Question

What is the slant height x of this square pyramid?

Answer 1

Since this is a square pyramid, the length of all the sides going up towards the top point of the pyramid (the apex) are all 4m. The triangular faces of the square pyramid are all equilateral triangles because we are given that one of the angles is 60°.

Since we are just looking for x, we can ignore all the other sides of the square pyramid except for the front face with x and 60°. That front face is an equilateral triangle. We see x splitting the equilateral triangle into two smaller triangles. These smaller triangles are all special 30-60-90 triangles (see picture), so you can use your rules for the side lengths of the triangles to figure out the length of x. 

In a 30-60-90 triangle, the sides are in a 1:2:√3 ratio for the short leg to the long leg to the hypotenuse. That means the long leg is the short leg times 2 and the hypotenuse is the short leg times √3. 

We know the hypotenuse is 4m and we are trying to find the long leg, x. We know that the ratio of the hypotenuse to the long leg is √3:2 and we can cross multiply and solve for x, then put x into simplest radical form:
$\frac{hypotenuse}{long \: leg} = \frac{ \sqrt{3}}{2} \\ \frac{4}{x} = \frac{ \sqrt{3}}{2} \\ 8 = x\sqrt{3}\\ x = \frac{8}{\sqrt{3}} \\ x = \frac{8}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\\ x= \frac{8\sqrt{3}}{3}$

Your final answer is $x= \frac{8\sqrt{3}}{3}$.

Answer 2

Answer:  The slant height of the given square pyramid is 2√3 m.

Step-by-step explanation:  We are given to find the slant height of the square pyramid shown in the figure.

As modified in the attached figure, the pyramid consists of four congruent equilateral triangles with side length 4 m and a square base of side length 4 m.

ΔABC is one of the four triangles.

We are to find the value of x.

In ΔABC, AB = BC = CA = 4 m  and  AD ⊥ BC, so D will be the mid-point of BC because any altitude of an equilateral triangle divides the opposite side into two equal parts.

That is,

$BD=DC=\dfrac{1}{2}\times BC=\dfrac{1}{2}\times4=2.$

Now, from the right-angled triangle ABD, we have

$AB^2=BD^2+AD^2~~~~~\textup{[Using Pythagoras theorem]}\\\\\Rightarrow AD^2=AB^2-BD^2\\\\\Rightarrow x^2=4^2-2^2\\\\\Rightarrow x=\sqrt{16-4}\\\\\Rightarrow x=\sqrt{12}\\\\\Rightarrow x=2\sqrt3.$

Thus, the slant height of the given square pyramid is 2√3 m.