CBSE Class 10 Mathematics Triangles, Trigonometry and Statistics Worksheet Set 02

Answer: If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, the triangles are similar.

Teacher's Note:
1. The equal angles must be the included angles.
2. Similar triangles have equal angles and proportional sides.

Answer: All congruent figures are similar because they have the same shape. Similar figures need not be congruent, since their sizes can differ.

Teacher's Note:
1. Congruent is similar plus equal size.
2. A map and its scale drawing are similar but not congruent.

Answer: The criteria are AA (or AAA), SSS (all three sides proportional) and SAS (one equal angle with proportional including sides).

Teacher's Note:
1. Two equal angles force the third to be equal as well.
2. SSS and SAS are checks using side ratios.

Answer: All equilateral triangles are similar. Each has three angles of 60° and sides in the ratio 1 : 1 : 1.

Teacher's Note:
1. Equal angles give similarity by the AA criterion.
2. Isosceles and scalene triangles can have different angles.

Answer: If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, it divides those two sides in the same ratio.

Teacher's Note:
1. The line must be parallel to the third side.
2. The converse also holds.

Answer: By the Basic Proportionality Theorem, \( \dfrac{AD}{DB} = \dfrac{AE}{EC} \), so \( \dfrac{2}{3} = \dfrac{2.4}{EC} \). EC = 3.6 cm.

Teacher's Note:
1. Set up the proportion with corresponding segments.
2. Cross-multiply to solve.

Answer: In a right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides. The hypotenuse, opposite the right angle, is the longest side.

Teacher's Note:
1. The theorem applies only to right-angled triangles.
2. Use it to test whether a triangle is right-angled.

Answer: Distance = \( \sqrt{10^2 - 8^2} = \sqrt{36} \) = 6 m.

Teacher's Note:
1. The ladder is the hypotenuse.
2. Subtract the squares, do not add.

Answer: The ratio of areas equals the square of the ratio of corresponding sides: \( \dfrac{BC^2}{EF^2} = \dfrac{64}{121} \). So \( \dfrac{BC}{15.4} = \dfrac{8}{11} \) and BC = 11.2 cm.

Teacher's Note:
1. Take the square root of the area ratio to get the side ratio.
2. BC and EF are corresponding sides.

Answer: sin A = \( \dfrac{BC}{AC} \), cos A = \( \dfrac{AB}{AC} \), tan A = \( \dfrac{BC}{AB} \).

Teacher's Note:
1. Opposite and adjacent are with respect to angle A.
2. AC is the hypotenuse.

Answer: sin 60° = \( \dfrac{\sqrt{3}}{2} \), cos 60° = \( \dfrac{1}{2} \), tan 60° = √3.

Teacher's Note:
1. Memorise the table of standard angles.
2. tan = sin ÷ cos.

Answer: cot 20° = tan (90° − 20°) = tan 70°. So the value is 1.

Teacher's Note:
1. Use cot θ = tan (90° − θ).
2. Complementary angles pair up.

Answer: sec (90° − A) = cosec A. The identity is sec²A − tan²A = 1.

Teacher's Note:
1. Co-function of secant is cosecant.
2. The identity follows from sin²A + cos²A = 1 divided by cos²A.

Answer: The expression equals \( \dfrac{1 - \sin^2 A}{1 - \cos^2 A} = \dfrac{\cos^2 A}{\sin^2 A} = \cot^2 A = \dfrac{49}{64} \).

Teacher's Note:
1. Use (1 + x)(1 − x) = 1 − x².
2. Then apply sin²A + cos²A = 1.

Answer: tan 45° = 1, so the value is \( \dfrac{1 - 1}{1 + 1} = 0 \).

Teacher's Note:
1. Substitute the standard value directly.
2. The numerator becomes zero.

Answer: sin²60° = \( \dfrac{3}{4} \), tan 45° = 1, cos²30° = \( \dfrac{3}{4} \). The value is \( \dfrac{3}{4} \) + 2 − \( \dfrac{3}{4} \) = 2.

Teacher's Note:
1. Two of the terms cancel.
2. Write each value before adding.

Answer: tan 60° = \( \dfrac{h}{20} \), so h = 20√3 ≈ 34.64 m.

Teacher's Note:
1. Tangent connects height and horizontal distance.
2. tan 60° = √3.

Answer: The three measures are the mean, the median and the mode.

Teacher's Note:
1. Each describes the centre of the data differently.
2. The mode is the most frequent value.

Answer: The class mark is the mid-value of a class, equal to (upper limit + lower limit) ÷ 2. For 10–25 it is (10 + 25) ÷ 2 = 17.5.

Teacher's Note:
1. Class marks are used as the representative value in grouped data.
2. Add the limits and halve.

Answer: Mean = \( \dfrac{\sum f_i x_i}{\sum f_i} \). Mode = \( l + \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h \). Median = \( l + \dfrac{\frac{n}{2} - cf}{f} \times h \).

Teacher's Note:
1. In each formula l is the lower limit of the relevant class and h is the class width.
2. For the median, cf is the cumulative frequency of the class before the median class.

Answer: Class marks: 5, 15, 25, 35, 45. Σfx = 25 + 120 + 300 + 350 + 225 = 1020 and Σf = 40. Mean = \( \dfrac{1020}{40} \) = 25.5.

Teacher's Note:
1. Multiply each frequency by its class mark.
2. Divide the total by the sum of frequencies.

Answer: n = 40, so n/2 = 20. Cumulative frequencies are 5, 13, 25, 35, 40, so the median class is 20–30 with cf = 13, f = 12 and h = 10. Median = 20 + \( \dfrac{20 - 13}{12} \times 10 \) ≈ 25.83.

Teacher's Note:
1. The median class is the first class whose cumulative frequency is at least n/2.
2. Use the lower limit of that class.

Answer: The modal class is 20–30 with f₁ = 12, f₀ = 8, f₂ = 10 and h = 10. Mode = 20 + \( \dfrac{12 - 8}{24 - 8 - 10} \times 10 \) = 20 + \( \dfrac{40}{6} \) ≈ 26.67.

Teacher's Note:
1. The modal class has the highest frequency.
2. f₀ and f₂ are the frequencies of the classes before and after it.