MSBSHSE Solutions For Class 8 Maths Part 1 Chapter 8- Quadrilateral: Constructions and Types

MSBSHSE Solutions For Class 8 Maths Part 1 Chapter 8 – Quadrilateral: Constructions and Types is available here. This Chapter mainly deals with the construction of quadrilaterals, rectangle, square, rhombus, parallelogram and trapezium. In order to enhance the performance of students in the Class 8 exam, expert tutors at BYJU’S have formulated the MSBSHSE Solutions. Students can easily download the MSBSHSE Solutions for Class 8 and can start practising offline, in order to gain that extra edge of knowledge. Solutions are provided in detail and a step-by-step guide to all the queries of the students. The exercises present in the chapter should be dealt with utmost sincerity, if one aims to score well in the examinations. Students are also advised to practice on a daily basis to excel in the board exams. To help gain interest among students, the links to this chapter are provided for free in the pdf format, which is readily available to download, from the below-mentioned links.

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Practice set 8.1 PAGE NO: 43

1. Construct the following quadrilaterals of given measures.

(1) In MORE, l(MO) = 5.8 cm, l(OR) = 4.4 cm, m∠M = 58°, m∠O = 105°, m∠R = 90°.

(2) Construct DEFG such that l(DE) = 4.5 cm, l(EF) = 6.5 cm, l(DG) = 5.5 cm, l(DF) = 7.2 cm, l(EG) = 7.8 cm.

(3) In ABCD, l(AB) = 6.4 cm, l(BC) = 4.8 cm, m∠A = 70°, m∠B = 50°, m∠C = 140°.

(4) Construct LMNO such that l(LM) = l(LO) = 6 cm, l(ON) = l(NM) = 4.5 cm, l(OM) = 7.5 cm.

Solution:

(1) In
MORE, l(MO) = 5.8 cm, l(OR) = 4.4 cm, m∠M = 58°, m∠O = 105°, m∠R = 90°.

Steps to construct a quadrilateral:

Step 1: Draw a line MO = 5.8cm.

Step 2: At point O, construct an angle of 105^o. Such that it forms a line OX.

Step 3: At point M, construct an angle of 58^o. Such that it forms a line MZ.

Step 4: With O as the centre and of radius 4.4cm, cut an arc on the line OX and mark that point as R.

Step 5: Now, at point R construct an angle of 90^o. Such that it forms a line RY and the point E is formed intersecting on the line MZ.

Here, is the required MORE quadrilateral.

(2) Construct
DEFG such that l(DE) = 4.5 cm, l(EF) = 6.5 cm, l(DG) = 5.5 cm, l(DF) = 7.2 cm, l(EG) = 7.8 cm.

Steps to construct a quadrilateral:

Step 1: Draw a line DE = 4.5cm.

Step 2: With D as the centre and radius 7.2cm, draw an arc.

Step 3: With E as the centre and radius 6.5cm, draw an arc cutting the previous arc at F.

Step 4: Join EF and DF.

Step 5: With D as the centre and radius 5.5cm, draw an arc.

Step 6: With E as the centre and radius 7.8cm, draw an arc cutting the previous ac at G.

Step 7: Join DG, EG and GF.

Here, is the required DEFG quadrilateral.

(3) In
ABCD, l(AB) = 6.4 cm, l(BC) = 4.8 cm, m∠A = 70°, m∠B = 50°, m∠C = 140°.

Steps to construct a quadrilateral:

Step 1: Draw a line AB = 6.4cm.

Step 2: construct an angle at point B of 50^o

Step 3: With B as the centre and radius of 4.8cm, draw an arc cutting the line BX at point C.

Step 4: construct an angle of 140^o at point C and name that line as Y.

Step 5: construct an angle of 70^o at point A, such that line AZ and line CY intersect at point D.

Here, is the required ABCD quadrilateral.

(4) Construct
LMNO such that l(LM) = l(LO) = 6 cm, l(ON) = l(NM) = 4.5 cm, l(OM) = 7.5 cm.

Steps to construct a quadrilateral:

Step 1: Draw a line LM = 6cm.

Step 2: With L as the centre and radius 6cm, draw an arc.

Step 3: With M as the centre and radius 7.5cm, draw an arc cutting the previous arc at O.

Step 4: Join OL and OL.

Step 5: With O as the centre and radius 4.5cm, draw an arc.

Step 6: With M as the centre and radius 4.5cm, draw an arc cutting the previous ac at N.

Step 7: Join ON, MN and OL.

Here, is the required LMNO quadrilateral.

Practice set 8.2 PAGE NO: 46

1. Draw a rectangle ABCD such that l(AB) = 6.0 cm and l(BC) = 4.5 cm.

Solution:

Steps to construct a rectangle:

Step 1: Draw a line AB = 6cm.

Step 2: Construct an angle of 90^o at point B.

Step 3: With B as centre and radius 4.5cm, draw an arc cutting the line BX at point C.

Step 4: With C as centre and radius 6cm, draw an arc.

Step 5: With A as the centre and radius 4.5cm, draw an arc cutting the previous arc at point D.

Step 6: Join AD and CD.

Here, is the required ABCD rectangle.

2. Draw a square WXYZ with side 5.2 cm.

Solution:

Steps to construct a square:

Step 1: Draw a line WX = 5.2cm.

Step 2: Construct an angle of 90^o at point X.

Step 3: With X as centre and radius 5.2cm, draw an arc cutting the line XP at point Y.

Step 4: With Y as centre and radius 5.2cm, draw an arc.

Step 5: With W as the centre and radius 5.2cm, draw an arc cutting the previous arc at point Z.

Step 6: Join YZ and WZ.

Here, is the required WXYZ square.

3. Draw a rhombus KLMN such that its side is 4 cm and m∠K = 75°.

Solution:

Steps to construct a rhombus:

Step 1: Draw a line KL = 4cm.

Step 2: Construct an angle of 75^o at point K.

Step 3: With K as centre and radius 4cm, draw an arc cutting the line KX at point N.

Step 4: With N as centre and radius 4cm, draw an arc.

Step 5: With L as the centre and radius 4cm, draw an arc cutting the previous arc at point M.

Step 6: Join LM and NM.

Here, is the required KLMN rhombus.

4. If diagonal of a rectangle is 26 cm and one side is 24 cm, find the other side.

Solution:

Let ABCD be the rectangle.

l(BC) = 24cm, l(AC) = 26cm

In ∆ABC,

m∠ABC = 90° [Angle of a rectangle]

By using Pythagoras theorem,

(26 )² = [l(AB)]² + (24)²

(26)² – (24)² = [l(AB)]²

(26 + 24) (26 – 24) = [l(AB)]² [Since, a² – b² = (a + b) (a – b)]

50 x 2 = [l(AB)]²

100 = [l(AB)]²

l(AB) = √100

By taking square root of both sides, we get

l(AB) =10 cm

∴The length of the other side is 10 cm.

5. Lengths of diagonals of a rhombus ABCD are 16 cm and 12 cm. Find the side and perimeter of the rhombus.

Solution:

In rhombus ABCD,

It is given that, l(AC) = 16 cm and l(BD) = 12 cm.

Let the diagonals of rhombus ABCD intersect at point O.

l(AO) = 12 l(AC) [Diagonals of a rhombus bisect each other]

l(AO) = 12 × 16

= 8 cm

Also, l(DO) = 12 l(BD) [Diagonals of a rhombus bisect each other]

l(DO) = 12 × 12

l(DO) = 6 cm

In ∆DOA,

m∠DOA = 90° [Diagonals of a rhombus are perpendicular to each other]

By using Pythagoras theorem,

= (8)² + (6)²

= 64 + 36

l(AD) = √100

By taking square root on both sides, we get

l(AD) = 10 cm

l(AB) = l(BC) = l(CD) = l(AD) = 10 cm [Since, sides of a rhombus are congruent]

Perimeter of rhombus ABCD

= l(AB) + l(BC) + l(CD) + l(AD)

= 10+10+10+10

= 40 cm

∴The side and perimeter of the rhombus are 10 cm and 40 cm respectively.

6. Find the length of diagonal of a square with side 8 cm.

Solution:

Let XYWZ be the square of side 8cm.

seg XW is a diagonal.

In ∆ XYW,

m∠XYW = 90° [Angle of a square]

By using Pythagoras theorem,

= (8)² + (8)²

= 64 + 64

By taking square root on both sides, we get

l(XW) = √128

= √64 × 2

= 8 √2 cm

∴ The length of the diagonal of the square is 8 √2 cm.

7. Measure of one angle of a rhombus is 50°, find the measures of remaining three angles.

Solution:

Let ABCD be the rhombus.

m∠A = 50°

m∠C = m∠A [Since, opposite angles of a rhombus are congruent]

∴ m∠C = 50°

Also, m∠D = m∠B …(i) [Opposite angles of a rhombus are congruent]

In rhombus ABCD, we know that sum of the measures of the angles of a quadrilateral is 360°.

m∠A + m∠B + m∠C + m∠D = 360°

50° + m∠B + 50° + m∠D = 360°

m∠B + m∠D + 100° = 360°

m∠B + m∠D = 360° – 100°

m∠B + m∠B = 260° [From (i)]

2m∠B = 260°

m∠B = 260/2

m∠B = 130°

m∠D = m∠B = 130° [From (i)]

∴ The measures of the remaining angles of the rhombus are 130°, 50° and 130°.

Practice set 8.3 PAGE NO: 49

1. Measures of opposite angles of a parallelogram are (3x – 2)° and (50 – x)°. Find the measure of its each angle.

Solution:

Let PQRS be the parallelogram.

m∠Q = (3x – 2)° and m∠S = (50 – x)°

m∠Q = m∠S …..(i) [Since, opposite angles of a parallelogram are congruent]

3x – 2 = 50 – x

3x + x = 50 + 2

4x = 52

x = 52/4

x = 13

Now, m∠Q = (3x – 2)°

(3 × 13 – 2)° = (39 – 2)° = 37°

m∠S = m∠Q = 37° [From(i)]

m∠P + m∠Q = 180° [Since, adjacent angles of a parallelogram are supplementary]

m∠P + 37° = 180°

m∠P = 180° – 37° = 143°

m∠R = m∠P = 143° [Since, opposite angles of a parallelogram are congruent]

∴ The measures of the angles of the parallelogram are 37°, 143°, 37° and 143°.

2. Referring the given figure of a parallelogram, write the answers of questions given below.
(1) If l(WZ) = 4.5 cm, then l(XY) = ?
(2) If l(YZ) = 8.2 cm, then l(XW) = ?
(3) If l(OX) = 2.5 cm, then l(OZ) = ?
(4) If l(WO) = 3.3 cm, then l(WY) = ?
(5) If m∠WZY = 120°, then m∠WXY = ? and m∠XWZ = ?

Solution:

(1) It is given that, l(WZ) = 4.5 cm

l(X Y) = l(WZ) [Since, opposite sides of a parallelogram are congruent ]

∴ l(X Y) = 4.5cm

(2) It is given that, l(YZ) = 8.2 cm

l(XW) = l(YZ) [Since, opposite sides of a parallelogram are congruent]

∴ l(XW) = 8.2cm

(3) It is given that, l(OX) = 2.5 cm

l(OZ) = l(OX) [Since, diagonals of a parallelogram bisect each other]

∴ l(OZ) = 2.5cm

(4) It is given that, l(WO) = 3.3 cm

l(WO) = 1/2 l(WY) [Since, diagonals of a parallelogram bisect each other]

3.3 = 1/2 l(WY)

3.3 × 2 = l(WY)

∴ l(WY) = 6.6cm

(5) It is given that, m∠WZY =120°

m∠WXY = m∠WZY [Since, opposite angles of a parallelogram are congruent]

So, m∠WXY = 120^o

Now,

m∠XWZ + m∠WXY = 180° [Since, adjacent angles of a parallelogram are supplementary]

m∠XWZ + 120° = 180°

m∠XWZ = 180°- 120°

∴ m∠XWZ = 60°

3. Construct a parallelogram ABCD such that l(BC) = 7 cm, m∠ABC = 40°, l(AB) = 3 cm.

Solution:

Steps to construct:

Step 1: Draw a line AB = 3cm.

Step 2: Construct an angle of 40^o at point B.

Step 3: With B as centre and radius 7cm, draw an arc cutting the line BX at point C.

Step 4: With C as centre and radius 3cm, draw an arc.

Step 5: With A as the centre and radius 7cm, draw an arc cutting the previous arc at point D.

Step 6: Join AD and CD.

Opposite sides of a parallelogram are congruent.

∴ l(AB) = l(CD) = 3cm

l(BC) = l(AD) = 7 cm

4. Ratio of consecutive angles of a quadrilateral is 1: 2: 3: 4. Find the measure of its each angle. Write with reason, what type of a quadrilateral it is.

Solution:

Let PQRS be the quadrilateral.

Ratio of consecutive angles of a quadrilateral is 1: 2: 3: 4.

Let us consider ‘x’ be the common multiple.

∴m∠P = x°, m∠Q = 2x°, m∠R = 3x° and m∠S = 4x°

In quadrilateral PQRS,

m∠P + m∠Q + m∠R + m∠S = 360° [Sum of the measures of the angles of a quadrilateral is 360°]

x° + 2x° + 3x° + 4x° = 360°

10 x° = 360°

x° = 360/10

x° = 36°

∴m∠P = x° = 36°

m∠Q = 2x° = 2 × 36° = 72°

m∠R = 3x° = 3 × 36° = 108° and

m∠S = 4x° = 4 × 36° = 144°

∴The measures of the angles of the quadrilateral are 36°, 72°, 108°, 144°.

Here, m∠P + m∠S = 36° + 144° = 180°

side PQ || side SR [Since, interior angles are supplementary]

m∠P + m∠Q = 36° + 72°

= 108°

≠ 180°

So, side PS is not parallel to side QR.

Since, one pair of opposite sides of the given quadrilateral is parallel.

∴The given quadrilateral is a trapezium.

5. Construct BARC such that l(BA) = l(BC) = 4.2 cm, l(AC) = 6.0 cm, l(AR) = l(CR) = 5.6 cm.

Solution:

Steps to construct:

Step 1: Draw a line BA = 4.2cm.

Step 2: With B as the centre and radius 4.2cm, draw an arc.

Step 3: With A as the centre and radius 6cm, draw an arc cutting the previous arc at C.

Step 4: Join BC and AC.

Step 5: With A as the centre and radius 5.6cm, draw an arc.

Step 6: With C as the centre and radius 5.6cm, draw an arc cutting the previous ac at D.

Step 7: Join AD, CD and BC.

6. Construct PQRS, such that l(PQ) = 3.5 cm, l(QR) = 5.6 cm, l(RS) = 3.5 cm, m∠Q = 110°, m∠R = 70°.

If it is given that PQRS is a parallelogram, which of the given information is unnecessary?

Solution:

Steps to construct:

Step 1: Draw a line PQ = 3.5cm.

Step 2: Construct an angle of 110^o at point Q.

Step 3: With Q as centre and radius 5.6cm, draw an arc cutting the line QX at point R.

Step 4: Construct an angle of 70^o at point R.

Step 5: With R as the centre and radius 3.5cm, draw an arc cutting the Line RY at point S.

Step 6: Join RS and PS.

• Since, the opposite sides of a parallelogram are congruent.Either l(PQ) or l(SR) is required.
• To construct a parallelogram, length of adjacent sides and measure of one angle is required.Either l(PQ) and m∠Q or l(SR) and m∠R is the unnecessary information given in the question.

In this chapter, we shall discuss concepts related to Construction of Quadrilaterals, with examples for better understanding. MSBSHSE Solutions contains all the solutions to the Maths problems provided in the textbook. Students can refer to these solutions and score good marks in the board exams.