If the side of a square is doubled, how does its area change?

Answer: Becomes 4 times. Thinking Process:Area $A = a^2$.New side $a' = 2a$.New Area $A' = (2a)^2 = 4a^2 = 4A$.The area becomes 4 times.

The ratio of the radii of two circles is 3:4. What is the ratio of their areas?

Answer: 9:16. Thinking Process:Area $\propto r^2$.Ratio of areas = $(r_1/r_2)^2 = (3/4)^2 = 9/16$.

If the length of a rectangle is increased by 10% and breadth is decreased by 10%, what is the percentage change in area?

Answer: 1% decrease. Thinking Process:Use Successive % Formula: $x + y + \frac{xy}{100}$.$x = +10, y = -10$.Change = $10 - 10 + \frac{10(-10)}{100} = -1\%$.Negative sign shows decrease.

Which triangle has the maximum area for a given perimeter?

Answer: Equilateral. For a fixed perimeter, the regular polygon (Equilateral Triangle in this case) encloses the maximum area.

If the circumference of a circle is equal to the perimeter of a square, which one has a larger area?

Answer: Circle. Thinking Process:For a given perimeter, a Circle encloses more area than a Square.Example: $Perimeter = 44$. Square Area = 121. Circle Area = 154.

Diagonals of a rhombus bisect each other at what angle?

Answer: 90°. A key property of a Rhombus is that its diagonals bisect each other at right angles ($90^\circ$).

The area of a triangle with sides 3, 4, and 5 is:

Answer: 6. Thinking Process:Check if it's a Right Angled Triangle: $3^2 + 4^2 = 9+16=25=5^2$. Yes.$Area = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 3 \times 4 = 6$.

What is the sum of interior angles of a pentagon?

Answer: 540°. Formula: $(n-2) \times 180^\circ$.For Pentagon, $n=5$.Sum = $(5-2) \times 180 = 3 \times 180 = 540^\circ$.

The diagonal of a square is $d$. What is its area?

Answer: d²/2. Area of Square = $\frac{1}{2} \times (\text{Diagonal})^2 = d^2/2$.

Which has a larger perimeter: a square of side 10 or a circle of diameter 10?

Answer: Square. Square Perimeter = $4 \times 10 = 40$.Circle Circumference = $\pi d = 3.14 \times 10 = 31.4$.Square is larger.

The length and breadth of a rectangular plot are in the ratio 5:3. If the perimeter is 320 m, find the area of the plot.

Answer: 6000 m². Thinking Process:1. Let length $L=5x$ and breadth $B=3x$.2. Perimeter $2(L+B) = 320 \implies 2(8x) = 320 \implies 16x = 320 \implies x=20$.3. $L = 100, B = 60$.4. Area = $100 \times 60 = 6000 m^2$.

The cost of fencing a circular field at the rate of ₹24 per meter is ₹5280. Find the cost of ploughing the field at ₹0.50 per m².

Answer: ₹1925. Thinking Process:1. Circumference = Total Cost / Rate = $5280/24 = 220$ m.2. $2\pi r = 220 \implies 2 \times \frac{22}{7} \times r = 220 \implies r = 35$ m.3. Area = $\frac{22}{7} \times 35 \times 35 = 22 \times 5 \times 35 = 3850 m^2$.4. Ploughing Cost = $3850 \times 0.50 = 1925$.

The area of a square is 1024 cm². Find the ratio of the length of its diagonal to the length of its side.

Answer: √2 : 1. Thinking Process:We don't need the area (Redundant data).For any square, Diagonal $d = a\sqrt{2}$.Ratio $d:a = a\sqrt{2} : a = \sqrt{2} : 1$.

The sides of a triangle are 13 cm, 14 cm, and 15 cm. The area of the triangle is:

Answer: 84 cm². Thinking Process:Use Heron's Formula.$s = \frac{13+14+15}{2} = \frac{42}{2} = 21$.$Area = \sqrt{21(21-13)(21-14)(21-15)}$$= \sqrt{21 \times 8 \times 7 \times 6}$$= \sqrt{(7 \times 3) \times (2^3) \times 7 \times (2 \times 3)}$$= \sqrt{7^2 \times 3^2 \times 2^4} = 7 \times 3 \times 4 = 84$.

The perimeter of a rhombus is 40 m and its height is 5 m. Finding its area.

Answer: 50 m². Thinking Process:1. Perimeter = $4a = 40 \implies Side a = 10$.2. A Rhombus is also a parallelogram.3. Area = Base $\times$ Height = $10 \times 5 = 50 m^2$.

The area of a field in the shape of a trapezium measures 1440 m². The perpendicular distance between its parallel sides is 24 m. If the ratio of the parallel sides is 5:3, the length of the longer parallel side is:

Answer: 75 m. Thinking Process:Let sides be $5x$ and $3x$. Height $h=24$.Area = $\frac{1}{2}(5x+3x) \times 24 = 1440$.$4x \times 24 = 1440$.$96x = 1440 \implies x = 15$.Longer side = $5x = 5(15) = 75$ m.

A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 m², then what is the width of the road?

Answer: 3 m. Thinking Process:Total Area = $60 \times 40 = 2400$.Road Area = Total - Lawn = $2400 - 2109 = 291 m^2$.Road Area Formula: $w(L+B-w)$.$w(60+40-w) = 291$.$100w - w^2 = 291 \implies w^2 - 100w + 291 = 0$.Factors of 291: 3 and 97.$(w-3)(w-97)=0$. Width must be 3.

The height of an equilateral triangle is 18 cm. Its area is:

Answer: 108√3 cm². Thinking Process:Height $h = \frac{\sqrt{3}}{2}a = 18 o a = \frac{36}{\sqrt{3}} = 12\sqrt{3}$.Area = $\frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} (12\sqrt{3})^2$$= \frac{\sqrt{3}}{4} (144 \times 3) = 108\sqrt{3}$.

A wheel makes 1000 revolutions in covering a distance of 88 km. The diameter of the wheel is:

Answer: 28 m. Thinking Process:Distance in 1 rev = Total Dist / Revs = $88000 / 1000 = 88$ meters.Circumference $\pi d = 88$.$\frac{22}{7} d = 88 \implies d = 28$ meters.

The perimeter of two squares are 24 cm and 32 cm. The perimeter (in cm) of a third square equal in area to the sum of the areas of these two squares is:

Answer: 40. Thinking Process:Square 1: Perim=24 $\to$ side=6 $\to$ Area=36.Square 2: Perim=32 $\to$ side=8 $\to$ Area=64.New Area = $36+64=100$.New Side = $\sqrt{100}=10$.New Perimeter = $4 \times 10 = 40$.

A circle is inscribed in a square of side 28 cm. Find the area of the remaining portion of the square which is not covered by the circle.

Answer: 168 cm². Thinking Process:Side of Square $a = 28$.Radius of Incircle $r = a/2 = 14$.Area Square = $28^2 = 784$.Area Circle = $\frac{22}{7} \times 14 \times 14 = 616$.Shaded Area = $784 - 616 = 168$.

Four circular cardboard pieces of radii 7 cm are placed on a paper in such a way that each piece touches two other pieces. The area of the portion enclosed between these pieces is:

Answer: 42 cm². Thinking Process:The centers form a square of side $2r = 14$.Area Enclosed = Area(Square) - $4 \times$ Area(Quadrant) = Area(Square) - Area(Circle).Square = $14^2 = 196$.Circle = $\frac{22}{7} \times 7^2 = 154$.Diff = $196 - 154 = 42$.

The perimeter of a triangular field is 240 m. Its two sides are 78 m and 50 m. Find the length of the altitude on the side of length 50 m.

Answer: 67.2 m. Thinking Process:1. Find third side: $240 - (78+50) = 112$ m.2. Sides: 50, 78, 112.3. Area (Heron's): $s=120$.$Area = \sqrt{120(120-50)(120-78)(120-112)}$$= \sqrt{120 \times 70 \times 42 \times 8} = 1680 m^2$.4. Altitude on 50m: $\frac{1}{2} \times 50 \times h = 1680 \implies h = 67.2$.

A cow is tied with a rope 14 m long at the corner of a rectangular field of dimensions 30m x 20m. The area of the field in which the cow can graze is:

Answer: 154 m². Thinking Process:The cow grazes a Quadrant (quarter circle) of radius 14 m.Area = $\frac{1}{4} \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 14 \times 14 = 154 m^2$.

Find the area of a regular hexagon of side 6 cm.

Answer: 54√3 cm². Thinking Process:Area = $6 \times \frac{\sqrt{3}}{4} a^2$.$= \frac{3\sqrt{3}}{2} (6)^2 = \frac{3\sqrt{3}}{2} \times 36 = 54\sqrt{3}$.

The ratio of the length of the equal sides and the third side of an isosceles triangle is 3:4. If the area of the triangle is $18\sqrt{5}$ units, the third side is:

Answer: 12 units. Thinking Process:Let sides be $3x, 3x, 4x$.It's isosceles. Height formula: $h=\sqrt{a^2 - (b/2)^2}$.$h = \sqrt{(3x)^2 - (2x)^2} = \sqrt{5x^2} = x\sqrt{5}$.Area = $\frac{1}{2} \times \text{Base} \times h = \frac{1}{2}(4x)(x\sqrt{5}) = 2x^2\sqrt{5}$.Given $2x^2\sqrt{5} = 18\sqrt{5} \implies x^2=9 \implies x=3$.Third side $4x = 4(3) = 12$.

If the area of an equilateral triangle is $A$ and its height is $h$, then $h^2/A = ?

Answer: √3. Thinking Process:$A = \frac{\sqrt{3}}{4}a^2$, $h = \frac{\sqrt{3}}{2}a$.$h^2 = \frac{3}{4}a^2$.Ratio $h^2/A = (\frac{3}{4}a^2) / (\frac{\sqrt{3}}{4}a^2) = \frac{3}{\sqrt{3}} = \sqrt{3}$.

Two small circles touch each other externally and the sum of their areas is $130\pi$ cm². If the distance between their centers is 14 cm, what are the radii?

Answer: 11, 3. Thinking Process:1. $r_1 + r_2 = 14$.2. $\pi r_1^2 + \pi r_2^2 = 130\pi \implies r_1^2 + r_2^2 = 130$.Check Options:A: $11+3=14$. $121+9=130$. Matches.

Two adjacent sides of a parallelogram are 15 cm and 18 cm. If the distance between the longer sides is 12 cm, find the distance between the smaller sides.

Answer: 14.4 cm. Thinking Process:Area = Base $\times$ Height.Case 1: Base 18, Height 12 $\to Area = 18 \times 12 = 216$.Case 2: Base 15, Height $h$ $\to Area = 15 \times h = 216$.$h = 216/15 = 14.4$ cm.

A path of uniform width runs round the inside of a rectangular field 38 m long and 32 m wide. If the path occupies 600 m², then the width of the path is:

Answer: 5 m. Thinking Process:Area of Park = $38 \times 32 = 1216$.Inner Area = $1216 - 600 = 616$.Sides of inner rectangle: $(38-2w)$ and $(32-2w)$.$(38-2w)(32-2w) = 616$.Option B (5): $28 \times 22 = 616$. Matches.

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Mensuration 2D (Area & Perimeter)

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1. Triangles (त्रिभुज)

Area:

A = \frac{1}{2} \times \text{Base} \times \text{Height}

Heron's Formula:

A = \sqrt{s(s-a)(s-b)(s-c)}

, where

s = \frac{a+b+c}{2}

(Semi-perimeter).

Equilateral Triangle:

Area: $A = \frac{\sqrt{3}}{4} a^2$
Height: $h = \frac{\sqrt{3}}{2} a$

Isosceles Triangle:

A = \frac{b}{4}\sqrt{4a^2 - b^2}

(where

a

is equal side).

Example:

Q: Find the area of an equilateral triangle with side 10 cm.

Solution:

A = \frac{\sqrt{3}}{4} (10)^2 = \frac{\sqrt{3}}{4} \times 100 = 25\sqrt{3} \text{ cm}^2

2. Quadrilaterals (चतुर्भुज)

Rectangle:

Area: $L \times B$
Perimeter: $2(L+B)$
Diagonal: $\sqrt{L^2 + B^2}$

Square:

Area: $a^2$ or $\frac{1}{2} \times (\text{Diagonal})^2$
Perimeter: $4a$
Diagonal: $a\sqrt{2}$

Parallelogram:

A = \text{Base} \times \text{Height}

Rhombus:

A = \frac{1}{2} \times d_1 \times d_2

(

d_1, d_2

are diagonals).
Side

a = \frac{1}{2}\sqrt{d_1^2 + d_2^2}

.

Trapezium:

A = \frac{1}{2} \times (\text{Sum of Parallel Sides}) \times \text{Height}

3. Circles (वृत्त)

Area:

A = \pi r^2

Circumference:

C = 2\pi r

Semi-Circle: Area =

\frac{\pi r^2}{2}

, Perimeter =

\pi r + 2r

.

Sector:

Area = $\frac{\theta}{360} \times \pi r^2$
Arc Length ( $l$ ) = $\frac{\theta}{360} \times 2\pi r$

Example:

Q: Find the circumference of a circle with radius 7 cm.

Solution:

C = 2 \times \frac{22}{7} \times 7 = 44 \text{ cm}

4. Polygons (बहुभुज)

Sum of Interior Angles:

(n-2) \times 180^\circ

Each Interior Angle (Regular):

\frac{(n-2) \times 180}{n}

Sum of Exterior Angles:

360^\circ

Regular Hexagon:

Area: $6 \times \frac{\sqrt{3}}{4} a^2 = \frac{3\sqrt{3}}{2} a^2$
It consists of 6 Equilateral Triangles.

5. Pathways (रास्ते)

Path outside a Rectangle:
Area =

(L + B + 2w) \times 2w

Path inside a Rectangle:
Area =

(L + B - 2w) \times 2w

Cross Paths:
Area =

w(L + B - w)

(where

w

is the width of the path).

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Mensuration 2D (Area & Perimeter)

Expert Answer & Key Takeaways

1. Triangles (त्रिभुज)

2. Quadrilaterals (चतुर्भुज)

3. Circles (वृत्त)

4. Polygons (बहुभुज)

5. Pathways (रास्ते)

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