Chapter 6 - Measuring space :perimeter and area

NCERT Solutions for Class 9 Maths Chapter 6 Measuring Space: Perimeter and Area

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1.0Download NCERT Class 9 Maths Chapter 6 Measuring Space : Perimeter and Area Solutions PDF

Class 9 Maths Chapter 6 Measuring Space: Perimeter and Area introduces students to the concepts of perimeter and area of plane figures, helping them understand how to measure two-dimensional shapes and apply these concepts in everyday situations. Download the NCERT Solutions for Class 9 Maths Chapter 6 with Detailed Answers for all the Exercises based on the latest NCERT syllabus 2026-27 to solve every textbook Problems with ease.

NCERT Solutions Class 9 Maths Chapter 6

2.0Learning Outcomes of Class 9 Maths Chapter 6 Measuring Space: Perimeter and Area

Calculate the perimeter of rectangles, parallelograms, triangles, and other common shapes.
Understand the relationship between a circle's circumference and diameter, and explain why π is irrational.
Find the length of an arc of a circle using the central angle.
Calculate the area of squares, rectangles, parallelograms, and triangles using standard formulas.
Apply Heron's formula to find the area of a triangle when its height is not known.
Use Brahmagupta's formula to calculate the area of a cyclic quadrilateral, and recognise Heron's formula as its special case.
Determine the circumcircle and incircle of a triangle.
Understand the ancient geometric method of squaring a rectangle.
Calculate the area of a circle, and the area of a sector and segment of a circle.

3.0Detailed NCERT Class 9 Maths Chapter 6 Measuring Space: Perimeter and Area Solutions

Exercise: 6.1

Unless stated otherwise, use the approximation $\frac{22}{7}$ for $π$ .

1.The perimeter of a circle is 44 cm . What is its radius?

Sol.

Given perimeter of a circle $= 44 cm$ radius $=$ ? As we know that Perimeter of circle $= 2 π r$ Where "r" is radius So, $2 π r = 44$ $2 \times \frac{22}{7} \times r = 44$ $r = \frac{44 \times 7}{2 \times 22}$ $r = 7 cm$ 2. Calculate, correct to 3 significant figures, the circumference of a circle with: (i) radius 7 cm (ii) radius 10 cm (iii) radius 12 cm

Sol. Circumference of a circle = ? (i) If $r = 7 cm$

Circumference of a circle $= 2 π r$ $= 2 \times \frac{22}{7} \times 7 = 44 cm$ (ii) Circumference of a circle $= 2 π r$ $= 2 \times \frac{22}{7} \times 10 = 62.857 cm$ (iii) Circumference of a circle $= 2 π r$

3.Calculate the length of the arc of a circle if: (i) The radius is 3.5 cm and the angle at the centre is $6 0^{\circ}$ , and (ii) The radius is 6.3 m and the angle at the centre is $12 0^{\circ}$ . Sol. As we know that Length of $arc = \frac{θ}{360} \times 2 π r$ (i)

When $r = 3.5 cm, θ = 6 0^{\circ}$ $ℓ = \frac{60}{360} \times 2 \times \frac{22}{7} \times 3.5$ $= \frac{1}{6} \times \frac{44}{7} \times \frac{7}{2} = \frac{11}{3} cm$ (ii)

When $r = 6.3 m, θ = 12 0^{\circ}$ $ℓ = \frac{120}{36 0 ^{\circ}} \times 2 \times \frac{22}{7} \times 6.3$ $= \frac{1}{3} \times \frac{44}{7} \times \frac{63}{10} = 13.2 m$ 4. Find the perimeter of a sector (i.e., the curved portion as well as the two straight portions) of a circle of radius 14 cm and sector angle $7 5^{\circ}$ .

Sol.

Since Length of arc $= \frac{7 5 ^{\circ}}{36 0 ^{\circ}} \times 2 \times \frac{22}{7} \times 14$ $= \frac{55}{3} = 18.33 cm$ Now, Total perimeter of sector

5.Find the perimeters of the following shapes (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate) (i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

Sol. (i) Total perimeter $= 2 π r + L + B$ As $2 r = 60 m$ $r = 30 m$ So, perimeter $= (2 \times \frac{22}{7} \times 30) + 60 + 80$ $= 188.57 + 140$ $= 328.57 m$ (ii) Total perimeter $= π r + π R + (12 - 8)$ $= π (r + R) + 4$ $= \frac{22}{7} (4 + 6) + 4$ $= (\frac{22}{7} \times 10) + 4 = 31.42 + 4$ $= 35.42 cm$ (iii) Total perimeter $= 4 \times$ perimeter of half circle

Since $2 r = 10, r = 5 cm$ $= 4 \times (\frac{1}{2} \times 2 π r) = 4 π r$ $= 4 \times \frac{22}{7} \times 5$ $= \frac{440}{7}$ $= 62.85 cm$ (iv) Here $2 r = 12 cm$ $r = 6 cm$ Total perimeter $= 3 \times$ (perimeter of half circle) $= 3 \times (\frac{1}{2} \times 2 π r) = 3 π r$ $= 3 \times \frac{22}{7} \times 6$ $= 56.57 cm$ (v) Total perimeter = circumference of 4 semi-circle (with radius 7 cm ) + circumference of 4 quadrants (with radius 14 cm ) $= 4 (π r_{1}) + 4 \times \frac{1}{4} \times 2 π r_{2}$ $= 4 π r_{1} + 2 π r_{2}$ $\Rightarrow 2 π (2 r_{1} + r_{2})$ $\Rightarrow 2 \times \frac{22}{7} (14 + 14)$ $= 2 \times \frac{22}{7} \times 28 = 44 \times 4$ $= 176 cm$ (vi) Total perimeter = (Perimeter of half larger circle) + (Perimeter of 4 smaller half circle)

Here radius of $(R)$ larger semicircle $= 14 cm$ Radius of smaller semicircle $(r) = \frac{7}{2} cm$ Now, total perimeter $= π R + 4 (π r)$ $= π (R + 4 r)$ $= \frac{22}{7} [14 + (4 \times \frac{7}{2})] = \frac{22}{7} [14 + 14]$ $= \frac{22}{7} \times 28 = 88 cm$ (vii) Here total perimeter can be find out by adding perimeter of all three semicircle formed sides of right angled triangle.

Here $2 r_{1} = 6; r_{1} = 3 cm$ Same as $r_{2} = 4 cm; r_{3} = 5 cm$ Total perimeter $= π r_{1} + π r_{2} + π r_{3}$ $= π (r_{1} + r_{2} + r_{3}) = \frac{22}{7} (3 + 4 + 5)$ $= \frac{22}{7} \times 12 = 37.71 cm$ (viii) Total perimeter = (Perimeter of semi circle) + (Perimeter of 3 smaller semi circle)

Here radius of (R) larger semicircle $= 6 cm$ Radius of smaller semicircle ( r ) = 2 cm Now, total perimeter $= π R + 3 (π r)$

$= π (R + 3 r) = \frac{22}{7} [6 + (3 \times 2)] = \frac{22}{7} [6 + 6] = \frac{22}{7} \times 12 = 37.71 cm$

(ix) Total perimeter = (Perimeter of half larger circle) + (Perimeter of 2 smaller half circle)

Now, total perimeter $= π R + 2 (π r)$ $= π (R + 2 r)$ $= \frac{22}{7} [10 + 2 (5)]$ $= \frac{22}{7} \times (20)$ $= \frac{440}{7}$ $= 62.85 cm$ 6. If the diameter of a car tyre is 56 cm , then: (i) How far does the car need to travel for the tyre to complete one revolution? (ii) How many revolutions does the tyre make if the car travels 10 km ?

Sol. Since diameter of car tyre ( 2 r ) $= 56 cm$ (Radius) $r = 28 cm$ As we know that Distance covered in one revolution $=$ circumference of tyre $= 2 π r$ $= 2 \times \frac{22}{7} \times 28 = 176 cm$ Total distance $= 10 km$ $= 10 \times 1000 \times 100$ $= 1000000 cm$ Total distance $=$ (Total number of revolutions) × (distance covered in 1 revolution)

So, total number of revolutions $= \frac{1000000}{176} = 5681.81$ (approx.) 7. Find the total perimeter of all the petals in each of the given flowers. (i)

(ii)

Sol. (i) Here total semicircle are 4 Each semicircle have diameter $= 14 cm$ Radius $= 7 cm$

Total perimeter = sum of arc length of all four semi-circles $= 4 \times π r$ $= 4 \times \frac{22}{7} \times 7 = 88 cm$ (ii) Here total semicircle are 6 Each semicircle have radius $= 42 cm$ Total perimeter = sum of arc length of all six semi-circles $= 6 \times π r$

8.The ratio of the perimeter of two circles is $5 : 4$ . What is the ratio of their radii? Sol. Given ratio of perimeter of two circles is $5 : 4$ Let their radii respectively $R, r$ Thus $\frac{2 π R}{2 π r} = \frac{5}{4}$ $\frac{R}{r} = \frac{5}{4}$ Ratio of their radii respectively $5 : 4$ .

Exercise: 6.2

1.Find the area of triangle ADE in figure.

Sol.

Construction: Draw perpendicular EF to AD Here $EF = CD$

Area of $△ AED = \frac{1}{2} \times$ Base × altitude $= \frac{1}{2} \times AD \times EF$ $= \frac{1}{2} \times 8 \times 10 = 40 cm^{2}$ 2. The parallel sides of a trapezium are 40 cm and 20 cm . If its non-parallel sides are both equal, each being 26 cm , find the area of the trapezium. Sol.

ABCD is trapezium in which $AB ∥ DC$ and $AD = BC = 26 cm$ $AB = 20 cm$ $CD = 40 cm$ Construction : Draw a parallel sides to AD, which BE so that ABED becomes a parallelogram ( $AB ∣∣ DE, AD ∣∣ BE$ ) Now $BE = 26 cm$ & $EC = 20 cm$ Now we can find out area of DEBC using Heron's formula.

Here $a = 26, b = 26, c = 20$ $S = \frac{a + b + c}{2} = \frac{26 + 26 + 20}{2} = \frac{72}{2} = 36$ Area of $△ EBC$ $= 36 (36 - 26) (36 - 26) (36 - 20)$ $= 36 \times 10 \times 10 \times 16$ $= 6 \times 10 \times 4$ $= 240 cm^{2}$ Draw ⟂ to EC such that BF ⟂ EC Now, area of $△ BEC$ $\frac{1}{2} \times EC \times BF = 240$ $\frac{1}{2} \times 20 \times BF = 240$ $BF = \frac{240 \times 2}{20}$ $BF = 24 cm$

Here BF will be also altitude of parallelogram ABED

Now area of ABED $=$ Base × altitude $= 20 \times 24 = 480 cm^{2}$ Total area of trapezium ABCD $= 480 + 240$ $= 720 cm^{2}$ 3. Find the area of a triangle, given that its sides are 8 cm and 11 cm long, and its perimeter is 32 cm .

Sol.

Let $PQ = 8 cm$ PR = 11 cm $∴$ Given perimeter $= 32 cm$

$QR = 32 - (8 + 11) = 14 cm$

Area of $△ PQR =$ ? $S = \frac{a + b + c}{2}$ $= \frac{8 + 11 + 14}{2}$ $= \frac{32}{2} = 16$ Area $= s (s - a) (s - b) (s - c)$ $= 16 (16 - 8) (16 - 11) (16 - 14)$ $= 4 \times 4 \times 4 \times 2 \times 5 \times 2$ $= 4 \times 2 \times 25$ $= 165 cm^{2}$ 4. The sides of a triangular plot are in the ratio $3 : 5 : 7$ ; its perimeter is 300 m . Find its area.

Sol. Let the sides of given triangle are $= 3 x, 5 x, 7 x$ Since perimeter is sum of all its sides

So, $3 x + 5 x + 7 x = 300$ $15 x = 300$ $x = 20$ sides are $a = 60, b = 100, c = 140$ $s = \frac{a + b + c}{2}$ $= \frac{60 + 100 + 140}{2} = \frac{300}{2} = 150$ Area of triangular plot $= 150 (150 - 60) (150 - 100) (150 - 140)$ $= 150 \times 90 \times 50 \times 10$ $= 10 \times 10 \times 5 \times 33$ $= 15003 m^{2}$ 5. One diagonal of a rhombus is twice as long as the other diagonal. If the rhombus has area $128 cm^{2}$ , find the length of the shorter diagonal.

Sol.

Let ABCD is rhombus & $AC = x cm BD = 2 x cm$

Since area of rhombus $= \frac{1}{2} \times$ Product of diagonals. So, $\frac{1}{2} \times x \times 2 x = 128 cm^{2}$ $x^{2} = 128$ $AC = x = 82 cm =$ smaller diagonal. 6. ABCD is a parallelogram. P and Q are any two points on side AB . What can you say about the ratio area ( $△ PCD$ ) : area $(△ QCD)?$ Sol.

Here ABCD is parallelogram As we know that "Triangles are on same base and between same parallels are equal in area's". Here $△ PCD$ & $△ QCD$ are on same base CD and between same parallels ( $AB ∥ CD$ ) So, area of $△ PCD$ : area of $△ QCD = 1 : 1$ . 7. $O$ is any point on the diagonal PR of a parallelogram PQRS. Prove that the areas of triangles PSO and PQO are equal. Sol.

Here PQRS is parallelogram ' O ' is any point on diagonal PR. T is intersection point of both diagonals. Here PT is the median of $△ PSQ$ , which will divide the $△ PSQ$ in two equal area. (Same for $△ OSQ$ also) Now, area of $△ PST =$ area of $△ PQT$

Same as area of $△ OST =$ area of $△ OQT$

Now, equation (1)-(2) area of $△ PSO =$ area of $△ PQO$ Hence proved. 8. If the mid-points of the sides of a 4-gon (also known as a quadrilateral, but we prefer to call it a '4-gon') are joined in order, prove that the area of the parallelogram thus formed will be half of the area of the given 4 -gon. (You may wonder whether the 4-gon thus formed is always a parallelogram, and if so, why? These questions will be tackled and answered in the chapter on quadrilaterals.) Sol.

ar (△ PBC) = \frac{1}{2} ar (△ ABC)

[Median of a triangle bisects the area of it]

ar (△ PBQ) = \frac{1}{2} ar (△ PBC)

\Rightarrow ar (△ PBQ) = \frac{1}{4} ar (△ ABC)

Similarly, $ar (△ ADR) = \frac{1}{2} ar (△ ADC)$ $ar (△ DSR) = \frac{1}{2} ar (△ ADR)$ $\Rightarrow ar (△ DSR) = \frac{1}{4} ar (△ ADC)$

Adding equation (i) and (ii) $ar (△ PBQ) + ar (△ DSR) = \frac{1}{4} [ar (△ ABC) +$ $ar (△ ADC)]$ $ar (△ PBQ) + ar (△ DSR) = \frac{1}{4} ar (ABCD)]$

Similarly, by joining diagonal BD it can to proved. $ar (△ ASP) + ar (△ QCR) = \frac{1}{4} ar (ABCD)$

Adding equation (iii) and (iv) $ar (△ PBQ) + ar (△ DSP) + ar (△ ASP) + ar (△ QCR) = \frac{1}{2} ar (ABCD)$ Subtracting both sides from ar(ABCD) $ar (ABCD) - [ar (△ PBQ) + ar (△ DSR) + ar (△ ASP) + ar (△ QCR)] = ar (ABCD) - \frac{1}{2}$ $ar (ABCD)$ $\Rightarrow ar (PQRS) = \frac{1}{2} ar (ABCD)$ 9. In $△ ABC$ , the midpoint of BC is D . Median AD is drawn. P is any point on AD . Show that area $(△ ABP) =$ area $(△ ACP)$ .

Sol. Since a median divides the triangle in two equal area. Here $A D$ is the Median of $△ A BC$ and $P D$ is the Median of $△ BPC$ . So, $ar (△ ABD) = ar (△ ACD)$

Same as, $ar (△ BPD) = ar (△ CPD)$

Equation (1)-(2)

So, $ar (△ ABP) = ar (△ ACP)$ Hence Proved 10. Given a square ABCD , let P be a point within it. Join PA, PB, PC, PD. What is the ratio of the areas of the red region ( $△ PAB$ and $△ PCD$ ) and the green region ( $△ PBC$ and $△ PDA$ )? Sol.

Let side of square ABCD be $= a$ Draw ⟂ from P to DC & AB at point M & N. Let $PM = x$ So $PN = a - x$ Now, area of $△ PAB +$ area of $△ PCD$ $= \frac{1}{2} \times a \times (a - x) + \frac{1}{2} \times a \times x$ $= \frac{1}{2} (a^{2} - ax + ax) = \frac{a ^{2}}{2}$ Same as, area of $△ PBC +$ area of $△ PDA$ $= a^{2} - \frac{a ^{2}}{2} = \frac{a ^{2}}{2}$ So, Ratio of Red Region to Green region will be: $= \frac{a ^{2}}{2} : \frac{a ^{2}}{2} = 1 : 1$ 11. In $△ ABC, D$ is the midpoint of AB . P is any point on BC , and Q is a point on AB such that $CQ ∥ PD . PQ$ is joined. Prove that Area $(△ BPQ) = \frac{1}{2}$ Area $(△ ABC)$ .

Sol.

Construction: Join D to C Since " $D$ " is mid point of $A B$ , thus $C D$ is the Median of $△ ABC$ .

And median divides the triangle in two equal area.

So, $ar (△ BDC) = \frac{1}{2} ar (△ ABC)$

Since $QC ∣ ∣ DP$ and $△ DPC$ & $△ DQP$ are on the same base and between the same parallel lines, so both will have equal area.

Thus, $ar (△ DPQ) = ar (△ DPC)$

Now, $ar (△ BDC) = ar (△ BDP) + ar (△ DPC)$ $= ar (△ BDP) + ar (△ DQP)$ $ar (△ BDC) = ar (△ BQP)$

So from equation (1), (2) & (3) $ar (△ BQP) = \frac{1}{2} ar (△ ABC)$ Hence Proved

Exercise : 6.3

1.Find the area of a sector of a circle with radius 7 cm if the angle of the sector is $6 0^{\circ}$ .

Sol. angle $(θ) = 6 0^{\circ}$ , radius $(r) = 7 cm$ . area of sector $= \frac{θ}{36 0 ^{\circ}} \times π r^{2}$ $= \frac{6 0 ^{\circ}}{36 0 ^{\circ}} \times \frac{22}{7} \times 7 \times 7$ $= \frac{1}{6} \times 154 = \frac{77}{3} cm^{2}$ 2. Find the area of a quadrant of a circle whose circumference is 44 cm .

Sol. Since $2 π r = 44$ $2 \times \frac{22}{7} \times r = 44$ $r = 7 cm$ area of quadrant $= \frac{1}{4} \times π r^{2}$ $= \frac{1}{4} \times \frac{22}{7} \times 7 \times 7 = \frac{77}{2} = 38.5 cm^{2}$ 3. The length of the minute hand of a clock is 7 cm . Find the area swept by the minute hand in 10 minutes.

Sol. For minute hand: $60 min = 36 0^{\circ}$ $10 min = 6 0^{\circ}$ radius $= 7 cm$ area swept by minute hand $= \frac{6 0 ^{\circ}}{36 0 ^{\circ}} \times \frac{22}{7} \times 7 \times 7$ $= \frac{1}{6} \times 154 = \frac{77}{3} cm^{2}$ 4. A chord of a circle of radius 10 cm subtends $9 0^{\circ}$ at the centre. Find the area of the corresponding: (i) minor sector (that subtends $9 0^{\circ}$ at the centre), and (ii) major sector (that subtends $27 0^{\circ}$ at the centre). (Use $π \approx 3.14$ )

Sol. (i) $θ = 9 0^{\circ}$ , radius $= 10 cm$ area of minor sector $= \frac{9 0 ^{\circ}}{36 0 ^{\circ}} \times 3.14 \times 10 \times 10$ $= \frac{157}{2} = 78.5 cm^{2}$ (ii) area of major sector $= \frac{27 0 ^{\circ}}{36 0 ^{\circ}} \times 3.14 \times 10 \times 10$ $= \frac{3}{4} \times 314 = 235.5 cm^{2}$ 5. A chord of a circle of radius 15 cm subtends an angle of $6 0^{\circ}$ at the centre of the circle. Find the areas of the corresponding minor and major segments of the circle. (Use $π \approx 3.14$ and $3 \approx 1.73$ )

Sol.

OA = OB =

radius

= 15 cm

θ = 6 0^{\circ}

Here AOB is an equilateral

Δ

. area of minor segment (APB)} = area of sector (AOBP) - (area of

△ AOB

)

\frac{6 0 ^{\circ}}{36 0 ^{\circ}} \times 3.14 \times 15 \times 15 - \frac{3}{4} (15)^{2}

= 1 5^{2} (\frac{1}{6} \times 3.14 - \frac{3}{4})

= (15)^{2} (\frac{157}{300} - \frac{1 7 ^{3}}{400})

= 225 \times \frac{628 - 519}{1200}

= \frac{225 \times 109}{1200}

= 20.43 cm^{2}

area of major sector

= π r^{2} - 20.43

= 3.14 \times 15 \times 15 - 20.43

= 706.5-20.43

= 686.07 cm^{2}

6. A car has two wipers which do not overlap. Each wiper has a blade of length 28 cm and sweeps through an angle of

12 0^{\circ}

. Find the total area cleaned at each sweep of the blades.

Sol. Area cleaned by 1 wiper $= \frac{12 0 ^{\circ}}{36 0 ^{\circ}} \times \frac{22}{7} \times 28 \times 28$ $= \frac{1}{3} \times 88 \times 28 = 821.33 cm^{2}$ area cleaned by both wiper $= 821.33 \times 2 = 1642.67 cm^{2}$ 7. A chord of a circle of radius $r$ subtends an angle of $6 0^{\circ}$ at the centre of the circle. Show that the area of the corresponding minor segment of the circle is equal to $π r^{2} (\frac{1}{6} - \frac{3}{4})$ .

Sol. Area of minor segment = area of minor sector (AOBP) - area of equilateral $△ AOB = \frac{6 0 ^{\circ}}{36 0 ^{\circ}} \times π r^{2} - \frac{3}{4} r^{2}$ $= π r^{2} (\frac{1}{6} - \frac{3}{4}) cm^{2}$ 8. An equilateral triangle is inscribed in a circle of radius $r$ . Show that the ratio of the area of the triangle to the area of the circle is equal to $\frac{3 3}{4 π} \approx 0.413$ .

Sol.

Let side of given triangle $= a$ Since inradius $(r) = \frac{Side of equilateral Δ}{2 3}$ $r = \frac{a}{3}$ so, $a = 3 r$

Now, $\frac{area of triangle}{area of circle} = \frac{\frac{3}{4} ( Side ) ^{2}}{π r ^{2}}$

9.A square is inscribed in a circle of radius r. Show that the ratio of the area of the square to the area of the circle is equal to $\frac{2}{π} \approx 0.637$ .

Sol.

Radius $= r$ where $a =$ side of square Here Diagonal of square $=$ diameter of circle $2 \cdot a = 2 r$ $a = 2 r$ area of square : area of circle $a^{2} : π r^{2}$ $2 r^{2} : π r^{2}$ 2 : $π$ $\Rightarrow \frac{2}{π}$ $\Rightarrow \frac{2 \times 7}{22} = \frac{7}{11} \approx 0.637$ 10. A hexagon is inscribed in a circle of radius r. Show that the ratio of the area of the hexagon to the area of the circle is equal to $\frac{3 3}{2 π} \approx 0.827$ .

Sol.

Side $=$ radius, Let $= a$ Here ABCDEF is hexagon whose area is defined by $= 6 \times \frac{3}{4} \times (side)^{2}$ $= \frac{3 3}{2} a^{2}$ Area of circle $= π (a)^{2}$ Ratio $\Rightarrow \frac{3 3}{2} a^{2} : π a^{2}$ $= \frac{3 3}{2} a^{2} : π$ $= \frac{3 3}{2 π} = \frac{3 \times 1.73 \times 7}{2 \times 22} \approx 0.827$ Here answer of this question is double of answer of question 8 because Hexagon is area is double of equilateral triangles area.

End of Chapter Exercise

In the problems below, unless stated otherwise use the approximation $\frac{22}{7}$ for $π$ .

1.Identities in algebra can sometimes be shown as area relationships. For example: The figure shown corresponds to the identity $(a + b)^{2} = a^{2} + 2 ab + b^{2}$ .

Do you see how?

Area model of an identity Draw figures corresponding to the identities $(a + b) (a - b) = a^{2} - b^{2}$ and $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 ab + 2 b c + 2 c a$ . Sol. For identify $a^{2} - b^{2}$ Here Unshaded region represent $a^{2} - b^{2}$ = region (I) + region (II) $= b (a - b) + a (a - b)$ $= ab - b^{2} + a^{2} - ab$ $= - b^{2} + a^{2}$ $= a^{2} - b^{2}$

For identify $(a + b + c)^{2}$ $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 ab + 2 bc + 2 ca$

2. An isosceles triangle has perimeter 40 cm; the equal sides are 15 cm each. Find the area of the triangle.

Sol.

Perimeter $= 40 cm$ $3^{rd}$ side $= 40 - (2 \times 15) = 10 cm$ $b = 10 cm$ $a = 15 cm$ area of isosceles triangle $= \frac{b}{4} (2 a)^{2} - (b)^{2}$ $= \frac{10}{4} 4 (15)^{2} - (10)^{2}$ $= \frac{10}{4} 800$ $= 502 cm^{2}$ 3. An isosceles triangle has base 10 cm , and its area is $60 cm^{2}$ . What are the lengths of the equal sides?

Sol. Here $b = 10, a =$ ? (Let equal side is a )

Δ PQR =

isosceles

Δ

Since

\frac{b}{4} (2 a)^{2} - (b)^{2} = 60

\frac{10}{4} 4 a^{2} - 100 = 60

4 a^{2} - 100 = (24)^{2}

4 a^{2} = 100 + 576

4 a^{2} = 676

a^{2} = 169

a = 13 cm

4. The area of a right-angled triangle is 54 sq. cm. One of its legs has length 12 cm . Find its perimeter.

Sol.

Area of equilateral triangle $= \frac{1}{2} \times$ Base × Height $\frac{1}{2} \times 12 \times x = 54$ $x = \frac{108}{12} = 9 cm$ $AC = (9)^{2} + (12)^{2}$ (using Pythagoras theorem) $AC = 81 + 144$ $AC = 225$ $= 15 cm$ Perimeter $= 9 + 12 + 15$ $= 36 cm$ 5. The sides of a triangle are in the ratio $2 : 3 : 4$ , and its perimeter is 45 cm . Find its area.

Sol. Let sides of triangle are $2 x, 3 x, 4 x$ So $2 x + 3 x + 4 x = 45$ $9 x = 45$ $x = 45$ Sides are $= 10, 15, 20 cm$ . Area of triangle we will find using Heron's formula. $S = \frac{45}{2} = 22.5$ area $= S (S - a) (S - b) (S - c)$ $= 22.5 \times 12.5 \times 7.5 \times 2.5$ $= \frac{225}{10} \times \frac{125}{10} \times \frac{75}{10} \times \frac{25}{10}$ $= \frac{15 \times 5 \times 5 \times 5 5 \times 3}{10 \times 10}$ $= \frac{1875 15}{100} cm^{2}$ 6. The sides of a triangle have lengths 7 cm , $24 cm, 25 cm$ . Find the area of the triangle in two different ways.

Sol. I Method Side lengths are $= 7 cm, 24 cm, 25 cm$ Area of triangle $= \frac{1}{2} \times 7 \times 24 = 84 cm^{2}$ II Method (Using Heron's formula) $S = \frac{7 + 24 + 25}{2} = \frac{56}{2} = 28$ area $= 28 \times (28 - 7) (28 - 24) \times (28 - 25)$ $= 7 \times 4 \times 7 \times 3 \times 2 \times 2 \times 3$ $= 7 \times 2 \times 3 \times 2$ $= 84 cm^{2}$ 7. If the wheel of a bicycle has a diameter of 60 cm , find how far a cyclist will have travelled after the wheel has rotated 100 times.

Sol. Diameter of wheel $= 60 cm$ . its radius $= 30 cm$ . Distance covered in 1 rotation = circumference of wheel

Now distance covered in 100 rotations $= 100 \times 2 π r$ $= 100 \times 2 \times \frac{22}{7} \times 30$ $= \frac{132000}{7} cm$ $= \frac{1320}{7}$ meter 8. Find the area of a quadrant of a circle whose circumference is 66 cm .

Sol. $2 π r = 56$ $2 \times \frac{22}{7} \times r = 66$ $r = \frac{66 \times 7}{22 \times 2}$ $r = \frac{21}{2} cm$ Now area of quadrant $= \frac{1}{4} π r^{2}$ $= \frac{1}{4} \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2}$ $= 86.625 cm^{2}$ 9. The wheel of a car has an outer radius of 28 cm . Calculate how far the car travels after one complete turn of the wheel, and how many times the wheel turns during a journey of 1 km .

Sol. Radius $= 28 cm$ Car travels after 1 complete turn $= 2 π r$ $= 2 \times \frac{22}{7} \times 28$ $= 176 cm$ Numbers of rotations of wheel $= \frac{total distance}{distance in onerotation}$ $= \frac{1 km}{176}$ $= \frac{1 \times 1000 \times 100}{176} \approx 568.18$ approx. 10. Two rectangles have the same area and the same perimeter. Does this mean that they are congruent to each other?

Sol. Yes, if two rectangle have both the same area and the same perimeter, they are congruent. Because fixing both the sum of their sides (Perimeter) and the product of their sides (area) restricts them to the same dimensions, ensuring that they are identical in size and shape. 11. You know that the area of a parallelogram is base × height. Using this and the figure, show that the area of a trapezium is half the sum of the parallel sides × height, i.e., $\frac{1}{2} (a + b) h$ .

Sol.

Since area of Parallelogram $= a \times h$ Now area of triangle $= \frac{1}{2} \times$ Base × Altitude $= \frac{1}{2} \times (b - a) \times h$ Now area of trapezium = area of parallelogram + area of triangle $= ah + \frac{1}{2} (b - a) h$ $= ah + \frac{bh}{2} - \frac{ah}{2}$ Area of trapezium $= \frac{ah}{2} + \frac{bh}{3} = \frac{1}{2} h (a + b)$ 12. By dividing a trapezium into two triangles show that its area is, half the sum of the parallel sides multiplied by the height (the same formula as the one given above).

Sol.

Area of trapezium $ABCD =$ area of both triangles $=$ or $(△ ABD) +$ or $(△ BDC)$ $= \frac{1}{2} \times ah + \frac{1}{2} \times b \times h$ $= \frac{1}{2} \times h (a + b)$ 13. Show how we can use two identical copies of a trapezium to make a parallelogram. How will this give us the formula for the area of a trapezium?

Sol. Let we have two identical copies of trapezium with parallel sides a, b & altitude as ' $h$ '

Since both trapeziums having same area. Area of parallelogram so formed $=$ Base × altitude $= (a + b) h$ Area of each trapezium $= \frac{1}{2}$ area of parallelogram $= \frac{1}{2} (a + b) h$ 14. Show that the area of a kite is half the product of its diagonals.

Show this: (i) using algebra, and (ii) using geometry.

Sol. As we know that diagonals of kite are ⟂ to each other and act as the base and height for triangles that compose the shape. Also longer diagonal will bisect smaller.

Let ABCD is kite and ' 0 ' is intersection point of its diagonals.

Let $BD = d_{1}$ and $AC = d_{2}$ Now or $△ ABD = \frac{1}{2} \times BD \times OA$ $= \frac{1}{2} \times BD \times \frac{d _{2}}{2}$ $= \frac{1}{4} d_{1} d_{2}$ Same as or $△ BC D = \frac{1}{2} \times B D \times OC$ $= \frac{1}{2} \times BD \times \frac{d _{2}}{2}$ $= \frac{1}{4} d_{1} d_{2}$ Now or $(□ ABCD) =$ or $(△ ABD) +$ or $(△ BCD)$ $= \frac{1}{4} d_{1} d_{2} + \frac{1}{4} d_{1} d_{2}$ $= \frac{1}{2} d_{1} d_{2}$ 15. Three problems about fitting congruent shapes together: (i) Rectangle ABCD has sides $a, b$ , and rectangle PQRS has sides 2a, 2b. Show that PQRS has 4 times the area of ABCD . Does this mean that 4 copies of rectangle ABCD will fit into rectangle PQRS ? (ii) $△ ABC$ has sides $a, b, c$ and $△ PQR$ has sides 2a, 2b, 2c. Show that $△ PQR$ has 4 times the area of $△ ABC$ . Does this mean that 4 copies of $△ ABC$ will fit into $△ PQR$ ? (ii) $△ ABC$ has sides $a, b, c$ and $△ PQR$ has sides 3a, 3b, 3c. Show that $△ PQR$ has 9 times the area of $△ ABC$ . Does this mean that 9 copies of $△ ABC$ will fit into $△ PQR$ ? Check and see! Sol. (i) area of $ABCD = a \times b$

area of

PQRS = 2 a \times 2 b = 4 ab = 4

(area of ABCD) (ii) area of

△ ABC

is given by

= S (S - a) (S - b) (S - c)

Where = its semi perimeter Area of

△ PQR

= S^{'} (S^{'} - 2 a) (S^{'} - 2 b) (S^{'} - 2 c)

= 2 S (2 S - 2 a) (2 S - 2 b) (2 S - 2 C)

= 4 S (S - a) (S - b) (S - c)

= 4 \times

area of

△ ABC

Note

S^{'} =

semi perimeter

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

S^{'} = a + b + c = 2 S

(iii) same as part (ii) we can find

16.What fraction of the triangle is shaded?

Sol. (i) Construction : Join Q and T

Since $△ PTS$ and $△ QTS$ are on same base ( $PS = SQ$ ) and having same altitude so

Let $x = ar Δ PTS = ar Δ QTS$

Same as ar $△ PQT = ar$ $Δ QT U = 2 x$ also again $ar Δ QT U = ar Δ RQ U = 2 x$

Total shaded area $= x + 2 x = 3 x$ required fraction $= \frac{3 x}{6 x} = \frac{1}{2}$ (ii) As given square is ABCD extend it to make such that it becomes PQRS square of size ( $3 \times 3$ ).

Now are $(△ ATD) = \frac{1}{2} ar (ASDT)$ $= \frac{1}{2}$ (area of 2 smaller square) = area of smaller square Same as Area of $(△ ADT) = ar (△ AWB) = ar (△ BVC) = ar (△ DUC) =$ area of smaller square ar(sq. ABCD) $= 5 \times$ area of smaller square shaded area fraction $= \frac{1}{5}$ 17. (i) What fraction of the rectangle is covered by the circles?

(ii) What fraction of the rectangle is covered by the circles?

Sol. (i) For figure

Let breadth is $= b$ Let length of Rectangle $= ℓ$

So $6 r = ℓ$ $r = \frac{ℓ}{6}$ same as $2 r = b$ $r = \frac{b}{2}$ area of 3 circles $= 3 π r^{2}$ area of rectangle $= ℓ \times b$ $= 6 r \times 2 r = 12 r^{2}$ Required fraction $= \frac{3 π r ^{2}}{12 r ^{2}} = \frac{π}{4} = \frac{11}{14}$ (ii)

Same as part (i) we can solve part (ii) Here $8 r = ℓ$ $2 r = b$ area of 4 circles $= 4 π r^{2}$ area of rectangle $= ℓ \times b$

Required fraction $= \frac{4 π r ^{2}}{ℓ \times b} = \frac{4 π r ^{2}}{16 r ^{2}}$ $= \frac{π}{4} = \frac{11}{14}$ 18. Use the above to make a conjecture about the area occupied by circles fitted into a rectangle in the manner shown. Test your conjecture for particular cases: 10 circles; 20 circles; 50 circles. Then prove your conjecture!

Sol. In previous Q. 17 (i) & (ii) the ratio of area of circles to the ratio of rectangle is always $\frac{11}{14}$ or $= 0.7857$ .

In case of 10 circles, 20 circles & 50 circles also ratio will be same.

Here we can take example of 10 circles, in which 10 circles are touches each other, also enclosed by a rectangle

Where radius of circle $= r$ , diameter $= 2 r$ Length and breadth of rectangle $= 20 r, 2 r$ Required fraction $= \frac{area of 10 circle}{area of Rectangle}$ $= \frac{10 \times \frac{22}{7} \times r ^{2}}{L \times B} = \frac{220 r ^{2}}{7 \times 20 r \times 2 r} = \frac{11}{14} = 0.7857$ Same we can prove for 20 circles and 50 circles also. 19. The figure shows nine identical rectangles fitted together to make a large rectangle whose area is $72 cm^{2}$ . Find the perimeter of each small rectangle.

Nine identical rectangles stacked together

Sol.

Let length of rectangle $= ℓ$ & breadth = b So, $4 ℓ = 5 b$ $ℓ = \frac{5 b}{4}$ Now area of each rectangle $= \frac{72}{9} = 8 cm^{2}$ $∵ ℓ \times b = 8$ $\frac{5 b ^{2}}{4} = 8 \Rightarrow b^{2} = \frac{32}{5} \Rightarrow b = \frac{4 10}{5} cm$ Now $ℓ = \frac{5}{4} \times \frac{4 10}{5} = 10 cm$ & $b = \frac{4 10}{5}$ Perimeter of Rectangle $= 2 (\frac{4 10}{5} + \frac{4 10}{5})$ $= \frac{2 \times 9 10}{5}$ $= \frac{18 10}{5} cm$ 20. Show that the areas of the shaded blue triangle and the shaded red triangle are equal.

Find a way of cutting up the blue triangle into some number of pieces and rearranging the pieces to cover the red triangle.

Sol.

Draw $AF ⊥ BC$ Let $AF = h$ Here $BD = DE = EC = b$ Since D, E trisect the base BC. & attitude of all triangles are same i.e. AF So $ar (△ ABD) = \frac{1}{2} \times AF \times BD = \frac{1}{2} bh$ [area of triangle $= \frac{1}{2} \times$ Base × Height] $ar (△ ADE) = \frac{1}{2} \times DE \times AF = \frac{1}{2} bh$ $ar (△ AEC) = \frac{1}{2} \times EC \times AF = \frac{1}{2} bh$ rearrangement Idea: The blue triangle can be we into smaller pieces parallel to one side and rearranged to exactly cover the red $Δ$ because both triangles have equal area and equal altitude. 21. The figure shows a quarter circle in a square. Its centre is at one vertex, and it passes through two adjacent vertices. There are two semicircles on two adjacent sides as diameters. They create the shaded regions A and B .

Show that $A$ and $B$ have equal area.

Sol. Let PQRS is a square with side $= 2 r$

Half area of region

A = \frac{1}{2} π r^{2} - \frac{1}{2} π r^{2} = \frac{r ^{2}}{2} (\frac{π}{2} - 1)

Now area of

A = r^{2} (\frac{π}{2} - 1)

area of region "B"

= \frac{1}{4} π (2 r)^{2} - (r^{2}) - 2 \times \frac{1}{4} π r^{2}

= \frac{π}{4} \times 4 r^{2} - r^{2} - \frac{π r ^{2}}{2}

= π r^{2} - \frac{π r ^{2}}{2} - r^{2}

= \frac{π r ^{2}}{2} - r^{2}

= r^{2} (\frac{π}{2} - 1)

So, area of region ' A ' = area of region ' B ' 22. In the figure, four semicircles have been drawn within the given square whose side is 2 units. The centres of these semicircles are the midpoints of the sides. They create a 4-petalled flower (shown in blue). Find the perimeter and the area of this flower.

Sol. Let mention blank part as mentioned in figure

Area of (I + III) $= (2)^{2} - π (1)^{2}$ $= 4 - π$

Same as area of (II + IV) = $4 - π$ Total shaded area $= (2)^{2} - 2 (4 - π)$ $= 4 - 8 + 2 π$ $= 2 π - 4$ $= \frac{44}{7} - 4$ $= \frac{44 - 28}{7}$ Area of 4 petalled flowers $= \frac{16}{7}$ unit Total Perimeter of 4 petalled flowers = 4 semicircle's perimeter $= 2 \times 2 π r$ $= 2 \times 2 \times \frac{22}{7} \times 1$ $= \frac{88}{7}$ unit 23. In the figure we see two concentric circles with a common centre O . A chord BC of the larger circle is drawn, touching the smaller circle at A . The length of BC is $ℓ$ .

Show that the area of the green region enclosed between the two circles is $\frac{1}{4} π r^{2}$ .

Sol. Here Let radii of smaller and larger circle are $= r & R$ respectively.

Shaded area $=$ area of larger circle - area of smaller circle. $= π R^{2} - π r^{2}$ Shaded area $= π (R^{2} - r^{2})$ Draw ⟂ form 'O' to BC

Using Pythagoras theorem $AB^{2} = R^{2} - r^{2}$ $(\frac{ℓ}{2})^{2} = R^{2} - r^{2}$ Now shaded area $= π (\frac{ℓ}{2})^{2} = \frac{π ℓ ^{2}}{2}$ 24. In the figure, semicircles have been drawn on all the sides of a right-angled triangle as shown. Show that Area (A) + Area (B) $= Area (C)$ .

Sol. Let $PQ = a$ $QR = b$ PR $= c = a^{2} + b^{2}$

Area of unshaded part in semi circle PQR is $\Rightarrow \frac{π}{2} (\frac{a ^{2} + b ^{2}}{2})^{2} - \frac{1}{2} ab$ $\Rightarrow \frac{π}{8} (a^{2} + b^{2}) - \frac{1}{2} ab$ area $(A) + area (B)$ $\Rightarrow \frac{π a ^{2}}{8} + \frac{π b ^{2}}{8} - (\frac{π}{8} (a^{2} + b^{2}) - \frac{1}{2} ab)$ $\Rightarrow \frac{π a ^{2}}{8} + \frac{π b ^{2}}{8} - \frac{π a ^{2}}{8} - \frac{π b ^{2}}{8} + \frac{1}{2} ab$ $\Rightarrow \frac{π r ^{2}}{4} - \frac{π r ^{2}}{4} + \frac{r ^{2}}{2}$ $\Rightarrow \frac{r ^{2}}{2}$ Thus, area of AEBF segment = area of $△ AOB = \frac{r ^{2}}{2}$ 25. In the figure shows two circles passing through each other's centres. Find the area of the region enclosed by the two circles in terms of the common radius r .

Sol. Let radius of circle $= r$

Here $△ ABD$ and $△ ABC$ are equilateral triangle and

Minor sector ABPC make $6 0^{\circ}$ angle at centre 'A of first circle

Here side of equilateral triangle $= r$ Area of segment (Minor ABPC) $=$ area of sector ABPC - area of $△ ABC$ $= \frac{6 0 ^{\circ}}{36 0 ^{\circ}} π (r)^{2} - \frac{3}{4} r^{2}$ $= \frac{π}{6} r^{2} - \frac{3 r ^{2}}{4}$ As shaded area consist of 4 such minor segments and 2 equilateral triangles.

So total shaded area $= 2 \times \frac{3}{4} (r)^{2} + 4 (\frac{π}{6} r^{2} - \frac{3}{4} r^{2})$ $= \frac{3}{4} r^{2} + \frac{44}{21} r^{2} - 3 r^{2}$ $= \frac{44}{21} r^{2} - \frac{3 r ^{2}}{2}$ 26. In the figure we see three triangles within a rectangle. The areas of the triangles are $A, B, C$ as marked. Show that the area of the rectangle is $\frac{2 ( A + C ) ( B + C )}{C}$ .

4.0Sol.

area of ' A '

= \frac{1}{2} \times y \times (ℓ - x)

area of ' B '

= \frac{1}{2} \times x (b - y)

area of '

C

= \frac{1}{2} xy

Now LHS

\frac{2 ( A + C ) ( B + C )}{C}

= \frac{2 ( \frac{1}{2} ( y ℓ - y x ) + \frac{1}{2} x y ) ( \frac{1}{2} ( x b - x y ) + \frac{1}{2} x y )}{\frac{1}{2} x y}

= \frac{y ℓ \cdot \frac{1}{2} \cdot xb}{\frac{1}{2} xy} = ℓ \cdot b

= Area of rectangle ABCD 27. In the figure we see two shaded regions formed by a quarter circle, a semicircle, and a triangle.

Sol.

Since $AB = r^{2} + r^{2} = 2 r^{2} = 2 r$ $BD = \frac{2 r}{2} = \frac{r}{2}$ Let $OA = r = OB =$ radius $ar (△ AOB) = \frac{1}{2} r^{2}$ Area of segment AFBO $= \frac{π r ^{2}}{4} - \frac{r ^{2}}{2}$ $= \frac{r ^{2}}{2} (\frac{π}{2} - 1)$ Now area of upper larger segment $= \frac{1}{2} π (\frac{r}{2})^{2}$ - are of segment AFBO $= \frac{π r ^{2}}{4} - \frac{r ^{2}}{2} (\frac{π}{2} - 1)$ $= \frac{1}{2} ab$ $=$ area of $△ PQR$ = area of region "C"

5.0Important Key Concepts of Class 9 Maths Measuring Space: Perimeter and Area - Chapter 6

Topic	What Students Learn
Perimeter of Shapes	Calculating the perimeter of rectangles, parallelograms, triangles, and other common figures
Circle's Circumference and π	The constant C/D ratio across all circles, and the proof that π is an irrational number
Length of an Arc	Using the central angle to calculate the length of an arc of a circle
Area of Square, Rectangle & Parallelogram	Standard area formulas for these shapes, along with their geometric derivation
Area of a Triangle	Calculating triangle area using the base-height formula
Heron's Formula	Finding the area of a triangle from its three sides alone, verified across equilateral, isosceles, and right-angled triangles
Circumcircle and Incircle of a Triangle	Locating and calculating the circumcircle and incircle associated with a triangle
Brahmagupta's Formula	Calculating the area of a cyclic quadrilateral, and understanding how Heron's formula emerges as its special case
Squaring a Rectangle	The ancient Baudhayana construction method for converting a rectangle into a square of equal area
Area of a Circle, Sector & Segment	Calculating the area of a full circle, along with the area of a sector and a segment
Historical Context of π and Area	The evolution of π and area formulas through the work of mathematicians such as Aryabhaṭa, Brahmagupta, and Baudhayana

6.0 Exercise-wise NCERT Class 9 Maths Chapter 6 Measuring Space: Perimeter and Area Detailed Solutions

Chapter 6 has three exercises plus an End-of-Chapter set, moving from perimeter and area of basic shapes to Heron's and Brahmagupta's formulas, and finally to circles. Here's what each one covers.

Exercise	Topics Covered
Exercise 6.1	Perimeter of shapes, the circumference of a circle, irrationality of π, and the length of an arc
Exercise 6.2	Area of squares, rectangles, parallelograms, and triangles, along with Heron's formula and Brahmagupta's formula for cyclic quadrilaterals
Exercise 6.3	Area of a circle, including problems involving sectors and segments
End-of-Chapter Exercises	Mixed, higher-order problems combining perimeter, area, Heron's formula, and Brahmagupta's formula

7.0Quick Revision on Class 9 Maths Chapter 6 - Measuring Space: Perimeter and Area

8.0Related Study Materials Class 9 Maths

Explore more learning resources in the form of ALLEN’s Class 9 Maths study materials according to the latest NCERT syllabus. You can refer to NCERT textbooks, chapter-wise notes, sample papers and previous years’ question papers along with NCERT Solutions to revise important concepts, strengthen your understanding and prepare well for school as well as CBSE examinations.

CBSE Class 9 Maths Syllabus

Class 9 Maths Revision Notes

NCERT Textbook for Class 9 Maths

CBSE Sample Papers for Class 9 Maths

9.0Advantages of Chapter 6 Maths Class 9 NCERT Solutions

Strengthens Perimeter Concepts: Helps students calculate the perimeter of different plane figures with ease.
Builds Area Calculation Skills: Explains methods to find the area of squares, rectangles, triangles, parallelograms, and other shapes.
Improves Mensuration Techniques: Develops the ability to apply appropriate formulas in different situations.
Enhances Problem-Solving Skills: Simplifies application-based questions involving perimeter and area.
Connects Maths to Real Life: Demonstrates the use of mensuration in practical situations such as land measurement and design.
Builds a Strong Foundation: Prepares students for advanced mensuration and geometry concepts in higher classes.

Do the NCERT Solutions for Class 9 Maths Chapter 6 explain every exercise in detail?

Why should I use NCERT Solutions for Class 9 Maths Chapter 6?

NCERT Solutions help students solve textbook questions accurately, revise important concepts, improve problem-solving skills, and prepare effectively for school examinations.

Why is the value of π (pi) important in this chapter?

What is the difference between the area of a circle and the area of a sector?

What types of shapes are covered in Chapter 6?

How does Chapter 6 help in solving real-life problems?

Join ALLEN!

Do the NCERT Solutions for Class 9 Maths Chapter 6 explain every exercise in detail?

Why should I use NCERT Solutions for Class 9 Maths Chapter 6?

NCERT Solutions help students solve textbook questions accurately, revise important concepts, improve problem-solving skills, and prepare effectively for school examinations.

Why is the value of π (pi) important in this chapter?

What is the difference between the area of a circle and the area of a sector?

What types of shapes are covered in Chapter 6?

How does Chapter 6 help in solving real-life problems?

Join ALLEN!

NCERT Solutions for Class 9 Maths Chapter 6 Measuring Space: Perimeter and Area

1.0Download NCERT Class 9 Maths Chapter 6 Measuring Space : Perimeter and Area Solutions PDF

2.0Learning Outcomes of Class 9 Maths Chapter 6 Measuring Space: Perimeter and Area

3.0Detailed NCERT Class 9 Maths Chapter 6 Measuring Space: Perimeter and Area Solutions

Exercise: 6.1

Exercise: 6.2

Exercise : 6.3

End of Chapter Exercise

4.0Sol.

5.0Important Key Concepts of Class 9 Maths Measuring Space: Perimeter and Area - Chapter 6

6.0 Exercise-wise NCERT Class 9 Maths Chapter 6 Measuring Space: Perimeter and Area Detailed Solutions

7.0Quick Revision on Class 9 Maths Chapter 6 - Measuring Space: Perimeter and Area

8.0Related Study Materials Class 9 Maths

9.0Advantages of Chapter 6 Maths Class 9 NCERT Solutions