A field is in the shape …

196 m s^2

Given: AB || CD AB = 25m CD = 10m BC = 14m AD = 13m From point C draw CE || DA. Hence ADCE is a llgrm having CD II AE & ADIICE AE= CD= 10m CE= AD = 13m BE= AB- AE =25-10-15 m BE= 15m In ABCE BC= 14m, CE= 13m, BE = 15rr Semiperimeter (s) = (a + b + c) / 2 Semiperimeter * (s) = (14 + 13 + 15) / 2 s= 42/2= 21m s = 21m Area of ABCE=√ s(s-a)(s-b)(s-c) Area of ABCE=√ 21(21-14)(21-13)(21-15) Area of ABCE=√ 21x7x 8×6 Area of ABCE= √ 7×3× 7×2 4×2×2×3 Area of ABCE=√7×7×3×3×2×2×4 Area of ABCE= 7×3×2×2= 21x 4= 84m² Area of ABCE= 84m² Area of ABCE=√ 7×3× 7× 4×2×2×3 Area of ABCE=√7x7×3×3×2×2×4 Area of ABCE= 7×3×2×2= 21x 4= 84m² Area of BCE = 84m ^ 2 Area of Delta*B * C * E = 1/2 * b * a × altitude Area of Delta*B * C * E = 1/2 * B * E * C * L 84 = 1/2 * 15 * C * l 84 * 2 = 15CL 168 = 15CL CL = 168/15 CL = 56/5 * m Height of trapezium= 56/ 5m Area of trapezium= 1/2( sum of Il sides) (height) Area of trapezium=1/2(25+10)(56/5) Area of trapezium= 1/2(35)(56/5) Area of trapezium= 7x28=196m² Hence the area of field is 196m²

Area of the field is 196m^2