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What is the height of an isosceles trapezoid, if the lengths of its bases are 5m and 11m, and the length of a leg 4m.

Please show your work. Thank You!

Respuesta :

frika

Let EB and CF be two heights. Then quadrilateral EBCF is rectangle, so BC=EF=5 m. It is given that bases are BC=5 m and AD=11 m. Triangles ABE and DCF are congruent, thus,

[tex]AE=FD=\dfrac{AD-EF}{2}=\dfrac{AD-BC}{2}=\dfrac{11-5}{2}=3.[/tex]

Consider right triangle ABE, by the Pythagorean theorem

[tex]AB^2=EB^2+AE^2,\\ \\4^2=3^2+EB^2,\\ \\EB^2=16-9=7,\\ \\EB=\sqrt{7}\ m.[/tex]

Answer: [tex]\sqrt{7}\ m.[/tex]

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MrRoyal

An isosceles trapezoid has 2 parallel sides and equal non-parallel legs. The height of the isosceles trapezoid is 2.65 m

Because the trapezoid is isosceles, the 2 legs of the trapezoid will be equal.

I have added an attachment that represents the given trapezoid.

From the attachment, we have:

[tex]l = 4[/tex] --- length of the leg

[tex]b =3[/tex] ---- base of the triangle formed by the leg of the trapezoid

The height (h) is calculated using the Pythagoras theorem as follows:

[tex]l^2 =h^2 + b^2[/tex]

So, we have:

[tex]4^2 = h^2 + 3^2[/tex]

[tex]16 = h^2 + 9[/tex]

Collect like terms

[tex]h^2 = 16 - 9[/tex]

[tex]h^2 = 7[/tex]

Take square roots of both sides

[tex]h = 2.65[/tex]

Hence, the height of the trapezoid is 2.65 m

Read more about isosceles trapezoid at:

https://brainly.com/question/14239662

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