Math, asked by Cholebhature10, 8 months ago

A field is in the shape of a trapezium whose parallel sides are 25m and 10m. The non-parallel sides are 14m and 13m. Find the area of field.

Answers

Answered by Uriyella

Given :–

A field is in the shape of a trapezium whose parallel sides = 25m and 10m.
And the non-parallel sides are 14m and 13m.

To Find :–

Area of the field.

Solution :–

Draw BE || AD

But AB || DC [Given]

So,

ABED is a parallelogram.

BE = AD = 14m
DE = AD = 25m

⟹ DC + CE = 25

⟹ 10 + CE = 25

⟹ CE = 25 – 10

⟹ CE = 15m

Now,

Area of ∆BCE = $\sqrt{s(s - a)(s-b)(s-c)}$

Where,

s = $\sf \dfrac{a + b + c}{2}$

⟹ $\sf \dfrac{13 + 15 + 14}{2}$

⟹ $\sf \dfrac{42}{2}$

Cut the denominator (2) and the numerator (42) by 2, we obtain

⟹ 21

s = 21

⟹ $\dfrac{1}{2}$ × b × h = $\sqrt{21(21 - 13)(21 - 15)(21-14)}$

⟹ $\dfrac{1}{2}$ × CE × BM = $\sqrt{21(8)(6)(7)}$

⟹ $\dfrac{1}{2}$ × 15 × BM = $\sqrt{(3 \times 7) (2 \times 2 \times 2)(2 \times 3) \times 7}$

⟹ $\dfrac{1}{2}$ × 15 × BM = 3 × 2 × 2 × 7

⟹ BM = $\dfrac{ 3 \times 7 \times 2 \times 2}{15}$

Cut the denominator (3) and the numerator (15) by 3, we obtain

⟹ BM = $\dfrac{56}{5}$

⟹ BM = 11.2m

Now, we need to find area of field (trapezium).

Area of trapezium ABCD,

⟹ $\dfrac{1}{2}$ × (sum of parallel lines) × height

⟹ $\dfrac{1}{2}$ × ( AB + DC) × $\dfrac{56}{5}$

⟹ $\dfrac{1}{2}$ × (12 + 10) × $\dfrac{56}{5}$

⟹ $\dfrac{1}{2}$ × 35 × $\dfrac{56}{5}$

Cut the denominator (2) and the numerator (56) by 2, we obtain

⟹ 1 × 35 × $\dfrac{28}{5}$

Now, cut the 35 and the denominator (5) by 5, we obtain

⟹ 28 × 7

⟹ 196

Hence,

The area of the field is 196m².

Answered by jevelin

Answer:

Draw a line BE parallel to AD and draw a perpendicular BF on CD

Draw a line BE parallel to AD and draw a perpendicular BF on CDit can be observe that ABED is a parallelogram

Draw a line BE parallel to AD and draw a perpendicular BF on CDit can be observe that ABED is a parallelogramBE=AD=13m

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

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

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

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

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

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

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

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