Measurement of Area Questions in English

(C) Area of square $= 121 \, cm^2$.
Side of square $= \sqrt{121} = 11 \, cm$.
Perimeter of square $= 4 \times \text{side} = 4 \times 11 = 44 \, cm$.
Since the same wire is used,the perimeter of the square equals the circumference of the circle.
Circumference of circle $= 2 \pi r = 44 \, cm$.
$2 \times \frac{22}{7} \times r = 44$.
$r = \frac{44 \times 7}{2 \times 22} = 7 \, cm$.
Area of circle $= \pi r^2 = \frac{22}{7} \times 7 \times 7 = 154 \, cm^2$.

(B) Given,the ratio of the area of the sector to the area of the circle is $1:4$.
Area of the circle $= 154 \, cm^2$.
Area of the sector $= \frac{1}{4} \times 154 = 38.5 \, cm^2$.
Since the ratio of the area of the sector to the area of the circle is equal to the ratio of the central angle $\theta$ to $360^{\circ}$,we have $\frac{\theta}{360^{\circ}} = \frac{1}{4}$,which gives $\theta = 90^{\circ}$.
Now,find the radius $r$ of the circle: $\pi r^2 = 154 \Rightarrow \frac{22}{7} \times r^2 = 154 \Rightarrow r^2 = 49 \Rightarrow r = 7 \, cm$.
The perimeter of a sector is given by the formula $P = \frac{\theta}{360^{\circ}} \times 2 \pi r + 2r$.
Substituting the values: $P = \frac{90^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 7 + 2 \times 7$.
$P = \frac{1}{4} \times 44 + 14 = 11 + 14 = 25 \, cm$.

(A) The sum of the areas of the squares is given by the sum of the squares of their side lengths:
Sum $= 20^2 + 21^2 + \ldots + 29^2$
Using the formula for the sum of squares of the first $n$ natural numbers,$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$,we can write:
Sum $= \sum_{k=1}^{29} k^2 - \sum_{k=1}^{19} k^2$
$= \frac{29(29+1)(2 \times 29 + 1)}{6} - \frac{19(19+1)(2 \times 19 + 1)}{6}$
$= \frac{29 \times 30 \times 59}{6} - \frac{19 \times 20 \times 39}{6}$
$= (29 \times 5 \times 59) - (19 \times 10 \times 13)$
$= 8555 - 2470 = 6085 \, cm^2$.

(B) The sum of the angles of a quadrilateral is $360^{\circ}$.
Let the angles be $2x, 4x, 7x,$ and $5x$.
$2x + 4x + 7x + 5x = 18x = 360^{\circ}$.
$x = 360^{\circ} / 18 = 20^{\circ}$.
The smallest angle of the quadrilateral is $2x = 2 \times 20^{\circ} = 40^{\circ}$.
Since the smallest angle of the triangle is equal to the smallest angle of the quadrilateral,the smallest angle of the triangle is $40^{\circ}$.
Another angle of the triangle is twice the smallest angle,which is $2 \times 40^{\circ} = 80^{\circ}$.
The sum of the angles of a triangle is $180^{\circ}$.
The third angle is $180^{\circ} - (40^{\circ} + 80^{\circ}) = 180^{\circ} - 120^{\circ} = 60^{\circ}$.
The angles of the triangle are $40^{\circ}, 60^{\circ},$ and $80^{\circ}$.
The second largest angle of the triangle is $60^{\circ}$.

(D) The side of the square is $\sqrt{1024\, cm^2} = 32\, cm$.
The length of the rectangle is $2 \times 32\, cm = 64\, cm$.
The breadth of the rectangle is $32\, cm - 12\, cm = 20\, cm$.
The required ratio of length to breadth is $64 : 20$.
Dividing both terms by $4$,we get $16 : 5$.

(A) Perimeter of the rectangle $= 2 \times (8 + 7) = 2 \times 15 = 30 \, cm$.
Perimeter of the square $= 2 \times 30 = 60 \, cm$.
Side of the square $= 60 / 4 = 15 \, cm$.
The diameter of the semicircle is equal to the side of the square,so $d = 15 \, cm$,which means the radius $r = 7.5 \, cm$.
The circumference of a semicircle is given by the formula $C = \pi r + d$.
$C = (22 / 7) \times 7.5 + 15$.
$C = 23.5714 + 15 = 38.5714 \, cm$.
Rounding to two decimal places,we get $38.57 \, cm$.

(C) The circumference of a circle is given by $C = 2\pi r$.
For the smaller circle,$2 \times \frac{22}{7} \times r_1 = 132 \implies r_1 = \frac{132 \times 7}{44} = 21 \text{ m}$.
For the larger circle,$2 \times \frac{22}{7} \times r_2 = 176 \implies r_2 = \frac{176 \times 7}{44} = 28 \text{ m}$.
The area of a circle is given by $A = \pi r^2$.
The difference in areas is $\pi(r_2^2 - r_1^2) = \frac{22}{7} \times (28^2 - 21^2)$.
Using the identity $a^2 - b^2 = (a - b)(a + b)$,we get $\frac{22}{7} \times (28 - 21)(28 + 21) = \frac{22}{7} \times 7 \times 49$.
Difference $= 22 \times 49 = 1078 \text{ m}^2$.