PQRS is a square whose side length is 20 cm. Both pairs of opposite vertices are joined (that is, both diagonals PR and QS are drawn), forming four congruent triangles inside the square. What is the sum of the perimeters of all four triangles?

Introduction / Context:
This problem explores basic properties of a square and the effect of drawing its diagonals. When both diagonals of a square are drawn, the square is divided into four congruent right isosceles triangles. The question asks for the sum of the perimeters of these four triangles. Such questions are useful for reinforcing understanding of symmetry, congruence, and the relationship between side lengths and diagonals.

Given Data / Assumptions:

• PQRS is a square.
• Each side of the square is 20 cm.
• Both diagonals PR and QS are drawn, meeting at the centre of the square.
• The diagonals divide the square into four congruent triangles.
• We must calculate the total of the perimeters of all four triangles.

Concept / Approach:
In a square with side length a, each diagonal has length a√2. When both diagonals are drawn, the square splits into four congruent right isosceles triangles. Each triangle has two legs equal to the side of the square (length a) and hypotenuse equal to half a diagonal. However, an easier way is to note that each small triangle still has legs along the sides of the square and hypotenuse equal to the full diagonal of the square. In fact, each triangle has sides 20 cm, 20 cm and 20√2 cm. Once we find the perimeter of one triangle, we can multiply by 4 to get the total perimeter of all four triangles.

Step-by-Step Solution:
Step 1: Note that the side length of the square PQRS is 20 cm. Step 2: Compute the diagonal of the square. For a square with side a, diagonal d = a√2. Here, d = 20√2 cm. Step 3: Each of the four triangles formed inside the square is right angled and congruent, with legs equal to the sides of the square. So, each triangle has sides 20 cm, 20 cm and 20√2 cm. Step 4: Find the perimeter of one triangle. Perimeter of one triangle = 20 + 20 + 20√2 = 40 + 20√2 cm. Step 5: Multiply by 4 to obtain the sum of the perimeters of all four triangles. Total perimeter = 4 * (40 + 20√2) = 160 + 80√2 cm. Step 6: Write it in the standard format: 80√2 + 160 cm.

Verification / Alternative check:
We can confirm the reasoning by visualizing or sketching the square. The diagonals of a square bisect each other at the centre and intersect at right angles, forming four identical right isosceles triangles. Each diagonal connects opposite vertices, and its length formula a√2 is standard. Since all four triangles share the same side lengths, multiplying the perimeter of one triangle by four is valid. If we approximate √2 ≈ 1.414, each triangle perimeter is approximately 40 + 20 * 1.414 ≈ 40 + 28.28 = 68.28 cm, and four times that is about 273.12 cm. Evaluating 80√2 + 160 ≈ 80 * 1.414 + 160 ≈ 113.12 + 160 = 273.12 cm, which matches and confirms our algebraic result.

Why Other Options Are Wrong:
Option 2: 80√2 + 80 cm is too small because it effectively uses half the correct constant part. Option 3: 40√2 + 160 cm is also smaller; it corresponds to only two triangle perimeters instead of four. Option 4: 40√2 + 80 cm is significantly less than the true total and mixes both parts incorrectly. Option 5: None of these is wrong because 80√2 + 160 cm appears among the options and is exactly the calculated total.

Common Pitfalls:
A common mistake is to treat each triangle as having a hypotenuse equal to half the diagonal or to mix up which sides are legs and which is the hypotenuse. Some students also forget to multiply the perimeter of one triangle by four and instead report only a single triangle perimeter. Others may miscalculate the length of the diagonal, using 2a instead of a√2. Carefully identifying side lengths, using the correct formula for the diagonal, and remembering that there are four congruent triangles avoids these issues.

Final Answer:
The sum of the perimeters of the four triangles is 80√2 + 160 cm.