In area of a trapezium we will discuss about the formula and the solved examples in area of a trapezium.
A trapezium is a quadrilateral having one pair of parallel opposite sides. In the given figure, ABCD is a trapezium in which AB ∥ DC.
Let ABCD be a trapezium in which AB ∥ DC, CE ⊥ AB, DF ⊥ AB and CE = DF = h.
Prove that:
Area of a trapezium ABCD = {¹/₂ × (AB + DC) × h} square units.
Proof: Area of a trapezium ABCD
= area (∆DFA) + area (rectangle DFEC) + area (∆CEB)
= (¹/₂ × AF × DF) + (FE × DF) + (¹/₂ × EB × CE)
= (¹/₂ × AF × h) + (FE × h) + (¹/₂ × EB × h)
= ¹/₂ × h × (AF + 2FE + EB)
= ¹/₂ × h × (AF + FE + EB + FE)
= ¹/₂ × h × (AB + FE)
= ¹/₂ × h × (AB + DC) square units.
= ¹/₂ × (sum of parallel sides) × (distance between them)
Formula of Area of a trapezium = ¹/₂ × (sum of parallel sides) × (distance between them)
1. Two parallel sides of a trapezium are of lengths 27 cm and 19 cm respectively, and the distance between them is 14 cm. Find the area of the trapezium.
Solution:
Area of the trapezium
= ¹/₂ × (sum of parallel sides) × (distance between them)
= {¹/₂ × (27 + 19) × 14} cm²
= 322 cm²
2. The area of a trapezium is 352 cm² and the distance between its parallel sides is 16 cm. If one of the parallel sides is of length 25 cm, find the length of the other.
Solution:
Let the length of the required side be x cm.
Then, area of the trapezium = {¹/₂ × (25 + x) × 16} cm²
= (200 + 8x) cm².
But, the area of the trapezium = 352 cm² (given)
Therefore, 200 + 8x = 352
⇒ 8x = (352 - 200)
⇒ 8x = 152
⇒ x = (152/8)
⇒ x = 19.
Hence, the length of the other side is 19 cm.
3. The parallel sides of a trapezium are 25 cm and 13 cm; its nonparallel sides are equal, each being 10 cm. Find the area of the trapezium.
Solution:
Let ABCD be the given trapezium in which AB = 25 cm, DC = 13 cm, BC = 10 cm and AD = 10 cm.
Through C, draw CE ∥ AD, meeting AB at E.
Also, draw CF ⊥ AB.
Now, EB = (AB - AE) = (AB - DC)
= (25 - 13) cm = 12 cm;
CE = AD = 10 cm; AE = DC = 13 cm.
Now, in ∆EBC, we have CE = BC = 10 cm.
So, it is an isosceles triangle.
Also, CF ⊥ AB
So, F is the midpoint of EB.
Therefore, EF = ¹/₂ × EB = 6cm.
Thus, in right-angled ∆CFE, we have CE = 10 cm, EF = 6 cm.
By Pythagoras’ theorem, we have
CF = [√CE² - EF²]
= √(10² - 6²)
= √64
= √(8 × 8)
= 8 cm.
Thus, the distance between the parallel sides is 8 cm.
Area of trapezium ABCD = ¹/₂ × (sum of parallel sides) × (distance between them)
= {¹/₂ × (25 + 13) × 8 cm²
= 152 cm²
4. ABCD is a trapezium in which AB ∥ DC, AB = 78 cm, CD = 52 cm, AD = 28 cm and BC = 30 cm. Find the area of the trapezium.
Solution:
Draw CE ∥ AD and CF ⊥ AB.
Now, EB = (AB - AE) = (AB - DC) = (78 - 52) cm = 26 cm,
CE = AD = 28 cm and BC = 30 cm.
Now, in ∆CEB, we have
S = ¹/₂ (28 + 26 + 30) cm = 42 cm.
(s - a) = (42 - 28) cm = 14 cm,
(s - b) = (42 - 26) cm = 16 cm, and
(s - c) = (42 - 30) cm = 12 cm.
area of ∆CEB = √{s(s - a)(s - b)(s - c)}
= √(42 × 14 × 16 × 12) cm²
= 336 cm²
Also, area of ∆CEB = ¹/₂ × EB × CF
= (¹/₂ × 26 × CF) cm²
= (13 × CF) cm²
Therefore, 13 × CF = 336
⇒ CF = 336/13 cm
Area of a trapezium ABCD
= {¹/₂ × (AB + CD) × CF} square units
= {¹/₂ × (78 + 52) × ³³⁶/₁₃} cm²
= 1680 cm²
● Area of a Trapezium
● Area of a Trapezium - Worksheet
Worksheet on Area of a Polygon
8th Grade Math Practice
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