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
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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