Given:
[tex]m\angle ACD=45^\circ[/tex]
Point B is the center of the base.
Each side of the base = 10 cm
To find:
The height of the pyramid.
Solution:
It is given that the measure of each side of the base is 10 cm. It means the base is square. The length of the diagonal of a square is:
[tex]d=a\sqrt{2}[/tex]
Where, a is the side length of the square.
Putting [tex]a=10[/tex] in the above formula, we get
[tex]d=10\sqrt{2}[/tex]
It is given that point B is the center of the base. It means point B bisect each diagonal of a base. So,
[tex]BC=\dfrac{d}{2}[/tex]
[tex]BC=\dfrac{10\sqrt{2}}{2}[/tex]
[tex]BC=5\sqrt{2}[/tex]
In a right angle triangle,
[tex]\tan \theta=\dfrac{Perpendicular}{Base}[/tex]
In triangle ABC,
[tex]\tan (45^\circ)=\dfrac{AB}{BC}[/tex]
[tex]1=\dfrac{h}{5\sqrt{2}}[/tex]
[tex]5\sqrt{2}=h[/tex]
[tex]7.0710678=h[/tex]
Round the value to the nearest integer.
[tex]h\approx 7[/tex]
The height of the pyramid is about 7 cm. Therefore, the correct option is A.
