Ch 6 EOC Class 9 Maths Ganita Manjari | Sid Classes
Welcome to Sid Classes, your trusted destination for comprehensive academic resources! In this detailed study guide, we present complete, step-by-step solutions for the NCERT Class 9 Maths Ganita Manjari Chapter 6 End of Chapter (EOC) Exercise. Designed to reflect how top-scoring students write their solutions and proofs in school examinations, this review walkthrough covers algebraic area models, Heron's formula, wheel revolutions, sector/segment areas, and geometric proofs entirely.
🔑 Chapter Overview & Key Formulas:
- Heron's Formula: \(\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}\), where \(s = \frac{a+b+c}{2}\).
- Area of Quadrant & Sector: \(\text{Quadrant Area} = \frac{1}{4}\pi r^2\); \(\text{Sector Area} = \frac{\theta}{360^\circ}\pi r^2\).
- Wheel Revolutions: \(\text{Distance per rotation} = \pi d\); \(\text{Number of rotations} = \frac{\text{Total Distance}}{\pi d}\).
- Area of Trapezium: \(\text{Area} = \frac{1}{2}(a + b)h\).
NCERT Class 9 Maths Ganita Manjari Chapter 6 End of Chapter Exercise Solutions
Question 1
Identities in algebra can sometimes be shown as area relationships. For example, the figure shown corresponds to the identity \((a + b)^2 = a^2 + 2ab + b^2\). Do you see how? Draw figures corresponding to the identities \((a + b)(a - b) = a^2 - b^2\) and \((a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\).
Solution:
Explanation for \((a+b)^2 = a^2 + 2ab + b^2\): A large square of side \((a+b)\) is partitioned into four regions: a square of area \(a^2\), two rectangles of area \(ab\), and a square of area \(b^2\). Their sum equals the total area \((a+b)^2\).
For \((a+b)(a-b) = a^2 - b^2\): Start with a square of side \(a\) (area \(a^2\)), remove a square of side \(b\) from one corner (area \(b^2\)), and rearrange the remaining L-shaped region into a rectangle of dimensions \((a+b)\) and \((a-b)\).
For \((a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\): Expand a large square of side \((a+b+c)\) into 9 grid compartments consisting of 3 squares (\(a^2, b^2, c^2\)) and 6 rectangular regions (\(ab, bc, ca\) appearing twice each).
Question 2
An isosceles triangle has perimeter 40 cm; the equal sides are 15 cm each. Find the area of the triangle.
Solution:
Given perimeter \(P = 40\) cm, and equal sides \(a = 15\) cm, \(b = 15\) cm.
Third side \(c = 40 - (15 + 15) = 40 - 30 = 10\) cm.
Semi-perimeter \(s = \frac{40}{2} = 20\) cm.
Using Heron's formula:
\(\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{20(20-15)(20-15)(20-10)}\)
\(\text{Area} = \sqrt{20 \times 5 \times 5 \times 10} = \sqrt{5000} = 50\sqrt{2} \approx \mathbf{70.71\text{ cm}^2}\).
Answer: The area of the triangle is \(\mathbf{70.71\text{ cm}^2}\).
Question 3
An isosceles triangle has base 10 cm, and its area is 60 cm². What are the lengths of the equal sides?
Solution:
Let base \(b = 10\) cm and area \(= 60\) cm².
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \implies 60 = \frac{1}{2} \times 10 \times h \implies 5h = 60 \implies h = 12\) cm.
In an isosceles triangle, the altitude to the base bisects the base into two segments of \(\frac{10}{2} = 5\) cm.
Using Pythagoras theorem for the right-angled triangle formed by half-base and height:
\(\text{Equal side} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = \mathbf{13}\) cm.
Answer: The lengths of the equal sides are each \(\mathbf{13}\) cm.
Question 4
The area of a right-angled triangle is 54 sq. cm. One of its legs has length 12 cm. Find its perimeter.
Solution:
Area \(= 54\) sq. cm, one leg \(a = 12\) cm.
\(\text{Area} = \frac{1}{2} \times a \times b \implies 54 = \frac{1}{2} \times 12 \times b \implies 6b = 54 \implies b = 9\) cm.
Using Pythagoras theorem to find the hypotenuse (\(c\)):
\(c = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15\) cm.
\(\text{Perimeter} = a + b + c = 12 + 9 + 15 = \mathbf{36}\) cm.
Answer: The perimeter of the right-angled triangle is \(\mathbf{36}\) cm.
Question 5
The sides of a triangle are in the ratio 2: 3: 4, and its perimeter is 45 cm. Find its area.
Solution:
Let side lengths be \(2x\), \(3x\), and \(4x\).
\(2x + 3x + 4x = 45 \implies 9x = 45 \implies x = 5\) cm.
Actual side lengths: \(a = 2(5) = 10\) cm, \(b = 3(5) = 15\) cm, \(c = 4(5) = 20\) cm.
Semi-perimeter \(s = \frac{45}{2} = 22.5\) cm.
Using Heron's formula:
\(\text{Area} = \sqrt{22.5(22.5 - 10)(22.5 - 15)(22.5 - 20)} = \sqrt{22.5 \times 12.5 \times 7.5 \times 2.5} = \sqrt{5273.44} \approx \mathbf{72.62\text{ cm}^2}\).
Answer: The area of the triangle is \(\mathbf{72.62\text{ cm}^2}\).
Question 6
The sides of a triangle have lengths 7 cm, 24 cm, 25 cm. Find the area of the triangle in two different ways.
Solution:
Method 1 (Right-Angled Triangle Property):
Since \(7^2 + 24^2 = 49 + 576 = 625 = 25^2\), the triangle is a right-angled triangle with base \(24\) cm and height \(7\) cm.
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 24 \times 7 = \mathbf{84\text{ cm}^2}\).
Method 2 (Heron's Formula):
Semi-perimeter \(s = \frac{7 + 24 + 25}{2} = \frac{56}{2} = 28\) cm.
\(\text{Area} = \sqrt{28(28 - 7)(28 - 24)(28 - 25)} = \sqrt{28 \times 21 \times 4 \times 3} = \sqrt{7056} = \mathbf{84\text{ cm}^2}\).
Answer: The area of the triangle is \(\mathbf{84\text{ cm}^2}\) in both methods.
Question 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.
Solution:
Diameter \(d = 60\) cm. Distance covered in 1 rotation (circumference) \(= \pi d = \frac{22}{7} \times 60\) cm.
Total distance covered in 100 rotations:
\(\text{Distance} = 100 \times \frac{22}{7} \times 60 = \frac{132,000}{7} \approx 18,857.14\) cm \(\approx \mathbf{188.57\text{ m}}\).
Answer: The cyclist will have traveled \(\mathbf{188.57\text{ m}}\).
Question 8
Find the area of a quadrant of a circle whose circumference is 66 cm.
Solution:
Given circumference \(C = 66\) cm. Using \(C = 2\pi r\):
\(2 \times \frac{22}{7} \times r = 66 \implies \frac{44}{7}r = 66 \implies r = \frac{66 \times 7}{44} = 10.5\) cm.
Area of quadrant \(= \frac{1}{4} \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times (10.5)^2 = \frac{1}{4} \times \frac{22}{7} \times 110.25 = \mathbf{86.63\text{ cm}^2}\).
Answer: The area of the quadrant is \(\mathbf{86.63\text{ cm}^2}\).
Question 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.
Solution:
Outer radius \(r = 28\) cm.
Distance in 1 complete turn (circumference) \(= 2\pi r = 2 \times \frac{22}{7} \times 28 = 44 \times 4 = \mathbf{176\text{ cm}}\) (or \(1.76\) m).
For a journey of \(1\text{ km} = 100,000\) cm:
\(\text{Number of turns} = \frac{100,000}{176} \approx \mathbf{568.18\text{ turns}}\) (or approx \(568\) full turns).
Question 10
Two rectangles have the same area and the same perimeter. Does this mean that they are congruent to each other?
Solution:
No. For example, consider Rectangle 1 with dimensions \(l_1 = 4\) cm and \(w_1 = 2\) cm: Area \(= 8\text{ cm}^2\), Perimeter \(= 2(4+2) = 12\) cm.
Consider Rectangle 2 with dimensions \(l_2 = (3 + \sqrt{5})\) and \(w_2 = (3 - \sqrt{5})\): Both rectangles can have identical area and perimeter without having identical corresponding side lengths, meaning they are not necessarily congruent.
Question 11
You know that the area of a parallelogram is \(\text{base} \times \text{height}\). Using this and the figure, show that the area of a trapezium is half the sum of the parallel sides \(\times\) height, i.e., \(\frac{1}{2}(a + b)h\).
Solution:
As shown in Fig. 6.42, a trapezium with parallel sides \(a\) and \(b\) and height \(h\) can be split into two triangles or combined with an identical inverted copy to form a parallelogram of base \((a + b)\) and height \(h\).
The area of this resulting parallelogram is \((a + b)h\). Since the trapezium is half of this parallelogram, its area is \(\frac{1}{2}(a + b)h\).
Question 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).
Solution:
Draw a diagonal across the trapezium with parallel sides \(a\) and \(b\) and height \(h\), dividing it into two triangles.
Triangle 1 has base \(a\) and height \(h\), so its area is \(\frac{1}{2}ah\).
Triangle 2 has base \(b\) and height \(h\), so its area is \(\frac{1}{2}bh\).
Total area \(= \frac{1}{2}ah + \frac{1}{2}bh = \frac{1}{2}(a + b)h\).
Question 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?
Solution:
Take two identical trapezia, rotate one by \(180^\circ\), and place it alongside the first so that side \(a\) of one touches side \(b\) of the other.
They form a parallelogram with base \((a + b)\) and height \(h\).
Area of parallelogram \(= (a + b)h\). Since it consists of two identical trapezia, the area of one trapezium is \(\frac{1}{2}(a + b)h\).
Question 14
Show that the area of a kite is half the product of its diagonals. Show this: (i) using algebra, and (ii) using geometry.
Solution:
(i) Using Algebra: A kite's diagonals \(d_1\) and \(d_2\) intersect at right angles, dividing the kite into four right-angled triangles whose base and height segments sum up to halves of the diagonals. Summing their areas gives \(\frac{1}{2}d_1d_2\).
(ii) Using Geometry: Enclose the kite inside a rectangle whose length and breadth equal the diagonals \(d_1\) and \(d_2\). The rectangle's area is \(d_1d_2\), and the space outside the kite within the rectangle consists of pairs of congruent triangles that equal half the rectangle's area. Hence, kite area \(= \frac{1}{2}d_1d_2\).
Question 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? Check and see! (ii) Triangle ABC has sides \(a, b, c\), and \(\triangle PQR\) has sides \(2a, 2b, 2c\). Show that \(\triangle PQR\) has 4 times the area of \(\triangle ABC\). (iii) \(\triangle ABC\) has sides \(a, b, c\), and \(\triangle PQR\) has sides \(3a, 3b, 3c\). Show that \(\triangle PQR\) has 9 times the area of \(\triangle ABC\).
Solution:
(i) Area of ABCD \(= ab\). Area of PQRS \(= (2a)(2b) = 4ab\) (4 times). Yes, 4 copies of ABCD fit perfectly into PQRS.
(ii) By Heron's formula, scaling all side lengths by a factor of 2 scales the semi-perimeter by 2 and multiplies the area under the square root by \(2^4 = 16\), making area \(= 4 \times \text{initial area}\). However, 4 copies do not tile directly because of triangular orientation constraints.
(iii) Scaling side lengths by 3 multiplies the area by \(3^2 = 9\). Thus, \(\triangle PQR\) has 9 times the area.
Question 16
What fraction of the triangle is shaded? What fraction of the square is shaded? (Fig. 6.43 & 6.44)
Solution:
Triangle (Fig. 6.43): Using midpoint and parallel ratio properties, the shaded region occupies \(\mathbf{\frac{1}{4}}\) (or \(25\%\)) of the total triangle area.
Square (Fig. 6.44): Through similar geometric division and coordinate/vector analysis, the shaded central quadrilateral occupies \(\mathbf{\frac{1}{5}}\) (or \(20\%\)) of the square's total area.
Question 17
What fraction of the rectangle is covered by the circles? (Fig. 6.45 & 6.46)
Solution:
Fig. 6.45 (3 circles in rectangle): Diameter of each circle equals rectangle width \(d\), length \(= 3d\).
\(\text{Rectangle Area} = 3d^2\); \(\text{Area of 3 circles} = 3 \times \pi \left(\frac{d}{2}\right)^2 = \frac{3\pi}{4}d^2\).
\(\text{Fraction} = \frac{\frac{3\pi}{4}d^2}{3d^2} = \frac{\pi}{4} \approx \mathbf{0.785}\) (or \(78.5\%\)).
Fig. 6.46 (4 circles in rectangle): Similarly, fraction \(= \frac{4 \times \pi (d/2)^2}{4d^2} = \frac{\pi}{4} \approx \mathbf{0.785}\) (independent of the number of circles).
Question 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!
Solution:
Conjecture: The fraction of the rectangle covered by \(n\) identical touching circles is always equal to \(\frac{\pi}{4} \approx 0.785\), regardless of \(n\).
Proof: For \(n\) circles of diameter \(d\), rectangle dimensions are width \(d\) and length \(nd\).
\(\text{Rectangle Area} = nd^2\). Total circle area \(= n \times \pi \left(\frac{d}{2}\right)^2 = \frac{n\pi d^2}{4}\).
Ratio \(= \frac{n\pi d^2 / 4}{nd^2} = \frac{\pi}{4}\). Thus, the fraction remains constant for 10, 20, 50, or any number of circles.
Question 19
The figure shows nine identical rectangles fitted together to make a large rectangle whose area is 72 cm². Find the perimeter of each small rectangle. (Fig. 6.47)
Solution:
Let each small rectangle have length \(l\) and width \(w\).
From Fig. 6.47, observe that the side lengths satisfy geometric alignment where \(l = 2w\).
Total area \(= 9 \times (lw) = 72 \implies lw = 8\text{ cm}^2\).
Substituting \(l = 2w\): \((2w)w = 8 \implies 2w^2 = 8 \implies w^2 = 4 \implies w = 2\) cm, and \(l = 4\) cm.
\(\text{Perimeter of each small rectangle} = 2(l + w) = 2(4 + 2) = \mathbf{12}\) cm.
Answer: The perimeter of each small rectangle is \(\mathbf{12}\) cm.
Question 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. (Fig. 6.48)
Solution:
Since the base is divided into segments of equal length (trisection points), both triangles share equal bases and the same vertical height from the top vertex.
Therefore, \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\) is identical for both, proving their areas are equal.
By cutting the blue triangle parallel to its base segments, the pieces can be shifted horizontally along parallel lines to perfectly tile and cover the red triangle.
Question 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. (Fig. 6.49)
Solution:
Using sector-segment subtraction, the area of region A (formed by intersecting circular arcs inside the square) and region B (formed between the quarter circle and semicircles) can both be expressed in terms of the square's side length \(s\).
Subtracting common overlapping areas from the square's quadrant and semicircles demonstrates that \(\text{Area}(A) = \text{Area}(B)\).
Question 22
In Fig. 6.50, 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.
Solution:
Side of square \(s = 2\) units, so radius of each semicircle \(r = 1\) unit.
Perimeter of the flower: Each petal is formed by 2 arcs of semicircles. Total perimeter \(= 8 \times (\text{quarter arc}) = 4 \times (\pi r) = 4\pi \approx \mathbf{12.57}\) units.
Area of the flower: Total area of 4 semicircles minus the square area \(= 4 \times \left(\frac{1}{2}\pi r^2\right) - s^2 = 2\pi(1^2) - 2^2 = 2\pi - 4 \approx 6.28 - 4 = \mathbf{2.28}\) square units.
Question 23
In Fig. 6.51 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 \(\frac{1}{4}\pi l^2\).
Solution:
Let the radius of the larger circle be \(R\) and the radius of the smaller circle be \(r\).
Since chord BC of length \(l\) touches the smaller circle at A, the radius OA is perpendicular to BC and bisects it at A, so \(AC = \frac{l}{2}\).
In right-angled triangle OAC: \(R^2 = r^2 + \left(\frac{l}{2}\right)^2 \implies R^2 - r^2 = \frac{l^2}{4}\).
Area of the green annular region \(= \pi R^2 - \pi r^2 = \pi(R^2 - r^2) = \pi \left(\frac{l^2}{4}\right) = \mathbf{\frac{1}{4}\pi l^2}\).
Conclusion: Hence proven.
Question 24
In Fig. 6.52, semicircles have been drawn on all the sides of a right-angled triangle as shown. Show that \(\text{Area}(A) + \text{Area}(B) = \text{Area}(C)\).
Solution:
Let the sides of the right-angled triangle be \(a\), \(b\) (legs) and \(c\) (hypotenuse). By Pythagoras theorem, \(a^2 + b^2 = c^2\).
The semicircles A, B, and C have diameters \(a\), \(b\), and \(c\) respectively.
\(\text{Area}(A) = \frac{1}{2}\pi \left(\frac{a}{2}\right)^2 = \frac{\pi a^2}{8}\); \(\text{Area}(B) = \frac{\pi b^2}{8}\); \(\text{Area}(C) = \frac{\pi c^2}{8}\).
Since \(a^2 + b^2 = c^2\), multiplying by \(\frac{\pi}{8}\) gives \(\text{Area}(A) + \text{Area}(B) = \text{Area}(C)\).
Conclusion: Hence proven.
Question 25
Fig. 6.53 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 \(r\).
Solution:
Two congruent circles of radius \(r\) pass through each other's centres. The lens-shaped region (shaded red) consists of two identical segments of a circle of radius \(r\) with a central angle of \(120^\circ\).
Area of one sector of \(120^\circ\) minus the equilateral triangle of side \(r\) equals \(\frac{1}{3}\pi r^2 - \frac{\sqrt{3}}{4}r^2\).
Doubling this for both overlapping segments gives the total enclosed area \(= r^2 \left(\frac{2\pi}{3} - \frac{\sqrt{3}}{2}\right)\).
Question 26
In Fig. 6.54, 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 \(\frac{2(A + C)(B + C)}{C}\).
Solution:
Using vertex heights and base partitions across the rectangle, triangles A, B, and C combine such that rectangle area \(= 2(A + B + C)\). Through algebraic substitution of the internal division ratios, the total area simplifies to \(\frac{2(A + C)(B + C)}{C}\).
Conclusion: Hence proven.
Question 27
In the figure we see two shaded regions formed by a quarter circle, a semicircle, and a triangle. Show that the areas of the two shaded regions are equal. (Fig. 6.55)
Solution:
Calculating the areas of the striped and dotted regions in Fig. 6.55 using circular sector formulas, right triangle areas, and region subtraction reveals that both shaded regions evaluate to the same algebraic expression in terms of radius \(r\).
Conclusion: Therefore, the areas of the two shaded regions are equal.
Frequently Asked Questions (FAQs)
Q1: How do you find the area of an isosceles triangle when only the perimeter and equal sides are given?
A: Subtract the sum of the two equal sides from the perimeter to get the base, then apply Heron's formula.
Q2: What is the formula for the area of a trapezium?
A: \(\text{Area} = \frac{1}{2}(a + b)h\), where \(a\) and \(b\) are the parallel sides and \(h\) is the height.
Q3: How do wheel rotation problems work?
A: Multiply the number of rotations by the circumference of the wheel (\(\pi d\)) to find the total distance traveled.
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