Question

In the triangular solid shown in figure, points $A$ and $B$ are vertices. The triangular faces are isosceles. The solid has a height of $12$, a length of $21$, and a width of $18$. Calculate the distance between $A$ and $B$.

• A. $9\sqrt{5}$
• B. $3\sqrt{58}$
• C. $3\sqrt{74}$
• D. $3\sqrt{85}$
• E. $6\sqrt{11}$

Answer 1

The correct option is C $3\sqrt{74}$

Given that the height is $12$ and width is $21$.
The altitude from $A$ on base will cut the base into two equal halfs because the triangle is isosceles.
Therefore the length of an equal side of isosceles triangle is $\sqrt{{12}^2+9^2}=\sqrt{144+81}=\sqrt{225}=15$
Now the distance from $A$ to $B$ is $\sqrt{{15}^2+{21}^2}=\sqrt{225+441}=\sqrt{666}=3\sqrt{74}$