CBSE Class 10 Mathematics Perimeter and Area of Plane Figures MCQs Set 06

Answer: (a) 4.5 m²

Explanation:
1. Let each equal side be a, so the hypotenuse is \( a\sqrt{2} \).
2. Perimeter: \( a(2 + \sqrt{2}) = 3(2 + \sqrt{2}) \), so a = 3 m.
3. Area = \( \frac{1}{2} \times 3 \times 3 \) = 4.5 m².

Teacher's Note:
1. The two equal sides and the hypotenuse of an isosceles right triangle are in the ratio 1 : 1 : √2.
2. Factor the perimeter before solving.

Answer: (c) 82%

Explanation:
1. New area = 1.4 × 1.3 × (side)² = 1.82 × (side)².
2. Increase = 82%.

Teacher's Note:
1. Successive percentage changes multiply as factors.
2. Do not add 40% and 30% for area.

Answer: (d) 44%

Explanation:
1. New area = 1.2 × 1.2 = 1.44 times the old area.
2. The increase is 44%.

Teacher's Note:
1. Area scales by the product of the two scale factors.
2. 1.2 × 1.2 = 1.44.

Answer: (d) 12 cm

Explanation:
1. Multiply by 12: the ratio becomes 6 : 4 : 3, with sum 13.
2. One part = 52 ÷ 13 = 4 cm, so the smallest side = 3 × 4 = 12 cm.

Teacher's Note:
1. Clear fractions in a ratio by multiplying by the LCM of the denominators.
2. The smallest side is the one with the fewest parts, here 3 parts.

Answer: (b) 200 m

Explanation:
1. Parallelogram area = base × 100; triangle area = \( \frac{1}{2} \) × base × h.
2. Equating, h = 200 m.

Teacher's Note:
1. A triangle on the same base and between the same parallels has half the parallelogram's area.
2. So an equal-area triangle must be twice as tall.

Answer: (c) 3x + 4

Explanation:
1. Perimeter = 4 × side = \( 4\left(\frac{3x}{4} + 1\right) \) = 3x + 4.

Teacher's Note:
1. Perimeter of a square is four times its side.
2. The last option is the area, not the perimeter.

Answer: (c) 69%

Explanation:
1. New area = (1.3)² = 1.69 times the old area.
2. Increase = 69%.

Teacher's Note:
1. Area changes as the square of the scale factor.
2. Do not quote 30% or 60%.

Answer: (a) \( 37\frac{1}{2} \)%

Explanation:
1. New width factor = \( \frac{1}{1.6} \) = 0.625.
2. Decrease = 1 − 0.625 = 0.375 = \( 37\frac{1}{2} \)%.

Teacher's Note:
1. With the area fixed, the two scale factors are reciprocals.
2. The decrease is measured against the original width.

Answer: (d) None of these

Explanation:
1. Poles around a closed boundary: 20 poles give 20 gaps of 5 m, so perimeter = 100 m.
2. Side = 25 m, area = 625 m², which is not among the first three options.

Teacher's Note:
1. For a closed figure, the number of poles equals the number of gaps.
2. Area = (perimeter ÷ 4)².

Answer: (b) 12 m

Explanation:
1. Let the sides be a (shorter) and b (longer): ab = 60 and \( \sqrt{a^2 + b^2} + b = 5a \).
2. For a = 5, b = 12: the diagonal is 13 and 13 + 12 = 25 = 5 × 5, so it works.
3. The length is 12 m.

Teacher's Note:
1. Try Pythagorean triples with product of legs 60.
2. Check both given conditions.

Answer: (b) 126, 32

Explanation:
1. d1 × d2 = 2 × 2016 = 4032.
2. Half-diagonals and side form a right triangle: d1² + d2² = 4 × 65² = 16900.
3. 126 × 32 = 4032 and 126² + 32² = 15876 + 1024 = 16900, so the diagonals are 126 cm and 32 cm.

Teacher's Note:
1. Diagonals of a rhombus bisect each other at right angles.
2. Check each option against both area and side.

Answer: (c) \( 64\sqrt{3} \) cm²

Explanation:
1. Diagonal = side × √2, so the side of the square is 12 cm and its perimeter is 48 cm.
2. The triangle has side 16 cm; area = \( \frac{\sqrt{3}}{4} \times 16^2 = 64\sqrt{3} \) cm².

Teacher's Note:
1. Diagonal of a square = side × √2.
2. Area of an equilateral triangle = (√3/4) × side².

Answer: (c) 27.75%

Explanation:
1. Area of a square = ½ × diagonal², so the area becomes (0.85)² = 0.7225 times.
2. Decrease = 100 − 72.25 = 27.75%.

Teacher's Note:
1. Area changes as the square of the diagonal.
2. Report the decrease, not the new value.

Answer: (d) 2

Explanation:
1. The diagonal of the square equals the diameter 2.
2. Area = ½ × diagonal² = ½ × 4 = 2 square units.

Teacher's Note:
1. The largest square in a circle has its diagonal equal to the diameter.
2. Area = d²/2.

Answer: (a) r² cm²

Explanation:
1. Best choice: base = the diameter 2r and the apex at the highest point, height r.
2. Area = ½ × 2r × r = r².

Teacher's Note:
1. Height is at most r for any point on the semicircle.
2. The base cannot exceed the diameter.

Answer: (b) 6r

Explanation:
1. The hexagon splits into six equilateral triangles of side r, so each side is r.
2. Perimeter = 6r.

Teacher's Note:
1. Side of a regular hexagon equals the radius of its circumcircle.
2. Six equilateral triangles make up the hexagon.

Answer: (a) \( 4\sqrt{3}r \)

Explanation:
1. The apothem equals r, and for a regular hexagon apothem = \( \frac{\sqrt{3}}{2} \) × side.
2. Side = \( \frac{2r}{\sqrt{3}} \); perimeter = \( 6 \times \frac{2r}{\sqrt{3}} = 4\sqrt{3}r \).

Teacher's Note:
1. The inradius of a regular hexagon is (√3/2) times the side.
2. Do not confuse the inscribed and circumscribed cases.

Answer: (d) 6 m

Explanation:
1. Let the sides be x and y: 29x² = 6y² − 1 and 9x − 4y = 1.
2. Substituting y = (9x − 1)/4 gives 11x² − 54x − 5 = 0, so x = 5 and y = 11.
3. Difference = 6 m.

Teacher's Note:
1. Use the second equation to express y in terms of x.
2. Check: 29 × 25 = 725 = 6 × 121 − 1.

Answer: (d) 1 : 2

Explanation:
1. For a square of side s, inradius = s/2 and circumradius = s/√2.
2. Ratio of areas = (s²/4) : (s²/2) = 1 : 2.

Teacher's Note:
1. Areas of circles are in the ratio of the squares of the radii.
2. Circumradius = half the diagonal.

Answer: (c) 2 : 5

Explanation:
1. Semicircle of radius r: a square of side s with a side on the diameter satisfies \( \left(\frac{s}{2}\right)^2 + s^2 = r^2 \), so s² = \( \frac{4r^2}{5} \).
2. In the full circle the square has area 2r².
3. Ratio = \( \frac{4}{5} : 2 \) = 2 : 5.

Teacher's Note:
1. A square in a semicircle has one side on the diameter and two vertices on the arc.
2. Use the Pythagoras theorem with half the side.