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.
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