Area Of Quadrilaterals

Area of the parallelogram:

The product of base of the parallelogram and the height of the parallelogram.

∴ area of the parallelogram = base × height

= b × h

Given that

Sides of the parallelogram are 5cm & 6cm

Area of the parallelogram = 25 cm²

We can write that as

Area = 5 cm × 5 cm

That means base of the parallelogram (b) = 5 cm

height of the parallelogram (h) = 5 cm

With height and base and side we will draw a parallelogram.

We need to find the value of X

In the above fig. triangle AOD is a right- angled triangle with AD as hypotenuse.

∴ AD² = A0² + OD²

6² = X² + 5²

X² = 36 – 25

X = √9

X = 3 cm

Using height, sides & X we can draw the parallelogram.

Given

Area = 25 cm²

Area = 5 cm × 5 cm

That means base of the parallelogram (b) = 5 cm

height of the parallelogram (h) = 5 cm

Perimeter = 24 cm

2 (A + B) = 24 cm

A + B = 12 cm

where A & B are sides of a parallelogram

since base is also a side of a parallelogram.

∴ B = 7 cm

With height and base and side we will draw a parallelogram.

We need to find the value of X

In the above fig. triangle POS is a right- angled triangle with PS as hypotenuse.

PS² = PO² + OS²

7² = X² + 5²

X = √(49-25)

= √24

Using height, sides & X we can draw the parallelogram.

From the given picture we can see that a triangle is drawn inside the parallelogram.

The area of the triangle is 5 cm²

We can see that the base and height of both the triangle and parallelogram is same

The area of the triangle =

5 = ( base × height)

10 cm² = base × height

The area of the parallelogram = base × height

= 10 cm²

Let ABCD is a parallelogram is formed by two set of parallel lines.

The distance between the two horizontal lines is 3 cm,

The distance between the vertically slanted lines is 2 cm.

Connect the two vertically slanted lines a right-angle triangle is formed.

From the right-angle triangle ABE

AB is the hypotenuse

AE is the opposite side

EB is the Adjacent side to θ

Sin θ = =

∴ θ = 30^o

∴ ∠ ABE is 30^o

From the fig. we can say that ∠ C = ∠ ABE = 30^o

In parallelogram opposite sides are equal

∴ ∠ BAD = ∠ C

∠ BAD = 30^o

∵ ∠ BAF = 90^o

∠ BAF = ∠ BAD + ∠ DAF

90^o = 30^o + ∠ DAF

∠ DAF = 60^o

From the diagram cos θ =

COS 60^o =

DA = 6 cm

We know that AD, AB are the sides of the parallelogram

Perimeter = 2 (AB + AD)

= 2 (4+6)

= 20 CM

Area = side × perpendicular distance to the opposite side

= AB × AF

= 4 × 3 = 12 cm²

Height of the triangle = 3.5 cm

Base of the triangle = 6 cm

The area of the triangle =

= 6 × 3.5

= 10.5 cm²

From the fig. we can say that the parallelogram is constituted of 2 triangles

∴ Area of the parallelogram = 2 × Area of triangle

= 2 × 10.5

= 21 cm²

Given area of the square = 4. cm²

We know that area of the square = × d²

d is the diagonal of the square

× d² = 4.

d² = 9

d = 3 cm

diagonal of the square = 3 cm

step – 1:

Draw the a horizonal line with the starting point as A and draw 45^o ray from point A.

(It is a square so, the angle between side and diagonal will be 90^o)

Step – 2:

We know the length of diagonal, so cut the ray with 3 cm arc

Step – 3:

Connect the intersecting point with horizontal line and name the intersection point as O

The above fig. forms a right-angle triangle

Sin 45^o =

BO =

In square all sides are equal we can draw rest of the fig. with the same measurement.

Thus, a square can be drawn when its area is given.

Area of the rhombus = 9

Area of the rhombus = × d₁ × d₂

× d₁ × d₂ = 9

d₁ × d₂ = 18

we can divide the diagonal like this.

d₁ × d₂ = 6 × 3

(you can also write it as 3 × 6 or 9 × 2)

∴ diagonal (d₁) = 6 cm

diagonal (d₂) = 3 cm

with these data we can draw a non- square rhombus.

Firstly, draw a diagonal and draw another diagonal passing through its mid points.

Note: the midpoint for both the diagonals should be same.

Connect all the adjacent points

Thus, a rhombus will be formed

Length of a diagonal (d₁) = 24 cm

Area of the rhombus = 216 cm²

Area of the rhombus = × d₁ × d₂

× d₁ × d₂ = 216

d₁ × d₂ = 432

24 × d₂ = 432

d₂ = 18 cm

(i) length of another diagonal = 18 cm

We can see that some part of the rhombus constitutes a right-angle triangle.

From that

H² = 9² + 12²

H = √225

= 15

(ii)∴ side of a rhombus is 15 cm

We know that sides are equal in a rhombus

(iii) So, perimeter = 4 × sides

= 4 × 15

= 60 cm

(iv) In rhombus distance between the side is also a side

∴ distance between the side is 15 cm

68-centimeter-long rope is used to make a rhombus

The above statement states that perimeter of the rhombus is 68 cm.

And they have given that diagonal (d₁) is 16 cm

(In question they have given that as meter that should be a meter)

Perimeter = 4 × a

Where,

a = side of the rhombus.

68 = 4a

a = 17 cm

From the triangle AOD

AD² = AO² + OD²

17² = 8² + X²

X = √(289-64)

= 15

(i) The distance between the other two corners (d₂) = 2X

= 2 × 15

= 30 cm

(ii)The area of the rhombus = × ×

= × ×

= 60 cm²

(i) Given that the small quadrilateral is formed by the midpoints of the diagonals of a rhombus.

In the small one the diagonals cut the other to the half of it.

So, this is a rhombus.

(ii) The area of the small rhombus = 3 cm²

× × = 3

Let, d₁, d₂ are the diagonals of smaller rhombus

d₃, d₄ are the diagonals of larger rhombus

d₁× d₂ = 24

d₁× d₂ = 4 × 6

d₁ = 4 cm

d₂ = 6 cm

They have given that diagonals of smaller rhombus is formed by midpoints of diagonals of larger rhombus.

d₃= 2 d₁ = 2 (4) = 8 cm

d₄= 2 d₂ = 2(6) = 12 cm

∴ the area of larger rhombus = × ×

= × ×

= 12 cm²

We have to draw largest rhombus inscribed in rectangle.

Rectangle breadth (b) = 6cm

Height (h) = 4 cm

We know that in rhombus is directly proportional to the diagonals of it.

In a rectangle we can’t draw diagonals more than its height and breadth.

So, the largest diagonals are height, breadth of rectangle

∴ the area of rhombus = × ×

= × ×

= × ×

= 3 cm²

A rhombus with area of 3 is largest possible in rectangle 6 × 4

As shown in the figure below:

Area of the quadrilateral = area of ADB + area of DCB

=

=

=

=

Hence, area of the quadrilateral is 40 square centimetres.

For this we choose a quadrilateral as shown below:

Clearly, diagonals of the quadrilateral are AC and BD, which are

perpendicular to each other.

Sum of areas of triangles ABD and CBD

Area of quadrilateral =

=

=

OA+OC=AC

=

Now,

Since, DB × AC = product of diagonals

Therefore, Area of quadrilateral =

Hence proved.

More elaborated diagram is as shown below:

Area of quadrilateral = area(ABC) + area(ACD) …..(1)

Triangle ABC is a right triangle.

Therefore, area(ABC) =

=

=

=

Also, using Pythagoras theorem, we have:

Now we find that all sides of triangle ACD are equal.

i.e. AC = CD = AD = 20 cm

Therefore, triangle ACD is an equilateral triangle.

Area of equilateral triangle ACD =

=

= 173.2 sq. cm

Therefore, from (1), we get:

Area of quadrilateral = area(ABC) + area(ACD)

= 96 sq. cm+ 173.2 sq. cm

= 269.2 sq. cm

Hence, area of the quadrilateral is 269.2 square centimetres.

From the figure below:

Area(ABCD) = area(ACD) + area(ABC)

=

= ……(1)

Again, Area(PQRS) = area(PRS) + area(PQR)

=

= ……(2)

Also, we know that the perpendicular distance between two parallel line are same everywhere. So, we have:

……(3)

Now, using (3) and taking ratio of (1) and (2), we get:

……(4)

Thus, we see that areas of quadrilateral area in the ratio of the lengths of the diagonals AC and PR. Hence proved.

(i) For the condition of quadrilaterals to have equal area the ratio in (4) should be one.

Therefore, the diagonals should be equal in when the areas of the two quadrilaterals are equal.

(ii) For this we draw a random quadrilateral which is neither a parallelogram nor trapezium, as shown in figure below:

Area of the above quadrilateral can be calculated to be

Area =

Case 1: we take AD = 10 cm, = 1 cm and = 2 cm so that area of quadrilateral becomes 15 square centimetres, as shown:

Area =

=

=

Case 2: we take AD = 6 cm, = 2 cm and = 3 cm so that area of quadrilateral becomes 15 square centimetres, as shown:

Area =

=

=

We know that a diagonal of a parallelogram divides it into two triangles of equal area.

Therefore, Area of quadrilateral =

=

=

= 192 sq. cm

Hence, area of the parallelogram is 192 square centimetres.