 and ∠B = 600.
Solution:
Steps of Construction:
(i) Construct AB = 5 cm.
(ii) At point A, construct ∠A = 900.
(iii) At point B, construct ∠B = 600.
(iv) Taking B as the centre and radius 4.2 cm, cut off ∠B at C.
(v) Taking A as the centre and radius 3.5 cm, cut off ∠A at D.
(vi) Now join CD.
Therefore, ABCD is the required quadrilateral.
8. Construct a quadrilateral PQRS where PQ = 4 cm, QR = 5 cm, RS = 4.5 cm, ∠Q = 600 and ∠R = 900.
Solution:
Steps of Construction:
(i) Construct a line segment QR = 5 cm.
(ii) At point Q, construct a ray QX making an angle of 600 and cut off QP = 4 cm.
(iii) At point R, construct a ray RY making an angle of 900 and cut off RS = 4.5 cm.
(iv) Now, join PS.
Therefore, PQRS is the required quadrilateral.
9. Construct a quadrilateral BEST where BE = 3.8 cm, ES = 3.4 cm, ST = 4.5 cm, TB = 5 cm and ∠E = 800.
Solution:
Steps of Construction:
(i) Construct a line segment BE = 3.8 cm.
(ii) At point E, construct a ray EX making an angle of 800 and cut off ES = 3.4 cm.
(iii) Taking B as the centre and 5 cm radius and S as the centre and 4.5 cm radius, construct arcs which intersect each other at T.
(iv) Now join TB and TS.
Therefore, BEST is the required quadrilateral.
10. Construct a quadrilateral ABCD where AB = 4.5 cm, BC = 4 cm, CD = 3.9 cm, AD = 3.2 cm and ∠B = 600.
Solution:
Steps of Construction:
(i) Construct AB = 4.5 cm.
(ii) At point B, construct ∠ABP = 600.
(iii) Cut off BC = 4 cm from BP.
(iv) Taking C as the centre and radius 3.9 cm, construct an arc.
(v) Taking A as the centre and radius 3.2 cm, construct an arc which meets the previous arc at D.
(vi) Now join AD and CD.
Exercise 14.2
1. Construct a parallelogram ABCD such that AB = 5 cm, BC = 3.2 cm and ∠B = 1200.
Solution:
Steps of Construction:
(i) Construct AB = 5 cm.
(ii) At point B, draw an angle 1200.
(iii) Taking B as the centre and radius 3.2 cm, cut off ∠B at C.
(iv) Taking C as the centre and radius AB, construct an arc.
(v) Taking A as the centre and radius 3.2 cm, construct an arc which meets the previous arc at D.
(vi) Now join AD and CD.
Therefore, ABCD is the required parallelogram.
2. Construct a parallelogram ABCD such that AB = 4.8 cm, BC = 4 cm and diagonal BD = 5.4 cm.
Solution:
Steps of Construction:
(i) Draw a triangle ABCD.
(ii) Taking B as the centre and radius 4 cm, construct an arc.
(iii) Taking D as the centre and radius 4.8 cm, construct an arc which meets the previous arc at C.
(iv) Now join CD, BC and AC.
Therefore, ABCD is the required parallelogram.
3. Construct a parallelogram ABCD such that BC = 4.5 cm, BD = 4 cm and AC = 5.6 cm.
Solution:
Steps of Construction:
(i) Draw a triangle BOC with BC = 4.5 cm
Here
BO = ½ × 4 = 2 cm
OC = ½ AC
= ½ × 5.6
= 2.8 cm
We know that the diagonals of a parallelogram bisect each other.
(ii) Produce OC to point A such that OC = OA.
(iii) Produce BO to point D such that OD = OB.
(iv) Now join AD.
Therefore, ABCD is the required parallelogram.
4. Construct a parallelogram ABCD such that AC = 6 cm, BD = 4.6 cm and angle between them is 450.
Solution:
Steps of Construction:
(i) Construct AO = ½ AC = 3 cm and produce AO to C such that OC = OA.
(ii) At point O, draw ∠COP = 450.
(iii) From OP, cut OD = ½ BD
= ½ × 4.6
= 2.3 cm
(iv) Produce OD to OB such that OB = OD.
(v) Now join AB, BC, CD and DA.
Therefore, ABCD is the required parallelogram.
5. Construct a rectangle whose adjacent sides are 5.6 cm and 4 cm.
Solution:
Steps of Construction:
(i) Construct AB = 5.6 cm.
(ii) At point B, draw ∠ABP = 900.
(iii) Cut off BC = 4 cm from BP.
(iv) Taking C as the centre and 5.6 cm radius, construct an arc.
(v) Taking A as the centre and 4 cm radius, construct an arc which meets the previous arc at point D.
(vi) Now join AD and CD.
Therefore, ABCD is the required rectangle.
6. Construct a rectangle such that one side is 5 cm and one diagonal is 6.8 cm.
Solution:
Steps of Construction:
(i) Construct AB = 5 cm.
(ii) At point A, draw ∠BAP = 900.
(iii) Taking B as the centre and 6.8 cm radius, construct an arc which meets AP at D.
(iv) Taking A as the centre and 6.8 cm radius, construct an arc.
(v) Taking D as the centre and 5 cm radius, construct an arc which meets the previous arc at C.
(vi) Now join BC and CD.
Therefore, ABCD is the required rectangle.
7. Construct a rectangle ABCD such that AB = 4 cm and ∠BAC = 600.
Solution:
Steps of Construction:
(i) Construct AB = 4 cm.
(ii) At point B, construct ∠ABP = 900.
(iii) At point A, draw ∠BAQ = 300. Let AQ meet BP at point D.
(iv) Taking D as the centre and 4 cm radius, construct an arc.
(v) Taking A as the centre and BD as the radius, construct an arc which meets the previous arc at point C.
(vi) Now join AC and CD.
Therefore, ABCD is the required rectangle.
8. Construct a rectangle such that one diagonal is 6.6 cm and the angle between two diagonals is 1200.
Solution:
Steps of Construction:
(i) Construct AO = ½ AC = (½ × 6.6) cm and produce AO to C such that OC = OA = 3.3 cm.
(ii) At point O, draw ∠COB = 1200.
(iii) Cut off OB = ½ AC = 3.3 cm from OB.
(iv) Produce BO to D such that OB = OD = 3.3 cm.
(v) Now join AB, BC, CD and DA.
Therefore, ABCD is the required rectangle.
9. Construct a rhombus whose one side is 5 cm, and one angle is 450.
Solution:
Steps of Construction:
(i) Construct AB = 5 cm.
(ii) At point A, draw ∠BAP = 450.
(iii) Cut off AD = 5 cm from AP.
(iv) Taking B as the centre and 5 cm radius, construct an arc.
(v) Taking D as the centre and 5 cm radius, construct an arc which meets the previous arc at point C.
(vi) Now join BC and CD.
Therefore, ABCD is the required rhombus.
10. Construct a rhombus whose one side is 4.5 cm and one diagonal is 5 cm.
Solution:
Steps of Construction:
(i) Construct AB = 4.5 cm.
(ii) Taking A as the centre and 4.5 cm radius, construct an arc.
(iii) Taking B as the centre and 5 cm radius, construct an arc which meets the previous arc at D.
(iv) Taking B as the centre and 4.5 cm radius, construct an arc.
(v) Taking D as the centre and 4.5 cm radius, construct an arc which meets the previous arc at point C.
(vi) Now join AD, BC and CD.
Therefore, ABCD is the required rhombus.
11. Construct a rhombus whose diagonals are 6.8 cm and 5.2 cm.
Solution:
Steps of Construction:
(i) Construct AC = 6.8 cm.
(ii) Construct one bisector PQ of AC to meet it at point O.
(iii) From POQ, cut off OB and OD such that
OB = OD = ½ BD = ½ × 5.2 = 2.6 cm
(iv) Now join AB, BC, CD and DA.
Therefore, ABCD is the required rhombus.
12. Construct a square whose one side is 4.3 cm.
Solution:
Steps of Construction:
(i) Construct BC = 4.3 cm.
(ii) At point B, draw ∠CBP = 900.
(iii) Cut off BA = 4.3 cm from BP.
(iv) Taking C as the centre and 4.3 cm radius, construct an arc.
(v) Taking A as the centre and 4.3 cm radius, construct an arc which meets the previous arc at D.
(vi) Now join AD and CD.
Therefore, ABCD is the required square.
13. Construct a square whose one diagonal is 6.2 cm.
Solution:
Steps of Construction:
(i) Construct AC = 6.2 cm.
(ii) Construct a perpendicular bisector PQ of AC to meet it at point O.
(iii) From POQ, cut off OB = OD = ½ AC = 3.1 cm.
(iv) Now join AB, BC, CD and DA.
Therefore, ABCD is the required square.
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