# Square and Its Properties
A square is a special type of quadrilateral, which is any four-sided figure. It is a regular polygon with four equal sides and four equal angles (90-degree angles, or right angles). Every square is a rectangle (because it has four right angles) and a rhombus (because it has four equal sides) and therefore it is a parallelogram (as its opposite sides are parallel and equal in length).
Here are the key properties of a square:
1. All four sides are of equal length, denoted typically as ‘a’.
2. All four interior angles are equal to 90 degrees, making the square a regular quadrilateral.
3. The diagonals of a square are equal in length and bisect each other at right angles, i.e., they are perpendicular bisectors of each other.
4. The diagonals also bisect the angles from which they are drawn, meaning they cut the square’s angles in half.
5. The diagonal of a square divides it into two congruent isosceles right-angled triangles.
6. The square has the largest area of all rectangles with a given perimeter.
7. The length of the diagonal ‘d’ can be found using the Pythagorean theorem, which in the case of a square is given by \( d = a\sqrt{2} \).
8. The perimeter ‘P’ of the square is given by \( P = 4a \).
9. The area ‘A’ of the square is given by \( A = a^2 \).
### Problems and Solutions about Square and Its Properties
**Problem 1:** Given a square with a side length of 5 cm, find its perimeter.
**Solution:** Using the formula for the perimeter of a square \( P = 4a \), we substitute the given side length to get \( P = 4 \times 5 cm = 20 cm\).
**Problem 2:** Calculate the area of a square whose side length is 8 inches.
**Solution:** The area of the square is given by \( A = a^2 \). Therefore, \( A = 8^2 = 64 \) square inches.
**Problem 3:** What is the length of the diagonal for a square with a side of 3 meters?
**Solution:** The length of the diagonal is given by \( d = a\sqrt{2} \). Thus, \( d = 3\sqrt{2} \approx 4.24 \) meters.
**Problem 4:** The perimeter of a square is 24 mm. What is the length of one side?
**Solution:** We use the perimeter formula \( P = 4a \) and solve for \( a \): \( a = \frac{P}{4} = \frac{24 mm}{4} = 6 mm \).
**Problem 5:** If the diagonal of a square is 10√2 cm, what is its area?
**Solution:** We first find the side using the diagonal formula inverted: \( a = \frac{d}{\sqrt{2}} = \frac{10\sqrt{2}}{\sqrt{2}} = 10 \) cm. Then \( A = a^2 = 10^2 = 100 \) square cm.
**Problem 6:** A square has an area of 121 sq ft. What is the length of its diagonal?
**Solution:** Find the side length: \( a = \sqrt{A} = \sqrt{121} = 11 \) ft. Then find the diagonal: \( d = a\sqrt{2} = 11\sqrt{2} \) ft.
**Problem 7:** The diagonal of a square is twice the length of its side. Is this statement true or false?
**Solution:** This statement is false. The diagonal is \( \sqrt{2} \) times the length of the side, not twice.
**Problem 8:** If the side of a square is doubled, by how much does the perimeter increase?
**Solution:** The original perimeter is \( P = 4a \). Doubling the side gives a new perimeter of \( P = 4(2a) = 8a \). This is twice the original perimeter.
**Problem 9:** Can a square have a diagonal length of 5 cm and a side length of 4 cm?
**Solution:** No, this is not possible due to the diagonal formula \( d = a\sqrt{2} \). If \( a = 4 \) cm, then \( d \) must be \( 4\sqrt{2} \), not 5 cm.
**Problem 10:** Given a square with a perimeter of 40 meters, find the length of the diagonal.
**Solution:** First, find the side length: \( a = \frac{P}{4} = \frac{40}{4} = 10 \) meters. Then find the diagonal: \( d = a\sqrt{2} = 10\sqrt{2} \) meters.
**Problem 11:** A square has a side length of ‘a’. Express the area of the square in terms of ‘a’.
**Solution:** The area is \( A = a^2 \).
**Problem 12:** If the area of a square is 169 cm², what is the perimeter?
**Solution:** Find the side length first: \( a = \sqrt{A} = \sqrt{169} = 13 \) cm. The perimeter is \( P = 4a = 4 \times 13 cm = 52 cm \).
**Problem 13:** The diagonal of a square is 16 cm long. What is the side length?
**Solution:** We have \( d = a\sqrt{2} \), so \( a = \frac{d}{\sqrt{2}} = \frac{16}{\sqrt{2}} = 8\sqrt{2} \) cm.
**Problem 14:** A square has the same area as a rectangle of 4 cm by 9 cm. What is the side length of the square?
**Solution:** Area of rectangle = Area of square, \( 4 \times 9 = a^2 \), so \( a^2 = 36 \) and \( a = 6 \) cm.
**Problem 15:** Calculate the side length of a square whose diagonal is 14 cm long.
**Solution:** Use \( a = \frac{d}{\sqrt{2}} = \frac{14}{\sqrt{2}} \approx 9.89 \) cm.
**Problem 16:** The ratio of the perimeter of two squares is 2:3. If the smaller square has a side of 5m, what is the side of the larger square?
**Solution:** Let the side of the larger square be \( x \). We have \( \frac{4 \times 5}{4 \times x} = \frac{2}{3} \), therefore, \( x = \frac{2}{3} \times 5 = \frac{10}{3} \) m.
**Problem 17:** Can a square have an area of 50 sq cm and a perimeter of 40 cm?
**Solution:** No. If \( A = a^2 = 50 \) then \( a \approx 7.07 \) cm. For this side length, perimeter would be \( P = 4a \approx 28.28 \) cm, not 40 cm.
**Problem 18:** A square and a circle have the same perimeter. If the square’s side length is 8 cm, what is the radius of the circle?
**Solution:** The perimeter of the square is \( P = 4 \times 8 = 32 \) cm. For the circle, \( P = 2\pi r \), so \( r = \frac{P}{2\pi} = \frac{32}{2\pi} \approx 5.09 \) cm.
**Problem 19:** How long is the diagonal of a square with a perimeter of 32 inches?
**Solution:** The side length is \( a = \frac{P}{4} = \frac{32}{4} = 8 \) inches. The diagonal is \( d = a\sqrt{2} = 8\sqrt{2} \approx 11.31 \) inches.
**Problem 20:** If a square’s side length is increased by 2 inches and its new area is 144 sq inches, find its original side length.
**Solution:** Let the original side length be \( x \). The new side length is \( x + 2 \). So \( (x + 2)^2 = 144 \), then \( x + 2 = 12 \) or \( x + 2 = -12 \) (rejecting the negative value since it is not meaningful for length). Thus, \( x = 12 – 2 = 10 \) inches.
These problems provide practice in using the properties of squares to solve a variety of algebraic and geometric problems. Remember, the key to solving these problems effectively is understanding the fundamental properties and characteristic equations for squares.
