GANITA MANJARI · GRADE 9 · PART I · CHAPTER 6

Chapter Summary

Think first. Use AI purposefully. Verify always.

Concept Map

Key Ideas

Perimeter is the total length of a closed boundary, counted once. It is a length, measured in units such as cm or m.
All circles share the ratio C/d = π. Diameter is twice radius. Physical measurements estimate this ratio; they do not give its exact value.
π cannot be written as a ratio of two integers. Its decimal expansion is non-terminating and non-repeating. This is a theorem, not a conclusion proved by inspecting a few digits.
An arc subtending θ degrees has θ/360 of the full circumference. On a stadium-shaped track the extra distance in outer lanes comes from the bends.
Semicircular arc length is proportional to diameter. Interior dotted arcs are not part of the outer boundary of a union of circles.
Rectangle area is ab. A parallelogram can be cut and rearranged to a rectangle with the same base and perpendicular height; its area is bh.
Two copies form a parallelogram, so triangle area is half base times perpendicular height. A median splits a triangle into equal areas because the two bases are equal and share an altitude.
Heron’s formula applies to valid triangles. Brahmagupta’s formula applies to cyclic quadrilaterals. A general quadrilateral’s sides alone do not determine its area.
Baudhāyana’s construction uses a right triangle and the difference of two squares to construct a square of area ab.
Rearranged sectors approach a parallelogram of base πr and height r. Thus circle area is π r². A sector uses two radii; a segment uses a chord and an arc.

Formula Bank

6.1 · Figures 6.2–6.4

P = 2 (a + b)

6.2 · Figures 6.5–6.7

C = πd = 2 πr

6.4 · Figures 6.8–6.11

l = \frac{θ}{360} \times 2 πr

6.6–6.7 · Figures 6.16–6.19

A = bh

6.8 · Figures 6.20–6.22

A = \frac{bh}{2}

6.8.1 · Examples 3–7, Figures 6.23–6.29

A = \sqrt{s (s - a) (s - b) (s - c)}

6.9 · Figure 6.30 and starred proof

\frac{{(a + b)}^{2}}{4} - \frac{{(a - b)}^{2}}{4} = ab

6.10–6.10.1 · Figures 6.36–6.40

A = \frac{θ}{360} π r^{2}

For Heron: s = (a+b+c)/2 and all triangle inequalities are strict. Cyclic quadrilaterals only:

s = \frac{a + b + c + d}{2}

A = \sqrt{(s - a) (s - b) (s - c) (s - d)}

A = {πr}^{2}

Visual Recall

Can a second congruent triangle complete a parallelogram?

The copy is a half-turn about the midpoint of the diagonal.

All 14 visual lessons →

Important Connections

Scaling lengths by k scales perimeter by k and area by k².
A rectangle is a special parallelogram; two congruent triangles form a parallelogram.
Heron’s expression is the zero-side limit of Brahmagupta’s cyclic-quadrilateral expression.
Arc length and sector area use the same fraction of a full turn, but different units.

Common Mistakes

Area counts square units inside a region; perimeter measures its boundary. Neither alone determines the other.
Using πr for a full circumference loses a factor of 2. The radius is not the diameter.
π = 22/7 is false. An observed lack of repetition in a finite decimal display is not a proof of irrationality.
Arc length is not sector perimeter. A one-lap stagger is not automatically the total stagger for every race format.
More pieces do not necessarily mean a longer path. Do not include internal arcs in an outer perimeter.
The sloping side is usually not the perpendicular height. Keeping side lengths fixed does not fix parallelogram area.
Equal areas do not imply congruent triangles. A median is not generally perpendicular to the base.
Do not use Brahmagupta for an arbitrary quadrilateral. Heron gives no non-degenerate triangle when the largest side equals the sum of the others.
The new square has the same area, not generally the same perimeter. Its side is √(ab), not (a+b)/2.
Finite sectors do not make an exact rectangle. A segment is not a sector. Areas need square units.

Mini Examples

TRACE · A rectangle has length 8 cm and width 3 cm.

Given

A rectangle has length 8 cm and width 3 cm.

To Find

Find its perimeter.

Formula / Property / Theorem / Reasoning

Opposite sides of a rectangle are equal.

Solution

P = 8 + 3 + 8 + 3 cm.
P = 2(8+3) cm.

Result

P = 22 cm. Original AIMS example.

DISCOVER · A circle has radius 5 cm.

Given

A circle has radius 5 cm.

To Find

Find its exact circumference.

Formula / Property / Theorem / Reasoning

C = 2πr.

Solution

The diameter is 10 cm.
C = π × 10 cm.

Result

10π cm. Original AIMS example.

APPROXIMATE · Compare 22/7 and 355/113 with a numerical reference for π.

Given

Compare $\frac{22}{7}$ and $\frac{355}{113}$ with a numerical reference for π.

To Find

Identify the closer approximation.

Formula / Property / Theorem / Reasoning

Compare absolute errors, not the number of digits written.

Solution

$\frac{22}{7}$ ≈ 3.14285714; its error is about 0.00126449.
$\frac{355}{113}$ ≈ 3.14159292; its error is about 0.00000027.

Result

$\frac{355}{113}$ is closer; neither is π. Based on the approximations discussed on pages 122–124.

ARC · Radius 3.5 cm; central angle 60°. Exercise 6.1 Q3(i) requests π ≈ 22/7.

Given

Radius 3.5 cm; central angle 60°. Exercise 6.1 Q3(i) requests π ≈ $\frac{22}{7}$ .

To Find

Find the arc length.

Formula / Property / Theorem / Reasoning

l = (θ/360) × 2πr.

Solution

l = ( $\frac{60}{360}$ ) × 7π = $\frac{7 π}{6}$ cm exactly.
Using the requested approximation gives l ≈ $\frac{11}{3}$ cm.

Result

About 3.67 cm (no final decimal-place requirement is stated in Q3).

PUZZLE · Two circles of radius r pass through one another’s centres (Example 1).

Given

Two circles of radius r pass through one another’s centres (Example 1).

To Find

Find the perimeter of their union.

Formula / Property / Theorem / Reasoning

The triangles from the two centres to either intersection are equilateral.

Solution

Each internal arc subtends 120°. Each outer arc therefore subtends 240°.
Each outer arc has length ( $\frac{240}{360}$ ) × 2πr = $\frac{4 πr}{3}$ .
Add the two outer arcs.

Result

$\frac{8 πr}{3}$ units.

RESHAPE · A parallelogram has base 9 cm and perpendicular height 4 cm.

Given

A parallelogram has base 9 cm and perpendicular height 4 cm.

To Find

Find its area.

Formula / Property / Theorem / Reasoning

Area = base × perpendicular height.

Solution

A = 9 × 4 ${cm}^{2}$ .
A horizontal shear keeps this base and height unchanged.

Result

36 ${cm}^{2}$ . Original AIMS example.

TRIANGLE · BC = 10 cm; altitude from A is 6 cm; D is the midpoint of BC.

Given

BC = 10 cm; altitude from A is 6 cm; D is the midpoint of BC.

To Find

Find areas ABC, ABD and ACD.

Formula / Property / Theorem / Reasoning

Equal bases BD and DC share the altitude from A.

Solution

Area ABC = $\frac{(10 \times 6)}{2}$ = 30 ${cm}^{2}$ .
BD = DC = 5 cm. Each smaller area is $\frac{(5 \times 6)}{2}$ .

Result

Area ABD = area ACD = 15 ${cm}^{2}$ . Original AIMS example.

SIDES · Sides 3, 4 and 5 units (Example 5).

Given

Sides 3, 4 and 5 units (Example 5).

To Find

Find the area in two ways.

Formula / Property / Theorem / Reasoning

Heron’s formula and the converse of the Baudhāyana–Pythagoras theorem.

Solution

s = $\frac{(3 + 4 + 5)}{2}$ = 6.
A = $\sqrt{6 \times 3 \times 2 \times 1}$ = 6 square units.
Because $3^{2}$ + $4^{2}$ = $5^{2}$ , the triangle is right-angled; $\frac{(3 \times 4)}{2}$ = 6.

Result

Both methods give 6 square units.

TRANSFORM · Rectangle sides 8 and 4 units.

Given

Rectangle sides 8 and 4 units.

To Find

Construct and verify the equal-area square.

Formula / Property / Theorem / Reasoning

${HP}^{2}$ = ${HK}^{2}$ − ${KP}^{2}$ .

Solution

HK = $\frac{(8 + 4)}{2}$ = 6.
KP = $\frac{(8 - 4)}{2}$ = 2.
${HP}^{2}$ = 36−4 = 32.

Result

Square side $\sqrt{32}$ = 4 $\sqrt{2}$ units; area 32 square units. Original numerical instance of Figure 6.30.

CIRCLE · Radius 7 cm, sector angle 60° (Exercise 6.3 Q1).

Given

Radius 7 cm, sector angle 60° (Exercise 6.3 Q1).

To Find

Find the area using the exercise approximation π ≈ $\frac{22}{7}$ .

Formula / Property / Theorem / Reasoning

Sector area = (θ/360)π $r^{2}$ .

Solution

Exact area = $\frac{49 π}{6}$ ${cm}^{2}$ .
Using π ≈ $\frac{22}{7}$ gives about $\frac{77}{3}$ ${cm}^{2}$ .

Result

About 25.67 ${cm}^{2}$ ; Q1 does not prescribe a final number of decimal places.

Concept Map

Key Ideas

Formula Bank

Visual Recall

Important Connections

Common Mistakes

Mini Examples

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