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

Answer: (d) more than 300

Explanation:
1. If 11 is the hypotenuse, the other two sides would need \( a^2+b^2 = 121 \) with integers, which has no solution. So 11 is a leg.
2. With 11 as a leg and hypotenuse c, other leg b: \( c^2-b^2 = 121 = (c-b)(c+b) = 1\times121 \) gives \( c = 61 \), \( b = 60 \). Area = \( \frac12\times11\times60 = 330 \).

Teacher's Note:
1. Difference of squares gives the integer triples.
2. Only the factor pair 1 × 121 works because both factors must have the same parity.

Answer: (c) \( \sqrt5+1 \)

Explanation:
1. Let AQ = x, AB = x + 2, AD = h and AP = t. Equal areas: \( \frac12xt = h \) and \( \frac12(x+2)(h-t) = h \).
2. From the first, \( t = \frac{2h}{x} \). Substituting, \( (x+2)\left(1-\frac2x\right) = 2 \), so \( x^2-2x-4 = 0 \) and \( x = 1+\sqrt5 \).

Teacher's Note:
1. Set up two equations from the equal areas.
2. The height h cancels.

Answer: (c) 20

Explanation:
1. Let the two kinds of corner triangles have legs a and b. Their total area is \( 2\cdot\frac{a^2}{2}+2\cdot\frac{b^2}{2} = a^2+b^2 = 200 \).
2. The rectangle has sides \( a\sqrt2 \) and \( b\sqrt2 \), so its diagonal is \( \sqrt{2a^2+2b^2} = \sqrt{400} = 20 \).

Teacher's Note:
1. The sides of the rectangle are the hypotenuses of the corner triangles.
2. Use Pythagoras for the diagonal.

Answer: (d) \( \frac{3\sqrt3\,S}{\pi} \)

Explanation:
1. Triangle side a: \( S = \frac{\sqrt3}{4}a^2 \), perimeter 3a.
2. Circle: \( 2\pi R = 3a \), area \( = \pi R^2 = \frac{9a^2}{4\pi} = \frac{9}{4\pi}\cdot\frac{4S}{\sqrt3} = \frac{3\sqrt3\,S}{\pi} \).

Teacher's Note:
1. The wire length (perimeter) is the same.
2. Eliminate a between the two areas.

Answer: (d) \( \frac32 \)

Explanation:
1. The largest hexagon is obtained by cutting an equilateral triangle of side a from each corner of a triangle of side 3a, leaving a hexagon of side a.
2. Triangle : hexagon \( = \frac{\sqrt3}{4}(3a)^2 : \frac{3\sqrt3}{2}a^2 = \frac94 : \frac32 = \frac32 \).

Teacher's Note:
1. Divide each side into three equal parts.
2. The cut-off corners are three small equilateral triangles.

Answer: (a) \( (\sqrt2+1)a \)

Explanation:
1. \( \pi(r+a)^2 = 2\pi r^2 \Rightarrow r+a = \sqrt2r \).
2. \( r = \frac{a}{\sqrt2-1} = (\sqrt2+1)a \).

Teacher's Note:
1. Area ratio 2 means radius ratio √2.
2. Rationalise the denominator.

Answer: (d) 18.75 cm

Explanation:
1. Semi-perimeter s = 48. Area \( = \sqrt{48\cdot18\cdot12\cdot18} = 432 \) cm².
2. \( R = \frac{abc}{4K} = \frac{30\times36\times30}{4\times432} = 18.75 \) cm.

Teacher's Note:
1. Use Heron's formula for the area.
2. Circumradius R = abc / 4K.

Answer: (a) 2 m

Explanation:
1. Area of the flag = 100 m², so the white cross has area 20 m².
2. Five squares: \( 5s^2 = 20 \Rightarrow s = 2 \) m.

Teacher's Note:
1. Area of the cross = five small squares.
2. Take 20% of the flag's area.

Answer: (d) Rs. 87

Explanation:
1. Hexagon area \( = \frac{3\sqrt3}{2}(4)^2 = 24\sqrt3 \approx 41.57 \) cm², circle area \( = 16\pi \approx 50.27 \) cm².
2. Design area \( \approx 8.70 \) cm², cost \( \approx 87 \) rupees.

Teacher's Note:
1. Designs = circle − hexagon.
2. Side of the inscribed hexagon equals the radius.