Question

1. Find the area of a trapezium shaped plot whose parallel sides are 25 m and 35 m and perpendicular distance
between them is 10 m.

​

Answer 1

Answer:

Construction:

⇒Draw BE∥AD such that D−E−C

⇒Draw BM⊥DC such that D−E−M−C

□ABED is parallelogram,

⇒AD=BE=13 m

⇒AB=DE=10 m

⇒BC=14 m

⇒DC=DE+EC ....(∵D−E−C)

∴EC=25−10

∴EC=15 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=

2

1

×BM×EC

⇒BM=

15

84×2

=11.2 cm

⇒Area of □ABED=11.2×10=112 m

2

⇒Area of the field □ABCD

=84+112 m

2

=196 m

2

solution

Answer 2

Step-by-step explanation:

Construction:

⇒Draw BE∥AD such that D−E−C

⇒Draw BM⊥DC such that D−E−M−C

□ABED is parallelogram,

⇒AD=BE=13 m

⇒AB=DE=10 m

⇒BC=14 m

⇒DC=DE+EC ....(∵D−E−C)

∴EC=25−10

∴EC=15 m

⇒In ΔBEC

⇒2S=13+14+15

⇒S=21 m

⇒Area of △BEC

$⇒a = \sqrt{s(s - a)(s - b )(s - c) }$

$⇒ \sqrt{21(21 - 13)(21 - 14)(21 - 15}$

$⇒ \sqrt{21 \times 8 \times 7 \times 6}$

=84m²

$⇒area \: of \: Δabce \: \frac{1}{2} \times bm \times ec$

$⇒bm = \frac{84 \times 2}{15}$

=11.2cm

⇒Area of □ ABED =11.2×10m²

⇒Area of the field □ABCD

=84+112 m²

=196m²

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