English, asked by kirankumari8336, 10 months ago

ABCD an isosceles trapezium its parallel side measure 13 cm and 35 cm its non parallel sides are equal and measure 10 centimetre find the area of the trapezium​

Answers

Answered by sachinBTS
2

Answer:

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Answered by TheProphet
5

S O L U T I O N : (Ques. Error)

\underline{\bf{Given\::}}

ABCD is an isosceles trapezium it's parallel side measure 13 cm & 25 cm, it's non - parallel sides are equal & measure 10 cm .

\underline{\bf{Explanation\::}}

Firstly, attachment a figure of Isosceles trapezium according to the question.

As we know that formula of the area of trapezium;

\boxed{\bf{Area=\frac{1}{2} \times (sum\:of\:base) \times height }}

A/q

We have,

  • AB = 13 cm
  • FC = 25 cm
  • AF = 10 cm
  • BC = 10 cm

So, ED = FC - AB

→ ED = 25 cm - 13 cm

ED = 12 cm

Therefore it's isosceles trapezium, so 2FE = ED

→ 2 FE =  12

→ FE = 12/2

→ FE = 6 cm

Now,

\underline{\underline{\tt{Using\:by\:Pythagoras\:theorem\::}}}

\mapsto\sf{(Hypotenuse)^{2} = (Base)^{2} + (Perpendicular)^{2}}

\mapsto\sf{(AF)^{2} = (FE)^{2} + (AE)^{2}}

\mapsto\sf{(10)^{2} = (6)^{2} + (AE)^{2}}

\mapsto\sf{100 = 36 + (AE)^{2}}

\mapsto\sf{(AE)^{2}  = 100 -36}

\mapsto\sf{(AE)^{2}  = 64}

\mapsto\sf{AE =\sqrt{64} }

\mapsto\bf{AE =8\:cm}

Now,

\mapsto\tt{Area \:of\:trapezium = \dfrac{1}{2} \times (Sum\:of\:base) \times height }

\mapsto\tt{Area \:of\:trapezium = \dfrac{1}{2} \times (13+25) \times 8}

\mapsto\tt{Area \:of\:trapezium = \dfrac{1}{\cancel{2}} \times 38 \times \cancel{8}}

\mapsto\tt{Area \:of\:trapezium = 38 \times 4}

\mapsto\bf{Area \:of\:trapezium = 152\:cm^{2}}

Thus,

The area of the trapezium will be 152 cm² .

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