Measurement of Area Questions in English

(C) In $\Delta OAB$,$E$ is the midpoint of $AO$ and $H$ is the midpoint of $BO$. By the Midpoint Theorem,$EH = \frac{1}{2} AB$.
Similarly,in $\Delta OBC$,$F$ is the midpoint of $DO$ and $G$ is the midpoint of $CO$. Note that the vertices are labeled such that $EFGH$ forms a quadrilateral inside. Applying the Midpoint Theorem to triangles $\Delta OAB, \Delta OBC, \Delta OCD$,and $\Delta ODA$:
$EH = \frac{1}{2} AB$
$HG = \frac{1}{2} CD$
$GF = \frac{1}{2} BC$
$FE = \frac{1}{2} DA$
Perimeter of $EFGH = EH + HG + GF + FE = \frac{1}{2} (AB + CD + BC + DA) = \frac{1}{2} (\text{Perimeter of } ABCD)$.
Therefore,the ratio of the perimeter of $EFGH$ to the perimeter of $ABCD$ is $\frac{1}{2} : 1$,which is $1:2$.

(D) The area of an equilateral triangle with side $a$ is given by $A = \frac{\sqrt{3}}{4} a^2$.
When each side is decreased by $2\, cm$,the new side becomes $(a - 2)$.
The new area is $A' = \frac{\sqrt{3}}{4} (a - 2)^2$.
According to the problem,the decrease in area is $4\sqrt{3}\, cm^2$,so $A - A' = 4\sqrt{3}$.
$\frac{\sqrt{3}}{4} a^2 - \frac{\sqrt{3}}{4} (a - 2)^2 = 4\sqrt{3}$.
Dividing both sides by $\frac{\sqrt{3}}{4}$,we get $a^2 - (a - 2)^2 = 16$.
Expanding the square: $a^2 - (a^2 - 4a + 4) = 16$.
$a^2 - a^2 + 4a - 4 = 16$.
$4a - 4 = 16$.
$4a = 20$.
$a = 5\, cm$.

(C) The circumference of the circular wire is equal to the perimeter of the rectangle.
Given diameter $d = 112 \, cm$.
Circumference of circle $= \pi \times d = \frac{22}{7} \times 112 = 22 \times 16 = 352 \, cm$.
Let the sides of the rectangle be $9x$ and $7x$.
The perimeter of the rectangle $= 2 \times (9x + 7x) = 2 \times 16x = 32x$.
Equating the perimeters: $32x = 352$.
$x = \frac{352}{32} = 11$.
The smaller side of the rectangle is $7x = 7 \times 11 = 77 \, cm$.

(D) The area of a triangle with sides $a, b, c$ is given by Heron's formula: $\text{Area} = \sqrt{S(S-a)(S-b)(S-c)}$,where $S = \frac{a+b+c}{2}$.
In $\Delta ABC$,the sides are $a = 60\, cm$,$b = 40\, cm$,and $c = 80\, cm$.
The semi-perimeter $S = \frac{60 + 40 + 80}{2} = \frac{180}{2} = 90\, cm$.
Area of $\Delta ABC = \sqrt{90(90-60)(90-40)(90-80)}$
$= \sqrt{90 \times 30 \times 50 \times 10} = \sqrt{1350000} = \sqrt{90000 \times 15} = 300 \sqrt{15}\, cm^2$.
$A$ diagonal divides a parallelogram into two triangles of equal area.
Therefore,the area of parallelogram $ABCD = 2 \times \text{Area of } \Delta ABC$.
Area $= 2 \times 300 \sqrt{15} = 600 \sqrt{15}\, cm^2$.

(A) Given,the ratio of breadth to length is $3:4$. Let the breadth be $3x$ and the length be $4x$.
The area of the lawn is $\frac{1}{12}$ hectare. Since $1 \text{ hectare} = 10000 \text{ m}^2$,the area is $\frac{1}{12} \times 10000 \text{ m}^2 = \frac{2500}{3} \text{ m}^2$.
The area of a rectangle is $\text{length} \times \text{breadth} = 4x \times 3x = 12x^2$.
Equating the two areas: $12x^2 = \frac{2500}{3}$.
$x^2 = \frac{2500}{3 \times 12} = \frac{2500}{36}$.
Taking the square root: $x = \sqrt{\frac{2500}{36}} = \frac{50}{6}$.
The breadth of the lawn is $3x = 3 \times \frac{50}{6} = \frac{50}{2} = 25 \text{ m}$.

(A) Let the side of the square be $x\,cm$.
According to the problem:
Area of the rectangle $= 3 \times$ Area of the square
The length of the rectangle is $20\,cm$ and the breadth is $\frac{3}{2}x$.
So,Area of the rectangle $= 20 \times \frac{3}{2}x = 30x$.
Area of the square $= x^2$.
Equating the areas:
$30x = 3 \times x^2$
Dividing both sides by $3x$ (since $x \neq 0$):
$10 = x$
Therefore,the side of the square is $10\,cm$.

(C) Let the diagonals of the rhombus be $d_1 = 12 \ cm$ and $d_2 = 16 \ cm$.
In a rhombus,the diagonals bisect each other at right angles $(90^{\circ})$.
This forms four right-angled triangles with legs equal to half the lengths of the diagonals.
Leg $1 = \frac{d_1}{2} = \frac{12}{2} = 6 \ cm$.
Leg $2 = \frac{d_2}{2} = \frac{16}{2} = 8 \ cm$.
The side of the rhombus acts as the hypotenuse of these right-angled triangles.
Using the Pythagorean theorem: $\text{Side} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \ cm$.

(D) The distance covered by the wheel in one revolution is equal to its circumference,which is given by the formula $C = \pi d$.
Given the diameter $d = 7\, m$,the circumference is $C = \frac{22}{7} \times 7 = 22\, m$.
The total distance to be covered is $22\, km = 22,000\, m$.
The number of revolutions is calculated by dividing the total distance by the distance covered in one revolution:
$\text{Number of revolutions} = \frac{\text{Total distance}}{\text{Circumference}} = \frac{22,000\, m}{22\, m} = 1000$.

(B) The area of an equilateral triangle is given by the formula $\frac{\sqrt{3}}{4} \times \text{side}^{2}$.
Given,$\frac{\sqrt{3}}{4} \times \text{side}^{2} = 9 \sqrt{3}$.
$\Rightarrow \text{side}^{2} = 9 \times 4 = 36$.
$\Rightarrow \text{side} = \sqrt{36} = 6 \ m$.
In an equilateral triangle,the median is also the altitude. Let $AD$ be the median to side $BC$. Since $D$ is the midpoint of $BC$,$BD = \frac{6}{2} = 3 \ m$.
Using the Pythagorean theorem in $\triangle ABD$:
$AD = \sqrt{AB^{2} - BD^{2}} = \sqrt{6^{2} - 3^{2}}$.
$AD = \sqrt{36 - 9} = \sqrt{27} = 3 \sqrt{3} \ m$.

(D) The length of the room is $8 \, m = 80 \, dm$ and the breadth is $6 \, m = 60 \, dm$.
Area of the floor $= 80 \, dm \times 60 \, dm = 4800 \, dm^2$.
Each tile is a square with a side of $4 \, dm$.
Area of one tile $= 4 \, dm \times 4 \, dm = 16 \, dm^2$.
Number of tiles required $= \frac{\text{Area of floor}}{\text{Area of one tile}} = \frac{4800}{16} = 300$.

(B) Let the side of the equilateral triangle be $a$.
The circumradius $R$ of an equilateral triangle is given by $R = \frac{a}{\sqrt{3}}$.
The area of the circumcircle is $\pi R^2 = 3 \pi$.
Substituting $R$,we get $\pi \left( \frac{a}{\sqrt{3}} \right)^2 = 3 \pi$.
$\frac{a^2}{3} = 3$.
$a^2 = 9$,which implies $a = 3 \, cm$.
The perimeter of the equilateral triangle is $3a = 3 \times 3 = 9 \, cm$.

(C) In $\triangle ABC$,$\angle BAC = 90^{\circ}$ and $AD \perp BC$.
Since $\triangle ABC \sim \triangle ACD$ (by $AA$ similarity criterion,as $\angle C = \angle C$ and $\angle ADC = \angle BAC = 90^{\circ}$),the ratio of their areas is equal to the square of the ratio of their corresponding sides.
Specifically,$\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle ACD)} = \left( \frac{BC}{AC} \right)^{2}$.
Given $BC = 8 \ cm$ and $AC = 6 \ cm$.
Therefore,$\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle ACD)} = \left( \frac{8}{6} \right)^{2} = \left( \frac{4}{3} \right)^{2} = \frac{16}{9}$.
Thus,the ratio is $16:9$.

(D) The ratio of the sides of the triangle is given as $\frac{1}{4}: \frac{1}{6}: \frac{1}{8}$.
To simplify this ratio,we multiply each term by the $LCM$ of $4, 6,$ and $8$,which is $24$.
Ratio $= (\frac{1}{4} \times 24) : (\frac{1}{6} \times 24) : (\frac{1}{8} \times 24) = 6: 4: 3$.
Let the sides be $6x, 4x,$ and $3x$.
The perimeter of the triangle is the sum of its sides,which is $91\, cm$.
So,$6x + 4x + 3x = 91$.
$13x = 91$.
$x = \frac{91}{13} = 7$.
The longest side is $6x = 6 \times 7 = 42\, cm$.
The shortest side is $3x = 3 \times 7 = 21\, cm$.
The difference between the longest and shortest side is $42\, cm - 21\, cm = 21\, cm$.

(B) Let the two equal sides of the isosceles right-angled triangle be $x \, cm$.
Then,the hypotenuse is $\sqrt{x^2 + x^2} = \sqrt{2}x \, cm$.
The perimeter is given as $x + x + \sqrt{2}x = 2p \, cm$.
$x(2 + \sqrt{2}) = 2p \implies x = \frac{2p}{2 + \sqrt{2}}$.
Rationalizing the denominator: $x = \frac{2p(2 - \sqrt{2})}{(2 + \sqrt{2})(2 - \sqrt{2})} = \frac{2p(2 - \sqrt{2})}{4 - 2} = p(2 - \sqrt{2})$.
The area of the triangle is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} x^2$.
Area $= \frac{1}{2} [p(2 - \sqrt{2})]^2 = \frac{1}{2} p^2 (4 + 2 - 4\sqrt{2}) = \frac{1}{2} p^2 (6 - 4\sqrt{2}) = (3 - 2\sqrt{2}) p^2 \, cm^2$.

(A) Let the radii of the two circles be $r_1$ and $r_2$ respectively.
The area of a circle is given by the formula $A = \pi r^2$.
According to the problem,the ratio of the areas is $\frac{\pi r_1^2}{\pi r_2^2} = \frac{4}{7}$.
By simplifying,we get $\frac{r_1^2}{r_2^2} = \frac{4}{7}$.
Taking the square root on both sides,we get $\frac{r_1}{r_2} = \sqrt{\frac{4}{7}} = \frac{2}{\sqrt{7}}$.
Therefore,the ratio of their radii is $2: \sqrt{7}$.

(D) Let the rhombus be $ABCD$ with diagonals $AC$ and $BD$ intersecting at $O$.
The perimeter of the rhombus is $20 \, cm$.
Side of the rhombus $(s) = \frac{20}{4} = 5 \, cm$.
Let diagonal $d_1 = BD = 8 \, cm$. Since diagonals of a rhombus bisect each other at right angles,$OB = \frac{d_1}{2} = 4 \, cm$.
In the right-angled triangle $\triangle AOB$,by Pythagoras theorem:
$OA^2 + OB^2 = AB^2$
$OA^2 + 4^2 = 5^2$
$OA^2 + 16 = 25$
$OA^2 = 9 \implies OA = 3 \, cm$.
Therefore,the other diagonal $d_2 = AC = 2 \times OA = 2 \times 3 = 6 \, cm$.
Area of the rhombus $= \frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 8 \times 6 = 24 \, cm^2$.

(A) In the given figure,there are four circles arranged in a $2 \times 2$ grid inside a square. The diameter of each circle is $14 \, cm$.
The side length of the square containing these circles is $2 \times 14 \, cm = 28 \, cm$. Thus,$DC = 28 \, cm$ and $BC = 28 \, cm$.
Given that $DC = CE$,we have $CE = 28 \, cm$.
The base of $\Delta BDE$ is $DE = DC + CE = 28 \, cm + 28 \, cm = 56 \, cm$.
The height of $\Delta BDE$ with respect to base $DE$ is $BC = 28 \, cm$.
Area of $\Delta BDE = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times DE \times BC$.
Area $= \frac{1}{2} \times 56 \times 28 = 28 \times 28 = 784 \, cm^2$.

(B) The side length of the square $ABCD$ is $28 \ cm$.
Area of the square $= \text{side} \times \text{side} = 28 \times 28 = 784 \ cm^2$.
Since there are four identical circles inscribed in the square,the diameter of each circle is half the side of the square.
Diameter of each circle $= 28 / 2 = 14 \ cm$.
Radius of each circle $(r) = 14 / 2 = 7 \ cm$.
Area of one circle $= \pi r^2 = (22 / 7) \times 7 \times 7 = 154 \ cm^2$.
Area of four such circles $= 4 \times 154 = 616 \ cm^2$.
Area of the shaded region $= \text{Area of square} - \text{Area of four circles} = 784 - 616 = 168 \ cm^2$.

(D) From statement $I$,the hypotenuse $b = 5\, cm$.
From statement $III$,one of the angles is $60^{\circ}$.
In a right-angled triangle $ABC$ (right-angled at $B$),if $\angle C = 60^{\circ}$,then $\angle A = 30^{\circ}$.
Using trigonometry,$\cos 60^{\circ} = \frac{BC}{AC} = \frac{a}{b}$.
Thus,$a = b \cos 60^{\circ} = 5 \times \frac{1}{2} = 2.5\, cm$.
Also,$\sin 60^{\circ} = \frac{AB}{AC} = \frac{c}{b}$.
Thus,$c = b \sin 60^{\circ} = 5 \times \frac{\sqrt{3}}{2} = 2.5\sqrt{3}\, cm$.
The area of the triangle is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times a \times c$.
Area $= \frac{1}{2} \times 2.5 \times 2.5\sqrt{3} = \frac{6.25\sqrt{3}}{2} = 3.125\sqrt{3}\, cm^{2} = \frac{25\sqrt{3}}{8}\, cm^{2}$.
Therefore,statements $I$ and $III$ together are sufficient to answer the question.

(B) The semi-perimeter of the triangle $(s)$ is calculated as:
$s = \frac{50 + 78 + 112}{2} = \frac{240}{2} = 120 \, cm$.
Using Heron's formula,the area of the triangle is:
$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$
$= \sqrt{120(120-50)(120-78)(120-112)}$
$= \sqrt{120 \times 70 \times 42 \times 8} = \sqrt{2822400} = 1680 \, cm^2$.
The altitude is smallest when the base is the largest side.
Let the base $b = 112 \, cm$ and the smallest altitude be $h$.
$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$
$1680 = \frac{1}{2} \times 112 \times h$
$1680 = 56 \times h$
$h = \frac{1680}{56} = 30 \, cm$.

(B) Given:
$AB + BC = 12 \text{ cm}$
$BC + CA = 14 \text{ cm}$
$CA + AB = 18 \text{ cm}$
Adding these three equations:
$(AB + BC) + (BC + CA) + (CA + AB) = 12 + 14 + 18$
$2(AB + BC + CA) = 44$
$AB + BC + CA = 22 \text{ cm}$
The perimeter of the triangle is $22 \text{ cm}$.
Let $r$ be the radius of the circle. The perimeter (circumference) of the circle is $2 \pi r$.
According to the problem,the perimeter of the circle is equal to the perimeter of the triangle:
$2 \pi r = 22$
$2 \times \frac{22}{7} \times r = 22$
$r = \frac{22 \times 7}{2 \times 22}$
$r = \frac{7}{2} \text{ cm}$

(A) The rate is $25$ $paisa$ per $m^2$,which is equal to $₹ \frac{25}{100} = ₹ \frac{1}{4}$ per $m^2$.
Area of the rectangular field = $\frac{\text{Total Expenditure}}{\text{Rate per } m^2} = \frac{1000}{1/4} = 4000\, m^2$.
Since the area of a rectangle is $\text{length} \times \text{breadth}$,we have $4000 = \text{length} \times 50$.
Therefore,the original length = $\frac{4000}{50} = 80\, m$.
If the length is increased by $20\, m$,the new length = $80 + 20 = 100\, m$.
The new area = $100\, m \times 50\, m = 5000\, m^2$.
The new expenditure = $\text{New Area} \times \text{Rate} = 5000 \times \frac{1}{4} = ₹ 1250$.

(B) Let the medians be $m_1 = 9 \, cm$,$m_2 = 12 \, cm$,and $m_3 = 15 \, cm$.
The area of a triangle with medians $m_1, m_2, m_3$ is given by the formula:
Area $= \frac{4}{3} \sqrt{s_m(s_m - m_1)(s_m - m_2)(s_m - m_3)}$,where $s_m = \frac{m_1 + m_2 + m_3}{2}$.
Here,$s_m = \frac{9 + 12 + 15}{2} = \frac{36}{2} = 18 \, cm$.
Area $= \frac{4}{3} \sqrt{18(18 - 9)(18 - 12)(18 - 15)}$
Area $= \frac{4}{3} \sqrt{18 \times 9 \times 6 \times 3}$
Area $= \frac{4}{3} \sqrt{2916}$
Area $= \frac{4}{3} \times 54 = 72 \, cm^2$.

(D) The perimeter of the rectangle is $P = 2(l + b) = 2(18 + 26) = 2(44) = 88 \text{ cm}$.
Since the perimeter of the circle is equal to the perimeter of the rectangle,we have $2 \pi r = 88 \text{ cm}$.
Substituting $\pi = \frac{22}{7}$,we get $2 \times \frac{22}{7} \times r = 88$.
$\frac{44}{7} \times r = 88 \Rightarrow r = 88 \times \frac{7}{44} = 14 \text{ cm}$.
The area of the circle is $A = \pi r^2 = \frac{22}{7} \times 14 \times 14$.
$A = 22 \times 2 \times 14 = 616 \text{ cm}^2$.

(A) Let the original radius of the circle be $r\, cm$.
The area of the original circle is $A_1 = \pi r^{2}$.
When the radius is increased by $1\, cm$,the new radius becomes $(r + 1)\, cm$.
The area of the new circle is $A_2 = \pi(r + 1)^{2}$.
According to the problem,the increase in area is $22\, cm^{2}$,so $A_2 - A_1 = 22$.
$\pi(r + 1)^{2} - \pi r^{2} = 22$
$\pi(r^{2} + 2r + 1 - r^{2}) = 22$
$\pi(2r + 1) = 22$
Substituting $\pi = \frac{22}{7}$,we get:
$\frac{22}{7}(2r + 1) = 22$
Dividing both sides by $22$:
$\frac{1}{7}(2r + 1) = 1$
$2r + 1 = 7$
$2r = 6$
$r = 3\, cm$.
Thus,the original radius of the circle is $3\, cm$.

(D) The sum of all interior angles of a polygon with $n$ sides is given by the formula $(n-2) \times 180^{\circ}$.
The sum of all exterior angles of any convex polygon is always $360^{\circ}$.
According to the problem,the sum of interior angles is twice the sum of exterior angles:
$(n-2) \times 180^{\circ} = 2 \times 360^{\circ}$
Divide both sides by $180^{\circ}$:
$n-2 = 2 \times 2$
$n-2 = 4$
$n = 6$
Therefore,the number of sides of the polygon is $6$.

(A) Let the rhombus be $ABCD$ with diagonals $AC = 8$ and $BD = 6.$ The diagonals of a rhombus bisect each other at right angles at point $O.$
Therefore,$BO = \frac{BD}{2} = \frac{6}{2} = 3$ units and $OC = \frac{AC}{2} = \frac{8}{2} = 4$ units.
In the right-angled triangle $\triangle BOC,$ by the Pythagorean theorem:
$BC^2 = BO^2 + OC^2$
$BC^2 = 3^2 + 4^2$
$BC^2 = 9 + 16 = 25$ sq. units.
Thus,the square of its side is $25.$

(C) The diagonal of a square inscribed in a circle is equal to the diameter of the circle.
Given radius $r = 8\, cm$.
Diameter $d = 2 \times r = 2 \times 8 = 16\, cm$.
Therefore,the diagonal of the square $BD = 16\, cm$.
The area of a square can be calculated using its diagonal as:
Area $= \frac{1}{2} \times (\text{diagonal})^2$
Area $= \frac{1}{2} \times (16)^2$
Area $= \frac{1}{2} \times 256 = 128\, cm^2$.

(D) Area of the square $= 1444 \ m^2$.
Let the side of the square be $a$.
So,$a^2 = 1444$.
$\therefore a = \sqrt{1444} = 38 \ m$.
Breadth of the rectangle $= \frac{1}{4} \times 38 = 9.5 \ m$.
Length of the rectangle $= 3 \times 9.5 = 28.5 \ m$.
Area of the rectangle $= \text{Length} \times \text{Breadth} = 28.5 \times 9.5 = 270.75 \ m^2$.
Difference between the area of the square and the area of the rectangle $= 1444 - 270.75 = 1173.25 \ m^2$.

(A) Let the length of the floor be $32x$ and the breadth be $21x$.
Given that the perimeter of the floor is $212$ feet.
The formula for the perimeter of a rectangle is $2 \times (\text{length} + \text{breadth})$.
So,$2(32x + 21x) = 212$.
$2(53x) = 212$.
$106x = 212$.
$x = 2$.
Therefore,length $= 32 \times 2 = 64$ feet and breadth $= 21 \times 2 = 42$ feet.
The area of the floor is $\text{length} \times \text{breadth} = 64 \times 42 = 2688$ square feet.
The cost of laying the carpet is $\text{Area} \times \text{cost per square foot} = 2688 \times 2.5 = ₹ 6720$.

(D) For the smaller circle,the circumference is $C_1 = 2 \pi r_1 = 88 \, m$.
$r_1 = \frac{88 \times 7}{2 \times 22} = 14 \, m$.
Area $A_1 = \pi r_1^2 = \frac{22}{7} \times 14 \times 14 = 616 \, m^2$.
For the larger circle,the circumference is $C_2 = 2 \pi r_2 = 220 \, m$.
$r_2 = \frac{220 \times 7}{2 \times 22} = 35 \, m$.
Area $A_2 = \pi r_2^2 = \frac{22}{7} \times 35 \times 35 = 3850 \, m^2$.
The difference between the areas is $A_2 - A_1 = 3850 - 616 = 3234 \, m^2$.

(D) The figure consists of a central rectangle and two semicircles at the ends.
Width of the rectangle $= 28 \ cm$,so the radius of the semicircles $r = 14 \ cm$.
The total length is $53 \ cm$. The length of the rectangular part $= 53 - 14 - 14 = 25 \ cm$.
Area of the rectangle $= \text{length} \times \text{width} = 25 \times 28 = 700 \ cm^2$.
Area of two semicircles $= 2 \times (\frac{1}{2} \pi r^2) = \pi r^2 = \frac{22}{7} \times 14 \times 14 = 22 \times 2 \times 14 = 616 \ cm^2$.
Total area $= 700 + 616 = 1316 \ cm^2$.
Note: The provided solution in the prompt was incorrect. Based on the geometry,the correct area is $1316 \ cm^2$. Since this is not in the options,the correct choice is $D$.

(B) Let the radius of the circle be $r$ and the side of the square be $a = 22\, cm$.
Given that the circumference of the circle is equal to the area of the square:
$2\pi r = a^2$
$2 \times \frac{22}{7} \times r = (22)^2$
$2 \times \frac{22}{7} \times r = 484$
$r = \frac{484 \times 7}{2 \times 22} = \frac{484 \times 7}{44} = 11 \times 7 = 77\, cm$.
The diameter of the circle is $d = 2r = 2 \times 77 = 154\, cm$.
The length of the rectangle $(l)$ is twice the diameter of the circle:
$l = 2 \times d = 2 \times 154 = 308\, cm$.
The perimeter of the rectangle is given as $668\, cm$:
$2(l + b) = 668$
$l + b = 334$
$308 + b = 334$
$b = 334 - 308 = 26\, cm$.
Thus,the breadth of the rectangle is $26\, cm$.

(C) The side of the first square is $\sqrt{\text{Area}} = \sqrt{200} = 10\sqrt{2}\, m$.
The diagonal of the first square is $\sqrt{2} \times \text{side} = \sqrt{2} \times 10\sqrt{2} = 20\, m$.
The diagonal of the new square is $\sqrt{2} \times 20 = 20\sqrt{2}\, m$.
The area of a square is given by $\frac{1}{2} \times (\text{diagonal})^2$.
Therefore,the area of the new square is $\frac{1}{2} \times (20\sqrt{2})^2 = \frac{1}{2} \times 400 \times 2 = 400\, m^2$.

(D) Let the original length be $L$ and the original breadth be $B.$ The original area is $A = L \times B.$
According to the problem,the new length $L' = L + 0.20L = 1.2L$ and the new breadth $B' = B - 0.20B = 0.8B.$
The new area is given as $192 \, m^2.$
Therefore,$L' \times B' = 192.$
Substituting the values,$(1.2L) \times (0.8B) = 192.$
$0.96 \times (L \times B) = 192.$
Since $L \times B = A,$ we have $0.96 \times A = 192.$
$A = \frac{192}{0.96} = \frac{19200}{96} = 200 \, m^2.$
Thus,the area of the original rectangle is $200 \, m^2.$

(D) The side of the square plot is $a = 28\, m$.
The area of the square plot is $a^2 = 28^2 = 784\, m^2$.
The diameter of the circular garden is $d = 28\, m$,so the radius is $r = \frac{d}{2} = 14\, m$.
The area of the circular garden is $\pi r^2 = \frac{22}{7} \times 14 \times 14 = 22 \times 2 \times 14 = 616\, m^2$.
The area of the space left out is the difference between the area of the square and the area of the circle:
Area left out $= 784\, m^2 - 616\, m^2 = 168\, m^2$.

(C) To find the cost of flooring,we need the total area of the hall and the rate of flooring per unit area.
Step $1$: From statement $I$,length $L = 6 \ m$.
Step $2$: From statement $II$,breadth $B = (2/3) \times 6 = 4 \ m$.
Step $3$: Area of the hall = $L \times B = 6 \ m \times 4 \ m = 24 \ m^2$.
Step $4$: Statement $III$ provides the rate of flooring. By combining all three statements,we can calculate the total cost = $\text{Area} \times \text{Rate}$.
Therefore,all three statements are required.

(C) Let the side of the square be $a \, cm$.
Given that the radius of the circle is equal to the side of the square,so radius $r = a$.
The circumference of the circle is $2 \pi r = 132 \, cm$.
Using $\pi = 22/7$,we have $2 \times (22/7) \times a = 132$.
$a = (132 \times 7) / (2 \times 22) = 66 \times 7 / 22 = 3 \times 7 = 21 \, cm$.
The length of the rectangle is $3/5$ of the side of the square,so length $l = (3/5) \times 21 = 63/5 = 12.6 \, cm$.
The breadth of the rectangle is given as $b = 8 \, cm$.
The area of the rectangle is $l \times b = 12.6 \times 8 = 100.8 \, cm^2$.

(C) Step $1$: Find the side of the square.
Perimeter of the square $= 56\, cm$.
Side of the square $= \frac{56}{4} = 14\, cm$.
Step $2$: Find the smallest side of the right-angled triangle.
Smallest side $= 14 - 8 = 6\, cm$.
Step $3$: Find the length of the rectangle.
Area of the rectangle $= 96\, cm^2$ and breadth $= 8\, cm$.
Length $= \frac{96}{8} = 12\, cm$.
Step $4$: Find the second largest side of the right-angled triangle.
Second largest side $= 12 - 4 = 8\, cm$.
Step $5$: Find the largest side (hypotenuse) of the right-angled triangle.
Using the Pythagorean theorem,$c = \sqrt{a^2 + b^2}$.
Largest side $= \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\, cm$.

(C) Circumference of the circle $= \pi \times \text{diameter} = \frac{22}{7} \times 56 = 176 \, cm$.
Perimeter of the square $= 272 - 176 = 96 \, cm$.
Side of the square $= \frac{96}{4} = 24 \, cm$.
Area of the square $= 24 \times 24 = 576 \, cm^2$.
Radius of the circle $= \frac{56}{2} = 28 \, cm$.
Area of the circle $= \pi r^2 = \frac{22}{7} \times 28 \times 28 = 2464 \, cm^2$.
Required sum of areas $= 576 + 2464 = 3040 \, cm^2$.

(C) Let the largest angle be $4x$ and the second largest angle be $3x$.
Given that the smallest angle is half the largest angle,so the smallest angle $= \frac{1}{2} \times 4x = 2x$.
The sum of the angles in a triangle is $180^{\circ}$.
Therefore,$4x + 3x + 2x = 180^{\circ}$.
$9x = 180^{\circ}$.
$x = 20^{\circ}$.
The largest angle $= 4x = 4 \times 20^{\circ} = 80^{\circ}$.
The smallest angle $= 2x = 2 \times 20^{\circ} = 40^{\circ}$.
The difference between the largest and the smallest angle $= 80^{\circ} - 40^{\circ} = 40^{\circ}$.

(D) Let the three angles of the quadrilateral be $13x^{\circ}$,$9x^{\circ}$,and $5x^{\circ}$ respectively.
The sum of all interior angles of a quadrilateral is $360^{\circ}$.
Given the $4^{th}$ angle is $36^{\circ}$,we have:
$13x + 9x + 5x + 36^{\circ} = 360^{\circ}$
$27x = 360^{\circ} - 36^{\circ}$
$27x = 324^{\circ}$
$x = \frac{324}{27} = 12$
The angles are:
$1^{st} = 13 \times 12 = 156^{\circ}$
$2^{nd} = 9 \times 12 = 108^{\circ}$
$3^{rd} = 5 \times 12 = 60^{\circ}$
$4^{th} = 36^{\circ}$
The largest angle is $156^{\circ}$ and the second smallest angle is $60^{\circ}$ (since $36 < 60 < 108 < 156$).
Difference $= 156^{\circ} - 60^{\circ} = 96^{\circ}$.

(D) For the smaller circle,the circumference $C_1 = 2 \pi r = 88 \text{ m}$.
Using $\pi = \frac{22}{7}$,we have $2 \times \frac{22}{7} \times r = 88$,which gives $r = \frac{88 \times 7}{44} = 14 \text{ m}$.
For the larger circle,the circumference $C_2 = 2 \pi R = 220 \text{ m}$.
$2 \times \frac{22}{7} \times R = 220$,which gives $R = \frac{220 \times 7}{44} = 35 \text{ m}$.
The difference between the areas is $\pi R^2 - \pi r^2 = \pi(R^2 - r^2)$.
Difference $= \frac{22}{7} \times (35^2 - 14^2) = \frac{22}{7} \times (35 - 14)(35 + 14)$.
Difference $= \frac{22}{7} \times 21 \times 49 = 22 \times 3 \times 49 = 3234 \text{ m}^2$.

(C) Area of the square $= a^2 = 196 \, cm^2$.
Therefore,the side $a = \sqrt{196} = 14 \, cm$.
Given that the side of the square is half the radius $(r)$ of the circle,so $a = r/2$.
Therefore,$r = 2 \times a = 2 \times 14 = 28 \, cm$.
The circumference of the circle $= 2 \pi r = 2 \times (22/7) \times 28 = 176 \, cm$.
According to the question,the breadth $(b)$ of the rectangle is equal to the circumference of the circle,so $b = 176 \, cm$.
The perimeter of the rectangle is given as $2(l + b) = 712 \, cm$.
Substituting the value of $b$: $2(l + 176) = 712$.
Dividing by $2$: $l + 176 = 356$.
Therefore,$l = 356 - 176 = 180 \, cm$.

(A) Let the perpendicular sides of the right-angled triangle be $5x$ and $12x$.
According to the question,the area of the triangle is given by:
$\frac{1}{2} \times \text{base} \times \text{height} = 270 \, cm^2$
$\frac{1}{2} \times 5x \times 12x = 270$
$30x^2 = 270$
$x^2 = \frac{270}{30} = 9$
$x = 3$
Now,the sides are $5 \times 3 = 15 \, cm$ and $12 \times 3 = 36 \, cm$.
The hypotenuse is given by the Pythagorean theorem:
$\text{Hypotenuse} = \sqrt{(15)^2 + (36)^2} = \sqrt{225 + 1296} = \sqrt{1521} = 39 \, cm$.
Alternatively,using the ratio $5:12:13$,the hypotenuse is $13x = 13 \times 3 = 39 \, cm$.

(B) Let the length and breadth of the rectangle be $a$ and $b$ respectively.
Given,diagonal $d = 25\, cm$ and area $A = 168\, cm^2$.
We know that $a^2 + b^2 = d^2 = 25^2 = 625$.
Also,$ab = 168$.
Using the identity $(a+b)^2 = a^2 + b^2 + 2ab$,we get $(a+b)^2 = 625 + 2(168) = 625 + 336 = 961$.
Therefore,$a+b = \sqrt{961} = 31\, cm$ (Equation $1$).
Using the identity $(a-b)^2 = a^2 + b^2 - 2ab$,we get $(a-b)^2 = 625 - 336 = 289$.
Therefore,$a-b = \sqrt{289} = 17\, cm$ (Equation $2$).
Adding Equation $1$ and Equation $2$: $2a = 31 + 17 = 48$,which gives $a = 24\, cm$.
Subtracting Equation $2$ from Equation $1$: $2b = 31 - 17 = 14$,which gives $b = 7\, cm$.
Thus,the length of the rectangle is $24\, cm$.

(B) Let the number of men standing in each row be $x$.
The total number of men arranged in a square formation is given by $x^2$.
According to the problem,the total number of men is $6000$,and $71$ men are left over after forming the square.
Therefore,the equation is: $x^2 + 71 = 6000$.
Subtracting $71$ from both sides,we get: $x^2 = 6000 - 71 = 5929$.
To find $x$,we calculate the square root of $5929$: $x = \sqrt{5929} = 77$.
Thus,there were $77$ men arranged in each row.

(A) Let the original length be $L$ and the original breadth be $B$. The original area is $A_1 = L \times B$.
After the change,the new length $L' = L + 0.10L = 1.10L$ and the new breadth $B' = B - 0.10B = 0.90B$.
The new area $A_2 = L' \times B' = (1.10L) \times (0.90B) = 0.99LB = 0.99A_1$.
The change in area is $A_2 - A_1 = 0.99A_1 - A_1 = -0.01A_1$.
This represents a decrease of $1 \%$.

(B) Let the original length be $L_1 = 6x$ and the original breadth be $B_1 = 5y$.
Original Area $A_1 = L_1 \times B_1 = 6x \times 5y = 30xy$.
After the change,the new length $L_2 = 7x$ and the new breadth $B_2 = 4y$.
New Area $A_2 = L_2 \times B_2 = 7x \times 4y = 28xy$.
The ratio of the new area to the original area is $A_2 : A_1 = 28xy : 30xy = 28 : 30$.
Simplifying the ratio by dividing by $2$,we get $14 : 15$.
Since the question asks for the ratio in which the area is diminished (original to new),the ratio is $30 : 28$,which simplifies to $15 : 14$.

(C) The horses are tethered at the vertices of the triangle with rope length $r = 7 \, m$. The area grazed by each horse is a sector of a circle with radius $r$ and central angle equal to the interior angle of the triangle at that vertex.
Let the angles at vertices $A, B,$ and $C$ be $\angle A, \angle B,$ and $\angle C$ respectively.
The sum of the interior angles of a triangle is $180^{\circ}$.
Total grazed area $= \text{Area of sector at } A + \text{Area of sector at } B + \text{Area of sector at } C$
$= \frac{\angle A}{360^{\circ}} \pi r^2 + \frac{\angle B}{360^{\circ}} \pi r^2 + \frac{\angle C}{360^{\circ}} \pi r^2$
$= \frac{(\angle A + \angle B + \angle C)}{360^{\circ}} \pi r^2$
Since $\angle A + \angle B + \angle C = 180^{\circ}$,
Total grazed area $= \frac{180^{\circ}}{360^{\circ}} \pi r^2 = \frac{1}{2} \pi r^2$
$= \frac{1}{2} \times \frac{22}{7} \times 7^2 = \frac{1}{2} \times \frac{22}{7} \times 49 = 11 \times 7 = 77 \, m^2$.