Math, asked by sushi7452, 1 month ago

A field is in the shape of a trapezium whose parallel sides are 60m and 70m. The non parallel
sides are 25m and 26m. Find the area of the field

Answers

Answered by shikshu1
0

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Step-by-step explanation:

Draw BE || AD such that D - E - C

Draw BM DC such that D - E - M - C

ABED is parallelogram

= AD = BE = 13m

= AB = DE = 10m

= BC = 14m

= DC = DE + EC .... ( D - E - C )

Therefore EC = 25 - 10 = 15 m

⇒In ΔBEC

⇒In ΔBEC⇒2S=13+14+15

⇒In ΔBEC⇒2S=13+14+15⇒S=21 m

⇒In ΔBEC⇒2S=13+14+15⇒S=21 m⇒Area of △BEC

⇒In ΔBEC⇒2S=13+14+15⇒S=21 m⇒Area of △BEC⇒A=

⇒In ΔBEC⇒2S=13+14+15⇒S=21 m⇒Area of △BEC⇒A= s(s−a)(s−b)(s−c)

⇒In ΔBEC⇒2S=13+14+15⇒S=21 m⇒Area of △BEC⇒A= s(s−a)(s−b)(s−c)

⇒In ΔBEC⇒2S=13+14+15⇒S=21 m⇒Area of △BEC⇒A= s(s−a)(s−b)(s−c)

⇒In ΔBEC⇒2S=13+14+15⇒S=21 m⇒Area of △BEC⇒A= s(s−a)(s−b)(s−c) =

⇒In ΔBEC⇒2S=13+14+15⇒S=21 m⇒Area of △BEC⇒A= s(s−a)(s−b)(s−c) = 21(21−13)(21−14)(21−15)

⇒In ΔBEC⇒2S=13+14+15⇒S=21 m⇒Area of △BEC⇒A= s(s−a)(s−b)(s−c) = 21(21−13)(21−14)(21−15)

⇒In ΔBEC⇒2S=13+14+15⇒S=21 m⇒Area of △BEC⇒A= s(s−a)(s−b)(s−c) = 21(21−13)(21−14)(21−15)

⇒In ΔBEC⇒2S=13+14+15⇒S=21 m⇒Area of △BEC⇒A= s(s−a)(s−b)(s−c) = 21(21−13)(21−14)(21−15) =

⇒In ΔBEC⇒2S=13+14+15⇒S=21 m⇒Area of △BEC⇒A= s(s−a)(s−b)(s−c) = 21(21−13)(21−14)(21−15) = 21×8×7×6

⇒In ΔBEC⇒2S=13+14+15⇒S=21 m⇒Area of △BEC⇒A= s(s−a)(s−b)(s−c) = 21(21−13)(21−14)(21−15) = 21×8×7×6

⇒In ΔBEC⇒2S=13+14+15⇒S=21 m⇒Area of △BEC⇒A= s(s−a)(s−b)(s−c) = 21(21−13)(21−14)(21−15) = 21×8×7×6

⇒In ΔBEC⇒2S=13+14+15⇒S=21 m⇒Area of △BEC⇒A= s(s−a)(s−b)(s−c) = 21(21−13)(21−14)(21−15) = 21×8×7×6 =84 m

⇒In ΔBEC⇒2S=13+14+15⇒S=21 m⇒Area of △BEC⇒A= s(s−a)(s−b)(s−c) = 21(21−13)(21−14)(21−15) = 21×8×7×6 =84 m 2

⇒Area of ΔBCE=

⇒Area of ΔBCE= 2

⇒Area of ΔBCE= 21

⇒Area of ΔBCE= 21

⇒Area of ΔBCE= 21 ×BM×EC

⇒Area of ΔBCE= 21 ×BM×EC⇒BM=

⇒Area of ΔBCE= 21 ×BM×EC⇒BM= 15

⇒Area of ΔBCE= 21 ×BM×EC⇒BM= 1584×2

⇒Area of ΔBCE= 21 ×BM×EC⇒BM= 1584×2

⇒Area of ΔBCE= 21 ×BM×EC⇒BM= 1584×2

⇒Area of ΔBCE= 21 ×BM×EC⇒BM= 1584×2 =11.2 cm

⇒Area of ΔBCE= 21 ×BM×EC⇒BM= 1584×2 =11.2 cm⇒Area of □ABED=11.2×10=112 m

⇒Area of ΔBCE= 21 ×BM×EC⇒BM= 1584×2 =11.2 cm⇒Area of □ABED=11.2×10=112 m 2

⇒Area of ΔBCE= 21 ×BM×EC⇒BM= 1584×2 =11.2 cm⇒Area of □ABED=11.2×10=112 m 2

⇒Area of ΔBCE= 21 ×BM×EC⇒BM= 1584×2 =11.2 cm⇒Area of □ABED=11.2×10=112 m 2 ⇒Area of the field □ABCD

⇒Area of ΔBCE= 21 ×BM×EC⇒BM= 1584×2 =11.2 cm⇒Area of □ABED=11.2×10=112 m 2 ⇒Area of the field □ABCD=84+112 m

⇒Area of ΔBCE= 21 ×BM×EC⇒BM= 1584×2 =11.2 cm⇒Area of □ABED=11.2×10=112 m 2 ⇒Area of the field □ABCD=84+112 m 2

⇒Area of ΔBCE= 21 ×BM×EC⇒BM= 1584×2 =11.2 cm⇒Area of □ABED=11.2×10=112 m 2 ⇒Area of the field □ABCD=84+112 m 2

⇒Area of ΔBCE= 21 ×BM×EC⇒BM= 1584×2 =11.2 cm⇒Area of □ABED=11.2×10=112 m 2 ⇒Area of the field □ABCD=84+112 m 2 =196 m

⇒Area of ΔBCE= 21 ×BM×EC⇒BM= 1584×2 =11.2 cm⇒Area of □ABED=11.2×10=112 m 2 ⇒Area of the field □ABCD=84+112 m 2 =196 m 2

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