Exercise 8.4 — Pythagoras Theorem

Exercise 8.4 — Pythagoras Theorem and Its Applications

Exercise 8.4 is the most extensive section of Chapter 8, Similar Triangles, in Class 10 Mathematics for students following the CBSE, Telangana, and Andhra Pradesh board syllabi. This exercise centers around the legendary Pythagoras Theorem — known in Indian mathematical tradition as the Baudhayana Theorem — and its converse, applying both to a wide variety of geometric shapes and real-life situations.

With 14 questions covering rhombuses, equilateral triangles, isosceles triangles, interior points, ladders, poles, and even equilateral triangles drawn on the sides of right triangles, this exercise is a complete workout in applying one central idea: the square of the hypotenuse equals the sum of squares of the other two sides.

The Two Supporting Theorems Behind This Exercise

Before the main exercise, the textbook proves two foundational results that almost every question in this exercise depends on:

• Altitude Theorem: If a perpendicular is drawn from the right-angle vertex to the hypotenuse, the two triangles formed are similar to the original triangle and to each other.
• Converse of Pythagoras Theorem: If the square of one side of a triangle equals the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

Question 1 — Sum of Squares of Sides of a Rhombus Equals Sum of Squares of Diagonals

This question asks us to prove that for any rhombus, the sum of the squares of its four sides equals the sum of the squares of its two diagonals. The key property used is that the diagonals of a rhombus bisect each other at right angles.

Question 2 — Proving AE² + CD² = AC² + DE² in a Right Triangle

ABC is right-angled at B, and D, E are arbitrary points on AB and BC respectively. We must show AE² + CD² = AC² + DE². The proof applies Pythagoras' Theorem to four separate right triangles inside the figure.

Question 3 — Relationship Between Side and Altitude of an Equilateral Triangle

This question proves that 3 times the square of a side of an equilateral triangle equals 4 times the square of its altitude. This relationship is extremely useful for quickly calculating the height or area of any equilateral triangle.

Question 4 — Geometric Mean Relation: PM² = QM × MR

PQR is right-angled at P, and M is a point on QR such that PM ⊥ QR. We must show PM² = QM · MR — a classic geometric mean relationship that arises from the altitude-on-hypotenuse theorem.

Question 5 — Three Geometric Mean Relations in a Right Triangle with Altitude

ABD is right-angled at A, with AC perpendicular to BD. This question asks for three separate relations, all following the same similar-triangles logic as Question 4, but applied to different pairs of corresponding sides.

Question 6 — Isosceles Right Triangle: AB² = 2AC²

ABC is isosceles and right-angled at C, meaning AC = BC. This is a direct, single-step application of the Pythagoras Theorem.

Question 7 — Interior Point Theorem in a Triangle

O is any point inside triangle ABC, with perpendiculars OD, OE, OF dropped onto BC, AC, AB respectively. This two-part question proves a beautiful symmetric relationship between the distances from O to the vertices and the segments created on each side.

Questions 8 & 9 — Real-Life Applications of Pythagoras Theorem

These two questions show how Pythagoras Theorem applies directly to everyday situations — a guy-wire attached to a pole, and the distance between the tops of two vertical poles.

Question 10 — Cevian Length in an Equilateral Triangle (9AD² = 7AB²)

In equilateral triangle ABC, point D lies on BC such that BD = (1/3)BC. We must prove 9AD² = 7AB². This question combines the altitude technique with careful algebraic manipulation.

Question 11 — Trisection Points on the Hypotenuse Side

ABC is right-angled at B, with D and E trisecting side BC (so BD = DE = EC = (1/3)BC). We must prove the relation 8AE² = 3AC² + 5AD².

Question 12 — Area Ratio of Similar Triangles Built on the Legs and Hypotenuse

ABC is isosceles and right-angled at B (so AB = BC). Similar triangles ACD and ABE are constructed externally on sides AC and AB. We must find the ratio of the areas of △ABE and △ACD — combining the Pythagoras Theorem with the area-ratio theorem from Exercise 8.3.

Question 13 — Equilateral Triangles Built on the Sides of a Right Triangle

Equilateral triangles are constructed on all three sides of a right-angled triangle. We must show that the area of the equilateral triangle built on the hypotenuse equals the sum of the areas of the equilateral triangles built on the other two sides — a beautiful generalization of the Pythagoras Theorem to areas of equilateral triangles instead of squares.

Question 14 — Equilateral Triangle on a Square's Side vs Its Diagonal

This question asks us to prove that the equilateral triangle built on the side of a square has exactly half the area of the equilateral triangle built on the square's diagonal.

Key Theorems Used in This Exercise

Common Mistakes to Avoid

• Misidentifying the hypotenuse: Always double check which side is opposite the right angle before writing the Pythagoras equation. Writing AB² = AC² + BC² instead of AC² = AB² + BC² (when ∠B = 90°) is a very common slip.
• Forgetting to prove similarity first: In questions like Q4 and Q5, you must explicitly establish the similar triangles using the altitude theorem before writing proportional side relationships — directly writing PM² = QM·MR without justification loses marks.
• Algebra errors when combining fractions: Questions like Q10 and Q11 require careful fraction subtraction (e.g., 1/2 − 1/3 = 1/6). Always find a common denominator carefully rather than estimating.
• Confusing "sum of squares of sides" with "square of sum": In Q1, the rhombus proof, students sometimes wrongly expand (OA + OB)² instead of keeping OA² and OB² as separate terms. Always apply Pythagoras Theorem to one triangle at a time.
• Using the wrong diagonal-to-side ratio: In Q14, remember the diagonal of a square is √2 times the side — not 2 times. Squaring √2 gives exactly 2, which is the key step in that proof.

Quick Reference — All Answers at a Glance

What This Exercise Prepares You For

Exercise 8.4 is the single most exam-critical section in Chapter 8 across Telangana SSC, AP SSC, and CBSE Class 10 boards, frequently contributing multiple 4-mark and 8-mark questions in a single paper. The Pythagoras Theorem proved here is foundational not just for geometry, but also for distance formulas in Coordinate Geometry and trigonometric identities in later chapters.

To strengthen the prerequisites used throughout this exercise, revisit Exercise 8.3 — Areas of Similar Triangles, since Q12 directly depends on the area-ratio theorem covered there. The Pythagoras Theorem itself also reappears in Coordinate Geometry, where it forms the basis of the distance formula between two points, and in Trigonometry, where it underlies the fundamental identity sin²θ + cos²θ = 1.