How to find the perimeter of a rectangle and an scalene triangle?

Question

The problem is as follows:

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&\textrm{21 cm}\\ 2.&\textrm{39 cm}\\ 3.&\textrm{28 cm}\\ 4.&\textrm{27 cm}\\ \end{array}$

I've attempted to follow a similar strategy of assigning values to the edges of the figures mentioned.

This was done as follows:

Since it was mentioned that the figure is a square then this means that the edges measure 6 cm.

Assuming that the inner sides of the scalene triangle are $a,\,b,\,c,\,d$ and $6$ (this starting from the left in clockwise direction).

This leads to:

$a+b+c+d+6=15$

The upper segment which has been cut by the triangle I'm assigning those as: $e,\,f,\,g$. (These starting from the left to the right).

Thus $e+f+g=6$

Up here the perimeter of the orange shaded region would be.

$3\times 6 + a+b+c+d = P$

Thus $a+b+c+d=9$

Therefore:

$P=18+9=27\,cm$

Therefore this should be the perimeter. But does it exist a better way than doing this?. Can it be obtained in a faster way?.

Answer 1

Note that the orange region's border contains three sides of the square (three quarters of the square's perimeter), plus the entire triangle's border, except for the bottom side of the square. Thus the answer is $\frac3424+15-\frac1424=\frac1224+15=12+15=27$.

Answer 2

You can simplify the calculations, at least, by thinking of it this way:

The perimeter of the orange shape consists of three sides of the square, along with two of the sides of the triangle. In particular, it consists of $$(\text{perimeter of triangle})+(\text{perimeter of square})-2\times (\text{side of square}).$$ So, the answer is $$15+24-2\cdot\frac{24}{4}=27\text{ cm}.$$