# Triangles inside a square

I have a question with a figure of Triangle inside a square. The base of the triangle is on the base of the square and the peak of the triangle touches the top of the square.It then asks the ratio of the area of triangle to the area of the square.

According to the book the answer is 1/2. But i am confused since i believe i needed more information. Would the ratio be any different if the alignment or (type) of the triangle is changed (ie currently it looks like an isosceles what happens if it was a right angle isosceles inside the square would the result be any different) I cant help but feel that i am missing some key concept here .

• A more interesting question is to find the triangle ABC having the optimal area, where A,B,C are points on the square or find the triangle ABC having the optimal perimeter, where A,B,C are points on the square. – Paracosmiste Aug 5 '13 at 21:54

I assume that the base of the triangle is the entire bottom side of the square. Then it doesn’t matter where the top vertex of the triangle is along the top edge of the square: the area of the triangle will be half the area of the square.

Suppose that the sides of the square have length $s$. Clearly the area of the square is $s^2$. The base of the triangle is $s$, and the height is also $s$: it’s the distance from the base of the square to the top edge of the square. Thus, the area of the triangle is $$\frac12\cdot\text{base}\cdot\text{height}=\frac12s^2\;,$$ half the area of the square.

In fact, the top vertex of the triangle could be anywhere on the straightline containing the top edge of the square, even beyond the edges of the square, and the triangle would still have base $s$, height $s$, and area $\frac12s^2$. In the figure below, for instance, triangles $\triangle ABC,\triangle ABD,\triangle ABE,\triangle ABF$, and $\triangle ABG$ all have area $\frac12s^2$, because they all have base $s$ and height $s$.

The following argument works if the top of the triangle is on the top side of the square. We reuse the nice picture by Brian M. Scott. Actually, $ABFD$ does not need to be a square, it can be any rectangle. Make a paper rectangle $ABFP$.

Drop a perpendicular from $E$ to $P$ on $AB$. Then $EB$ splits rectangle $EPBF$ into two congruent parts.

Similarly, $EA$ splits rectangle $EPAD$ into two congruent parts.

Use scissors to cut along lines $EB$ and $EA$. The triangles $EBF$ and $EAD$ can be assembled to make a copy of $\triangle ABE$. So the area of $ABFD$ is twice the area of $\triangle ABE$.

Remark: We do not even need to assume that $ABFD$ is a rectangle. Suppose it is a parallelogram. Draw a line through $E$ parallel to $BF$, meeting $AB$ at $P$, and use the same argument as in the solution above.

• However correct me if i am wrong the tip of the triangle has to touch somewhere on the parallel side of the base. right ? for the height – Rajeshwar Jun 8 '12 at 14:47
• @Rajeswahr: The base of the triangle has to be $AB$. The tip $E$ of the triangle has to lie on the opposite side $DF$. For my argument, $E$ has to be between $D$ and $F$. For the argument of Brian M. Scott, it can be on the line $DF$, even to the right of $F$ or the left of $D$. But you were only asked for $E$ between $D$ and $F$. (My argument can be adapted to $E$ not between $D$ and $F$, but it becomes less simple.) – André Nicolas Jun 8 '12 at 15:08