Question

the parallel sides of a trapezium is 77m and 60m and its non parallel sides are 25m find the area of trapezium ?

Answer 1

$\red{\large\underline{\underline\mathtt{Question:}}}$

The parallel sides of a trapezium is 77m and 60m and its non parallel sides are 25m . Find the area of trapezium ?

$\purple{\large\underline{\underline\mathtt{Given:}}}$

• Parallel side $\rightarrow p_{1} = 60m$
• Parallel side $\rightarrow p_{2} = 77m$
• Non - parallel equal side = $25 m$

$\green{\large\underline{\underline\mathtt{We\: know:}}}$

• Pythagoras theorem

$\blue{\mathrm{\boxed{\blue{h^{2} = b^{2} + p^{2}}}}}$

where,

1. h = Hypotenuse
2. b = base
3. p = height

• Area of a trapezium:-

$\purple{\mathrm{\boxed{\blue{\rightarrow \dfrac{1}{2} \times (sum\:of\:parallel\:sides) \times height}}}}$

$\blue{\large\underline{\underline\mathtt{Concept:}}}$

By seeing the above figure ∆ADE , we can conclude that the figure formed is as right-angled triangle .

Since it is a right-angled triangle ,we can use the

Pythagoras theorem to find its height.

By Pythagoras theorem,

$\mathrm{h^{2} = b^{2} + p^{2}}$

$\mathrm{\Rightarrow 25^{2} = \bigg({\dfrac{17}{2}\bigg)^{2} + p^{2}}}$

$\mathrm{\Rightarrow 25^{2} - \bigg({\dfrac{17}{2}\bigg)^{2} = p^{2}}}$

$\mathrm{\Rightarrow \sqrt{25^{2} - \bigg({\dfrac{17}{2}\bigg)^{2}} = p}}$

$\mathrm{\Rightarrow \sqrt{625 - \dfrac{289}{4}} = p}$

$\mathrm{\Rightarrow \sqrt{552.75} = p}$

$\mathrm{\purple{\Rightarrow 23.5= p}}$

$\red{{\boxed{\purple{\therefore Height = 23.5m}}}}$

$\red{\large\underline{\underline\mathtt{Solution:}}}$

Given,

• Parallel side $\rightarrow p_{1} = 60m$
• Parallel side $\rightarrow p_{2} = 77m$
• Height $\rightarrow h = 23.5$

By the formula ,

$\mathrm{\dfrac{1}{2} \times (sum\:of\:parallel\:sides) \times height}$

$\mathrm{\Rightarrow \dfrac{1}{2} \times (p_{1} + p_{2}) \times h}$

$\mathrm{\Rightarrow \dfrac{1}{2} \times (60 + 77) \times 23.5}$

$\mathrm{\Rightarrow \dfrac{1}{2} \times 137 \times 23.5}$

$\mathrm{\Rightarrow \dfrac{1}{\cancel{2}} \times 137 \times \cancel{23.5}}$

$\mathrm{\Rightarrow 137 \times 11.75}$

$\mathrm{\purple{\Rightarrow 1610.5m^{2}}}$

$\red{{\boxed{\purple{\therefore Area = 1610.75 m^{2}}}}}$

$\blue{\large\underline{\underline\mathtt{Extra\: Information:}}}$

• Area of square = side × side
• Area of reactangle = length × breadth
• Area of Equilateral triangle = $\dfrac{\sqrt{3}a^{2}}{4}$

______________________________________

Answer 2

$\pink{\large\underline{\underline\mathtt{Question:-}}}$

The parallel sides of a trapezium is 77m and 60m and its non parallel sides are 25m . Find the area of trapezium ?

$\purple{\large\underline{\underline\mathtt{Given:-}}}$

Parallel side$\rightarrow p_{1} = 60m→p$

Parallel side$\rightarrow p_{2} = 77m→p$

Non - parallel equal side = 25 m25m

$\green{\large\underline{\underline\mathtt{We\: know:}}}$

Pythagoras theorem

$\blue{\mathrm{\boxed{\blue{h^{2} = b^{2} + p^{2}}}}}$

where,

h = Hypotenuse

b = base

p = height

Area of a trapezium:-

$\purple{\mathrm{\boxed{\blue{\rightarrow \dfrac{1}{2} \times (sum\:of\:parallel\:sides) \times height}}}}$

→

2

1

×(sumofparallelsides)×height

$\blue{\large\underline{\underline\mathtt{Concept:}}}$

By seeing the above figure ∆ADE , we can conclude that the figure formed is as right-angled triangle .

Since it is a right-angled triangle ,we can use the

Pythagoras theorem to find its height.

By Pythagoras theorem,

$\mathrm{h^{2} = b^{2} + p^{2}}$

$\mathrm{\Rightarrow 25^{2} = \bigg({\dfrac{17}{2}\bigg)^{2} + p^{2}}}$

$\mathrm{\Rightarrow 25^{2} - \bigg({\dfrac{17}{2}\bigg)^{2} = p^{2}}}$

$\mathrm{\Rightarrow \sqrt{25^{2} - \bigg({\dfrac{17}{2}\bigg)^{2}} = p}}$

$\mathrm{\Rightarrow \sqrt{625 - \dfrac{289}{4}} = p}$

$\mathrm{\Rightarrow \sqrt{552.75} = p}$

$\mathrm{\purple{\Rightarrow 23.5= p}}$

$\red{{\boxed{\purple{\therefore Height = 23.5m}}}}$

$\red{\large\underline{\underline\mathtt{Solution:}}}$

Given,

Parallel sid$\rightarrow p_{1} = 60m→p$

=60m

Parallel side$\rightarrow p_{2} = 77m→p$

=77m

Height$\rightarrow h = 23.5→h=23.5$

By the formula ,

$\mathrm{\dfrac{1}{2} \times (sum\:of\:parallel\:sides) \times height}$

$\mathrm{\Rightarrow \dfrac{1}{2} \times (p_{1} + p_{2}) \times h}$

$\mathrm{\Rightarrow \dfrac{1}{2} \times (60 + 77) \times 23.5}$

$\mathrm{\Rightarrow \dfrac{1}{2} \times 137 \times 23.5}$

$\mathrm{\Rightarrow \dfrac{1}{\cancel{2}} \times 137 \times \cancel{23.5}}$

$\mathrm{\Rightarrow 137 \times 11.75}$

$\mathrm{\purple{\Rightarrow 1610.5m^{2}}}$

$\red{{\boxed{\purple{\therefore Area = 1610.75 m^{2}}}}}$

$\blue{\large\underline{\underline\mathtt{Extra\: Information:}}}$

Area of square = side × side

Area of reactangle = length × breadth

Area of Equilateral triangle = $\dfrac{\sqrt{3}a^{2}}{4}$

______________________________________