Math, asked by pashajahangeer1975, 10 months ago

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

Answered by Anonymous
6

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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Answered by jevelin
2

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

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=15m

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 ΔBEC

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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by \: eric \: taeyong

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