Finding the sum of the areas of three squares from the sum of three other quadrilaterals.

Question

The sum of the areas of squares $X,$ $Y,$ and $Z$ is 112. Find the sum of the areas of squares $P,$ $Q,$ and $R.$

How can I form a system of equations to solve for the sum of the areas of squares P, Q, and R.Is there a better way to solve for the areas without using algebraic manipulations?

Answer 1

$$\text{Let}\quad p^2=P, q^2=Q, r^2= R;$$ $$ x^2= X, y^2=Y, z^2= Z\;; X+Y+Z=Q1=112; $$ where the smaller case letter represents side of the area of square with bigger case letter.

Let $\alpha $ is angle between $z$ and $y$ and by Law of Cosines

$$ X= Z+Y -2 zy \cos \alpha$$ $$ P= Z+Y +2 zy \cos \alpha$$

because between $(Z,Y)$ the vertically opposite angle is supplementary $( \pi- \alpha).$

Adding

$$ P+X = 2(Z+Y)\tag1$$ $$ X+Y+Z= Q1 \tag2 $$ From (1) and (2)

$$ P = 3(Y+Z) -Q1 \tag3 $$ By cyclic symmetry $$ Q= 3(Z+X )-Q1 \tag4 $$ $$ R= 3(X+Y)-Q1 \tag5 $$ Add (3), (4),(5) and simplify $$ P+Q+R= 6\; Q1- 3\; Q1 =3\; Q1= 3 (112) =336$$

Answer 2

1. Using the cosine rule (and some work), conclude that $ \sum x^2 = \sum 2 xy \cos \alpha$.
2. Show that $p^2 = y^2 + z^2 + 2 yz \cos \alpha$.
3. Hence conclude that $ \sum p^2 = 3 \sum x^2 $.