A field is in the shape of a trapezium whose parallelsides are 25 m and 10 m. The non-parallel sidesare 14 m and 13 m. Find the area of the field.
Answers
Solution :-
( Refer to Image First .)
From image we have :-
→ AB = 10m. (Given).
→ DC = 25m . (Given).
→ AD = 14m. (Given).
→ BC = 13m. (Given).
Construction :-
→ Draw AE & BF that are Perpendicular to DC.
now,
→ Let DE = x m.
→ EF = AB = 10m
→ FC = 25 - (10 + x)
→ FC = (15 - x) m.
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Than, in Right ∆AED, we have ,
→ AE² = AD² - DE² (By Pythagoras Theoram)
→ AE² = (14)² - (x)² ------ Equation (1).
Similarly,
in Right ∆BFC, we have ,
→ BF² = BC² - FC² (By Pythagoras Theoram)
→ AE² = (13)² - (15-x)² ------ Equation (2).
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Since, AE = BF ( Height of Trapezium.)
Comparing Both Equations we get ,
→ (14)² - x² = (13)² - (15 - x)²
→ 196 - x² = 169 - (225 + x² - 30x)
→ 196 - x² = 169 - 225 - x² + 30x
→ 30x = 196 + 225 - 169
→ 30x = 252
→ x = 8.4 m.
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Therefore,
→ AE² = (14)² - x²
→ AE² = (14)² - (8.4)²
→ AE² = (14+8.4)(14 - 8.4)
→ AE² = 22.4 * 5.6
→ AE² = 4 * 5.6 * 5.6
→ AE² = 2² * 5.6²
→ AE² = (2*5.6)²
→ AE = 2*5.6
→ AE = 11.2 m.
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Hence,
→ Area of Trapezium ABCD = (1/2) * (sum of Parallel sides) * Height
→ Area of Trapezium ABCD = (1/2) * ( AB + DC) * AE
→ Area of Trapezium ABCD = (1/2) * ( 10 + 25) * 11.2
→ Area of Trapezium ABCD = (1/2) * 35 * 11.2
→ Area of Trapezium ABCD = 5.6 * 35
→ Area of Trapezium ABCD = 196m² (Ans.)
Hence, Area of Required Trapezium is 196m².
Step-by-step explanation:
Given:
AB || CD
AB =25 m
CD= 10m
BC=14m
AD= 13m
From point C draw CE || DA . Hence ADCE is a ||
grm having CD || AE & AD||CE
AE= CD= 10m
CE= AD= 13m
BE= AB- AE =25- 10=15 m
BE= 15m
In ∆BCE
BC= 14m, CE= 13m, BE= 15m
Semiperimeter (s)= (a+b+c)/2
Semiperimeter(s) =( 14+13+15)/2
s= 42/2= 21m
s= 21m
Area of ∆BCE= √ s(s-a)(s-b)(s-c)
Area of ∆BCE=√ 21(21-14)(21-13)(21-15)
Area of ∆BCE= √ 21×7× 8×6
Area of ∆BCE= √ 7×3× 7× 4×2×2×3
Area of ∆BCE=√7×7×3×3×2×2×4
Area of ∆BCE= 7×3×2×2= 21× 4= 84m²
Area of ∆BCE= 84m²
Area of ∆BCE= 1/2 × base × altitude
Area of ∆BCE= 1/2 × BE ×CL
84= 1/2×15×CL
84×2= 15CL
168= 15CL
CL= 168/15
CL= 56 /5m
Height of trapezium= 56/ 5m
Area of trapezium= 1/2( sum of || sides)×( height)
Area of trapezium=1/2(25+10)(56/5)
Area of trapezium= 1/2(35)(56/5)
Area of trapezium= 7×28= 196m²
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Hence the area of field is 196m².