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, with the help of 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 helps students 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 Download PDF

## 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 centre and 4.5 cm radius and R as centre and 6 cm radius construct arcs which intersect each other at Q.

(iii) Now join PQ and QR.

(iv) Taking P as centre and 5 cm radius and R as 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 centre and 4 cm radius construct and arc and with B as centre and 5.4 cm construct an arc which meets the previous arc at the point D.

Now join AD and BD.

(iii) Taking B as centre and 5 cm radius, construct an arc

Taking D as centre and 5.6 cm radius, construct an arc which meets the previous arc at the 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 centre and 4 cm radius, construct an arc

Taking Q as 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 centre and 3.5 cm radius, construct an arc

Taking Q as 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 centre and 5.5 cm radius and taking D as centre and 7 cm radius construct arcs which intersect each other at B.

(iii) Join BC and BD.

(iv) Taking C as centre and 5.5 cm radius and taking D as 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, âˆ B = 45 ^{0}, âˆ C = 90^{0} and âˆ D = 120^{0}.**

**Solution:**

Steps of Construction:

(i) Construct BC = 6 cm.

(ii) At the point B, draw âˆ CBP = 45^{0}.

(iii) At the point C, draw âˆ BCQ = 90^{0}.

(iv) Cut off CD = 4 cm from CQ.

(v) At the point D, draw âˆ CDR = 120^{0}.

(iv) Let BP and DR meet at the point A.

Therefore, ABCD is the required quadrilateral.

**6. Construct a quadrilateral PQRS where PQ = 4 cm, QR = 6 cm, âˆ P = 60 ^{0}, âˆ Q = 90^{0} and âˆ R = 120^{0}.**

**Solution:**

Steps of Construction:

(i) Construct a line segment QR = 6 cm.

(ii) At the point Q, construct a ray QX making an angle of 90^{0} and cut off QP = 4 cm.

(iii) At the point P, construct a ray making an angle of 60^{0} and at R, a ray making an angle 120^{0} which meets each other at the 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, âˆ A = 90 ^{0} and âˆ B = 60^{0}.**

**Solution:**

Steps of Construction:

(i) Construct AB = 5 cm.

(ii) At the point A, construct âˆ A = 90^{0}.

(iii) At the point B, construct âˆ B = 60^{0}.

(iv) Taking B as centre and radius 4.2 cm cut off âˆ B at C.

(v) Taking A as centre and radius 3.5 cm cut off âˆ 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, âˆ Q = 60 ^{0} and âˆ R = 90^{0}.**

**Solution:**

Steps of Construction:

(i) Construct a line segment QR = 5 cm.

(ii) At the point Q, construct a ray QX making an angle of 60^{0} and cut off QP = 4 cm.

(iii) At the point R, construct a ray RY making an angle of 90^{0} 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 âˆ E = 80 ^{0}.**

**Solution:**

Steps of Construction:

(i) Construct a line segment BE = 3.8 cm.

(ii) At the point E, construct a ray EX making an angle of 80^{0} and cut off ES = 3.4 cm.

(iii) Taking B as centre and 5 cm radius and S as 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 âˆ B = 60 ^{0}.**

**Solution:**

Steps of Construction:

(i) Construct AB = 4.5 cm.

(ii) At point B, construct âˆ ABP = 60^{0}.

(iii) Cut off âˆ BC = 4 cm from BP.

(iv) Taking C as centre and radius 3.9 cm construct an arc.

(v) Taking A as 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 âˆ B = 120 ^{0}.**

**Solution:**

Steps of Construction:

(i) Construct AB = 5 cm.

(ii) At point B, draw angle 120^{0}.

(iii) Taking B as centre and radius 3.2 cm cut off âˆ B at C.

(iv) Taking C as centre and radius AB construct an arc.

(v) Taking A as 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 centre and radius 4 cm, construct an arc.

(iii) Taking D as 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 = Â½ Ã— 4 = 2 cm

OC = Â½ AC

= Â½ Ã— 5.6

= 2.8 cm

We know that the diagonals of 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 45 ^{0}.**

**Solution:**

Steps of Construction:

(i) Construct AO = Â½ AC = 3 cm and produce AO to C such that OC = OA.

(ii) At the point O, draw âˆ COP = 45^{0}.

(iii) From OP

Cut OD = Â½ BD

= Â½ Ã— 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 the point B, draw âˆ ABP = 90^{0}.

(iii) Cut off BC = 4 cm from BP.

(iv) Taking C as centre and 5.6 cm radius, construct an arc.

(v) Taking A as 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 the point A, draw âˆ BAP = 90^{0}.

(iii) Taking B as centre and 6.8 cm radius, construct an arc which meets AP at D.

(iv) Taking A as centre and 6.8 cm radius, construct an arc.

(v) Taking D as 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 âˆ BAC = 60 ^{0}.**

**Solution:**

Steps of Construction:

(i) Construct AB = 4 cm.

(ii) At the point B, construct âˆ ABP = 90^{0}.

(iii) At the point A, draw âˆ BAQ = 30^{0}. Let AQ meet BP at the point D.

(iv) Taking D as centre and 4 cm radius construct an arc.

(v) Taking A as centre and BD as radius, construct an arc which meets the previous arc at the 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 angle between two diagonals is 120 ^{0}.**

**Solution:**

Steps of Construction:

(i) Construct AO = Â½ AC = (Â½ Ã— 6.6) cm and produce AO to C such that OC = OA = 3.3 cm.

(ii) At the point O, draw âˆ COB = 120^{0}.

(iii) Cut off OB = Â½ 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 45 ^{0}.**

**Solution:**

Steps of Construction:

(i) Construct AB = 5 cm.

(ii) At the point A, draw âˆ BAP = 45^{0}.

(iii) Cut off AD = 5 cm from AP.

(iv) Taking B as centre and 5 cm radius, construct an arc.

(v) Taking D as centre and 5 cm radius, construct an arc which meets the previous arc at the 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 centre and 4.5 cm radius, construct an arc.

(iii) Taking B as centre and 5 cm radius, construct an arc which meets the previous arc at D.

(iv) Taking B as centre and 4.5 cm radius, construct an arc.

(v) Taking D as 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 the point O.

(iii) From POQ, cut off OB and OD such that

OB = OD = Â½ BD = Â½ Ã— 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 the point B, draw âˆ CBP = 90^{0}.

(iii) Cut off BA = 4.3 cm from BP.

(iv) Taking C as centre and 4.3 cm radius, construct an arc.

(v) Taking A as 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 = Â½ AC = 3.1 cm.

(iv) Now join AB, BC, CD and DA.

Therefore, ABCD is the required square.