Question

29. The area of a trapezium with equal non-parallel sides is 168 m². If the lengths of
the parallel sides are 36m and 20m,find the length of each non-parallel side.​

Answer 1

Step-by-step explanation:

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Answer 2

Given :-

• Area of the trapezium = 168 m²
• Length of one parallel side = a = 36 m
• Length of other parallel side = b = 20 m

To Find :-

• Length of each non - parallel sides.

Solution :-

Draw perpendicular lines AE ⊥ FC and BD ⊥ FC.

We know that,

$\large\boxed{\underline{{\sf Area\:of\:the\:trapezium=\dfrac{1}{2}\times h\times(a+b) }}}$

Putting the values together,

$:\implies\sf{168=\dfrac{1}{2} \times h \times (36+20)}$

$:\implies\sf{168 \times \dfrac{1}{2} \times h \times 56}$

$:\implies\sf{168\times 2 =h \times 56}$

$:\implies\sf{336 =h \times 56}$

$:\implies\sf{h=\dfrac{336}{50} }$

$:\implies\sf{h=6 }$

∴ Height of the trapezium = 6 m.

→  AE = BD

Now,

➲  AB = ED = 20 m.

➲  FC = AE = 36 m.

➲  FE = DC = ?

Value of FE :-

$:\implies\rm{FE=\dfrac{FC-ED}{2} }$

$:\implies\rm{FE=\dfrac{36-20}{2} }$

$:\implies\rm{FE=\cancel\dfrac{16}{2} }$

$:\implies\rm{FE=8 }$

Therefore, FE = DC = 8.

Now,

We have to find the length of the non parallel sides.

According to the Question :-

Δ AEF is a right angled triangle.

By Pythagoras theorem :-

$:\implies\rm{AF^2=AE^2+FE^2 }$

$:\implies\rm{AF^2=6^2+8^2 }$

$:\implies\rm{AF^2=36+64}$

$:\implies\rm{AF^2=100 }$

$:\implies\rm{AF=\sqrt{100} }$

$:\implies\rm{AF=10\:m }$

Length of one parallel side of the trapezium, AF = 10 m.

∴ AF = BC