Question

Find the area of a field in the shape of a trapezium whose parallel sides are of lengths
48 m and 160 m and non-parallel sides of lengths 50 m and 78m.
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solve the question step by step​

Answer 1

$\underline{\sf{\green{ Question}}}$

• Find the area of a field in the shape of a trapezium whose parallel sides are of lengths 48 m and 160 m and non -parallel sides of lengths 50 m and 78m.

$\underline{\sf{\green{Step\ by\ step\ Explanation}}}$

Given

• PQRS is a trapezium
• Parallel sides = lengths 48 m and 160 m
• Non - parallel sides= lengths 50 m and 78m

To Find

• Area Of Field

Construction

• QA||PS
• PQ||SA

Opposite side are equal

Therefore PQAS Is a parallelogram

Now in parallelogram PQAS

PS=QA=50m

PQ=SA=48m

Now,

• AR=SR-SA
• AR=100m-48m
• AR=52m

Now in triangle AQR

• AQ=50m
• AR=52m
• QR=72m

$\sf{Semi- perimeter= \cfrac{Sum\ of\ all\ sides}{2}}$

$\sf{Semi- perimeter= \cfrac{50+78+52}{2}}$

$\sf{Semi- perimeter= \cfrac{180}{2}}$

$\sf{Semi- perimeter= 90}$

By using Heron's formula

$\pink{\sf Area\ of\ triangle\ AQR= \sqrt{S(S-A)(S-B)(S-C)}}$

S= Semi-perimeter

A,B and C are the sides of triangle AQR

Here A=50m , B= 52m and C=78m

$\sf Area\ of\ triangle\ AQR= \sqrt{90(90-50)(90-52)(90-78)}$

$\sf Area\ of\ triangle\ AQR= \sqrt{90×40×38×12}$

$\sf Area\ of\ triangle\ AQR= \sqrt{90×18240}$

$\sf Area\ of\ triangle\ AQR= \sqrt{1641600}$

$\sf Area\ of\ triangle\ AQR= 1281cm$

Now

$\sf Area of triangle AQR= 1/2 base × height$

$\sf 1281=1/2×52×Height$

$\sf Height=1281×2/52$

$\sf Height=2562/52$

$\sf ~~~Height=49 m$

$\\\\\sf{Area\ of\ parallelogram\ PQRS = 1/2 Sum\ of\ parallelogram\ Side× Height}$

$\sf{Area\ of\ parallelogram\ PQRS = 1/2 (100+48)× 49}$

$\sf{Area\ of\ parallelogram\ PQRS = 1/2 ×148× 49}$

$\sf{Area\ of\ parallelogram\ PQRS = 74× 49}$

$\sf{\red{Area\ of\ parallelogram\ PQRS = 3626m²}}$