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The Baudhayana-Pythagoras Theorem - Halving a Square

Grade 8CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Baudhayana-Pythagoras theorem provides a method to construct a square whose area is exactly half of a given square by connecting the midpoints of the adjacent sides.

A large square with a smaller square inscribed inside it by joining the midpoints of its sides.
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If the side of the original square is ss, its area is s2s^2. The side of the new square formed by joining the midpoints is s2\frac{s}{\sqrt{2}}, resulting in an area of (s2)2=s22(\frac{s}{\sqrt{2}})^2 = \frac{s^2}{2}.

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According to Baudhayana's Sulba Sutras, the diagonal of a square produces an area twice as large as the original square. Conversely, the side of a square is the diagonal of a square with half the area.

A square showing side s and diagonal d.
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This geometric halving is a practical application of the Baudhayana-Pythagoras theorem (a2+b2=c2a^2 + b^2 = c^2) where aa and bb are the half-lengths of the original square's sides.

📐Formulae

a2+b2=c2a^2 + b^2 = c^2

Areaoriginal=s2Area_{original} = s^2

Areahalved=s22Area_{halved} = \frac{s^2}{2}

Diagonal=s2Diagonal = s\sqrt{2}

Sidenew=Areaoriginal2Side_{new} = \sqrt{\frac{Area_{original}}{2}}

💡Examples

Problem 1:

A square has a side length of 10 cm10\text{ cm}. Find the area of the square formed by joining the midpoints of its sides.

Solution:

  1. Area of the original square: Area=102=100 cm2Area = 10^2 = 100\text{ cm}^2.
  2. According to the property of halving a square, the area of the inner square is half of the original.
  3. Areainner=1002=50 cm2Area_{inner} = \frac{100}{2} = 50\text{ cm}^2.

Explanation:

By joining the midpoints, we effectively create four right-angled triangles at the corners. Each triangle has a base and height of 5 cm5\text{ cm} (half the side). Using the Baudhayana-Pythagoras theorem, the side of the inner square is 52+52=50\sqrt{5^2 + 5^2} = \sqrt{50}. The area is thus (50)2=50(\sqrt{50})^2 = 50.

Problem 2:

Calculate the difference in area between a square of side 12 cm12\text{ cm} and a square formed by halving it through its midpoints.

Solution:

Area of outer square: 12×12=144 cm212 \times 12 = 144\text{ cm}^2 Area of inner square: 1442=72 cm2\frac{144}{2} = 72\text{ cm}^2 Difference in area: 144−7272\begin{array}{r} 144 \\ -72 \\ \hline 72 \end{array} The difference is 72 cm272\text{ cm}^2.

Explanation:

The area of the square formed by joining midpoints is always half of the parent square. Subtracting half from the whole leaves the other half as the difference.

Problem 3:

Find the length of the diagonal of a square whose side is 5 cm5\text{ cm}.

Solution:

Using the Baudhayana-Pythagoras Theorem for the diagonal dd: d2=52+52d^2 = 5^2 + 5^2 d2=25+25d^2 = 25 + 25 d2=50d^2 = 50 d=50=52 cmd = \sqrt{50} = 5\sqrt{2}\text{ cm} Using 2≈1.414\sqrt{2} \approx 1.414: d≈5×1.414=7.07 cmd \approx 5 \times 1.414 = 7.07\text{ cm}

Explanation:

The diagonal of a square acts as the hypotenuse of a right-angled triangle where the legs are the sides of the square.

Problem 4:

A square plot has a side of 14 m14\text{ m}. A smaller square garden is designed inside it by joining the midpoints of the sides of the plot. Calculate the side length of this garden.

Square with side 14 and inner square with vertices at midpoints.

Solution:

Side of original square (s)=14 m\text{Side of original square (s)} = 14\text{ m} Side of new square=s2\text{Side of new square} = \frac{s}{\sqrt{2}} Side=142=14×22=72 m\text{Side} = \frac{14}{\sqrt{2}} = \frac{14 \times \sqrt{2}}{2} = 7\sqrt{2}\text{ m} Using 2≈1.414:\text{Using } \sqrt{2} \approx 1.414: Side≈7×1.414=9.898 m\text{Side} \approx 7 \times 1.414 = 9.898\text{ m}

Explanation:

When midpoints of a square with side ss are joined, the side of the inner square forms the hypotenuse of a right triangle with legs s2\frac{s}{2}. Using a2+b2=c2a^2 + b^2 = c^2, we get (142)2+(142)2=Side2(\frac{14}{2})^2 + (\frac{14}{2})^2 = Side^2, which simplifies to Side=49+49=98=72Side = \sqrt{49+49} = \sqrt{98} = 7\sqrt{2}.

Problem 5:

The area of a square formed by joining the midpoints of a larger square is 32 cm232\text{ cm}^2. What is the side length of the larger square?

Diagram showing an inner square with area 32 and an outer square with side s.

Solution:

Area of inner square=32 cm2\text{Area of inner square} = 32\text{ cm}^2 Area of larger square=2×Area of inner square\text{Area of larger square} = 2 \times \text{Area of inner square} Area of larger square=2×32=64 cm2\text{Area of larger square} = 2 \times 32 = 64\text{ cm}^2 Side of larger square=64=8 cm\text{Side of larger square} = \sqrt{64} = 8\text{ cm}

Explanation:

Since the inner square (joining midpoints) has exactly half the area of the outer square, we multiply the inner area by 2 to find the outer area. The side length is then the square root of that area.