# Perimeter and Area of Trapezium

Here we will discuss about the perimeter and area of a trapezium and some of its geometrical properties.

Area of a trapezium (A) = $$\frac{1}{2}$$ (sum of parallel sides) × height

= $$\frac{1}{2}$$ (a + b) × h

Perimeter of a trapezium (P) = sum of parallel sides + sum of oblique sides

Some geometrical properties of a trapezium:

In a trapezium PQRS in which sides PQ and RS are parallel, and X and Y are respectively the middle points of PS and QR,

XY = $$\frac{1}{2}$$ (PQ + SR)

Area of ∆QSR = area of ∆PSR

Area of ∆PQS = area of ∆PQR

Solved example problem on finding the perimeter and area of a trapezium:

1. In the trapezium PQRS, PQ ∥ RS and ∠PSR = 90°. If PQ = 15 cm, SR = 40 cm and the diagonal PR = 41 cm then find the area of a trapezium.

Solution:

In the right-angled ∆PSR,

PR$$^{2}$$ = PS$$^{2}$$ + SR$$^{2}$$

Therefore, 41$$^{2}$$ cm $$^{2}$$ = PS$$^{2}$$ + 40$$^{2}$$ cm$$^{2}$$

⟹ PS$$^{2}$$ = (41$$^{2}$$ - 40$$^{2}$$) cm$$^{2}$$

= (41 + 40) (41 – 40) cm$$^{2}$$

= 81 × 1 cm$$^{2}$$

= 81 cm$$^{2}$$

Therefore, PS = 9 cm

Therefore, area of the trapezium PQRS = $$\frac{1}{2}$$ (sum of the parallel sides) × height

= $$\frac{1}{2}$$ (PQ + SR) × PS

= $$\frac{1}{2}$$ (15 + 40) × 9 cm$$^{2}$$

= $$\frac{1}{2}$$ × 55 × 9 cm$$^{2}$$

= $$\frac{495}{2}$$ cm$$^{2}$$

= 247.5 cm$$^{2}$$

2. The parallel sides of a trapezium measure 46 cm and 25 cm. Its other sides are 20 cm and 13 cm. Find the distance between the parallel sides and the area of the trapezium.

Solution:

PQRS is a trapezium in which RS ∥PQ, RS = 25 cm and PQ = 46 cm.

Also, PS = 20 cm and QR = 13 cm

Draw RT ∥ SP and RU ⊥ PQ

Then RSPT is a parallelogram.

So, RT = SP = 20 cm and PT = SR = 25 cm

Therefore, TQ = PQ – PT = 46 cm – 25 cm = 21 cm

Area of the ∆RTQ = $$\sqrt{s(s - a)(s - b)(s - c)}$$

where s = $$\frac{\textrm{RT + TQ + QR}}{2}$$

= $$\frac{\textrm{20 + 21 + 13}}{2}$$ cm

= 27 cm

Now, plug the values in $$\sqrt{s(s - a)(s - b)(s - c)}$$.

= $$\sqrt{27(27 - 20)(27 - 21)(27 - 13)}$$ cm$$^{2}$$

= $$\sqrt{27 ∙ 7 ∙ 6 ∙ 14}$$ cm$$^{2}$$

= $$\sqrt{3 ∙ 3 ∙ 3 ∙ 7 ∙ 3 ∙ 2 ∙ 7 ∙ 2}$$ cm$$^{2}$$

= $$\sqrt{3^{2} ∙ 3^{2} ∙ 7^{2} ∙ 2^{2}}$$ cm$$^{2}$$

= 3 ∙ 3 ∙ 7 ∙ 2 cm$$^{2}$$

= 126 cm$$^{2}$$

Also, the area of the ∆RTQ = $$\frac{1}{2}$$ TQ × RU = $$\frac{1}{2}$$ × 21 cm × RU cm$$^{2}$$

Therefore, 126 cm$$^{2}$$ = $$\frac{1}{2}$$ × 21 cm × RU

or, RU = $$\frac{126 × 2}{21}$$ cm

or, RU = 12 cm

Therefore, the distance between the parallel sides = 12 cm

Therefore, area of the trapezium PQRS = $$\frac{1}{2}$$ × (SR + PQ) × RU

= $$\frac{1}{2}$$ × (25 + 46) × 12 cm$$^{2}$$

= $$\frac{1}{2}$$ × (25 + 46) × 12 cm$$^{2}$$

= $$\frac{1}{2}$$ × 71 × 12 cm$$^{2}$$

= $$\frac{852}{2}$$ cm$$^{2}$$

= 426 cm$$^{2}$$

Application on Perimeter and Area of Trapezium:

3. The shape of the cross section of a canal is a trapezium. If the canal is 10 m wide on the top and 6 m wide at the bottom, and the area of its cross section is 72 m2 then find the depth of the canal.

Solution:

The cross section is the trapezium PQRS in which PQ ∥ RS. Here PQ = 10 m, RS = 6 m, and area of the trapezium PQRS = 72 m2.

Let d be the depth of the canal.

Then, area of the trapezium PQRS = $$\frac{1}{2}$$(PQ + RS)d

⟹ 72 m= $$\frac{1}{2}$$(10 + 6) × d

⟹ d = $$\frac{72 × 2}{16}$$ m = 9 m

Therefore, the depth of the canal = 9 m.