Chapter 7 Area (Class 8 - Latest Maths NCERT (Ganita Prakash II) Solutions)

Solution:

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(i) Finding the missing side length in the first figure:

Let us label the relevant rectangles and vertices to find the missing side length.

In rectangle ABCD (area = $21\text{ in}^2$):

$\text{Area} = \text{Length} \times \text{Breadth}$

$7 \times \text{BC} = 21$

$\text{BC} = 3\text{ in}$

Now, considering the vertical dimension involving AF:

$\text{AF} = \text{AD} + \text{DF}$

$\text{AF} = 3\text{ in} + 4\text{ in} = 7\text{ in}$

In the rectangle EFAG (area = $28\text{ in}^2$):

$\text{EF} \times \text{AF} = 28\text{ in}^2$

$\text{EF} \times 7\text{ in} = 28\text{ in}^2$

$\text{EF} = 4\text{ in}$

Now, considering the horizontal dimension involving HA:

$\text{HA} = \text{HG} + \text{GA}$

$\text{HA} = 3\text{ in} + 4\text{ in} = 7\text{ in}$

In rectangle HIJA (area = $35\text{ in}^2$):

$\text{Area} = \text{HA} \times \text{AJ}$

$35\text{ in}^2 = 7\text{ in} \times \text{AJ}$

$\text{AJ} = 5\text{ in}$

Now, considering the vertical dimension involving AK:

$\text{AK} = \text{AJ} + \text{JK}$

$\text{AK} = 5\text{ in} + 2\text{ in} = 7\text{ in}$

In rectangle KLMA (area = $14\text{ in}^2$):

$\text{Area} = \text{KL} \times \text{LM}$

$x\text{ in} \times 7\text{ in} = 14\text{ in}^2$

$x = 2\text{ in}$

Thus, the missing side length = $2\text{ in}$.

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(ii) Finding the missing side lengths in the second figure:

Let us label the relevant rectangles and regions to find the missing side lengths.

In rectangle ABGH (area = $29\text{ m}^2$):

$\text{Area} = \text{AB} \times \text{AH}$

$\text{AB} \times 4\text{ m} = 29\text{ m}^2$

$\text{AB} = \frac{29}{4}\text{ m} = 7.25\text{ m}$

Area of rectangle HGDC = Area of total lower-left structure ($\text{Area} = 50\text{ m}^2$) – Area of rectangle ABGH ($29\text{ m}^2$)

$\text{Area of HGDC} = 50\text{ m}^2 - 29\text{ m}^2 = 21\text{ m}^2$

In rectangle HGDC:

$\text{CD} \times \text{GD} = 21\text{ m}^2$

Since $\text{CD} = \text{AB} = \frac{29}{4}\text{ m}$:

$\frac{29}{4}\text{ m} \times \text{GD} = 21\text{ m}^2$

$\text{GD} = \frac{84}{29}\text{ m} \approx 2.9\text{ m}$

In rectangle BEFG (area = $11\text{ m}^2$):

$\text{BG} \times \text{BE} = 11\text{ m}^2$

$4\text{ m} \times \text{BE} = 11\text{ m}^2$

$\text{BE} = \frac{11}{4}\text{ m} = 2.75\text{ m}$

Thus, the missing side lengths are $\text{AB} = \frac{29}{4}\text{ m}$, $\text{BE} = \frac{11}{4}\text{ m}$, and $\text{GD} = \frac{84}{29}\text{ m}$.

Solution:

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(i) Measurements needed:

Length of outer rectangle ABCD = $A$

Width of outer rectangle ABCD = $B$

Length of inner rectangle EFGH = $a$

Width of inner rectangle EFGH = $b$

Let $A = 10\text{ m}$, $B = 8\text{ m}$, $a = 6\text{ m}$, $b = 4\text{ m}$.

Calculation:

$\text{Area of path} = \text{Area of ABCD} - \text{Area of EFGH}$

$= 10 \times 8 - 6 \times 4$

$= 80 - 24$

$= 56\text{ m}^2$

Formula:

$\text{Area of path} = (A \times B) - (a \times b)$

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(ii) Finding area when path width is given:

Yes, if the width of the path is uniform along each side, we can find its area, but we also need the dimensions of either the outer or inner rectangle.

Let the width of the path $d = 2\text{ m}$ (uniform on all sides).

Length of inner park EFGH = $l = 16\text{ m}$

Width of inner park EFGH = $w = 11\text{ m}$

Length of outer rectangle = $l + 2d$

$= 16 + 2(2) = 20\text{ m}$

Width of outer rectangle = $w + 2d$

$= 11 + 2(2) = 15\text{ m}$

Breaking the path into rectangles:

Now there are $4$ rectangular sections:

Left rectangle area = $w \times d = 11 \times 2 = 22\text{ m}^2$

Right rectangle area = $w \times d = 11 \times 2 = 22\text{ m}^2$

Top rectangle area = $(l + 2d) \times d = 20 \times 2 = 40\text{ m}^2$

Bottom rectangle area = $(l + 2d) \times d = 20 \times 2 = 40\text{ m}^2$

Total area of path = $22 + 22 + 40 + 40 = 124\text{ m}^2$

Formula:

$\text{Area of path} = 2d(l + w) + 4d^2$

where $d$ = width of path, $l$ = length of inner park, and $w$ = width of inner park.

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(iii) Does the area change when the outer rectangle is shifted?

No, the area of the path does not change.

Reason: The area of the path depends only on the total area of the outer rectangle ABCD and the area of the inner rectangle EFGH. Shifting the position of the inner rectangle inside the outer rectangle alters the widths of the surrounding path on opposite sides, but the sum of their areas remains exactly the same.

Solution:

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Measurements needed:

Length of plot = $14\text{ m}$

Width of plot = $12\text{ m}$

Width of horizontal path = $x_1$

Width of vertical path = $x_2$

Assign values:

Let $x_1 = 2\text{ m}$

$x_2 = 2\text{ m}$

Calculation:

Area of horizontal path = $14 \times 2 = 28\text{ m}^2$

Area of vertical path = $12 \times 2 = 24\text{ m}^2$

Area of overlapping square in the center = $2 \times 2 = 4\text{ m}^2$

$\therefore \text{Area of crosspath} = 28 + 24 - 4 = 48\text{ m}^2$

Formula:

$\text{Area of crosspath} = (L \times w_1) + (W \times w_2) - (w_1 \times w_2)$

where $L$ = length of plot, $W$ = width of plot, $w_1$ = width of horizontal path, and $w_2$ = width of vertical path.

Solution:

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After naming the figure,

The area of the spiral tube = Area of rectangle ABEC + Area of rectangle DEGF + Area of rectangle GHIJ + Area of rectangle JKML + Area of rectangle NOPL + Area of rectangle PQRS + Area of rectangle STUV + Area of rectangle VWYX + Area of rectangle XZA1B1

$= \text{AC} \times \text{AB} + \text{EG} \times \text{DE} + \text{IH} \times \text{JI} + \text{LJ} \times \text{LM} + \text{NO} \times \text{NL} + $$ \ \text{PQ} \times \text{PS} + \text{UT} \times \text{ST} + \text{VX} \times \text{VW} + \text{ZA}_1 \times \text{A}_1\text{B}_1$

$= 20 \times 1 + 18 \times 1 + 20 \times 1 + 13 \times 1 + 15 \times 1 + 8 \times 1 + 10 \times 1 + $$ \ 3 \times 1 + 5 \times 1$

$= 20 + 18 + 20 + 13 + 15 + 8 + 10 + 3 + 5$

$= 112\text{ sq. units}$

Thus, the area of the spiral tube is $112\text{ sq. units}$.

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Let the length of the straight tube be $x$.

The area of the bent tube on the left = Area of rectangle BACD + Area of rectangle BGFE

$= \text{AC} \times \text{CD} + \text{BG} \times \text{BE}$

$= 5 \times 1 + 4 \times 1$

$= 9\text{ sq. units}$

Area of straight tube = $x \times 1$

Area of bent tube = $9\text{ sq. units}$

According to the question, both have the same area.

So $x \times 1 = 9$

$\Rightarrow x = 9\text{ units}$.

Solution:

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Let '$a$' be the side of the square.

Area of triangle DCB (region 3):

$= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \text{DC} \times \text{BC}$

$= \frac{1}{2} \times a \times a = \frac{a^2}{2}\text{ sq. units}$

In right-angled triangle DCB, by Pythagoras theorem:

$\text{DB} = \sqrt{\text{DC}^2 + \text{BC}^2} = \sqrt{a^2 + a^2} = \sqrt{2}a\text{ units}$

$\therefore \text{AE} = \text{DE} = \text{EB} = \frac{\text{DB}}{2} = \frac{\sqrt{2}a}{2} = \frac{\sqrt{2} \times \sqrt{2}}{2 \times \sqrt{2}}a = \frac{2}{2\sqrt{2}}a = \frac{a}{\sqrt{2}}\text{ units}$

Area of triangle AED (region 1):

$= \frac{1}{2} \times \text{DE} \times \text{AE} = \frac{1}{2} \times \frac{a}{\sqrt{2}} \times \frac{a}{\sqrt{2}}$

$= \frac{a^2}{4}\text{ sq. units}$

Area of triangle AEB (region 2):

$= \frac{1}{2} \times \text{EB} \times \text{AE} = \frac{1}{2} \times \frac{a}{\sqrt{2}} \times \frac{a}{\sqrt{2}} = \frac{a^2}{2} \times \frac{1}{2}$

$= \frac{a^2}{4}\text{ sq. units}$

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If the side of the square is doubled $= 2a$

Area of triangle BCD (region 3):

$= \frac{1}{2} \times \text{DC} \times \text{BC}$

$= \frac{1}{2} \times 2a \times 2a = 2a^2\text{ sq. units}$

In right-angled triangle BCD, by Pythagoras theorem:

$\text{BD} = \sqrt{\text{DC}^2 + \text{BC}^2}$

$= \sqrt{(2a)^2 + (2a)^2}$

$= \sqrt{4a^2 + 4a^2} = 2\sqrt{2}a\text{ units}$

$\therefore \text{AE} = \text{ED} = \text{EB} = \frac{\text{BD}}{2} = \frac{2\sqrt{2}a}{2} = \sqrt{2}a\text{ units}$

Area of triangle AED (region 1):

$= \frac{1}{2} \times \text{DE} \times \text{AE}$

$= \frac{1}{2} \times \sqrt{2}a \times \sqrt{2}a = \frac{1}{2} \times 2a^2 = a^2\text{ sq. units}$

Area of triangle AEB (region 2):

$= \frac{1}{2} \times \text{EB} \times \text{AE} = \frac{1}{2} \times \sqrt{2}a \times \sqrt{2}a$

$= a^2\text{ sq. units}$

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The increase in area of region 1:

$= \frac{a^2}{\frac{a^2}{4}} = 4\text{ times}$

The increase in area of region 2:

$= \frac{a^2}{\frac{a^2}{4}} = 4\text{ times}$

The increase in area of region 3:

$= \frac{2a^2}{\frac{a^2}{2}} = \frac{2a^2}{a^2} \times 2 = 4\text{ times}$

Thus, the increase in the areas of regions 1, 2, and 3 is $4$ times.

Reason: If the sidelength of the square is doubled, then the area becomes $4$ times.

Solution:

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This is a geometric dissection and rearrangement puzzle. Follow the step-by-step instructions below to perform this activity:

Step 1: Dissection of the Square

• Draw a square on a piece of thick chart paper or cardboard.

• Draw two perpendicular intersecting line segments inside the square from the edges such that they divide the square into $4$ distinct polygonal pieces (two trapezoids and two right-angled or asymmetric four-sided pieces meeting at the perpendicular intersection point).

• Carefully cut along these lines using a pair of scissors or a craft knife to obtain the $4$ pieces.

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Step 2: Rearrangement into a Larger Square with a Hole

• Take the $4$ cut-out pieces and flip or rotate them outward.

• Align the straight outer boundaries of the pieces along the new extended perimeter.

• By fitting the angled edges together, the pieces form a larger outer square frame.

• Because the pieces expand outward during this specific rearrangement, a symmetrical square-shaped "hole" or open space is automatically created right in the center.

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Conclusion:

By dissecting a square using two perpendicular lines and sliding the resulting pieces outward, we successfully rearrange them into a larger square enclosing a central hole.

Formula Used:

(i) For Triangle $ABC$:

Base ($BC$) $= 4\text{ cm}$ and Height ($AE$) $= 3\text{ cm}$.

(ii) For Triangle $DEF$:

Base ($EF$) $= 5\text{ cm}$ and Height ($DN$) $= 3.2\text{ cm}$.

(iii) For Triangle $NAT$:

It is a right-angled triangle where Base ($AT$) $= 3\text{ cm}$ and Height ($NA$) $= 4\text{ cm}$.

Given:

In $\Delta ABC$:

Case 1: Base ($BC$) $= 6\text{ units}$ and corresponding Height ($AX$) $= 4\text{ units}$.

Case 2: Base ($AC$) $= 8\text{ units}$ and corresponding Height ($BY$) $= ?$.

To Find:

The length of the altitude $BY$.

Solution:

The area of a triangle remains the same regardless of which side is taken as the base.

First, calculate the area using base $BC$ and altitude $AX$:

Now, we use the same area to find altitude $BY$ using base $AC$:

Given:

1. $\Delta SUB$ is an isosceles triangle with $SU = SB$.

2. $SE \perp UB$ (Altitude from the vertex to the base).

3. Area of $\Delta SEB = 24\text{ sq. units}$.

To Find:

Area of $\Delta SUB$.

Solution:

In an isosceles triangle, the altitude drawn from the vertex connecting the equal sides to the base ($SE$) acts as a line of symmetry. This means it bisects the base $UB$ and divides the triangle into two congruent triangles of equal area.

The total area of $\Delta SUB$ is the sum of the areas of these two smaller triangles:

Solution:

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According to the ancient geometric methods given in the Śulba-Sūtras, a rectangle can be transformed into a triangle of equal area through the following steps:

1. Let us take a rectangle ABCD, with length $a$ and breadth $b$.

2. Now mark the midpoint $E$ of side CD.

Draw a line perpendicular to CD passing through $E$.

Mark a point $M$ on it such that $\text{ME} = b$.

3. Draw a triangle using base = AB (same as the rectangle’s length) and join $M$ to $A$ and $B$ to form $\triangle MAB$.

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Verification of Area:

• Area of the rectangle ABCD = $\text{length} \times \text{breadth} = a \times b$

• Area of the constructed triangle MAB = $\frac{1}{2} \times \text{base} \times \text{height} $$ \ = \frac{1}{2} \times a \times b = \frac{1}{2}ab$

Wait, to ensure the triangle has the equal area (same area $ab$), the height of the triangle must be $2b$ (or the base must be $2a$). Following the classical Śulba-Sūtra construction where the height is taken as twice the breadth (i.e., $\text{ME} = 2b$), the area becomes:

$\text{Area of } \triangle MAB = \frac{1}{2} \times a \times (2b) = a \times b$

Thus, the triangle has an area exactly equal to the area of the given rectangle.

Solution:

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According to the ancient geometric methods given in the Śulba-Sūtras, a triangle can be transformed into a rectangle of equal area through the following steps:

1. Take a triangle ABC with base $b$ and height $h$.

2. Find the midpoint $M$ of the height.

3. Draw a line parallel to the base through $M$.

This line intersects the sides of the triangle.

4. Create a rectangle using:

Length = Same as the triangle’s base = $b$

Width = Half of the triangle’s height = $\frac{h}{2}$

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Verification of Area:

• Area of the triangle ABC = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times b \times h$

• Area of the constructed rectangle = $\text{length} \times \text{width} = b \times \frac{h}{2} = \frac{1}{2}bh$

Thus, the rectangle has an area exactly equal to the area of the given triangle.

Solution:

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Given that $ABCD$, $BCEF$, and $BFGH$ are identical squares.

Let us assume that the intersection of line segments $AB$ and $DH$ is point $I$.

Let the side of each square be $a$.

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(i) Finding the area of the blue region when the red region is $49$ sq. units:

Area of the red region ($\triangle HDC$) = $\frac{1}{2} \times \text{DC} \times \text{HC}$

(Since $\text{HC} = \text{HB} + \text{BC} = a + a = 2a$)

$= \frac{1}{2} \times a \times 2a = a^2 = 49$

$\Rightarrow a = 7\text{ units}$

Now, the area of the blue region ($\triangle IAD$):

$\text{Area} = \frac{1}{2} \times \text{AI} \times \text{AD}$

$= \frac{1}{2} \times \left(\frac{a}{2}\right) \times a$

$= \frac{1}{2} \times \left(\frac{7}{2}\right) \times 7 = \frac{49}{4}\text{ sq. units} = 12.25\text{ sq. units}$

Thus, the area of the blue region is $12.25\text{ square units}$.

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(ii) Finding the area of each square when the total enclosed area is $180$ sq. units:

Given, the total area enclosed by the blue and red regions = $\text{Area of } \triangle HDC + \text{Area of } \triangle AID = 180\text{ sq. units}$

$\text{Area of } \triangle HDC = a^2$

$\text{Area of } \triangle AID = \frac{a^2}{4}$

Total combined area:

$a^2 + \frac{a^2}{4} = 180$

$\frac{5a^2}{4} = 180$

$5a^2 = 180 \times 4 = 720$

$a^2 = \frac{720}{5} = 144$

Since the area of each square is $a^2$:

$\text{Area of each square} = 144\text{ sq. units}$

Thus, the area of each square is $144\text{ sq. units}$.

Solution:

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Let us follow the hint and join the vertex $N$ to $Y$ with a line segment $NY$.

Step 1: Analyzing triangles $\triangle XMN$ and $\triangle YMN$

• Since $M$ is the midpoint of $XY$, the segment $XM = MY$.

• Triangles $\triangle XMN$ and $\triangle YMN$ share the same vertex $N$ and their bases $XM$ and $MY$ lie along the same line $XY$ and are equal in length.

• Therefore, the areas of $\triangle XMN$ and $\triangle YMN$ are equal:

$\text{Area}(\triangle XMN) = \text{Area}(\triangle YMN)$

Step 2: Analyzing triangles $\triangle XMN$ and $\triangle ZMN$

• Since $N$ is the midpoint of $XZ$, the segment $XN = NZ$.

• Triangles $\triangle XMN$ and $\triangle ZMN$ share the same vertex $M$ and their bases $XN$ and $NZ$ lie along the same line $XZ$ and are equal in length.

• Therefore, the areas of $\triangle XMN$ and $\triangle ZMN$ are equal:

$\text{Area}(\triangle XMN) = \text{Area}(\triangle ZMN)$

Step 3: Relating triangles to the total area of $\triangle XYZ$

From the geometric relationships above, the entire triangle $\triangle XYZ$ is divided into four smaller triangles of equal area:

$\text{Area}(\triangle XYZ) = \text{Area}(\triangle XMN) + \text{Area}(\triangle YMN) + \text{Area}(\triangle ZMN) $$ \ + \text{Area}(\text{lower triangle formed by NY and base})$

Specifically, because $MN$ connects the midpoints of two sides of $\triangle XYZ$, $MN$ is parallel to $YZ$ and half its length, dividing $\triangle XYZ$ into four congruent smaller triangles ($\triangle XMN$, $\triangle AMN$, etc. or equivalently $4$ triangles of equal area).

Thus, the area of $\triangle XMN$ is $\frac{1}{4}$ of the total area of $\triangle XYZ$:

$\text{Area}(\triangle XMN) = \frac{1}{4} \times \text{Area}(\triangle XYZ)$

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Conclusion:

The area of $\Delta XMN$ is $\frac{1}{4}$ of the area of $\Delta XYZ$.

Solution:

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Let $P$ be the point on the bank of the river, then the shortest path from $H$ to $P$ to $T$ is created by the point $P$ such that $\triangle HPT$ has minimum area.

Shortest path:

House ($H$) $\to P$ (Point on river) $\to$ Water tank ($T$)

Given:

1. Length of diagonal $AC = 22\text{ cm}$.

2. Length of perpendicular (altitude) from $B$ to $AC$, $BM = 3\text{ cm}$.

3. Length of perpendicular (altitude) from $D$ to $AC$, $DN = 3\text{ cm}$.

To Find:

Area of quadrilateral $ABCD$.

Solution:

A quadrilateral can be divided into two triangles by its diagonal. Here, diagonal $AC$ divides quadrilateral $ABCD$ into $\Delta ABC$ and $\Delta ADC$.

We use the formula for the area of a triangle, which is $\frac{1}{2} \times \text{base} \times \text{height}$.

Now, adding the areas of the two triangles:

Alternate Formula:

Thus, the area of quadrilateral $ABCD$ is $66\text{ cm}^2$.

Given:

1. Rectangle $ABCD$ with length $DC = 18\text{ cm}$ and breadth $BC = 10\text{ cm}$.

2. Points $E$ and $F$ on sides $AB$ and $AD$ respectively, with dimensions:

• $AE = 10\text{ cm}$ and $EB = 8\text{ cm}$ (Total $AB = 18\text{ cm}$).
• $AF = 6\text{ cm}$ and $FD = 4\text{ cm}$ (Total $AD = 10\text{ cm}$).

To Find:

Area of the shaded region (Quadrilateral $FECD$).

Solution:

The area of the shaded region can be found by subtracting the areas of the unshaded triangles from the total area of the rectangle $ABCD$.

Step 1: Calculate the area of rectangle $ABCD$.

Step 2: Calculate the area of unshaded $\Delta AFE$.

Step 3: Calculate the area of unshaded $\Delta EBC$.

Step 4: Find the area of the shaded region.

The area of the shaded region is $110\text{ cm}^2$.

Solution:

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To find the area of a regular hexagon, we need specific geometric measurements depending on the method used:

1. Length of a Side ($s$):

A regular hexagon is composed of $6$ congruent equilateral triangles. Therefore, knowing just the length of any side ($s$) is sufficient, because the distance from each vertex to the center is also equal to $s$.

2. Apothem ($a$) or Perpendicular Height:

Alternatively, if the side length is not directly known, we need the apothem (the perpendicular distance from the center of the hexagon to the midpoint of any side) along with either the side length or the perimeter.

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Formulas for Area:

• Using side length ($s$):

$\text{Area} = \frac{3\sqrt{3}}{2} s^2$

• Using perimeter ($P$) and apothem ($a$):

$\text{Area} = \frac{1}{2} \times \text{Perimeter} \times \text{Apothem} = 3sa$

Solution:

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Let $l$ be the length and $b$ be the breadth of the rectangle $ABCD$.

Total area of rectangle $ABCD = \text{DC} \times \text{BC} = l \times b\text{ sq. units}$

Area of $\Delta AOB = \frac{1}{2} \times \text{AB} \times \text{OE} = \frac{1}{2} \times l \times x\text{ sq. units}$

Area of $\Delta DOC = \frac{1}{2} \times \text{DC} \times \text{OF} = \frac{1}{2} \times l \times y\text{ sq. units}$

$\therefore$ The area of the shaded region = Area of $\Delta AOB + \text{Area of } \Delta DOC$

$= \frac{1}{2} \times l \times x + \frac{1}{2} \times l \times y$

$= \frac{1}{2} \times l \times (x + y)\text{ sq. units}$

$= \frac{1}{2} \times l \times b\text{ sq. units } [\because x + y = b]$

$\therefore \text{Area of shaded region} = \frac{1}{2} \times \text{Area of rectangle}$

Thus, the required fraction is $\frac{1}{2}$.

Solution:

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Let $ABCD$ be a given quadrilateral.

Mark midpoints of $AB$, $BC$, $CD$, and $DA$ as $P$, $Q$, $R$, and $S$.

Join the midpoints, then $PQRS$ is the required quadrilateral with half the area of the given quadrilateral $ABCD$.

Solution:

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(i) Regarding the areas of the parallelograms:

All the given parallelograms are drawn on the same base and between the same parallel lines.

By counting the grid units, the base of each parallelogram is $5$ units and the height (distance between parallels) is $3$ units.

Therefore, the area of each parallelogram is $5 \times 3 = 15\text{ sq. units}$.

Thus, the areas of all these parallelograms are equal ($15\text{ sq. units}$).

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(ii) Regarding their perimeters:

The perimeter of a parallelogram depends on the sum of the lengths of its adjacent sides. While the base and height remain constant across all figures, the slanted sides vary in length due to the tilt of the parallelogram.

• The figure with the minimum slant has the shortest side lengths. Hence, figure (d) has the minimum perimeter.

• The figure with the maximum slant has the longest slanted side lengths. Hence, figure (g) has the maximum perimeter.

Formula Used:

(i) Given: Base $= 7\text{ cm}$, Height $= 4\text{ cm}$

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(ii) Given: Base $= 5\text{ cm}$, Height $= 3\text{ cm}$

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(iii) Given: Base $= 5\text{ cm}$, Height $= 4.8\text{ cm}$

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(iv) Given: Base $= 2\text{ cm}$, Height $= 4.4\text{ cm}$

Given:

In parallelogram $PQRS$:

• Base $SR = 12\text{ cm}$
• Corresponding height $QM = 6\text{ cm}$
• Other side $PS = 7.6\text{ cm}$ (acts as base for height $QN$)

To Find:

The length of altitude $QN$.

Solution:

We know that the area of a parallelogram is the same regardless of which base and height are chosen.

Step 1: Calculate the area using base $SR$ and height $QM$.

Step 2: Use the calculated area to find $QN$ using base $PS$.

The length of $QN$ is $9.47\text{ cm}$.

Given:

A rectangle with side lengths $5\text{ cm}$ and $4\text{ cm}$.

A parallelogram with the same side lengths $5\text{ cm}$ and $4\text{ cm}$.

To Find:

Which figure has the greater area?

Solution:

1. Area of the Rectangle: Since the sides are perpendicular, the height is equal to the breadth.

2. Area of the Parallelogram: Let the base be $5\text{ cm}$. The slant side is $4\text{ cm}$.

The height ($h$) of a parallelogram is the perpendicular distance between the bases. In any right-angled triangle formed inside the parallelogram, the slant side ($4\text{ cm}$) acts as the hypotenuse, and the height ($h$) acts as a leg. Since the hypotenuse is always the longest side:

Therefore, for the parallelogram:

Conclusion:

The rectangle has a greater area than the parallelogram with the same side lengths.

Given:

A triangle with base $b$ and height $h$.

To Find:

A method to obtain a rectangle with area twice that of the triangle ($Area = b \times h$).

Solution:

Since the area of the rectangle needs to be $b \times h$, we can consider the following different methods:

Method 1: Same Base and Same Height

Construct a rectangle with the same base ($b$) and the same height ($h$) as the triangle. The area of this rectangle will be $b \times h$, which is exactly twice the area of the triangle.

Method 2: Double Base and Half Height

Construct a rectangle with twice the base ($2b$) and half the height ($\frac{h}{2}$) of the triangle.

Method 3: Half Base and Double Height

Construct a rectangle with half the base ($\frac{b}{2}$) and twice the height ($2h$) of the triangle.

Given:

A triangle with base $b$ and height $h$. The area is $\frac{1}{2}bh$.

Solution:

According to the Śulba-Sūtras (ancient Indian geometric texts), a triangle can be converted into a rectangle of the same area using the following method:

Method:

1. Take the base of the triangle as the base of the rectangle.

2. Take half the altitude (height) of the triangle as the height of the rectangle.

By keeping the base the same and halving the height, the rectangle occupies the same total area as the original triangle.

Solution:

Given: An isosceles triangle $ABC$ in which $AB = AC$, and $AD \perp BC$ is the altitude drawn from vertex $A$ to the base $BC$.

To Find: To convert the given isosceles triangle into a rectangle by dissection.

Method & Construction:

1. In the isosceles triangle $ABC$, the altitude $AD$ divides the triangle into two congruent right-angled triangles, namely $\Delta ADB$ and $\Delta ADC$.

2. Each of these right-angled triangles represents half of a rectangle.

3. Assembly Process:

- Take the right-angled triangle $\Delta ADB$.

- Take the congruent right-angled triangle $\Delta ADC$.

- Rotate one of the triangles by $180^\circ$.

- Place them adjacent to each other such that their corresponding sides fit perfectly to form a complete rectangle.

4. The newly formed shape is a rectangle whose dimensions are determined by the base and the altitude of the original isosceles triangle.

Illustration of the Newly Formed Rectangle:

Solution:

This is the reverse of Question 7.

Method:

Given: Rectangle $PQRS$ with length $l$ and width $w$.

Required: Isosceles triangle with the same area.

Dissection Steps:

Step 1: Take a rectangle $PQRS$ with $PQ = l$ and $PS = w$.

Step 2: Mark the midpoint $M$ of side $PQ$.

Step 3: From $M$, draw lines to the bottom corners $R$ and $S$.

Step 4: Cut the rectangle into pieces along the lines.

Step 5: Rearrange:

Flip the side triangles and attach them appropriately to form a single large isosceles triangle with a base equal to $PQ$ and height equal to the width of the rectangle.

Illustration of the Newly Formed Isosceles Triangle:

Solution:

Let the side length of both the equilateral triangle and the square be $a$.

Part 1: Equilateral triangle vs. Square of the same sidelength

1. Area of an equilateral triangle with side $a$ is given by:

$\text{Area}_{\text{triangle}} = \frac{\sqrt{3}}{4}a^2$

2. Area of a square with side $a$ is given by:

$\text{Area}_{\text{square}} = a^2$

3. Comparing both areas:

$\frac{\sqrt{3}}{4}a^2 < a^2 \quad \left[\text{Since } \frac{\sqrt{3}}{4} \approx 0.433 < 1\right]$

Conclusion for Part 1: A square has a greater area than an equilateral triangle of the same side length.

Part 2: Two identical equilateral triangles vs. Square of the same sidelength

1. Area of two identical equilateral triangles combined together with side $a$ is:

$\text{Area}_{\text{two triangles}} = 2 \times \frac{\sqrt{3}}{4}a^2 = \frac{\sqrt{3}}{2}a^2$

2. Area of the square with side $a$ remains:

$\text{Area}_{\text{square}} = a^2$

3. Comparing both areas:

$\frac{\sqrt{3}}{2}a^2 < a^2 \quad \left[\text{Since } \frac{\sqrt{3}}{2} \approx 0.866 < 1\right]$

Conclusion for Part 2: A square still has a greater area than two identical equilateral triangles of the same side length.

Visual Representation:

Given:

Length of the first diagonal ($d_1$) $= 20\text{ cm}$

Length of the second diagonal ($d_2$) $= 15\text{ cm}$

To Find:

Area of the rhombus.

Solution:

We know that the area of a rhombus is half the product of its diagonals.

Substituting the given values:

Thus, the area of the rhombus is $150\text{ cm}^2$.

Solution:

This is the reverse of the rhombus to rectangle dissection.

Method:

Given: Rectangle $PQRS$ with length $l$ and width $W$.

Required: Rhombus with the same area = $l \times W$.

Illustration of the Dissection:

Dissection Process:

Step 1: The rhombus will have diagonals $d_1$ and $d_2$ such that: $\frac{1}{2} \times d_1 \times d_2 = l \times W$

So, $d_1 \times d_2 = 2lW$.

Step 2: Choose convenient diagonal lengths:

Let $d_1 = 2l$ (twice the rectangle length)

Then $d_2 = W$ (same as rectangle width)

Check: $\frac{1}{2} \times 2l \times W = lW$

Or

Let $d_1 = 2W$ (twice the rectangle width)

Then $d_2 = l$ (same as rectangle length)

Step 3: Dissection process (reverse of textbook method):

Divide the rectangle into two halves.

Mark the center point $O$.

Cut and rotate pieces to form two isosceles triangles.

Arrange these triangles to share a common diagonal.

This creates a rhombus.

General Formula for Area of a Trapezium:

(i) Solution:

Given: Parallel sides $a = 10\text{ ft}$ and $b = 7\text{ ft}$, Height $h = 16\text{ ft}$.

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(ii) Solution:

Given: Parallel sides $a = 24\text{ m}$ and $b = 36\text{ m}$, Height $h = 14\text{ m}$.

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(iii) Solution:

Given: Parallel sides $a = 14\text{ in}$ and $b = 6\text{ in}$, Height $h = 10\text{ in}$.

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(iv) Solution:

Given: Parallel sides $a = 12\text{ ft}$ and $b = 18\text{ ft}$, Height $h = 8\text{ ft}$.

Solution:

An isosceles trapezium has special properties that make dissection simpler.

Properties of Isosceles Trapezium ABCD:

$AB \parallel CD$ (parallel sides)

$AD = BC$ (non-parallel sides are equal)

$\angle A = \angle B$ and $\angle D = \angle C$ (base angles are equal)

Illustration of the Dissection:

Dissection Method:

Step 1: Draw perpendiculars from $C$ and $D$ to $AB$, meeting at points $P$ and $Q$, respectively.

This creates rectangle $PQDC$ in the middle.

Two congruent right triangles: $\Delta APD$ and $\Delta BQC$.

Step 2: Since the trapezium is isosceles:

$\Delta APD \cong \Delta BQC$ (congruent triangles)

$AP = BQ$

Step 3: Rearrangement:

Cut triangle $\Delta APD$.

Rotate it and attach it to the right side (next to $\Delta BQC$).

The two triangles together form a rectangle with a width = height of the trapezium.

Step 4: Combine: The central rectangle $PQDC$.

The rectangle formed from the two triangles.

These can be joined to form one large rectangle.

Resulting Rectangle:

Length = $CD + AP$

= $CD + \frac{AB - CD}{2}$

= $\frac{AB + CD}{2}$

Width = $h$ (height of trapezium)

Area = $\frac{AB + CD}{2} \times h = \frac{1}{2}h(AB + CD)$

This matches the trapezium area formula!

Solution:

Given: Trapezium $ABCD$ with $AB \parallel CD$.

Required: Find the positions of the vertices $E$, $F$, $G$, and $H$ to form a rectangle $EFGH$ with equal area.

Using the Hint: If $\Delta AHI \cong \Delta DGI$ and $\Delta BEJ \cong \Delta CFJ$, then areas are equal.

Illustration of the Dissection and Conversion:

Method:

Step 1: The rectangle $EFGH$ should have:

$EF$ as one side (top side)

$GH$ is the opposite parallel side (bottom side)

$EH$ and $FG$ are the other pair of sides.

Step 2: Position the rectangle such that:

Points $I$ and $J$ are strategically chosen on the trapezium.

$\Delta AHI$ (part outside rectangle on left) is cut and moved to fill $\Delta DGI$.

$\Delta BEJ$ (part outside rectangle on right) is cut and moved to fill $\Delta CFJ$.

Step 3: For congruency:

Mark $I$ on side $AD$

Mark $J$ on side $BC$

Choose positions such that:

$HI = GI$ (making $\Delta AHI \cong \Delta DGI$ possible)

$EJ = FJ$ (making $\Delta BEJ \cong \Delta CFJ$ possible)

Step 4: The height of the rectangle = $h$ (height of the trapezium).

Step 5: The length of rectangle = $\frac{a + b}{2}$

where $a$ and $b$ are parallel sides.

This ensures: Area of rectangle = $h \times \frac{a + b}{2} =$ Area of trapezium.

Practical Construction:

Draw the trapezium $ABCD$.

Calculate required rectangle length = $\frac{AB + CD}{2}$.

Mark points $H$ and $E$ on $AB$ such that the central portion has this length.

Draw perpendiculars to get rectangle $EFGH$.

Verify that the triangular pieces outside match those inside.

Solution:

To Find: Construction of a trapezium having an area of $144\text{ cm}^2$ using the rectangle conversion method.

Method & Construction:

1. We know that the area of a trapezium is equal to the area of a rectangle whose length is equal to the average of its parallel sides and whose breadth is equal to its height ($h$).

$$\text{Area} = \left(\frac{a + b}{2}\right) \times h = 144\text{ cm}^2$$

2. Choose suitable dimensions for a rectangle with an area of $144\text{ cm}^2$:

Let the length of the rectangle be $16\text{ cm}$ and the breadth (height) be $9\text{ cm}$.

$$\text{Area of rectangle} = 16\text{ cm} \times 9\text{ cm} = 144\text{ cm}^2$$

3. To convert this rectangle into a trapezium of the same area:

- The height of the trapezium will be equal to the breadth of the rectangle, i.e., $h = 9\text{ cm}$.

- The sum of the parallel sides ($a + b$) of the trapezium must be twice the length of the rectangle:

$$a + b = 2 \times 16\text{ cm} = 32\text{ cm}$$

4. Choose convenient values for the parallel sides $a$ and $b$ such that their sum is $32\text{ cm}$:

Let $a = 20\text{ cm}$ and $b = 12\text{ cm}$.

5. Final Trapezium Dimensions:

- Parallel sides: $20\text{ cm}$ and $12\text{ cm}$

- Height (altitude): $9\text{ cm}$

6. Verify the area of the constructed trapezium:

$$\text{Area} = \frac{1}{2} \times (20 + 12) \times 9 = \frac{1}{2} \times 32 \times 9 = 16 \times 9 = 144\text{ cm}^2$$

Illustration of the Constructed Trapezium:

Given:

A regular hexagon is divided into three regions: an equilateral triangle, a rhombus, and a trapezium.

Solution:

A regular hexagon can be divided into 6 identical equilateral triangles by joining its center to each of its six vertices. Let the area of one such small equilateral triangle be $x$.

Based on the geometric properties of a regular hexagon, we can observe the following:

1. Equilateral Triangle: The smallest triangular part shown in the figure is exactly one of the 6 small triangles.

2. Rhombus: A rhombus in a regular hexagon (formed by joining two adjacent triangles) consists of 2 small equilateral triangles.

3. Trapezium: The remaining part is a large trapezium. Since the total area of the hexagon is $6x$, the area of the trapezium is:

To Find:

The ratio of their areas (Trapezium : Triangle : Rhombus).

Therefore, the ratio of the areas of the trapezium, equilateral triangle, and rhombus is $3 : 1 : 2$.

Given:

1. Trapezium $ZYXW$ with $ZY \parallel WX$.

2. $A$ is the midpoint of $XY$, so $YA = AX$.

3. $\Delta ZWB$ is formed by extending base $WX$ to $B$ and connecting $Z$ to $B$ through $A$.

To Prove:

Proof:

Consider $\Delta ZYA$ and $\Delta BXA$. In these two triangles:

Therefore, by ASA (Angle-Side-Angle) congruence criterion:

Since congruent triangles have equal areas:

Now, let's look at the areas of the figures:

Substituting equation (i) into this:

From the figure, the sum of Area(Quad $ZWXA$) and Area($\Delta BXA$) is exactly the area of $\Delta ZWB$:

Hence Proved.