Question

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

Answer 1

$\large\bf\underline{Given:-}$

• Parellel sides of trapezium are 25m and 10m
• non parellel sides are 14m and 10m

$\large\bf\underline {To \: find:-}$

• Area of trapezium.

$\huge\bf\underline{Solution:-}$

Let PQRS be the given trapezium in which:-

• PQ = 25m
• RS = 10m
• PS = 14m
• QR = 10m

• PH is perpendicular to RS

Now,

• PA = SR

»» AQ = PQ - PA

»» AQ = 25 - 10

»» AQ = 15m

AND,

»» AB = 1/2 × AQ

»» AB = 1/2 × 15

»» AB = 7.5m

IN right angled triangle ∆RAB

• AR = 14m
• AB = 7.5

• BY PYTHAGORAS THEOREM

»» RA² = AB² + RB²

»» 196 = 56.25 + RB²

»» RB² = 196 - 56.25

»» RB² = 139.75

»» RB = √139.75

»» RB = 11.82

So the distance between parellel sides is 11.82m

Area of trapezium = 1/2(sum of || sides )× height

Area of trapezium = 1/2(25+10)×11.82

Area of trapezium = 1/2 × 35 × 11.82

Area of trapezium = 17.5 × 11.82

• »★ Area of trapezium = 206.85m²

Answer 2

$\sf\red{\underline{\underline{Answer:}}}$

$\sf{Area \ of \ trapezium \ is \ 157.5 \ m^{2}}$

$\sf\orange{Given:}$

$\sf{In \ a \ rectangular \ field,}$

$\sf{\implies{Parallel \ sides \ are \ 25 \ m \ and \ 10 \ m}}$

$\sf{\implies{Non-parallel \ sides \ are \ 14 \ m \ 10 \ m}}$

$\sf\pink{To \ find:}$

$\sf{Area \ of \ the \ field.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{Construction:}$

$\sf{In \ trapezium \ ABCD,}$

$\sf{1. \ Draw \ BE \ parallel \ to \ AD.}$

$\sf{2. \ Draw \ BF \ perpendicular \ to \ CD.}$

__________________________________

$\sf{In \ \triangle \ ECB,}$

$\sf{By \ heron's \ formula}$

$\sf{s=\frac{14+15+10}{2}}$

$\sf{\therefore{s=\frac{39}{2}=19.5}}$

$\sf{Area \ of \ triangle=\sqrt{s(s-a)(s-b)(s-c)}}$

$\sf{=\sqrt{19.5(19.5-14)(19.5-15)(19.5-10)}}$

$\sf{=\sqrt{19.5\times5.5\times4.5\times9.5}}$

$\sf{=\sqrt{4584.937}(approx)}$

$\sf{\therefore{Area \ of \ triangle=67.7 \ m^{2} (approx)}}$

$\sf{Area \ of \ triangle=\frac{1}{2}\times \ base\times \ height}$

$\sf{\therefore{67.7=\frac{1}{2}\times15\times \ BF}}$

$\sf{\therefore{BF=\frac{67.7}{7.5}}}$

$\sf{\therefore{BF=9 \ m(approx)}}$

$\boxed{\sf{Area \ of \ trapezium=\frac{1}{2}\times(Sum \ of \ \parallel \ sides)\times \ (Height)}}$

$\sf{\therefore{Area \ of \ trapezium=\frac{1}{2}\times(10+25)\times9}}$

$\sf{\therefore{Area \ of \ trapezium=35\times4.5}}$

$\sf{\therefore{Area \ of \ trapezium=157.5 \ m^{2}}}$

$\sf\purple{\tt{\therefore{Area \ of \ trapezium \ is \ 157.5 \ m^{2}}}}$