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 m² = $\frac{1}{2}$(10 + 6) × d

⟹ d = $\frac{72 × 2}{16}$ m = 9 m

Therefore, the depth of the canal = 9 m.

9th Grade Math

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