Quadrilaterals

We know that the sum of the angles in a Quadrilateral is 360⁰.

Let us consider Quadrilateral ABCD,

Let ∠A = 70⁰ and ∠B = 130⁰ ,

According to the problem, it is told that the remaining two angles are equal,

i.e, ∠C =∠D

∴∠A +∠B +∠C +∠D = 360⁰

Now substituting the values and conditions we get,

⇒ 70⁰ + 130⁰ + ∠C + ∠C = 360⁰

⇒ 200⁰ + 2 × ∠C = 360⁰

⇒ 2 × ∠C = 360⁰ - 200⁰

⇒ 2 × ∠C = 160⁰

⇒ ∠C = 80⁰

The values of the other two angles is70⁰, 70⁰.

According to the problem it is given that,

∠R = 125⁰ and ∠P and ∠Q are supplementary angles.

We know that, the sum of supplementary angles equals to 180⁰.

So, ∠P +∠Q = 180⁰.

We know that the sum of all angles in a quadrilateral is 360⁰.

So, ∠P +∠Q +∠R +∠S = 360⁰.

Substituting the values and conditions we obtain,

⇒ 180⁰ + 125⁰ + ∠S = 360⁰

⇒ 305⁰ + ∠S = 360⁰

⇒ ∠S = 360⁰ - 305⁰

⇒ ∠S = 55⁰.

The value of ∠S is55⁰.

Let us assume a quadrilateral ABCD.

We know that the sum of angles in a quadrilateral is 360⁰.

i.e, ∠A +∠B +∠C +∠D = 360⁰. ...... (1)

It is also given that the first three angles are in the ratio of 2:3:5, and the fourth angle is 90⁰.

Let us take the first three angles to be 2x, 3x and 5x.

Substituting the values in the eq(1) we get,

⇒ 2x + 3x + 5x + 90⁰ = 360⁰

⇒ 10x = 360⁰ - 90⁰

⇒ 10x = 270⁰

⇒ x =

⇒ x = 27⁰.

Now we find the values of the angles using the value of x.

⇒ ∠A = 2x

⇒ ∠A = 2 × 27⁰

⇒ ∠A = 54⁰.

⇒ ∠B = 3x

⇒ ∠B = 3 × 27⁰

⇒ ∠B = 81⁰.

⇒ ∠C = 5x

⇒ ∠C = 5 × 27⁰

⇒ ∠C = 135⁰.

The three angles are 54⁰, 81⁰, 135⁰.

We know that the sum of the angles in a quadrilateral is 360⁰.

i.e, ∠A +∠B +∠C +∠D = 360⁰. ...... (1)

We also know that the sum of angles in a triangle is 180⁰.

According to the problem, it is given that ∠C +∠D = 100⁰.

Substituting the condition in the eq(1) we get,

⇒ ∠A + ∠B + 100⁰ = 360⁰

⇒ ∠A + ∠B = 360⁰ - 100⁰

⇒ ∠A +∠B = 260⁰.

⇒

⇒ . ...... (2)

According to the problem, it is given that the ΔABP is formed by the intersection of angular bisectors of ∠A and ∠B.

From ΔABP, We can write that,

⇒

⇒ 130⁰ + ∠P = 180⁰ (From eq(2))

⇒ ∠P = 180⁰ - 130⁰

⇒ ∠P = 50⁰.

The value of the ∠P is 50⁰.

Let us consider trapezium PQRS with PQ||RS.

According to the problem, it is given that ∠P = 70⁰and ∠Q = 80⁰.

We know that the sum of adjacent angles between two parallel lines is 180⁰.

i.e, ∠P +∠S = 180⁰ ...... - - (1)

∠Q +∠R = 180⁰. ...... - (2)

Substituting the value of ∠P in eq(1) we get,

⇒ 70⁰ + ∠S = 180⁰

⇒ ∠S = 180⁰ - 70⁰

⇒ ∠S = 110⁰.

Substituting the value of ∠Q in eq(2) we get,

⇒ 80⁰ + ∠R = 180⁰

⇒ ∠R = 180⁰ - 80⁰

⇒ ∠R = 100⁰.

The values of ∠S and ∠R is110⁰ and 100⁰.

Consider a trapezium ABCD, it is given that

AB||CD and the AD is not parallel to BC.

From the figure, we can say the similarities of ΔABC and ΔADC.

⇒ AC = AC (Common side)

⇒ ∠A =∠C (Alternate angles on opposite sides of traversal line are equal)

⇒ ∠B≠∠D (One angle is obtuse and other is acute)

⇒ AB≠CD (From the figure)

From the above conclusions, we can say that any of the property like SSS, SAS, ASA is not satisfied. So, we can conclude that,

ΔABC is not congruent toΔADC.

Given:

PQRS is an Isosceles Trapezium.

∠SRP = 30⁰

∠PQS = 40⁰

From the figure, we can say that,

⇒ ∠PQS =∠RSQ (alternate angles on both sides of traversal line)

So, ∠RSQ = 40⁰.

Also,

⇒ ∠SRP =∠RPQ (alternate angles on both sides of traversal line)

So, ∠RPQ = 30⁰.

The values of ∠RSQ and ∠RPQ is40⁰ and 30⁰.

Let us assume a parallelogram ABCD,

We know that the sum of the adjacent angles in a parallelogram is 180⁰.

We also know that the angles at the opposite vertices are equal.

i.e., ∠A +∠B = 180⁰. ...... - (1)

∠A =∠C ...... ...... (2)

∠B =∠D ...... ...... (3)

According to the problem, it is given that the adjacent angles are in the ratio of 2:1.

Let us assume that the adjacent angles be A and B.

Let’s take the values of ∠B = 2x and ∠A = x

From eq(1) we get,

⇒ x + 2x = 180⁰

⇒ 3x = 180⁰

⇒

⇒ x = 60⁰.

Now find the value of ∠A and ∠B,

⇒ ∠A = x

⇒ ∠A = 60⁰

⇒ ∠B = 2x

⇒ ∠B = 2 × 60⁰

⇒ ∠B = 120⁰

From (2) and (3) we get,

∠C = 60⁰ and ∠D = 120⁰

The values of ∠A,∠B,∠C and ∠D is 60⁰, 120⁰, 60⁰, 120⁰.

Let us consider a parallelogram ABCD,

We know that the lengths of opposite sides of a parallelogram are equal.

i.e, AD = BC ...... (1)

AB = CD ...... (2)

We also know that Perimeter of Parallelogram is the sum of the lengths of its sides.

i.e, Perimeter of Parallelogram(P) = AB + BC + CD + DA - - (3)

According to the problem it is given that,

The perimeter of Parallelogram(P) is 450m and

One side is 75m larger than the other side.

Let us consider the length of smaller be ‘x’m

Then from the figure,

BC = AD = xm ...... - - (4)

Then, AB = CD = (x + 75)m ...... (5)

From eq(1),

⇒ x + x + 75 + x + x + 75 = 450 (since P = 450m)

⇒ 4x + 150 = 450

⇒ 4x = 450 - 150

⇒ 4x = 300

⇒

⇒ x = 75m.

From Eq(4) we can say that,

BC = DA = 75m.

AB = CD = (75 + 75) = 150m.

The lengths of sides AB, BC, CD and DA is 150m, 75m, 150m, 75m.

Consider a Parallelogram ABCD,

Diagonals AC and BD intersect at O.

According to the problem it is given that ∠DAC = 40⁰,∠CAB = 35⁰ and ∠DOC = 110⁰.

Since AC is a straight line it’s angle ∠AOC is 180⁰.

So, ∠AOC can be written as

⇒ ∠AOC =∠AOD +∠COD ...... (1)

⇒ ∠AOC =∠AOB +∠BOC ...... (2)

Substituting the values in the eq(1) we get,

⇒ 180⁰ = ∠AOD + 110⁰

⇒ ∠AOD = 180⁰ - 110⁰

⇒ ∠AOD = 70⁰ ...... (3)

Since BD is a straight line it’s angle ∠BOD is 180⁰.

So, ∠BOD can be written as

⇒ ∠BOD =∠DOC +∠BOC ...... - (4)

Substituting the values in eq(4) we get,

⇒ 180⁰ = ∠BOC + 110⁰

⇒ ∠BOC = 180⁰ - 110⁰

⇒ ∠BOC = 70⁰ ...... - (5)

Using eq. (2) and (5) we get,

⇒ 180⁰ = ∠AOB + 70⁰

⇒ ∠AOB = 180⁰ - 70⁰

⇒ ∠AOB = 110⁰ ...... - - (6)

From the alternative angles property of traversal line between two Parallel lines, we can write

∠OCD =∠OAB = 35⁰

∠BCO =∠DAO = 40⁰

We sum that sum of all angles in a triangle is 180⁰.

From the ΔAOD, ΔDOC, ΔCOB, ΔAOB we can write,

⇒ ∠DAO +∠AOD +∠ODA = 180⁰

⇒ ∠DOC +∠OCD +∠CDO = 180⁰

⇒ ∠COB +∠OBC +∠BCO = 180⁰

⇒ ∠AOB +∠OBA +∠BAO = 180⁰

On substituting the known values in these equations we get,

⇒ 40⁰ + 70⁰ + ∠ODA = 180⁰

⇒ 110⁰ + ∠ODA = 180⁰

⇒ ∠ODA = 180⁰ - 110⁰

⇒ ∠ODA = 70⁰

⇒ 110⁰ + 35⁰ + ∠CDO = 180⁰

⇒ 145⁰ + ∠CDO = 180⁰

⇒ ∠CDO = 180⁰ - 145⁰

⇒ ∠CDO = 35⁰.

⇒ 40⁰ + ∠OBC + 70⁰ = 180⁰

⇒ ∠OBC + 110⁰ = 180⁰

⇒ ∠OBC = 180⁰ - 110⁰

⇒ ∠OBC = 70⁰.

⇒ 110⁰ + ∠OBA + 35⁰ = 180⁰

⇒ ∠OBA + 145⁰ = 180⁰

⇒ ∠OBA = 180⁰ - 145⁰

⇒ ∠OBA = 35⁰

We know that ∠CBO =∠CBD = 70⁰ and ∠ACB =∠OCB = 40⁰

And also ∠ADC =∠ADO +∠ODC

⇒ ∠ADC = 70⁰ + 35⁰

⇒ ∠ADC = 105⁰.

The values of ∠ABO,∠ADC,∠ACB,∠CBD is 35⁰, 105⁰, 40⁰, 70⁰.

Consider a Parallelogram ABCD,

It is told that the side BC is produced to E and ∠BCE = 105⁰.

From the figure, we can say that

∠BCE +∠BCD = 180⁰ (Since ∠DCE is the straight angle)

By substituting the values we get,

⇒ 105⁰ + ∠BCD = 180⁰

⇒ ∠BCD = 180⁰ - 105⁰

⇒ ∠BCD = 75⁰

We know in a parallelogram the opposite angles are equal and the sum of the adjacent angles is 180⁰.

i.e, ∠C =∠A ...... (1)

∠B =∠D ...... (2)

∠C +∠D = 180⁰ ...... (3)

From Eq(1) we can write

∠A = 75⁰.

From Eq(3), we can write

⇒ 75⁰ + ∠D = 180⁰

⇒ ∠D = 180⁰ - 75⁰

⇒ ∠D = 105⁰.

From Eq(2), we can write

∠B = 105⁰.

The angles ∠A,∠B,∠C,∠D are75⁰, 105⁰, 75⁰, 105⁰.

Consider a Parallelogram KLMN,

According to the problem, it is given that ∠K = 60⁰.

We know that in a parallelogram,

⇒ The adjacent angles are supplementary.

⇒ The opposite angles are equal.

From the figure, we can say that,

∠K =∠M ...... - (1)

∠L =∠N ...... - (2)

∠K +∠L = 180⁰ (3)

From(1) we can write that

∠M = 60⁰

From(3) we can write that

⇒ 60⁰ + ∠L = 180⁰

⇒ ∠L = 180⁰ - 60⁰

⇒ ∠L = 120⁰

From(2), we can say that

∠N = 120⁰.

The angles ∠K,∠L,∠M,∠N are 60⁰, 120⁰, 60⁰, 120⁰.

Consider a triangle ABCD,

According to the problem, it is given that the sides are in the ratio 2:1.

The perimeter of the rectangle 30cm.

We know that the sides of the rectangles are represented by Length(L) and Breadth(B).

We also know that,

Perimeter(P) of the rectangle = 2 × (L + B).

From the problem, we came to know that the ratio L: B = 2:1.

Let us assume the values of Length and Breadth be 2x and.

From the problem, we have P = 30cm.

Substituting the values we get,

⇒ 30 = 2 × (2x + x)

⇒ 30 = 2 × (3x)

⇒ 30 = 6x

⇒

⇒ x = 5cm.

Using the value of ‘x’ we find the value of length and breadth of Rectangle.

Now,

⇒ L = 2x

⇒ L = 2 × 5

⇒ L = 10cm.

⇒ B = 5cm.

From this, we can say that AB and CD are lengths, while BC and DA are breadths.

The lengths of AB, BC, CD, DA are 10cm, 5cm, 10cm, 5cm.

Consider Rectangle ABCD,

According to the problem, it is given that ∠OCD = 30⁰.

We know that in a rectangle The sides are perpendicular to each other. So, we can write

⇒ ∠OCD +∠OCB = 180⁰.

⇒ 30⁰ + ∠OCB = 90⁰

⇒ ∠OCB = 90⁰ - 30⁰

⇒ ∠OCB = 60⁰.

We know that the alternate angles along the traversal line between two parallel lines are equal.

So,

∠OCD =∠OAB = 30⁰.

∠OCB =∠OAD = 60⁰.

We know that the diagonals in a rectangle bisect each other.

So, we can say that,

AO = OB = OC = DO.

We also the angles opposite to the equal sides are also equal.

So, From the figure, we can say that

∠OBC =∠OCB = 60⁰.

From ΔOBC, we can say that,

∠O +∠B +∠C = 180⁰. (Sum of angles in a triangle is 180⁰)

By substituting the values we get,

⇒ ∠O + 60⁰ + 60⁰ = 180⁰.

⇒ ∠O + 120⁰ = 180⁰

⇒ ∠O = 180⁰ - 120⁰

⇒ ∠O = 60⁰.

The value of ∠BOC is 60⁰ and the ΔBOC is Equilateral Triangle.

The Criteria required for a quadrilateral to become Rectangle is:

⇒ The Lengths of opposite sides are to be equal.

⇒ The opposite sides are to be parallel.

⇒ The adjacent angles are need to be supplementary.

⇒ The sides are to be perpendicular to each other.

⇒ The diagonals bisect each other.

⇒ Each diagonal bisects the Rectangle into congruent triangles.

The criteria required for a quadrilateral to become Parallelogram is:

⇒ The lengths of opposite sides are to be equal.

⇒ The opposite sides need to be parallel.

⇒ The adjacent angles are need to be supplementary.

⇒ The diagonals bisect each other.

⇒ Each diagonal bisects the Parallelogram into congruent triangles.

⇒ The angles of the opposite vertices are needed to be equal.

From the above statements, we can clearly say that the rectangle satisfies the requires criteria of Parallelogram.

But, the parallelogram doesn’t satisfy the criteria of sides to be perpendicular.

So, from these, we can conclude that All Rectangles are parallelograms, but all parallelograms are not rectangles.

Let us consider a Rectangle ABCD,

According to the problem, it is given that

The sides are in the ratio 4:3.

The area of Rectangle is 1728m².

The cost of fencing is Rs.2.50/m.

Let us consider the length(L) of the Rectangle be ‘4x’m and the breadth(B) of the rectangle be ‘3x’m.

We know that Area(A) of a Rectangle is

A = L × B

From the given data we can write,

⇒ 4x × 3x = 1728

⇒ 12x² = 1728

⇒

⇒ x² = 144

⇒ x = √144

⇒ x = 12

Now we find the length and breadth of the triangle.

⇒ Length(L) = 4x

⇒ L = 4 × 12

⇒ L = 48m

⇒ Breadth(B) = 3x

⇒ B = 3 × 12

⇒ B = 36m

To find the total cost for fencing, we need to find the perimeter first.

As we know that the fencing the sides of the rectangle but not total internal part of the rectangle.

Now, let’s find the Perimeter (P) of the rectangle.

We know that Perimeter(P) of the Rectangle is

P = 2 × (L + B)m

Now let’s substitute the values of length and breadth of the rectangle.

⇒ P = 2 × (48 + 36)

⇒ P = 2 × (84)

⇒ P = 168m

The total cost required for fencing(F_c) = Perimeter*cost per meter

⇒ F_c = 168 × 2.50

⇒ F_c = 420

The total fencing cost required is Rs.420.

Let us consider a Rectangular yard ABCD,

According to the problem, it is given that

ΔADF, ΔBCE are a congruent isosceles right triangle.

It is also given that,

Length of AB is 25m.

Length of BC is 15m.

From the figure, we can clearly say that,

AB = DF + FE + EC and

DF = EC.

From the above equations and data are given in the problem,

⇒ 25 = (2 × DF) + 15

⇒ 2 × DF = 25 - 15

⇒ 2 × DF = 10

⇒

⇒ DF = 5m

From isosceles right triangle ADF,

We can see that AD = DF,

So breath of Rectangle ABCD is 5m.

We know that Area of Rectangle = Length × Breadth

From the figure and calculations, we can clearly say,

Length of Rectangle ABCD is 25m.

The breadth of Rectangle ABCD is 5m.

The base of Triangle ADF is 5m.

The height of Triangle ADF is 5m.

Area of Rectangle ABCD = 25 × 5

Area of Rectangle ABCD = 125m².

Area of Triangle

Area of Triangle

Area of Triangle ADF = 12.5m²

The area in the Rectangular yard that containing flower beds is the area of two Triangles ADF and BCE.

But according to the problem, it is told that ΔADF and ΔBCE are congruent.

So, the area of both the triangles is the same.

Area of ΔADF = Area of ΔBCE = 12.5m².

The area occupies by flower beds = Area of ΔADF + Area of ΔBCE.

The area occupies by flower beds = 12.5 + 12.5

The area occupies by flower beds = 25m².

The fraction(f) of the yard that is occupied by the flower bed is given by

⇒

⇒

The fraction is .

Given a rhombus ABCD,

According to the problem, it is given that ∠C = 70⁰.

We know that in a rhombus, the sum of adjacent angles is 180⁰.

The opposite angles are equal.

So, we can say that;

∠A =∠C.

∠B =∠D.

∠C +∠D = 180⁰.

From the values and equations, we can write that,

⇒ ∠A = 70⁰.

⇒ 70⁰ + ∠D = 180⁰

⇒ ∠D = 180⁰ - 70⁰

⇒ ∠D = 110⁰.

⇒ ∠B = 110⁰.

The angles ∠A, ∠B,∠D is 70⁰, 110⁰, 110⁰.

Consider a Rhombus PQRS,

According to the problem, it is given that:

PQ = 3x - 7.

QR = x + 3.

We know that the lengths of sides of a rhombus are equal.

So, we can clearly say that,

PQ = QR

⇒ 3x - 7 = x + 3

⇒ 3x - x = 3 + 7

⇒ 2x = 10

⇒

⇒ x = 5 units

Now, let's find the length of each side,

⇒ PQ = 3x - 7

⇒ PQ = (3 × 5) - 7

⇒ PQ = 15 - 7

⇒ PQ = 8 units

Since all the sides of a rhombus are equal.

Length of side PS is 8units.

The criteria required for a quadrilateral to become Rhombus is:

⇒ The lengths of all sides are equal.

⇒ The opposite sides need to be parallel.

⇒ The adjacent angles are need to be complimentary.

⇒ The diagonals bisect each other.

⇒ Each diagonal bisects the Rhombus into congruent triangles.

⇒ The opposite angles are need to be equal.

The criteria required for a quadrilateral to become Parallelogram is:

⇒ The lengths of opposite sides are to be equal.

⇒ The opposite sides need to be parallel.

⇒ The adjacent angles are need to be supplementary.

⇒ The diagonals bisect each other.

⇒ Each diagonal bisects the Parallelogram into congruent triangles.

⇒ The angles of the opposite vertices are needed to be equal.

From the above statements, we can clearly say that the Rhombus meets the criteria of Parallelogram. So, we can clearly say that

A rhombusis a Parallelogram.

Consider a square ABCD,

According to the problem, it is given that the area of ΔABD is 36cm².

We know that the diagonal of a square bisect into congruent triangles.

So, from the figure we can clearly say that

Area of ΔABD = Area of ΔBCD

So, Area of ΔBCD = 36cm².

From the figure, we can clearly say that,

Area of square ABCD = Area of ΔABD + Area of ΔBCD

Area of square ABCD = 36 + 36cm²

Area of square ABCD = 72cm².

The Area of ΔBCD = 36cm².

The Area of square ABCD = 72cm².

Given:

Side(a) of square ABCD = 5cm.

Perimeter(P₂) of square PQRS = 40cm.

We know that Perimeter of a square having side length ‘x’ is ‘4x’.

We also know that Area of a square having side length ‘x’ is ‘x²’.

So perimeter(P₁) of square ABCD is 4a

i.e, P₁ = 4a

⇒ P₁ = 4 × 5

⇒ P₁ = 20cm.

Area(A₁) of square ABCD is a²

i.e, A₁ = a²

⇒ A₁ = 5²

⇒ A₁ = 25cm².

Let's find the side length(b) of square PQRS

We have P₂ = 4b

⇒ 4b = 40

⇒ b =

⇒ b = 10cm.

Area(A₂) of square PQRS is b²

i.e, A₂ = b²

⇒ A₂ = 10²

⇒ A₂ = 100cm².

Let's find the ratio of Perimeters(RP) of both squares

i.e, r_p =

⇒ r_p =

⇒ r_p =

Let's find the ratio of Areas(r_A) of both squares

i.e., r_ag =

⇒ r_ag =

⇒ r_ag =

The ratio of perimeters is.

The ratio of area is.

Consider a square field ABCD

According to the problem, it is given that the side length (a) of the square field is 20m.

i.e, a = 20m.

We need to find the Perimeter of the square as the fencing covers only the surface of the square.

We know that the Perimeter(P) of the square of side length ‘x’ is given by 4x.

So, The perimeter of the square ABCD is,

⇒ P = 4 × 20

⇒ P = 80m.

We need them to fence the square four times.

The required length(L) of the fence is given by:

L = 4P

⇒ L = 4 × 80

⇒ L = 320m.

The required length of the fence is 320m.

The differences between Square and Rhombus are:

AB = 4 cm, BC = 6 cm

Construction:

1) Draw a line segment ‘AB’ of length ‘4’cm.

2) Now with the help of protractor at point ‘B’ draw a perpendicular line to the segment ‘AB’.

3) Now with the help of compass draw an arc of radius ‘6’cm taking Point ‘B’ as centre.

4) The arc and line intersect at a point, this point is ‘C’.

5) Join points ‘B’, ’C’ to form line segment ‘BC’.

6) Now draw a line perpendicular to BC from point ‘C’ with the help of protractor.

7) Now draw a line perpendicular to ‘AB’ from point ‘A’ with the help of protractor.

8) Now, these two lines intersect at a point, this point is ‘D’.

9) Join points ‘C’, ’D’ to form line segment ‘CD’ and points ‘A’, ’D’ to form line segment ‘AD’.

10) Thus Rectangle ‘ABCD’ is formed.

AB = 6 cm, AC = 7.2 cm

Construction:

1) Draw a line segment ‘AB’ of length ‘6’cm.

2) Draw a perpendicular line to ‘AB’ using Protractor at point ‘B’.

3) Now, taking ‘A’ as centre draw an arc of length ‘7.2’cm.

4) The arc will intersect the line at a point, this is the point ‘C’. Join the points ‘A’, ’C’ to form line segment ‘AC’ and ‘B’, ’C’ to form line segment ‘BC’.

5) Now draw perpendicular line to line segment ‘BC’ using protractor at point ‘C’.

6) Now draw perpendicular line to line segment ‘AB’ using protractor at point ‘A’.

7) These two perpendicular lines intersect at a point, this point is ‘D’

8) Join points ‘A’, ’D’ to form line segment ‘AD’ and points ‘B’, ’D’ to form line segment ‘BD’.

9) Thus Rectangle ‘ABCD’ is formed.

which has side - length 2 cm

Construction:

1) Draw a line segment ‘AB’ of length 2cm.

2) Now, draw a perpendicular line to the line segment ‘AB’ using protractor at ‘B’.

3) Now taking point ‘B’ as centre draw an arc of length ‘2’cm.

4) The arc intersects the line at point, this point is ‘C’.

5) Join points ‘B’ and ‘C’ to form line segment ‘BC’.

6) Now draw a perpendicular to the line segment ‘BC’ using protractor at point ‘C’.

7) Now draw a perpendicular to the line segment ‘AB’ using protractor at point ‘A’.

8) Join points ‘A’, ’D’ to form line segment ‘AD’ and points ‘C’, ’D’ to form line segment ‘CD’.

9) Thus Square ‘ABCD’ is formed.

which has a diagonal 6 cm.

Construction:

1) Find the side - length of the square using the length of the diagonal.

2) We know that for a square of side - length ‘a’, the length of diagonal is ‘a√2’.

3) From this, we can the length of the side is i.e, ‘3√2’cm.

4) So, now we draw a line segment ‘AB’ of length ‘3√2’cm.

5) Draw a line perpendicular to line segment ‘AB’ using protractor at point ‘B’.

6) Now, taking point ‘A’ as centre draw an arc of radius ‘6’cm.

7) The arc intersects the line at a point, this point is ‘C’.

8) Join points ‘A’, ’C’ to form the line segment ‘AC’ and points ‘B’, ’C’ to form line segment ‘BC’.

9) Now draw a perpendicular to the line segment ‘BC’ using protractor at point ‘C’.

10) Now draw a perpendicular to the line segment ‘AB’ using protractor at point ‘A’.

11) These two perpendiculars intersect at a point, this point is ‘D’.

12) Join points ‘C’, ’D’ to form line segment ‘CD’ and points ‘A’, ’D’ to form the line segment ‘AD’.

13) Thus square ‘ABCD’ is formed.

Allister:

1) Draw a horizontal line of the length of diagonal.

2) Mark a point as ‘A’ and from this point draw a line of length ‘6’ cm at an inclination of ‘45⁰’ with the horizontal line.

3) Now from the end of diagonal (i.e, Point ‘C’) draw a line of inclination of ‘45⁰’ with the diagonal in the direction of the horizontal line.

4) Now, the horizontal line and the line drawn towards the horizontal line will intersect at a point, this point is ‘B’.

5) Now draw a perpendicular to the line segment ‘BC’ using protractor at point ‘C’.

6) Now draw a perpendicular to the line segment ‘AB’ using protractor at point ‘A’.

7) Join points ‘A’, ’D’ to form the line segment ‘AD’ and points ‘C’, ’D’ to form line segment ‘CD’.

8) Thus square ‘ABCD’ is formed.