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\(^{2}\)

9th Grade Math

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