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The Baudhayana-Pythagoras Theorem - Further Applications of the Baudhāyana-Pythagoras Theorem

Grade 8CBSE

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

🔑Concepts

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The Baudhāyana-Pythagoras theorem states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. This can be visualized as the area of a square built on the longest side being equal to the combined areas of squares built on the legs.

Right-angled triangle with sides a, b and hypotenuse c.
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In a rectangle, the diagonal divides the rectangle into two identical right-angled triangles. The length of the diagonal dd can be calculated using the length ll and breadth ww of the rectangle as d=l2+w2d = \sqrt{l^2 + w^2}.

Rectangle with length l, width w, and diagonal d.
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For an isosceles triangle, the altitude (height) drawn from the vertex between equal sides to the base bisects the base. This creates two right-angled triangles where the hypotenuse is the equal side ss, one leg is the height hh, and the other leg is half the base b2\frac{b}{2}.

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Pythagorean triplets are sets of three positive integers (a,b,c)(a, b, c) that satisfy a2+b2=c2a^2 + b^2 = c^2. A common way to generate these for any integer m>1m > 1 is using the forms 2m2m, m2−1m^2 - 1, and m2+1m^2 + 1.

📐Formulae

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

d=l2+w2d = \sqrt{l^2 + w^2}

Altitude of isosceles triangle h=s2−(b2)2\text{Altitude of isosceles triangle } h = \sqrt{s^2 - \left(\frac{b}{2}\right)^2}

Pythagorean Triplet for m>1:(2m,m2−1,m2+1)\text{Pythagorean Triplet for } m > 1: (2m, m^2 - 1, m^2 + 1)

💡Examples

Problem 1:

A ladder 13 m13\text{ m} long reaches a window which is 12 m12\text{ m} above the ground on side of a wall. Find the distance of the foot of the ladder from the wall.

Solution:

Let the distance of the foot of the ladder from the wall be xx. The ladder, the wall, and the ground form a right-angled triangle where the ladder is the hypotenuse. According to the Baudhāyana-Pythagoras theorem: x2+122=132x^2 + 12^2 = 13^2 x2+144=169x^2 + 144 = 169 x2=169−144x^2 = 169 - 144 169−14425\begin{array}{r} 169 \\ -144 \\ \hline 25 \end{array} x2=25x^2 = 25 x=25=5 mx = \sqrt{25} = 5\text{ m}.

Explanation:

We used the theorem a2+b2=c2a^2 + b^2 = c^2 where c=13c=13 (ladder) and b=12b=12 (wall height). Solving for aa gives the distance on the ground.

Problem 2:

Find the length of the diagonal of a rectangle whose length is 15 cm15\text{ cm} and breadth is 8 cm8\text{ cm}.

Solution:

In a rectangle, the diagonal dd, length ll, and breadth ww satisfy the relation: d2=l2+w2d^2 = l^2 + w^2 d2=152+82d^2 = 15^2 + 8^2 d2=225+64d^2 = 225 + 64 225+64289\begin{array}{r} 225 \\ +64 \\ \hline 289 \end{array} d=289d = \sqrt{289} d=17 cmd = 17\text{ cm}.

Explanation:

The diagonal divides the rectangle into two right-angled triangles. By applying the theorem to one of these triangles, we find the diagonal length.

Problem 3:

Write a Pythagorean triplet whose smallest member is 66.

Solution:

Using the general form (2m,m2−1,m2+1)(2m, m^2 - 1, m^2 + 1): Let 2m=62m = 6 m=3m = 3 Then, m2−1=32−1=9−1=8m^2 - 1 = 3^2 - 1 = 9 - 1 = 8 And, m2+1=32+1=9+1=10m^2 + 1 = 3^2 + 1 = 9 + 1 = 10 Check: 62+82=36+64=100=1026^2 + 8^2 = 36 + 64 = 100 = 10^2. The triplet is (6,8,10)(6, 8, 10).

Explanation:

We identify mm from the given even number and use the standard algebraic identities for Pythagorean triplets to find the other two numbers.

Problem 4:

A wire of length 25 m25\text{ m} is attached to the top of a vertical pole of height 24 m24\text{ m}. How far from the base of the pole should the other end of the wire be anchored to the ground so that the wire is taut?

A right triangle representing a pole, a wire, and the ground distance.

Solution:

Let hh be the height of the pole, LL be the length of the wire (hypotenuse), and xx be the distance from the base. Given: h=24 mh = 24\text{ m}, L=25 mL = 25\text{ m}. Using the Baudhāyana-Pythagoras theorem: x2+h2=L2x^2 + h^2 = L^2 x2+242=252x^2 + 24^2 = 25^2 x2+576=625x^2 + 576 = 625 x2=625−576x^2 = 625 - 576 x2=49x^2 = 49 x=49=7 mx = \sqrt{49} = 7\text{ m} The wire should be anchored 7 m7\text{ m} from the base.

Explanation:

The pole, the ground, and the wire form a right-angled triangle where the wire is the hypotenuse. We solve for the unknown base using the square root of the difference between the squares of the hypotenuse and the height.

Problem 5:

An isosceles triangle has equal sides of 10 cm10\text{ cm} each and a base of 12 cm12\text{ cm}. Calculate the altitude of the triangle.

Isosceles triangle with an altitude dividing the base into two equal parts.

Solution:

In an isosceles triangle, the altitude hh bisects the base bb. Base b=12 cmb = 12\text{ cm}, so half the base is 122=6 cm\frac{12}{2} = 6\text{ cm}. The equal side s=10 cms = 10\text{ cm} acts as the hypotenuse for the right triangle formed by the altitude. h2+62=102h^2 + 6^2 = 10^2 h2+36=100h^2 + 36 = 100 h2=100−36h^2 = 100 - 36 h2=64h^2 = 64 h=64=8 cmh = \sqrt{64} = 8\text{ cm} The altitude of the triangle is 8 cm8\text{ cm}.

Explanation:

By drawing an altitude, we split the isosceles triangle into two right-angled triangles. We then use the Pythagoras theorem on one of these triangles with a base of 6 cm6\text{ cm} and hypotenuse of 10 cm10\text{ cm}.