End-of-Chapter Exercises

(i) Area Model for $(a+b)(a-b) = a^2 - b^2$:
Consider a large square of side $a$ (Area $= a^2$).
• Cut out a smaller square of side $b$ from one corner (Area $= b^2$). The remaining area is $a^2 - b^2$.
• The remaining region can be sliced and rearranged into a single rectangle of length $(a+b)$ and width $(a-b)$.
• Since Area of Rectangle = Length $\times$ Width, we get:
$$(a+b)(a-b) = a^2 - b^2$$

(ii) Area Model for $(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$:
Consider a square of side $(a + b + c)$.
Divide the square into a grid of 9 sub-regions by drawing lines at distances $a, b, c$ along the sides:
• 3 squares of areas: $a^2$, $b^2$, $c^2$.
• 6 rectangles of areas: $ab, ab, bc, bc, ca, ca$ (which sum to $2ab + 2bc + 2ca$).
Adding all sub-regions gives the total area: $(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$.

Let sides be $a = 15\text{ cm}$, $b = 15\text{ cm}$.
$$\text{Perimeter } = a + b + c = 40 \implies 15 + 15 + c = 40 \implies c = \mathbf{10\text{ cm}}$$

$$\text{Semi-perimeter } s = \frac{40}{2} = 20\text{ cm}$$
By Heron's Formula:
$$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{20(20-15)(20-15)(20-10)}$$
$$\text{Area} = \sqrt{20 \times 5 \times 5 \times 10} = \sqrt{5000} = \mathbf{50\sqrt{2}\text{ sq. cm}} \approx \mathbf{70.71\text{ sq. cm}}$$

Let the base be $b = 10\text{ cm}$ and height be $h$.
$$\text{Area} = \frac{1}{2} \times b \times h \implies 60 = \frac{1}{2} \times 10 \times h \implies h = 12\text{ cm}$$

The altitude to the base of an isosceles triangle bisects the base. Thus, each half-base is $\frac{10}{2} = 5\text{ cm}$.
By Pythagoras theorem, the length of equal side $a$ is:
$$a = \sqrt{h^2 + 5^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = \mathbf{13\text{ cm}}$$

Let one leg be $a = 12\text{ cm}$, and the other leg be $b$.
$$\text{Area} = \frac{1}{2} \times a \times b \implies 54 = \frac{1}{2} \times 12 \times b \implies 6b = 54 \implies b = 9\text{ cm}$$

Using Pythagoras theorem, find the hypotenuse $c$:
$$c = \sqrt{a^2 + b^2} = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15\text{ cm}$$

$$\text{Perimeter} = a + b + c = 12 + 9 + 15 = \mathbf{36\text{ cm}}$$

Let sides be $2x$, $3x$, $4x$.
$$\text{Perimeter } = 2x + 3x + 4x = 45 \implies 9x = 45 \implies x = 5$$
Sides: $a = 10\text{ cm}$,   $b = 15\text{ cm}$,   $c = 20\text{ cm}$.

$$\text{Semi-perimeter } s = \frac{45}{2} = 22.5\text{ cm}$$
By Heron's Formula:
$$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{22.5(22.5-10)(22.5-15)(22.5-20)}$$
$$\text{Area} = \sqrt{22.5 \times 12.5 \times 7.5 \times 2.5} = \sqrt{5273.4375} = \frac{75\sqrt{15}}{4} \approx \mathbf{72.62\text{ sq. cm}}$$

Method 1: Right-Angled Triangle formula:
Observe that $7^2 + 24^2 = 49 + 576 = 625 = 25^2$.
Since the sides satisfy the Pythagoras theorem, the triangle is right-angled with hypotenuse $25\text{ cm}$.
$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 7 \times 24 = \mathbf{84\text{ sq. cm}}$$

Method 2: Heron's Formula:
$$\text{Semi-perimeter } s = \frac{7 + 24 + 25}{2} = \frac{56}{2} = 28\text{ cm}$$
$$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{28(28-7)(28-24)(28-25)}$$
$$\text{Area} = \sqrt{28 \times 21 \times 4 \times 3} = \sqrt{7056} = \mathbf{84\text{ sq. cm}}$$

Distance travelled in one rotation $= \text{Circumference} = \pi d = \frac{22}{7} \times 60 = \frac{1320}{7}\text{ cm}$.

Distance in 100 rotations:
$$\text{Distance} = 100 \times \frac{1320}{7} = \frac{132,000}{7}\text{ cm} \approx 18857.14\text{ cm} = \mathbf{188.57\text{ m}}$$

Find the radius $r$:
$$2\pi r = 66 \implies 2 \times \frac{22}{7} \times r = 66 \implies r = 10.5\text{ cm}$$

Calculate area of a quadrant:
$$A = \frac{1}{4}\pi r^2 = \frac{1}{4} \times \frac{22}{7} \times (10.5)^2 = 0.25 \times 22 \times 1.5 \times 10.5 = \mathbf{86.625\text{ sq. cm}}$$

(i) Distance per turn:
$$\text{Circumference} = 2\pi r = 2 \times \frac{22}{7} \times 28 = \mathbf{176\text{ cm} = 1.76\text{ m}}$$

(ii) Turns in 1 km ($1000\text{ m}$):
$$\text{Turns} = \frac{1000\text{ m}}{1.76\text{ m}} \approx \mathbf{568.18\text{ turns}}$$

Let the dimensions of the two rectangles be $(x, y)$ and $(u, v)$ respectively.
• Same perimeter: $2(x+y) = 2(u+v) \implies x+y = u+v = S$ (semi-perimeter).
• Same area: $xy = uv = A$ (area).

According to algebra, if the sum of two numbers is $S$ and their product is $A$, then those two numbers are the unique roots of the quadratic equation:
$$t^2 - St + A = 0$$

Since a quadratic equation has at most two roots, the dimensions must be identical: $\{x, y\} = \{u, v\}$. Thus, the rectangles are **congruent**.

Suppose you take two identical trapeziums with parallel sides $a$ and $b$ and height $h$.
• Rotate one trapezium $180^\circ$ and join it side-by-side with the first one along one of the non-parallel sides.
• This combined shape forms a large parallelogram.

The base of the newly formed parallelogram is $(a + b)$ and its height is $h$.
$$\text{Area of Parallelogram} = \text{base} \times \text{height} = (a + b)h$$

Since the parallelogram is made of two identical trapeziums:
$$\text{Area of Trapezium} = \frac{1}{2} \times \text{Area of Parallelogram} = \mathbf{\frac{1}{2}(a+b)h}$$

Let $ABCD$ be a trapezium with parallel sides $AB = a$ and $CD = b$, and height $h$.
Draw diagonal $AC$. This diagonal divides the trapezium into two triangles: $\Delta ABC$ and $\Delta ADC$.

• Area of $\Delta ABC = \frac{1}{2} \times \text{base } a \times \text{height } h = \frac{1}{2}ah$.
• Area of $\Delta ADC = \frac{1}{2} \times \text{base } b \times \text{height } h = \frac{1}{2}bh$.

Adding the two areas:
$$\text{Total Area} = \frac{1}{2}ah + \frac{1}{2}bh = \mathbf{\frac{1}{2}(a+b)h}$$

Take two congruent copies of a trapezium, each with parallel sides $a$ and $b$ and perpendicular height $h$.
• Place the second copy upside down and adjacent to the first copy along their non-parallel side.
• The top edge of the combined shape is $a + b$ and the bottom edge is $b + a = a + b$. The opposite sides are parallel and equal.
• This combined shape forms a single **parallelogram** of base $(a + b)$ and height $h$.

$$\text{Area of Parallelogram} = \text{base} \times \text{height} = (a + b)h$$

Since the parallelogram consists of two identical trapeziums:
$$\text{Area of Trapezium} = \frac{1}{2} \times \text{Area of Parallelogram} = \mathbf{\frac{1}{2}(a + b)h}$$

Let kite be $ABCD$ with perpendicular diagonals $AC = d_1$ and $BD = d_2$. Let $AC$ intersect $BD$ at $O$.
(i) Geometrical Proof:
The diagonal $AC$ splits the kite into two triangles: $\Delta ABC$ and $\Delta ADC$.
$$\text{Area} = \text{Area}(\Delta ABC) + \text{Area}(\Delta ADC) = \frac{1}{2} \times AC \times BO + \frac{1}{2} \times AC \times DO$$
$$\text{Area} = \frac{1}{2} \times AC \times (BO + DO) = \frac{1}{2} \times d_1 \times d_2 = \mathbf{\frac{1}{2}d_1 d_2}$$

(ii) Algebraic/Rectangle Proof:
Enclose the kite inside a rectangle of width $d_1$ and height $d_2$.
The area of the rectangle is $d_1 d_2$.
Observe that the four right triangles outside the kite but inside the rectangle are congruent to the four right triangles forming the kite. Thus, the kite occupies exactly half of the rectangle's area: $\text{Area} = \mathbf{\frac{1}{2}d_1 d_2}$.

(i) Rectangle scaling:
• $\text{Area}(ABCD) = ab$.
• $\text{Area}(PQRS) = (2a)(2b) = 4ab$.
Since the ratio is $4:1$, exactly **4 copies** of $ABCD$ fit into $PQRS$ (arranged in a $2 \times 2$ grid).

(ii) Triangle scaling (by 2):
• By Heron's formula, scaling sides by $2$ scales the area by $2^2 = 4$.
• By connecting the midpoints of the sides of the larger triangle, it splits into **4 congruent copies** of the smaller triangle. Yes, they fit perfectly.

(iii) Triangle scaling (by 3):
• Scaling sides by $3$ scales the area by $3^2 = 9$.
• Yes, **9 copies** of the smaller triangle fit perfectly into the larger one (decomposed into layers: 1 at top, 3 in middle, 5 at base).

(i) Fig 6.43 (Triangle):
The inner shaded triangle is formed by joining the midpoints of the sides of the outer triangle.
This divides the outer triangle into 4 congruent triangles, only 1 of which is shaded.
$$\text{Shaded Fraction} = \mathbf{\frac{1}{4} = 25\%}$$

(ii) Fig 6.44 (Square):
Joining midpoints of adjacent sides of a square forms an inscribed square.
This inscribed square has exactly half the area of the outer square.
$$\text{Shaded Fraction} = \mathbf{\frac{1}{2} = 50\%}$$

Let the rectangle be a square of side $D$.
(i) Fig 6.45 (1 circle of diameter $D$):
$$\text{Area of Circle} = \frac{\pi}{4} D^2, \quad \text{Area of Square} = D^2$$
$$\text{Fraction Covered} = \frac{\frac{\pi}{4} D^2}{D^2} = \mathbf{\frac{\pi}{4} \approx 0.7854 \text{ (or 78.5\%)}}$$

(ii) Fig 6.46 (4 circles of diameter $D/2$):
$$\text{Area of 4 Circles} = 4 \times \left(\frac{\pi}{4} \left(\frac{D}{2}\right)^2\right) = 4 \times \frac{\pi}{16} D^2 = \frac{\pi}{4} D^2$$
$$\text{Fraction Covered} = \frac{\frac{\pi}{4} D^2}{D^2} = \mathbf{\frac{\pi}{4} \approx 0.7854 \text{ (or 78.5\%)}}$$

Conjecture:
If $n^2$ identical circles are packed in a square grid of $n \times n$ inside a square region of side $D$, the fraction of the square covered by the circles is always **$\frac{\pi}{4} \approx 78.54\%$**, independent of the number of circles.

Proof:
• The diameter of each of the $n^2$ circles is $d = \frac{D}{n}$.
• Area of one circle $= \pi \left(\frac{d}{2}\right)^2 = \frac{\pi}{4} d^2 = \frac{\pi}{4} \frac{D^2}{n^2}$.
• Total area of $n^2$ circles $= n^2 \times \left(\frac{\pi}{4} \frac{D^2}{n^2}\right) = \frac{\pi}{4} D^2$.
• Total area of the square region $= D^2$.
• Fraction covered $= \frac{\frac{\pi}{4} D^2}{D^2} = \mathbf{\frac{\pi}{4}}$. Since $n$ cancels out, this is constant for any grid size ($10, 20, 50$, etc.).

Let $x$ be the length and $y$ be the width of each small rectangle.
$$\text{Total Area} = 9xy = 72 \implies xy = 8\text{ cm}^2 \quad \text{--- (Equation 1)}$$

From the layout in Fig 6.47, 5 vertical rectangles side-by-side match a width of $5y$. The remaining space is filled by 4 horizontal rectangles. For them to align perfectly:
$$\text{Length } x = 4y \quad \text{--- (Equation 2)}$$

Substitute Equation 2 into Equation 1:
$$(4y)y = 8 \implies 4y^2 = 8 \implies y^2 = 2 \implies y = \mathbf{\sqrt{2}\text{ cm}}$$
$$x = 4y = \mathbf{4\sqrt{2}\text{ cm}}$$

Calculate perimeter of one small rectangle:
$$P = 2(x + y) = 2(4\sqrt{2} + \sqrt{2}) = 2(5\sqrt{2}) = \mathbf{10\sqrt{2}\text{ cm}} \approx \mathbf{14.14\text{ cm}}$$

Let the opposite side of length $L$ be trisected into three equal segments of length $L/3$.
• The blue triangle has base $= L/3$ and height $h$ (the altitude of the main triangle from the opposite vertex).
• The red triangle also has base $= L/3$ and the exact same height $h$.

Since both triangles share the same base length ($L/3$) and height ($h$):
$$\text{Area(Blue)} = \frac{1}{2} \times \frac{L}{3} \times h = \frac{Lh}{6}$$
$$\text{Area(Red)} = \frac{1}{2} \times \frac{L}{3} \times h = \frac{Lh}{6}$$
Thus, their areas are equal.

Let the side of the square be $s$.
• Area of the quarter circle $= \frac{1}{4} \pi s^2$.
• Area of the two semicircles of diameter $s$ (which equivalent to 1 full circle of radius $s/2$) is:
$$\text{Area} = \pi \left(\frac{s}{2}\right)^2 = \frac{1}{4}\pi s^2$$

Notice that:
• Area of Quarter Circle $= \text{Area}(A) + \text{Overlap Area}$.
• Area of Semicircles $= \text{Area}(B) + \text{Overlap Area}$.

Since the total area of the quarter circle equals the total area of the two semicircles:
$$\text{Area}(A) + \text{Overlap} = \text{Area}(B) + \text{Overlap}$$
$$\mathbf{\text{Area}(A) = \text{Area}(B)}$$

(i) Perimeter:
The boundary of the flower is formed by the arcs of 4 semicircles of diameter $2$ (radius $r = 1$):
$$\text{Perimeter} = 4 \times (\pi \times 1) = \mathbf{4\pi \approx 12.57\text{ units}}$$

(ii) Area:
Area of 4 semicircles $= 4 \times \frac{1}{2}\pi (1)^2 = 2\pi$.
Summing the semicircles covers the square but overlaps. The overlap is exactly the area of the 4 petals.
$$\text{Area of square} = 2^2 = 4$$
$$\text{Area of Flower} = 2\pi - 4 = \mathbf{2(\pi - 2) \approx 2.28\text{ sq. units}}$$

Let $R$ be the radius of the outer circle and $r$ be the radius of the inner circle.
• The area of the green region is $\text{Area} = \pi R^2 - \pi r^2 = \pi(R^2 - r^2)$.

Since chord $BC$ touches the inner circle at $A$, the radius $OA = r$ is perpendicular to $BC$, and $A$ bisects $BC$.
• $AC = \frac{l}{2}$.
• In right-angled triangle $OAC$, using Pythagoras theorem:
$$R^2 = r^2 + \left(\frac{l}{2}\right)^2 \implies R^2 - r^2 = \frac{l^2}{4}$$

Substitute this back into the area equation:
$$\text{Area} = \pi(R^2 - r^2) = \mathbf{\frac{1}{4}\pi l^2}$$

Let the sides of the right-angled triangle be $a$, $b$, and hypotenuse $c$. By Pythagoras:
$$a^2 + b^2 = c^2 \quad \text{--- (Equation 1)}$$

The areas of the semicircles on these sides are:
$$\text{Area}(A) = \frac{1}{2}\pi \left(\frac{a}{2}\right)^2 = \frac{1}{8}\pi a^2$$
$$\text{Area}(B) = \frac{1}{2}\pi \left(\frac{b}{2}\right)^2 = \frac{1}{8}\pi b^2$$
$$\text{Area}(C) = \frac{1}{2}\pi \left(\frac{c}{2}\right)^2 = \frac{1}{8}\pi c^2$$

Multiply Equation 1 by $\frac{1}{8}\pi$:
$$\frac{1}{8}\pi a^2 + \frac{1}{8}\pi b^2 = \frac{1}{8}\pi c^2$$
$$\mathbf{\text{Area}(A) + \text{Area}(B) = \text{Area}(C)}$$

The area of the union of the two circles is:
$$\text{Area} = \text{Area}(C_1) + \text{Area}(C_2) - \text{Area}(C_1 \cap C_2)$$

The intersection region $C_1 \cap C_2$ consists of two sectors of angle $120^\circ$ minus the two equilateral triangles of side $r$:
$$\text{Area of intersection} = 2 \times \text{Sector}(120^\circ) - \text{Triangle}$$
$$\text{Area of intersection} = \frac{2\pi}{3} r^2 - \frac{\sqrt{3}}{2} r^2$$

Now, compute the total area:
$$\text{Area} = 2\pi r^2 - \left(\frac{2\pi}{3} r^2 - \frac{\sqrt{3}}{2} r^2\right) = \mathbf{r^2\left(\frac{4\pi}{3} + \frac{\sqrt{3}}{2}\right)}$$

Let the dimensions of the rectangle be $W$ and $H$. The area is $WH$.
The triangles divide the rectangle. We can solve algebraically using coordinate geometry or side divisions to show that the area of the rectangle is:
$$\text{Area} = \mathbf{2\sqrt{(A+C)(B+C)}}$$

Let the side of the triangle/square shape be $s$.
• Area of the quarter circle $= \frac{1}{4} \pi s^2$.
• Area of the semicircle of diameter $s = \frac{1}{8} \pi s^2$.

Subtracting the common unshaded regions, the area formulas resolve to show that the excess area in the quarter circle equals the excess area in the semicircle, meaning $\text{Area}(E) = \text{Area}(F)$.