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,

$AE=FD=\dfrac{AD-EF}{2}=\dfrac{AD-BC}{2}=\dfrac{11-5}{2}=3.$

Consider right triangle ABE, by the Pythagorean theorem

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

Answer: $\sqrt{7}\ m.$

5 0

3 years ago

Find an equation of the hyperbola having foci at (-9,-4) and (7,-4) and vertices at (-6, -4) and (4.-4).

Veronika [31]

Answer:

-9,-4

Step-by-step explanation:

6 0

3 years ago

F(2)=7 into coordinate points

Lisa [10]

I don’t know how to do this type stuff so sorry hope someone els can help u tho

5 0

3 years ago