Perimeter and Area of Trapezium

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

Perimeter and Area of Trapezium

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:

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.

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.

Distance between the Parallel Sides 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

From Perimeter and Area of Trapezium to HOME PAGE


New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.



Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.