Squares and Square Roots | A

Question 1

Cut out two identical squares of paper. Draw, label, and cut as follows:

Now place the pieces 5, 6, 7, and 8 around Square 1 to get a square with double the area.

Question diagram 1
Check your answer with HomiSolve it yourself, then let Homi check your steps and spot mistakes.
Solution
Understand the Question
  • Let Square 1 have side length ss and area s2s^2.
  • An identical square (Square 2) is divided along its diagonals into 44 congruent right-angled isosceles triangles numbered 5,6,7,5, 6, 7, and 88.
  • Each triangle has an area equal to s24\dfrac{s^2}{4} and a hypotenuse equal to ss.
  • Placing these 44 triangles along the four edges of Square 1 adds an area of 4×s24=s24 \times \dfrac{s^2}{4} = s^2, resulting in a combined area of s2+s2=2s2s^2 + s^2 = 2s^2, which is exactly double the area of Square 1.

Step 1 · Calculate the Area and Dimensions of the Pieces

Let ss be the side length of Square 1. Area of Square 1=s2\text{Area of Square 1} = s^2

Square 2 is identical to Square 1 and cut into four identical right-angled isosceles triangles (5,6,7,85, 6, 7, 8). Each triangle has legs of length s2\dfrac{s}{\sqrt{2}} and hypotenuse ss.

Area of one triangle=12×(s2)×(s2)=12×s22=s24\begin{aligned} \text{Area of one triangle} &= \dfrac{1}{2} \times \left(\dfrac{s}{\sqrt{2}}\right) \times \left(\dfrac{s}{\sqrt{2}}\right) \\[0.6em] &= \dfrac{1}{2} \times \dfrac{s^2}{2} \\[0.6em] &= \dfrac{s^2}{4} \end{aligned}

Total area when pieces 5,6,7,5, 6, 7, and 88 are added to Square 1:

Total Area=Area(Square 1)+Area(5)+Area(6)+Area(7)+Area(8)=s2+4×(s24)=s2+s2=2s2\begin{aligned} \text{Total Area} &= \text{Area(Square 1)} + \text{Area}(5) + \text{Area}(6) + \text{Area}(7) + \text{Area}(8) \\[0.6em] &= s^2 + 4 \times \left(\dfrac{s^2}{4}\right) \\[0.6em] &= s^2 + s^2 \\[0.6em] &= 2s^2 \end{aligned}

Side length of the new square (SnewS_{\text{new}}):

Snew2=2s2Snew=2s2=s2\begin{aligned} S_{\text{new}}^2 &= 2s^2 \\[0.6em] S_{\text{new}} &= \sqrt{2s^2} = s\sqrt{2} \end{aligned}

Step 2 · Arrange the Pieces to Form the Larger Square

Diagram 1

Place the hypotenuse (length ss) of each triangle (5,6,7,85, 6, 7, 8) along the four outer edges of Square 1.

The legs of the adjacent triangles align along the outer boundary to form each side of the new square:

Side length of large square=s2+s2=2s2=s2\begin{aligned} \text{Side length of large square} &= \dfrac{s}{\sqrt{2}} + \dfrac{s}{\sqrt{2}} \\[0.6em] &= \dfrac{2s}{\sqrt{2}} \\[0.6em] &= s\sqrt{2} \end{aligned}

Since all four sides are equal to s2s\sqrt{2} and meet at right angles, the resulting figure is a square of area (s2)2=2s2(s\sqrt{2})^2 = 2s^2.

Answer

Placing the hypotenuse of each of the 44 triangular pieces (5,6,7,85, 6, 7, 8) along the four edges of Square 1 forms a new square with side length s2s\sqrt{2} and area 2s22s^2 (double the area of Square 1).

Common Mistakes
  • Side-Matching Error: Attempting to attach the legs of the triangles to Square 1 instead of their hypotenuses. Since the side of Square 1 is ss, only the hypotenuse (length ss) matches its edges.
  • Doubling Dimensions vs. Doubling Area: Assuming that doubling the area doubles the side length. Doubling the side length quadruples the area (4s24s^2), whereas doubling the area scales the side length by a factor of 2\sqrt{2} (s2s\sqrt{2}).

More questions in A

Q1

Cut out two identical squares of paper. Draw, label, and cut as follows:

Now place the pieces 5, 6, 7, and 8 around Square 1 to get a square with double the area.

Q2

Halving a Square Using Paper

Cut out a square from a piece of paper. Now make a square whose area is half the area of the first square.

Q3

Will the square having half the sidelength have half the area? Why not? How many such squares will fill the original square?

Fold the square paper inward, as shown, such that the crease lines pass through the midpoints of the sides. PQRSPQRS is the required square with half the area.

Q4

There are 3 closed boxes—one containing only red balls, the second containing only blue balls and the third containing only green balls. The boxes are labelled RED, BLUE and GREEN such that ‘no’ box has the correct label. We need to find which label goes with which box. How can this be done if we are allowed to open only one box?

← Back to Squares and Square Roots