NCERT Solutions for Class 10 Maths: Triangles Exercise 6.6

Core Theorems for Solving Triangles Ex 6.6

The problems in Exercise 6.6 are not based on a single new concept. Instead, they are application-based questions that require you to strategically use theorems from the entire chapter. A strong foundation is key. Here are the most important tools you'll need:

1. Theorem on Areas of Similar Triangles: If two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides. So, if ΔABC ~ ΔPQR, then Ar(ABC)/Ar(PQR) = (AB/PQ)² = (BC/QR)² = (AC/PR)². This is crucial for problems comparing areas.

1. Pythagoras' Theorem and its Converse: In a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides. Many proofs in this exercise require you to find or construct right-angled triangles to apply this theorem. You will often need to create multiple equations using this theorem and then add or subtract them to reach the desired result.

1. Angle Bisector Theorem: The internal bisector of an angle of a triangle divides the opposite side in the ratio of the sides containing the angle. For a ΔABC where AD is the angle bisector of ∠A, this means BD/DC = AB/AC. This theorem is directly applied in a few specific problems.

1. Properties of Medians and Altitudes: Problems often involve medians and altitudes. Remember that a median connects a vertex to the midpoint of the opposite side. An altitude is a perpendicular from a vertex to the opposite side. Combining these with Pythagoras' Theorem is a common strategy.

How to Solve Proofs using Pythagoras' Theorem and Constructions

1. Step 1: Understand the Goal and Draw a Diagram — Read the problem carefully to identify the 'Given' information and the 'To Prove' statement. Draw a neat, labeled diagram. For example, let's prove that in ΔABC, if AD is a median to side BC, then AB² + AC² = 2(AD² + BD²). Your diagram should show ΔABC with median AD.
2. Step 2: Make a Construction — Notice that there are no right angles given. To use Pythagoras' Theorem, we must create them. Draw an altitude AE ⊥ BC. Now you have two right-angled triangles, ΔAEB and ΔAEC.
3. Step 3: Apply Pythagoras' Theorem — In right ΔABE, AB² = AE² + BE². (Equation 1) In right ΔACE, AC² = AE² + CE². (Equation 2)
4. Step 4: Add the Equations and Substitute — Add Equation 1 and Equation 2: AB² + AC² = 2AE² + BE² + CE². Now, express BE and CE in terms of BD, CD, and DE. Since AD is the median, BD = CD. BE = BD - DE CE = CD + DE = BD + DE Substitute these into the equation: AB² + AC² = 2AE² + (BD - DE)² + (BD + DE)².
5. Step 5: Simplify to Get the Final Result — Expand the terms: AB² + AC² = 2AE² + (BD² - 2·BD·DE + DE²) + (BD² + 2·BD·DE + DE²) AB² + AC² = 2AE² + 2BD² + 2DE² AB² + AC² = 2(AE² + DE²) + 2BD². Now, look at the right-angled triangle ΔADE. Here, AD² = AE² + DE². Substitute this back: AB² + AC² = 2(AD²) + 2BD². This proves the result.

Exam Alert: Common Pitfalls in Complex Triangle Proofs

These problems are tricky, and it's easy to make small mistakes. Watch out for these common errors:

• Forgetting to Square the Ratio: When using the theorem on areas of similar triangles, a very common mistake is to write Ar(Δ1)/Ar(Δ2) = Side1/Side2 instead of (Side1/Side2)². Always remember to square the ratio of the sides.

• Incorrect Correspondence in Similarity: When you state that two triangles are similar (e.g., ΔABC ~ ΔPQR), the order of vertices matters. AB corresponds to PQ, BC to QR, and ∠A to ∠P. A wrong correspondence will lead to incorrect ratios and a failed proof.

• Algebraic Mistakes: The proofs often involve adding or subtracting multiple equations derived from Pythagoras' theorem. Be very careful with signs and algebraic manipulations, especially when expanding expressions like (a+b)² or (a-b)².

• Hesitation to Construct: Some proofs are impossible without an auxiliary construction, like drawing a perpendicular (altitude) or extending a line. If you're stuck, ask yourself, "Can I create a right-angled triangle to use Pythagoras' Theorem?"

Practice Questions with Solutions

• Q: In ΔABC, AD is the bisector of ∠A, meeting BC at D. If AB = 6 cm, AC = 8 cm, and BD = 3 cm, find the length of CD. A: Step 1: By the Angle Bisector Theorem, the ratio of the sides containing the angle is equal to the ratio of the segments of the opposite side. So, AB/AC = BD/CD. Step 2: Substitute the given values: 6/8 = 3/CD. Step 3: Cross-multiply to solve for CD: 6 CD = 8 3 => 6 * CD = 24. Step 4: CD = 24/6 = 4 cm. Final answer: CD = 4 cm.
• Q: The areas of two similar triangles are 81 cm² and 49 cm² respectively. If the altitude of the first triangle is 4.5 cm, find the corresponding altitude of the second triangle. A: Step 1: For similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding altitudes. So, Area(Δ1)/Area(Δ2) = (h1/h2)². Step 2: Substitute the given values: 81/49 = (4.5/h2)². Step 3: Take the square root of both sides: √(81/49) = 4.5/h2 => 9/7 = 4.5/h2. Step 4: Cross-multiply: 9 h2 = 7 4.5 => 9 * h2 = 31.5. Step 5: h2 = 31.5/9 = 3.5 cm. Final answer: The altitude of the second triangle is 3.5 cm.
• Q: In an equilateral triangle ABC, D is a point on side BC such that BD = (1/3)BC. Prove that 9AD² = 7AB². A: Step 1: Draw altitude AE ⊥ BC. In an equilateral triangle, E is the midpoint of BC. Let side length be 'a'. So, AB = BC = AC = a. BE = a/2. Step 2: Given BD = (1/3)BC = a/3. Then DE = BE - BD = a/2 - a/3 = a/6. Step 3: In right ΔABE, AE² = AB² - BE² = a² - (a/2)² = a² - a²/4 = 3a²/4. Step 4: In right ΔADE, AD² = AE² + DE² = 3a²/4 + (a/6)² = 3a²/4 + a²/36. Step 5: Find common denominator: AD² = (27a² + a²)/36 = 28a²/36 = 7a²/9. Step 6: Multiply by 9: 9AD² = 7a². Since AB = a, 9AD² = 7AB². Final answer: 9AD² = 7AB².
• Q: D and E are points on the sides AB and AC respectively of ΔABC such that DE || BC. If AD:DB = 2:3 and area of ΔADE = 16 cm², find the area of trapezium DBCE. A: Step 1: Since DE || BC, ΔADE is similar to ΔABC (by AA similarity criterion). Step 2: The ratio AD:DB = 2:3 implies AD/AB = AD/(AD+DB) = 2/(2+3) = 2/5. Step 3: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. So, Area(ΔADE)/Area(ΔABC) = (AD/AB)² = (2/5)² = 4/25. Step 4: Substitute Area(ΔADE) = 16 cm²: 16/Area(ΔABC) = 4/25. Area(ΔABC) = (16 25) / 4 = 4 25 = 100 cm². Step 5: Area of trapezium DBCE = Area(ΔABC) - Area(ΔADE) = 100 cm² - 16 cm² = 84 cm². Final answer: Area of trapezium DBCE = 84 cm².

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