Question

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​

Answer 1

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Answer 2

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 .