A field is in shape of trapezium whose parallel sides are 25m and 10m. The non parallel sides are 14m and 13m . Find the area of field.
Answers
Answer:
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
Your answer is 196m^2.
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Answer:
Answer:Draw a line BE parallel to AD and draw a perpendicular BF on CD
Answer:Draw a line BE parallel to AD and draw a perpendicular BF on CDit can be observe that ABED is a parallelogram
Answer:Draw a line BE parallel to AD and draw a perpendicular BF on CDit can be observe that ABED is a parallelogramBE=AD=13m
Answer:Draw a line BE parallel to AD and draw a perpendicular BF on CDit can be observe that ABED is a parallelogramBE=AD=13mED=AB=10m
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Answer:Draw a line BE parallel to AD and draw a perpendicular BF on CDit can be observe that ABED is a parallelogramBE=AD=13mED=AB=10mEC=25-ED=15mfor ΔBECsemi perimeter S=13+14+15/2m
Answer:Draw a line BE parallel to AD and draw a perpendicular BF on CDit can be observe that ABED is a parallelogramBE=AD=13mED=AB=10mEC=25-ED=15mfor ΔBECsemi perimeter S=13+14+15/2m42/2=21
Answer:Draw a line BE parallel to AD and draw a perpendicular BF on CDit can be observe that ABED is a parallelogramBE=AD=13mED=AB=10mEC=25-ED=15mfor ΔBECsemi perimeter S=13+14+15/2m42/2=21By heron formula
Answer:Draw a line BE parallel to AD and draw a perpendicular BF on CDit can be observe that ABED is a parallelogramBE=AD=13mED=AB=10mEC=25-ED=15mfor ΔBECsemi perimeter S=13+14+15/2m42/2=21By heron formulaArea of Δ √s(s-a) (s-b) (s-c)
Answer:Draw a line BE parallel to AD and draw a perpendicular BF on CDit can be observe that ABED is a parallelogramBE=AD=13mED=AB=10mEC=25-ED=15mfor ΔBECsemi perimeter S=13+14+15/2m42/2=21By heron formulaArea of Δ √s(s-a) (s-b) (s-c)Area of ΔABEC= √21 (21-13)(21-14)(21-15)m²
Answer:Draw a line BE parallel to AD and draw a perpendicular BF on CDit can be observe that ABED is a parallelogramBE=AD=13mED=AB=10mEC=25-ED=15mfor ΔBECsemi perimeter S=13+14+15/2m42/2=21By heron formulaArea of Δ √s(s-a) (s-b) (s-c)Area of ΔABEC= √21 (21-13)(21-14)(21-15)m²[√21(8)(7)(6)]m²
Answer:Draw a line BE parallel to AD and draw a perpendicular BF on CDit can be observe that ABED is a parallelogramBE=AD=13mED=AB=10mEC=25-ED=15mfor ΔBECsemi perimeter S=13+14+15/2m42/2=21By heron formulaArea of Δ √s(s-a) (s-b) (s-c)Area of ΔABEC= √21 (21-13)(21-14)(21-15)m²[√21(8)(7)(6)]m²=84m²
Answer:Draw a line BE parallel to AD and draw a perpendicular BF on CDit can be observe that ABED is a parallelogramBE=AD=13mED=AB=10mEC=25-ED=15mfor ΔBECsemi perimeter S=13+14+15/2m42/2=21By heron formulaArea of Δ √s(s-a) (s-b) (s-c)Area of ΔABEC= √21 (21-13)(21-14)(21-15)m²[√21(8)(7)(6)]m²=84m²Area of Δ BEC =½x CE x BF
Answer:Draw a line BE parallel to AD and draw a perpendicular BF on CDit can be observe that ABED is a parallelogramBE=AD=13mED=AB=10mEC=25-ED=15mfor ΔBECsemi perimeter S=13+14+15/2m42/2=21By heron formulaArea of Δ √s(s-a) (s-b) (s-c)Area of ΔABEC= √21 (21-13)(21-14)(21-15)m²[√21(8)(7)(6)]m²=84m²Area of Δ BEC =½x CE x BF=84=12x15xBF
Answer:Draw a line BE parallel to AD and draw a perpendicular BF on CDit can be observe that ABED is a parallelogramBE=AD=13mED=AB=10mEC=25-ED=15mfor ΔBECsemi perimeter S=13+14+15/2m42/2=21By heron formulaArea of Δ √s(s-a) (s-b) (s-c)Area of ΔABEC= √21 (21-13)(21-14)(21-15)m²[√21(8)(7)(6)]m²=84m²Area of Δ BEC =½x CE x BF=84=12x15xBFBF=16815=11.2m
Answer:Draw a line BE parallel to AD and draw a perpendicular BF on CDit can be observe that ABED is a parallelogramBE=AD=13mED=AB=10mEC=25-ED=15mfor ΔBECsemi perimeter S=13+14+15/2m42/2=21By heron formulaArea of Δ √s(s-a) (s-b) (s-c)Area of ΔABEC= √21 (21-13)(21-14)(21-15)m²[√21(8)(7)(6)]m²=84m²Area of Δ BEC =½x CE x BF=84=12x15xBFBF=16815=11.2mArea of the field = 84+112=196m2
Answer:Draw a line BE parallel to AD and draw a perpendicular BF on CDit can be observe that ABED is a parallelogramBE=AD=13mED=AB=10mEC=25-ED=15mfor ΔBECsemi perimeter S=13+14+15/2m42/2=21By heron formulaArea of Δ √s(s-a) (s-b) (s-c)Area of ΔABEC= √21 (21-13)(21-14)(21-15)m²[√21(8)(7)(6)]m²=84m²Area of Δ BEC =½x CE x BF=84=12x15xBFBF=16815=11.2mArea of the field = 84+112=196m2Step-by-step explanation:
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