Area of some geometrical figures Questions in English

(B) Given sides of the triangle are $a = 4 \, cm, b = 5 \, cm, c = 6 \, cm$.
Semi-perimeter $s = \frac{a + b + c}{2} = \frac{4 + 5 + 6}{2} = \frac{15}{2} \, cm$.
Using Heron's formula,Area $= \sqrt{s(s - a)(s - b)(s - c)}$.
Area $= \sqrt{\frac{15}{2} \left( \frac{15}{2} - 4 \right) \left( \frac{15}{2} - 5 \right) \left( \frac{15}{2} - 6 \right)}$.
Area $= \sqrt{\frac{15}{2} \times \frac{7}{2} \times \frac{5}{2} \times \frac{3}{2}}$.
Area $= \sqrt{\frac{15 \times 7 \times 15}{16}} = \frac{15}{4} \sqrt{7} \, cm^2$.

(D) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Substituting the given vertices:
$x_1 = a \cos \theta, y_1 = b \sin \theta$
$x_2 = -a \sin \theta, y_2 = b \cos \theta$
$x_3 = -a \cos \theta, y_3 = -b \sin \theta$
Area $= \frac{1}{2} |a \cos \theta (b \cos \theta - (-b \sin \theta)) + (-a \sin \theta)(-b \sin \theta - b \sin \theta) + (-a \cos \theta)(b \sin \theta - b \cos \theta)|$
$= \frac{1}{2} |a \cos \theta (b \cos \theta + b \sin \theta) - a \sin \theta (-2b \sin \theta) - a \cos \theta (b \sin \theta - b \cos \theta)|$
$= \frac{ab}{2} |\cos^2 \theta + \cos \theta \sin \theta + 2 \sin^2 \theta - \sin \theta \cos \theta + \cos^2 \theta|$
$= \frac{ab}{2} |2 \cos^2 \theta + 2 \sin^2 \theta|$
$= \frac{ab}{2} \times 2(\cos^2 \theta + \sin^2 \theta) = ab(1) = ab$.

(A) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by the formula:
Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Substituting the given vertices $(-4, 1), (1, 2), (4, -3)$:
Area $= \frac{1}{2} |-4(2 - (-3)) + 1(-3 - 1) + 4(1 - 2)|$
Area $= \frac{1}{2} |-4(5) + 1(-4) + 4(-1)|$
Area $= \frac{1}{2} |-20 - 4 - 4|$
Area $= \frac{1}{2} |-28| = 14$ square units.

(B) The quadrilateral $PQRS$ formed by drawing lines through the vertices parallel to the opposite sides is a parallelogram.
The area of this parallelogram $PQRS$ is twice the area of the triangle $\Delta PQR$.
The area of $\Delta PQR$ with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by:
$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Substituting the coordinates $P(2, 1), Q(4, -1), R(3, 2)$:
$\text{Area}(\Delta PQR) = \frac{1}{2} |2(-1 - 2) + 4(2 - 1) + 3(1 - (-1))|$
$= \frac{1}{2} |2(-3) + 4(1) + 3(2)|$
$= \frac{1}{2} |-6 + 4 + 6| = \frac{1}{2} |4| = 2$
Therefore,the area of the parallelogram $PQRS = 2 \times \text{Area}(\Delta PQR) = 2 \times 2 = 4$.

(A) The vertices of triangle $ABC$ are $A(2, 1)$,$B(4, 3)$,and $C(2, 5)$.
The mid-points $D$,$E$,and $F$ are calculated as follows:
$D = \left( \frac{2+4}{2}, \frac{1+3}{2} \right) = (3, 2)$
$E = \left( \frac{4+2}{2}, \frac{3+5}{2} \right) = (3, 4)$
$F = \left( \frac{2+2}{2}, \frac{1+5}{2} \right) = (2, 3)$
The area of triangle $DEF$ is given by the determinant formula:
Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Area $= \frac{1}{2} |3(4 - 3) + 3(3 - 2) + 2(2 - 4)|$
Area $= \frac{1}{2} |3(1) + 3(1) + 2(-2)|$
Area $= \frac{1}{2} |3 + 3 - 4| = \frac{1}{2} |2| = 1$ sq. units.
Alternatively,the area of the triangle formed by joining the mid-points is $\frac{1}{4}$ of the area of the original triangle $ABC$.

(C) The equation $|x| + |y| = 1$ represents a square with vertices at $(1, 0)$,$(0, 1)$,$(-1, 0)$,and $(0, -1)$.
The distance between consecutive vertices (e.g.,$(1, 0)$ and $(0, 1)$) is the side length $s = \sqrt{(1-0)^2 + (0-1)^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2}$.
The area of a square is given by $s^2$.
Therefore,the area $= (\sqrt{2})^2 = 2$ square units.
Alternatively,the area of the square formed by $|x| + |y| = a$ is $2a^2$. Here $a = 1$,so the area is $2(1)^2 = 2$.

(B) Let the midpoints be $D(0, 0), E(1, 2),$ and $F(-3, 4)$.
The area of triangle $DEF$ is given by the formula:
Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Area of $\Delta DEF = \frac{1}{2} |0(2 - 4) + 1(4 - 0) + (-3)(0 - 2)|$
$= \frac{1}{2} |0 + 4 + 6| = \frac{1}{2} \times 10 = 5$.
The area of triangle $ABC$ is $4$ times the area of the triangle formed by joining the midpoints.
Area of $\Delta ABC = 4 \times \text{Area of } \Delta DEF = 4 \times 5 = 20$.

(B) The area of a quadrilateral with vertices $(x_1, y_1)$,$(x_2, y_2)$,$(x_3, y_3)$,and $(x_4, y_4)$ is given by the formula: $\text{Area} = \frac{1}{2} |(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)|$.
Substituting the given coordinates $(1, 1)$,$(3, 4)$,$(5, -2)$,and $(4, -7)$:
$\text{Area} = \frac{1}{2} |(1 \times 4 + 3 \times -2 + 5 \times -7 + 4 \times 1) - (1 \times 3 + 4 \times 5 + -2 \times 4 + -7 \times 1)|$.
$\text{Area} = \frac{1}{2} |(4 - 6 - 35 + 4) - (3 + 20 - 8 - 7)|$.
$\text{Area} = \frac{1}{2} |(-33) - (8)|$.
$\text{Area} = \frac{1}{2} |-41| = 20.5 \text{ square units}$.

(C) The vertices of the square $OPQR$ are $O(0, 0)$,$P(\alpha, 0)$,$Q(\alpha, \alpha)$,and $R(0, \alpha)$.
Area of square $OPQR = \alpha^2$.
$M$ is the midpoint of $PQ$,so $M = (\frac{\alpha + \alpha}{2}, \frac{0 + \alpha}{2}) = (\alpha, \frac{\alpha}{2})$.
$N$ is the midpoint of $QR$,so $N = (\frac{0 + \alpha}{2}, \frac{\alpha + \alpha}{2}) = (\frac{\alpha}{2}, \alpha)$.
The coordinates of the vertices of $\triangle OMN$ are $O(0, 0)$,$M(\alpha, \frac{\alpha}{2})$,and $N(\frac{\alpha}{2}, \alpha)$.
Area of $\triangle OMN = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
$= \frac{1}{2} |0(\frac{\alpha}{2} - \alpha) + \alpha(\alpha - 0) + \frac{\alpha}{2}(0 - \frac{\alpha}{2})|$
$= \frac{1}{2} |\alpha^2 - \frac{\alpha^2}{4}| = \frac{1}{2} |\frac{3\alpha^2}{4}| = \frac{3\alpha^2}{8}$.
Ratio = $\frac{\text{Area of square}}{\text{Area of } \triangle OMN} = \frac{\alpha^2}{\frac{3\alpha^2}{8}} = \frac{8}{3} = 8 : 3$.

(C) The area of a polygon with vertices $(x_1, y_1), (x_2, y_2), \dots, (x_n, y_n)$ is given by the formula: $\text{Area} = \frac{1}{2} |(x_1y_2 - y_1x_2) + (x_2y_3 - y_2x_3) + (x_3y_4 - y_3x_4) + (x_4y_5 - y_4x_5) + (x_5y_1 - y_5x_1)|$. \\ Substituting the given points $A(1, 1), B(7, 21), C(7, -3), D(12, 2), E(0, -3)$: \\ $\text{Area} = \frac{1}{2} |(1 \times 21 - 1 \times 7) + (7 \times -3 - 21 \times 7) + (7 \times 2 - (-3) \times 12) + (12 \times -3 - 2 \times 0) + (0 \times 1 - (-3) \times 1)|$ \\ $\text{Area} = \frac{1}{2} |(21 - 7) + (-21 - 147) + (14 + 36) + (-36 - 0) + (0 + 3)|$ \\ $\text{Area} = \frac{1}{2} |14 - 168 + 50 - 36 + 3| = \frac{1}{2} |-137| = \frac{137}{2} \text{ square units}$.

(B) The area of the triangle formed by joining the midpoints of the sides of a triangle is one-fourth of the area of the original triangle.
First,calculate the area of $\Delta ABC$ using the coordinates $A(5, -1), B(-7, 6),$ and $C(1, 3)$.
Area of $\Delta ABC = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
$= \frac{1}{2} |5(6 - 3) + (-7)(3 - (-1)) + 1(-1 - 6)|$
$= \frac{1}{2} |5(3) - 7(4) + 1(-7)|$
$= \frac{1}{2} |15 - 28 - 7|$
$= \frac{1}{2} |-20| = 10 \text{ sq. units}$.
Since the area of $\Delta DEF = \frac{1}{4} \times \text{Area of } \Delta ABC$,
Area of $\Delta DEF = \frac{1}{4} \times 10 = 2.5 \text{ sq. units}$.

(D) Let the midpoints be $M_1(1, 2)$,$M_2(-1, 1)$,and $M_3(0, 3)$.
The area of the triangle formed by the midpoints of the sides of a triangle is one-fourth the area of the original triangle.
Area of the triangle formed by midpoints $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Area $= \frac{1}{2} |1(1 - 3) + (-1)(3 - 2) + 0(2 - 1)|$
Area $= \frac{1}{2} |1(-2) - 1(1) + 0| = \frac{1}{2} |-2 - 1| = \frac{1}{2} |-3| = 1.5 \text{ sq. units}$.
The area of the original triangle $= 4 \times (\text{Area of the triangle formed by midpoints}) = 4 \times 1.5 = 6 \text{ sq. units}$.

(A) The given equation is $|x + y| + |x - y| = 11$.
Let $u = x + y$ and $v = x - y$. The equation becomes $|u| + |v| = 11$.
This represents a square in the $uv$-plane with vertices at $(11, 0), (0, 11), (-11, 0), (0, -11)$.
The area of this square in the $uv$-plane is $2 \times (11)^2 = 242$.
The transformation from $(x, y)$ to $(u, v)$ is given by $u = x + y$ and $v = x - y$.
The Jacobian of this transformation is $J = \begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{vmatrix} = \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} = -1 - 1 = -2$.
The area in the $xy$-plane is given by $\text{Area}_{xy} = \frac{\text{Area}_{uv}}{|J|} = \frac{242}{|-2|} = \frac{242}{2} = 121$.
Alternatively,the equation $|x + y| + |x - y| = a$ represents a square with vertices at $(\pm a/2, \pm a/2)$. Here $a = 11$,so vertices are $(\pm 5.5, \pm 5.5)$.
The side length of this square is the distance between $(5.5, 5.5)$ and $(-5.5, 5.5)$,which is $11$.
Thus,the area is $11^2 = 121$.

(C) Let the side length of the square be $a$. Let the coordinates of vertices $O, P, Q, R$ be $(0,0), (a, 0), (a, a), (0, a)$ respectively.
The area of the square $OPQR$ is $a^2$.
$M$ is the midpoint of $PQ$,so its coordinates are $\left(a, \frac{a}{2}\right)$.
$N$ is the midpoint of $QR$,so its coordinates are $\left(\frac{a}{2}, a\right)$.
The area of $\Delta OMN$ with vertices $(0,0), (a, \frac{a}{2}), (\frac{a}{2}, a)$ is given by:
Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Area $= \frac{1}{2} |0(\frac{a}{2} - a) + a(a - 0) + \frac{a}{2}(0 - \frac{a}{2})|$
Area $= \frac{1}{2} |a^2 - \frac{a^2}{4}| = \frac{1}{2} |\frac{3a^2}{4}| = \frac{3a^2}{8}$.
The ratio of the area of the square to the area of $\Delta OMN$ is $a^2 : \frac{3a^2}{8} = 1 : \frac{3}{8} = 8 : 3$.

(A) Let $ABCD$ be the given quadrilateral with vertices $A(-4, 5), B(0, 7), C(5, -5),$ and $D(-4, -2).$
To find the area of quadrilateral $ABCD$,we divide it into two triangles by drawing the diagonal $AC$.
Area $(ABCD) = \text{Area}(\Delta ABC) + \text{Area}(\Delta ACD).$
The area of a triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|.$
For $\Delta ABC$ with vertices $A(-4, 5), B(0, 7), C(5, -5):$
Area $(\Delta ABC) = \frac{1}{2} |-4(7 - (-5)) + 0(-5 - 5) + 5(5 - 7)|$
$= \frac{1}{2} |-4(12) + 0(-10) + 5(-2)|$
$= \frac{1}{2} |-48 - 10| = \frac{1}{2} |-58| = 29 \text{ unit}^2.$
For $\Delta ACD$ with vertices $A(-4, 5), C(5, -5), D(-4, -2):$
Area $(\Delta ACD) = \frac{1}{2} |-4(-5 - (-2)) + 5(-2 - 5) + (-4)(5 - (-5))|$
$= \frac{1}{2} |-4(-3) + 5(-7) - 4(10)|$
$= \frac{1}{2} |12 - 35 - 40| = \frac{1}{2} |-63| = 31.5 \text{ unit}^2.$
Total Area $(ABCD) = 29 + 31.5 = 60.5 = \frac{121}{2} \text{ unit}^2.$

(D) From the figure,the rectangle has dimensions $12 \times 9$. The area of the rectangle is $12 \times 9 = 108 \text{ sq units}$.
The triangular corner cut from the rectangle has legs of length $3$ and $4$,and a hypotenuse of length $5$.
The area of this right-angled triangle is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 3 = 6 \text{ sq units}$.
The area of the resulting pentagon is the area of the rectangle minus the area of the triangle: $108 - 6 = 102 \text{ sq units}$.
The ratio of the area of the pentagon to the area of the rectangle is $\frac{102}{108}$.
Dividing both numerator and denominator by $6$,we get $\frac{17}{18}$.

(D) Let the widths of the columns be $x_1, x_2, x_3, x_4$ and the heights of the rows be $y_1, y_2, y_3, y_4$.
From the given areas,we have:
$x_1 y_1 = 10$,$x_2 y_1 = 4$,$x_3 y_2 = 12$,$x_4 y_2 = 15$,$x_4 y_3 = 25$.
We want to find the area of rectangle $KLMN$,which is $x_1 y_3$.
From the relations:
$x_1 = \frac{10}{y_1}$,$x_2 = \frac{4}{y_1}$,$x_3 = \frac{12}{y_2}$,$x_4 = \frac{15}{y_2}$,$y_3 = \frac{25}{x_4} = \frac{25}{15/y_2} = \frac{25 y_2}{15} = \frac{5}{3} y_2$.
Thus,the area $KLMN = x_1 y_3 = \left(\frac{10}{y_1}\right) \left(\frac{5}{3} y_2\right) = \frac{50}{3} \frac{y_2}{y_1}$.
Looking at the grid,the area $KLMN$ is the product of the width of the first column and the height of the third row.
Using the property that for any four rectangles in a $2 \times 2$ grid,the product of the areas of diagonally opposite rectangles is equal:
Let $A_{i,j}$ be the area of the rectangle in row $i$ and column $j$.
$A_{1,1} \cdot A_{2,2} = A_{1,2} \cdot A_{2,1}$.
Using the given values: $10 \cdot A_{2,2} = 4 \cdot A_{2,1} \Rightarrow A_{2,1} = 2.5 A_{2,2}$.
Also,$A_{2,2} \cdot A_{3,3} = A_{2,3} \cdot A_{3,2} \Rightarrow A_{2,2} \cdot A_{3,3} = 12 \cdot 15 = 180$.
$A_{3,3} \cdot A_{4,4} = A_{3,4} \cdot A_{4,3} \Rightarrow A_{3,3} \cdot A_{4,4} = 25 \cdot A_{4,3}$.
By observing the grid,the area of $KLMN$ is $A_{4,1} = \frac{A_{4,3} \cdot A_{2,1}}{A_{2,3}} = \frac{A_{4,3} \cdot 2.5 A_{2,2}}{12}$.
Following the grid proportions,the area is $50$.

(A) The given pair of lines $xy=0$ represents the coordinate axes,i.e.,$x=0$ and $y=0$.
The other lines are $x-4=0$ (which is $x=4$) and $y+5=0$ (which is $y=-5$).
These four lines $x=0, x=4, y=0$,and $y=-5$ enclose a rectangular region in the Cartesian plane.
The vertices of this rectangle are $(0, 0), (4, 0), (4, -5)$,and $(0, -5)$.
The length of the rectangle is the distance between $x=0$ and $x=4$,which is $|4-0| = 4 \text{ units}$.
The breadth of the rectangle is the distance between $y=0$ and $y=-5$,which is $|0-(-5)| = 5 \text{ units}$.
Therefore,the area of the rectangle is $\text{length} \times \text{breadth} = 4 \times 5 = 20 \text{ sq units}$.

(B) The area of a triangle with vertices in polar coordinates $(r_1, \theta_1)$,$(r_2, \theta_2)$,and $(r_3, \theta_3)$ is given by the formula:
$\text{Area} = \frac{1}{2} |r_1 r_2 \sin(\theta_2 - \theta_1) + r_2 r_3 \sin(\theta_3 - \theta_2) + r_3 r_1 \sin(\theta_1 - \theta_3)|$
Substituting the given values $(1, 0)$,$(2, \frac{\pi}{3})$,and $(3, \frac{2\pi}{3})$:
$\text{Area} = \frac{1}{2} |1 \cdot 2 \sin(\frac{\pi}{3} - 0) + 2 \cdot 3 \sin(\frac{2\pi}{3} - \frac{\pi}{3}) + 3 \cdot 1 \sin(0 - \frac{2\pi}{3})|$
$\text{Area} = \frac{1}{2} |2 \sin(\frac{\pi}{3}) + 6 \sin(\frac{\pi}{3}) + 3 \sin(-\frac{2\pi}{3})|$
$\text{Area} = \frac{1}{2} |2(\frac{\sqrt{3}}{2}) + 6(\frac{\sqrt{3}}{2}) + 3(-\frac{\sqrt{3}}{2})|$
$\text{Area} = \frac{1}{2} |\sqrt{3} + 3\sqrt{3} - \frac{3\sqrt{3}}{2}|$
$\text{Area} = \frac{1}{2} |4\sqrt{3} - 1.5\sqrt{3}| = \frac{1}{2} |2.5\sqrt{3}| = \frac{5\sqrt{3}}{4} \text{ square units}$

(C) Let the vertices of the triangle be $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$.
The midpoints of the sides are given as $M_1(2, 7)$,$M_2(1, 1)$,and $M_3(10, 8)$.
The area of the triangle formed by the midpoints of the sides of a triangle is $\frac{1}{4}$ of the area of the original triangle.
First,calculate the area of the triangle formed by the midpoints using the formula: $\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
$\text{Area}_{\text{mid}} = \frac{1}{2} |2(1 - 8) + 1(8 - 7) + 10(7 - 1)|$.
$\text{Area}_{\text{mid}} = \frac{1}{2} |2(-7) + 1(1) + 10(6)|$.
$\text{Area}_{\text{mid}} = \frac{1}{2} |-14 + 1 + 60| = \frac{1}{2} |47| = 23.5$.
Since $\text{Area}_{\text{original}} = 4 \times \text{Area}_{\text{mid}}$,we have $\text{Area}_{\text{original}} = 4 \times 23.5 = 94$.