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sarkhan2018

In this attached picture according to the conditions of the problem we have an isosceles trapezoid and since we know that legs are equal (AD=BC=5 cm), we have to calculate bases and height in order to find the area. Working with the triangle BCD, we apply Pythagoras theorem and find that CD = [tex] \sqrt{75+25} [/tex] = 10 cm. Since BDC is a right triangle, applying theorem for the area of triangles, we find that [tex] \frac{1}{2} * BF = \frac{1}{2} * 5 * \sqrt{75} [/tex] and BF= [tex] 0.5\sqrt{75} [/tex]. Since ABCD is an isosceles trapezoid, triangles ADE and BFC are congruent with Angle Side Angle theorem. Then, DE=FC and with the help of Pythagoras theorem, DE=FC=2.5 cm. Then, AB=EF=5 cm and the area of the trapezoid is  [tex]A= BF * \frac{AB+CD}{2} = 0.5 \sqrt{75} * \frac{5+10}{2} = 18.75 \sqrt{3} cm^{2} [/tex]

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ashishdwivedilVT

The area of the given trapezoid will be [tex]\frac{75}{4} \sqrt{3}[/tex].

It is given that

In trapezoid ABCD

AE ⊥ CD

AC = √75

AD=BC =5

AB =5

In the triangle ADC

From Pythagoras theorem

AC²+AD²=DC²

(√75)² +5² = DC²

DC=10

Let DE=CF=x (Isosceles trapezoid)

What is isosceles Trapezoid?

Trapezoid in which the base angles are equal and therefore the left and right side lengths are also equal.

AB=EF= DC-2x

5 = DC-2x

x=5/2

AE = [tex]\sqrt({5^{2} } -(5/2)^2)\\\\[/tex]

AE = [tex]\sqrt{75} /2[/tex]

So, the area of the trapezoid ABCD = [tex]\frac{1}{2} *(AB+CD)*AE[/tex]

Area of the trapezoid ABCD = [tex]\frac{1}{2} *(5+10)*\frac{\sqrt{75} }{2}[/tex]

Area of the trapezoid ABCD = [tex]\frac{75}{4} \sqrt{3}[/tex]

Therefore, The area of the given trapezoid ABCD will be [tex]\frac{75}{4} \sqrt{3}[/tex].

To get more about trapezoid visit:

https://brainly.com/question/1410008

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