Question

What is the height of the pyramid ???

Answer 1

Given:

$m\angle ACD=45^\circ$

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:

$d=a\sqrt{2}$

Where, a is the side length of the square.

Putting $a=10$ in the above formula, we get

$d=10\sqrt{2}$

It is given that point B is the center of the base. It means point B bisect each diagonal of a base. So,

$BC=\dfrac{d}{2}$

$BC=\dfrac{10\sqrt{2}}{2}$

$BC=5\sqrt{2}$

In a right angle triangle,

$\tan \theta=\dfrac{Perpendicular}{Base}$

In triangle ABC,

$\tan (45^\circ)=\dfrac{AB}{BC}$

$1=\dfrac{h}{5\sqrt{2}}$

$5\sqrt{2}=h$

$7.0710678=h$

Round the value to the nearest integer.

$h\approx 7$

The height of the pyramid is about 7 cm. Therefore, the correct option is A.