**Problem 1 :**

Find the area of the trapezoid ABCD shown below.

**Problem 2 :**

In a trapezoid the measurement of one parallel side two more than the other parallel side and the height is 4 cm. The area of the trapezoid is 64 cm^{2}. Find the lengths of the two parallel sides.

**Problem 3 :**

The shape of the top surface of a table is a trapezoid. Find its area, if its parallel sides are 1 m and 1.2 m and perpendicular distance between 0.8 m.

**Problem 4 :**

The wall is in the shape as shown below has to be painted. If one can of paint covers 0.5 m^{2}, how many cans of paint will be needed, if only one coat of paint is applied ?

**Problem 1 :**

Find the area of the trapezoid ABCD shown below.

**Solution :**

Area of the trapezoid ABCD is

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

Substitute a = 5, b = 12 and h = 4.

= (1/2)(5 + 12)4

= (1/2)(17)4

= 34 cm^{2}

**Problem 2 :**

In
a trapezoid the measurement of one parallel side two more than the
other parallel side and the height is 4 cm. The area of the trapezoid is
64 cm^{2}. Find the lengths of the two parallel sides.

**Solution :**

Let 'a' and 'b' be the two parallel sides.

One parallel side is two more than the other parallel side.

Then,

a = b + 2

Area of the trapezium = 64 cm^{2}

(1/2)(a + b)h = 64

Substitute h = 4 and a = b + 2.

(1/2)(b + 2 + a)4 = 64

(2b + 2)2 = 64

Divide each side by 2.

2b + 2 = 32

Subtract 2 from each side.

2b = 30

Divide each side by 2.

b = 15

Then,

a = b + 2

a = 15 + 2

a = 17

So, the lengths of the two parallel sides are 15 cm and 17 cm.

**Problem 3 :**

The shape of the top surface of a table is a trapezoid. Find its area, if its parallel sides are 1 m and 1.2 m and perpendicular distance between 0.8 m.

**Solution :**

Area of a the top of surface of the table (trapezoid) is

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

= (1/2)(1.2 + 1)0.8

= (1/2) (2.2)0.8

= 1.1 (0.8)

= 8.8 m^{2}

**Problem 4 :**

The wall is in the shape as shown below has to be painted. If one can of paint
covers 0.5 m^{2}, how many cans
of paint will be needed, if only one coat
of paint is applied ?

**Solution : **

In the figure shown above, the perpendicular distance between the two sides BC and AB at any point is same, which is 5 cm.

Then, the sides BC and AD are parallel.

In the quadrilateral above, because the two sides BC and AD are parallel, ABCD is a trapezoid.

Let 'a' and 'b' be the lengths of two parallel sides and.

a = BC = 5cm

b = AD = AF + FE + ED = 3 + 5 + 4 = 12 cm

Height (h) = 5 cm

Area of the trapezoid ABCD is

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

Substitute a = 5, b = 12 and h = 5.

= (1/2)(5 + 12)5

= 42.5 m^{2}

One can of paint covers 0.5 m^{2}.

Number of cans of paint required 42.5 m^{2} is

= 42.5 / 0.5

= 425 / 5

= 85

So, 85 cans of paint required to cover the above wall.

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