AREA OF RHOMBUS TRAPEZIUM AND PARALLELOGRAM

Problem 1 :

a)  Find the area of rhombus which has diagonals of length 12 cm and 8 cm.

b)  One diagonal of a rhombus is twice as long as the other diagonal. If the rhombus has area 32 cm², find the length of the shorter diagonal ?

Solution :

a)  

Area of rhombus = (1/2) x d₁ x d₂

diagonal 1 = 12 cm and diagonal 2 = 8 cm

= (1/2) x 12 x 8

= 6 x 8

= 48 cm²

So, the area of rhombus is 48 cm².

b) Diagonal 1 = 2 diagonal 2

Area of rhombus = 32 cm²

(1/2) x d₁ x d₂ = 32

(1/2) x 2d₂ x d₂ = 32

(d₂)² = 32

d₂ = √32

d₂ = √(2 x 2x 2 x2 x 2)

= 2 x 2 √2 

= 4√2 cm

d₁ = 2(4√2)

= 8√2

So, the length of the shorter diagonal is 4√2 cm.

Problem 2 :

The area of the trapezium ABCD is 204 cm². Find the area of triangle DBC.

Solution :

The area of the trapezium ABCD = 204 cm²

(1/2) h (a + b) = 204

Here a = 15 cm and b = 36 cm

(1/2) x h (15 + 36) = 204

(1/2) x h x 51 = 204

h = (204 x 2)/51

h = 4 x 2

h = 8 cm

Height of the trapezium = 8 cm

After connecting the D and B, we see a triangle.

= (1/2) x base x height

base = DC = 36 cm

height = 8 cm

= (1/2) x 8 x 36

= 4 x 36

= 144 cm²

Problem 3 :

a) A kite has diagonals of length 16 cm and 10 cm. Find its area.

b) Find the area of a kite with diagonals of length a cm and b cm.

Solution :

Area of kite = (1/2) x d₁ x d₂

d₁ = 16 cm and d₂ = 10 cm

= (1/2) x 16 x 10

= 80 cm²

Area of kite with diagonals a cm and b cm.

Area of kite = (1/2) x a x b

Problem 4 :

Parallelogram ABCD has AB = 10 cm and diagonal DB = 15 cm. If the shortest distance from C to line AB is 8 cm, find the shortest distance from A to DB.

Solution :

Area of parallelogram = base x height

= 10 x 8

= 80 cm²

Area of triangle ADB = (1/2) x 80

= 40 cm²

(1/2) x base x height = 40

(1/2) x 15 x height = 40

height = (40 x 2)/15

= 16/3 cm

Problem 5 :

Area of this trapezium  :

Solution :

EC = 10 cm (since it is parallelogram)

AB = 24 cm

AE + EB = 24

10 + EB = 24

EB = 24 - 10

EB = 14 cm

In triangle DEB,

area of triangle DEB = √s(s - a)(s - b)(s - c)

s = (a + b + c)/2

s = (EC + EB + BC)/2

= (10 + 14 + 15)/2

= 39/2

= 19.5

Applying the values of s, a, b and c in the formula, we get

= √19.5(19.5 - 10)(19.5 - 14) (19.5 - 15)

= √19.5 x 9.5 x 5.5 x 4.5

= √4584.9375

= 67.71

Area of triangle CEB = (1/2) x EB x CF

(1/2) x 14 x CF = 67.71

7 x CF = 67.71

CF = 67.71/7

CF = 9.63 cm

Problem 6 :

In the figure below AB = 9√2 meters, ED = 6 meters and <BAE = 45°. What is the area of the kite

A) 90 m²         B) 108 m²        C) 135 m²  

D) 216 m²       E) Not enough information

Solution :

Let AE = BE = x

AB² = AE² + BE²

(9√2)² = x² + x²

81(2) = 2x²

x² = 81

x = 9 cm

Area of kite = (1/2) x 18 x 12

= 6 x 18

= 108 cm²

Problem 7 :

Find the area of the shaded region :

Solution :

Area of shaded region = area of rectangle - area of kite

= length x width - (1/2) x d₁ x d₂

= 24 x 20 - (1/2) x (20 x 24)

= 480 - 240

= 240 cm²

So, area of the shaded region is 240 cm²

Problem 8 :

Find the area of the right trapezoid.

Solution :

Area of trapezoid = (1/2) x 4 x (10 + 6)

= (1/2) x 4 x 16

= 2 x 16

32 cm²

Problem 9 :

Find the missing measurement of the trapezoid.

Solution :

a = ? ft, b = 10.9 ft and height = 7.9 ft

area of trapezoid = (1/2) x 7.9 x (a + 10.9)

59.3 = (1/2) X 7.9 x (a + 10.9)

59.3(2) / 7.9 = a + 10.9

15.01 = a + 10.9

a = 15.01 - 10.9

a = 4.11

So, the missing measure is 4.11 ft.

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