In one of my tutorial classes, when I was studdying in 9th class (I am in 10th now), our tutor gave us a problem saying it’s a difficult one, and to him, it was incomplete.

This is that problem:

In a rectangle ABCD, point B is joined to the points P and Q on the sides AD and DC respectively. And P and Q are joined to form three distinct triangle; namely PDQ or x, QCB or y, BAP or z and BPQ or f.

Given the area of x = 3m^2, y = 4m^2 and z = 5m^2, find the area of f.

Note: all the information regarding this problem has been provided. The dimensions of any of the line segments are not known. This also means that the exact positions of P and Q on their respective line segments are also not known. diagram

Because of the lack of information (See Note), my tutor declared it as an incomplete problem (he is just a high-school Science and Mathematics teacher, please don’t expect him to think beyond the limits of a typical high-school student, that is, in case he is wrong).

After a few hours of brainstorming, I came up with what I called an ingenious method. But then I felt that my method actually was not accurate.

So, can anybody tell me if this problem is complete and solvable, if yes, is my method correct (see below), if no, what is the best/better method to solve it.


As all the triangles with their areas given are on the edge of the rectangle, we know that they all are right triangles. So we can say that one of the arms of the right angle is the altitude and the other is the base.

Now, we can use the formula (1/2) * base * altitude to calculate their areas. Lets take a as altitude and b as base.

(Here’s the first part of the trick) We now try all possible _a_s and _b_s for the three triangles, where a and b are natural numbers only. (This is also where the major weakness of my method is. More numbers, i.e., any real numbers should have been tried, but that is infeasible). Note that all possible values are those values that form a triangle of that specified area.

For triangle x, we have the following dimensions: -a = 1, b = 6 -a = 2, b = 3 -a = 3, b = 2 -a = 6, b = 1 #

For triangle y, we have the following dimensions: -a = 1, b = 8 -a = 2, b = 4 -a = 4, b = 2# -a = 8, b = 1

For triangle z, we have the following dimensions: -a = 1, b = 10 -a = 2, b = 5 -a = 5, b = 2 -a = 10, b = 1#

(Here’s the other part of the trick) Then, we select those specific dimensions of triangles, which when placed together form a rectangle.

The condition we require: When the triangles are placed together to form a quadrilateral the opposite sides of it must be equal.

If this quadrilateral’s opposite sides are equal, we know that it is a parallelogram. And as all of the triangles that form the edges are right angled, we have at least one of the angles of the parallelogram right angled_, so we know that this parallelogram is a rectangle.

The correct set of dimensions (that we will be using here) are high-lighted in bold, while the once suffixed with a ‘#’ can also be used.

So, in triangle x we take PD as altitude and DQ as base. In triangle y we take BC as altitude and CQ as base. In triangle z we take PA as altitude and AB as base.

Henceforth, we now have: (PA + PD) = BC (DQ + CQ) = AB Therefore, the condition is matched.


Now we can either find the area of f by finding the area of the rectangle and subtracting it by the areas of the three triangles, or by finding the three triangle’s hypotenuse using the Pythagorean formula and then apply the _Herron’s formula _on the hypotenuses. I prefer the former.

>After calculating, we have, _f = 8m^2_

Thank you =)


6 Answers 6


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Let BC = 1 unit. Then, $4 = y = [\triangle CBQ] = \frac {1.QC}{2}$ yields QC = 8 units

Let DQ = s units. Then AB = s + 8 units

$5 = z = [\triangle ABP] = \frac {(s+8).AP}{2}$ yields $AP = \frac {10}{s + 8}$ units

$DP = 1 – AP = … = \frac {s – 2}{s + 8}$

Similarly, $3 = (\frac {1}{2}) {s}{\frac{ s – 2}{s + 8}}$.

The above result is a solvable quadratic with roots $s = 12$ or $–4$ (rejected)

Therefore, the rectangle occupies 20 square units and x, y, z occupy …. Giving f = ...

  • $\begingroup$ How do you justify the BC = 1 assertion? With that you show there is one solution but how do you know this is the only one? $\endgroup$ Apr 8, 2015 at 14:53
  • $\begingroup$ @GertBeukema This is called unitisation. I will use length as an example. I have the right to declare the length of an object (say 3.52 cm) as an equivalent to 1 (special) unit [defined by me]. Thus, another object which is 26.048 cm long is equal to 7.4 times of my (special) units. Also, I think the solution is unique (as the proof shows). Other possible candidate is the rejected one. $\endgroup$
    – Mick
    Apr 8, 2015 at 15:44
  • $\begingroup$ But there are other solutions, e.g. AB = 5, BC = 4 or 10 and 2 or 6 2/3 and 3. $\endgroup$ Apr 8, 2015 at 16:06
  • $\begingroup$ The problem I see with your unitization here is that you still keep using 3,4 and 5 as the areas, so these are now not unitized. With BC = 1 you are only looking at one specific case, which luckily for you has a solution. $\endgroup$ Apr 8, 2015 at 16:11
  • $\begingroup$ @GertBeukema Note that the question is asking what f is (rather than the dimension of the rectangle). Thus, your suggested candidates need not be distinguished. For you latest comment:- Choosing one particular portion to be unitized can do the job already. Hence, there is no need to unitize the area too. Besides, one has to be very careful in unitizing more than one dimension because conflict might occur. $\endgroup$
    – Mick
    Apr 8, 2015 at 16:28

Let BC = h and AB = k

QC = 8/h (since area BCQ = 4)

AP = 10/k (since area ABP = 5)

(h - 10/k)(k - 8/h) = 6 (since area DPQ = 3)

Now: k = 20/h or k = 4/h (h <> 0)

Since the area of the rectangle is k*h and larger than 12 we have to disregard the second solution.

Using the first solution we get 20 as the area of the rectangle and f = 8.

Since there only is one solution, I'd call the problem complete.

  • 1
    $\begingroup$ You should include the computation: $$ (h - 10/k)(k - 8/h) = 6 $$ $$ (hk-10)(hk-8)=6hk $$ $$ (hk)^2-24 hk+80=0 $$ $$ (hk-20)(hk-4)=0 $$ $\endgroup$
    – san
    Jun 26, 2015 at 21:06

Fantastic problem! Thanks for sharing. I solved it with Euler Math Toolbox (see below). However, I'd like to see a more geometric solution. Probably, I am just blind right now.

The method of the solution is straightforward. $xh$ is $x$ depending on $a$ and $b$ (computing the area as side times side divided by 2). $fh$ is $f$ depending on $a$, $b$, $z$ and $y$. Then, surprisingly, we can get rid of the total area $a*b=x+y+z+f$

>xh &= (b-2*z/a)*(a-2*y/b)/2

$${ (a - {{2y} \over b} ) (b - {{2z} \over a} )} \over 2$$

>fh &= expand(a*b-(xh+y+z))

$$ {{ab} \over 2} - {{2yz} \over {ab}}$$

>fhh &= subst(x+y+z+f,a*b,fh)

$$ {{z + y + x + f} \over 2} - {{ 2yz} \over {z + y + x + f}} $$

>&solve(f=fhh,f), function f(x,y,z) &= f with %[2]

$$ [ {f = -\sqrt {z^2 - 2yz + 2xz + y^2 + 2xy + x^2}, f = \sqrt {z^2 - 2yz + 2xz + y^2 + 2xy + x^2}} ] $$

$$ \sqrt {x^2 - 2yz +2xz + y^2 + 2xy + x^2} $$


$$ 8 $$


This problem is complete in that we are given all of the required information to solve it.

Let a = |AB| = |CD| and b = |BC| = |DA| since ABCD is a rectangle

Let |CQ| = qa so |QD| = (1 - q)a

Let |DP| = pb so |PA| = (1 - p)b

Given the areas of the 3 triangles we then have (after multiplying both sides by 2)

(1) 10 = (1 - p)ab from area of z is 5

(2) 8 = qab from area of y is 4

(3) 6 = (1-q)pab from area of x is 6

From (2) we have q = 8/ab and substituting into (3) we have p = 6(ab - 8)

Substituting into (1) we have 10 = [1 - (6/(ab - 8))]ab which can be rearranged to give the quadratic equation

(ab)^2 - 24ab + 80 = 0 or (ab - 4)(ab - 20) = 0

We can reject the root ab = 4 since this gives a negative area for triangle f, so we have [area of f] = 20 - 3 - 4 - 5 = 8

For extra points we could go on the solve for p and q, getting p = 1/2 and q = 2/5

  • $\begingroup$ Sorry, I missed a divide by sign in my answer, it should be p = 6/(ab - 8) not 6(ab - 8) $\endgroup$ Sep 27, 2016 at 17:03

My solution:

Let B = the length of line BC.
Let H = the length of line AB.
Then line AP = 4/z.
Then Line CQ = 8/y.
Let R = B*H = the area of the rectangle ABCD.
Then H = R/B.

Let R = B*H (area of the rectangle = Base * Height).

If you mirror the 3 outside triangles (z,x,y) about their hypotenuse to create smaller rectangles inside R and sum the areas of these 3 rectangles, you get the area of the rectangle (R), plus some overlap. The area of overlap is (z/H)*(y/B).

2*(z+x+y) - (z/H)(y/B) = BH


2*z+2*x+2*y-z*y/R = R

We how have 1 equation and 1 unknown R (where R = the area of ABCD). At this point the problem is effectively solved.

To continue to the final solution, convert the equation into a more conventional form by multiplying both sides by -R, then adding R^2 to both sides, then use the quadratic formula to solve for R.

R^2-2*R*x-2*R*y-2*Rz+yz = 0

Only the larger of the two solutions for R has physical significance. After simplification you get:

f := sqrt( (z+x+y)^2 - 4*(z*y) ) <==============================

Plug in any values you want for x, y, and z; and get the area f.

Note: This solution does not require the knowledge of any lengths.


Let $P = (0, 2), Q = (3, 0)$ and $B = (a,b)$ to force $x = 3$. Using the shoelace formula, the area of $\Delta PQB$ is $\frac{1}{2}(3b+2a-6)$, which is equal to the area of the rectangle minus the three smaller triangles, or $ab - (3+4+5)$.

Thus $3b+2a-6 = 2ab - 24$, or $2ab - 2a - 3b - 18 = 0$. This factors as $(2a-3)(b-1) = 21$, and as $B$ must have a larger $x$-coordinate than $Q$, and a larger $y$-coordinate than $P$, the only option is the pair $(7, 3)$ which gives $a = 5, b = 4$. Thus the area is $\frac{1}{2}(3b+2a-6) = \boxed{8}$.


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