CBSE Class 10 Maths: Triangles Exercise 6.3 - Understanding Similarity Criteria

What are Similar Triangles and Why Do They Matter?

Similar triangles are triangles that have the same shape but not necessarily the same size. Think of a photograph and its enlargement – the objects in both are the same shape but one is bigger than the other. This concept is fundamental in geometry, allowing us to find unknown lengths or angles in figures that are related. Unlike congruent triangles, where all corresponding sides and angles are equal, similar triangles only require corresponding angles to be equal and corresponding sides to be in proportion.

Understanding similar triangles is vital for various real-world applications, from architecture and engineering (scaling models) to optics (how lenses focus light) and even map-making. In CBSE Class 10, mastering the criteria for similarity is crucial for solving problems in Chapter 6 and other geometry-related chapters. We will explore how to confidently apply these criteria to prove similarity and then use the property of proportional sides to find missing values.

The Three Key Similarity Criteria

AA (Angle-Angle) Similarity Criterion

If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar. The third angle will automatically be equal. This is the most frequently used criterion.

SSS (Side-Side-Side) Similarity Criterion

If the corresponding sides of two triangles are proportional, then their corresponding angles are equal, and hence the two triangles are similar. Remember to check that the ratios of all three pairs of corresponding sides are equal.

SAS (Side-Angle-Side) Similarity Criterion

If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar. The 'included angle' part is critical here – the angle must be between the two proportional sides.

Applying Similarity Criteria: Worked Examples

• Example 1: Using AA Similarity Problem: In the given figure, if ΔOAB ~ ΔOCD, ∠A = 75° and ∠C = 45°, find ∠B and ∠D. Solution: Step 1: Understand the given information. We are given that ΔOAB ~ ΔOCD. This means their corresponding angles are equal and corresponding sides are proportional. Given ∠A = 75° and ∠C = 45°. Step 2: Identify corresponding angles. Since ΔOAB ~ ΔOCD, the corresponding angles are: ∠A = ∠C (This is where we need to be careful. The similarity statement ΔOAB ~ ΔOCD implies A corresponds to C, B to D, and O to O.) Wait, let's re-read the problem or adjust the example. If ΔOAB ~ ΔOCD, then ∠A = ∠C, ∠B = ∠D, and ∠AOB = ∠COD. Let's assume the question meant ∠OAB and ∠OCD are given, or it's a specific configuration where A and C are not corresponding, but ∠OAB and ∠ODC are. Let's rephrase the problem for clarity, common in Ex 6.3: Problem: In ΔABC and ΔPQR, if ∠A = ∠P and ∠B = ∠Q, prove ΔABC ~ ΔPQR and find ∠R if ∠C = 60°. Solution: Step 1: Identify given conditions. We are given ∠A = ∠P and ∠B = ∠Q. Step 2: Apply AA Similarity Criterion. Since two angles of ΔABC are equal to two corresponding angles of ΔPQR, by AA Similarity Criterion, ΔABC ~ ΔPQR. Step 3: Use properties of similar triangles. Because the triangles are similar, their corresponding angles are equal. So, ∠C = ∠R. Given ∠C = 60°. Step 4: Conclude the unknown angle. Therefore, ∠R = 60°. Final Answer: ΔABC ~ ΔPQR by AA Similarity, and ∠R = 60°.
• Example 2: Using SSS Similarity Problem: Are the triangles ΔABC and ΔDEF similar if AB = 3 cm, BC = 4 cm, AC = 5 cm, and DE = 6 cm, EF = 8 cm, DF = 10 cm? Solution: Step 1: List the side lengths of both triangles. ΔABC: AB = 3, BC = 4, AC = 5 ΔDEF: DE = 6, EF = 8, DF = 10 Step 2: Check the ratios of corresponding sides. We need to compare the ratios of sides. Let's try pairing the smallest, middle, and largest sides. Ratio of smallest sides: AB/DE = 3/6 = 1/2 Ratio of middle sides: BC/EF = 4/8 = 1/2 Ratio of largest sides: AC/DF = 5/10 = 1/2 Step 3: Apply SSS Similarity Criterion. Since AB/DE = BC/EF = AC/DF = 1/2, all corresponding sides are proportional. By SSS Similarity Criterion, ΔABC ~ ΔDEF. Final Answer: Yes, ΔABC and ΔDEF are similar by SSS similarity.
• Example 3: Using SAS Similarity Problem: In ΔPQR and ΔXYZ, PQ = 4 cm, PR = 6 cm, ∠P = 70°. Also, XY = 6 cm, XZ = 9 cm, ∠X = 70°. Are ΔPQR and ΔXYZ similar? Solution: Step 1: Identify the given information for both triangles. ΔPQR: PQ = 4, PR = 6, ∠P = 70° ΔXYZ: XY = 6, XZ = 9, ∠X = 70° Step 2: Check for equality of included angles. We have ∠P = 70° and ∠X = 70°. So, ∠P = ∠X. Step 3: Check the proportionality of the sides including these angles. The sides including ∠P are PQ and PR. The sides including ∠X are XY and XZ. Calculate the ratios: PQ/XY = 4/6 = 2/3 PR/XZ = 6/9 = 2/3 Step 4: Apply SAS Similarity Criterion. Since ∠P = ∠X and the sides including these angles are proportional (PQ/XY = PR/XZ), by SAS Similarity Criterion, ΔPQR ~ ΔXYZ. Final Answer: Yes, ΔPQR and ΔXYZ are similar by SAS similarity.

Common Mistakes to Avoid in Triangles Ex 6.3

Students often make crucial mistakes when dealing with similar triangles. Here's how to steer clear of them:

1. Incorrect Correspondence: The most common error is matching non-corresponding vertices. If ΔABC ~ ΔPQR, it means A corresponds to P, B to Q, and C to R. Ensure you write side ratios and angle equalities according to this correct correspondence (AB/PQ = BC/QR = AC/PR). A good practice is to redraw the triangles, orienting them similarly.
2. Confusing Congruence with Similarity: Remember, congruent triangles are a special case of similar triangles where the ratio of corresponding sides is 1 (i.e., they are equal). Don't assume that if shapes look similar, they must also be congruent, or vice-versa.
3. Applying SAS Incorrectly: For SAS, the angle must be the included angle (the angle between the two proportional sides). If the angle is not included, you cannot use SAS similarity.
4. Arithmetic Errors in Ratios: Be careful with your calculations when simplifying fractions for side ratios. A single mistake can lead to an incorrect conclusion about similarity.

Practice Questions with Solutions

• Q: In the figure, OD ⋅ OB = OC ⋅ OA. Show that ∠A = ∠C and ∠B = ∠D. A: Step 1: Rearrange the given proportion. Given OD ⋅ OB = OC ⋅ OA. We can write this as OA/OD = OB/OC (or OA/OC = OB/OD). Let's use OA/OC = OB/OD. Step 2: Identify included angles. Observe the vertically opposite angles ∠AOB and ∠COD. They are equal. Step 3: Apply SAS Similarity Criterion. In ΔAOB and ΔCOD, we have OA/OC = OB/OD and ∠AOB = ∠COD (vertically opposite angles). Therefore, by SAS Similarity Criterion, ΔAOB ~ ΔCOD. Step 4: Conclude corresponding angles. Since the triangles are similar, their corresponding angles are equal. Thus, ∠A = ∠C and ∠B = ∠D. Final answer: Shown that ∠A = ∠C and ∠B = ∠D.
• Q: ΔABC is an isosceles triangle with AC = BC. If AB² = 2AC², prove that ΔABC is a right-angled triangle. A: Step 1: Analyze the given information. Given ΔABC is isosceles with AC = BC. Also, AB² = 2AC². Step 2: Substitute AC for BC in the equation. Since AC = BC, we can write AB² = AC² + AC². This means AB² = AC² + BC². Step 3: Apply the converse of the Pythagorean Theorem. The equation AB² = AC² + BC² is exactly the form of the Pythagorean theorem. According to the converse of the Pythagorean Theorem, if the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle. Step 4: Conclude the type of triangle. Here, AB is the hypotenuse, and the angle opposite to AB is ∠C. Therefore, ∠C = 90°. Final answer: ΔABC is a right-angled triangle with the right angle at C.
• Q: A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds. A: Step 1: Convert units and calculate distance covered. Girl's height h_g = 90 cm = 0.9 m. Lamp-post height h_l = 3.6 m. Speed v = 1.2 m/s. Time t = 4 s. Distance covered by girl = v × t = 1.2 m/s × 4 s = 4.8 m. Let this be BC. Step 2: Draw a diagram and identify similar triangles. Let AB be the lamp-post, CD be the girl, and CE be her shadow. The light ray from A passes over the girl's head D to form the shadow at E. We have ΔABE and ΔCDE. Step 3: Prove similarity. In ΔABE and ΔCDE: ∠B = ∠C = 90° (Lamp-post and girl are perpendicular to the ground). ∠E = ∠E (Common angle). By AA Similarity Criterion, ΔABE ~ ΔCDE. Step 4: Use proportionality of sides. Since ΔABE ~ ΔCDE, we have AB/CD = BE/CE. Let CE = x (length of shadow). Then BE = BC + CE = 4.8 + x. Substitute the values: 3.6/0.9 = (4.8 + x)/x. Step 5: Solve for x. 4 = (4.8 + x)/x 4x = 4.8 + x 3x = 4.8 x = 4.8 / 3 x = 1.6 m. Final answer: The length of her shadow after 4 seconds is 1.6 meters.
• Q: E is a point on the side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF. A: Step 1: Identify given information. ΔABC is isosceles with AB = AC. This implies ∠B = ∠C (angles opposite to equal sides). AD ⊥ BC means ∠ADB = 90°. EF ⊥ AC means ∠EFC = 90°. Step 2: Consider the triangles to be proved similar: ΔABD and ΔECF. Step 3: Look for equal angles. From AB = AC, we know ∠B = ∠C. In ΔABD, ∠B is present. In ΔECF, ∠C is present (which is the same as ∠B). So, ∠ABD = ∠ECF (as ∠ABC = ∠ECF). We also have ∠ADB = 90° (given AD ⊥ BC). And ∠EFC = 90° (given EF ⊥ AC). So, ∠ADB = ∠EFC. Step 4: Apply AA Similarity Criterion. Since ∠ABD = ∠ECF and ∠ADB = ∠EFC, by AA Similarity Criterion, ΔABD ~ ΔECF. Final answer: ΔABD ~ ΔECF is proven by AA Similarity.

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