Class 9 Maths Chapter 6 End-of-Chapter Exercises Solutions

Consider a square of side a.

Its area is:

Now remove a smaller square of side b from one corner.

Its area is:

As shown in the figure, divide the remaining L-shaped region into two parts:

Now rearrange the region by rotating Piece II and attaching it adjacent to Piece I.

This forms a single rectangle whose dimensions are:

Since both configurations represent the exact same area:

(a + b)(a − b) = a² − b²

Consider a square whose side length is:

Divide each side into three parts of lengths a, b, and c.

This partitions the large square into 9 smaller regions:

= a² + b² + c² + 2ab + 2bc + 2ca

Equating the two expressions for the total area of the large square:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Equal sides of the isosceles triangle:

15 cm, 15 cm

40 cm

Let the base be b cm.

b = 40 − 30 = 10 cm

Since the triangle is isosceles, the perpendicular from the vertex to the base bisects the base.

Half of base = 10 / 2 = 5 cm

Let the perpendicular height be h.

Using √2 ≈ 1.4142:

Base:

Area:

60 cm²

Let the height of the triangle be h.

Since the triangle is isosceles, the perpendicular from the vertex to the base bisects the base.

Half of base = 10 / 2 = 5 cm

Now consider either of the two right-angled triangles formed by the height.

The two equal sides are 13 cm each.

Area:

54 cm²

One leg:

12 cm

Let the other leg be a cm.

For a right-angled triangle,

12 cm and 9 cm

Now find the hypotenuse using Pythagoras' theorem:

Ratio of sides:

2:3:4

45 cm

Let the sides be 2x, 3x, and 4x.

Since the perimeter is 45 cm:

Now find the semi-perimeter:

s = (10 + 15 + 20) / 2

= √[(45/2) × (25/2) × (15/2) × (5/2)]

The sides are:

7 cm, 24 cm, 25 cm

We can find the area in two ways.

Method 1: Using Heron's formula

The semi-perimeter is:

s = (7 + 24 + 25) / 2

= √[28(28 − 7)(28 − 24)(28 − 25)]

= √(28 × 21 × 4 × 3)

Method 2: Using Pythagoras' theorem

Check whether the triangle is right-angled:

and

Diameter of the wheel:

r = d / 2 = 60 / 2 = 30 cm

Using π = 22/7, the distance travelled in one revolution is:

Distance in 1 revolution = 2 × (22/7) × 30

Converting to metres (1 m = 100 cm):

Using π = 22/7, we get:

A quadrant is one-fourth of a circle.

Radius of the wheel:

Distance travelled in one complete turn:

One complete turn of the wheel covers a distance equal to its circumference.

Using π = 22/7:

Distance in one turn = 2 × (22/7) × 28

Converting into metres:

Distance in one complete turn = 176 cm = 1.76 m

Number of turns during a journey of 1 km:

1 km = 1000 m = 100000 cm

Number of turns = Total distance / Distance in one turn

Number of turns = 6250/11 ≈ 568.18 ≈ 568 turns (rounded to nearest whole number)

Yes.

Let the sides of the first rectangle be a and b, and the sides of the second rectangle be c and d.

Since their perimeters are equal:

a + b = c + d --- (1)

Since their areas are equal:

Now square equation (1):

Expanding both sides:

a² + 2ab + b² = c² + 2cd + d²

Using ab = cd, subtract 4ab from the left and 4cd from the right:

a² − 2ab + b² = c² − 2cd + d²

Taking positive side lengths and assuming without loss of generality that a ≥ b and c ≥ d:

a − b = c − d --- (3)

We now have equations (1) and (3):

a + b = c + d --- (1)

a − b = c − d --- (3)

The two rectangles are congruent.

Let the parallel sides of the trapezium be a and b, and its height be h.

Using the construction shown in Fig. 6.42, the trapezium is divided into:

Area of the parallelogram:

Area of the triangle:

The remaining base is b − a.

Area of the trapezium:

The trapezium consists of these two parts.

Taking 1/2 × h common:

= 1/2 × h × [2a + (b − a)]

= 1/2 × h × (a + b)

Let the parallel sides of the trapezium be a and b, and its height be h.

Draw diagonal BD to divide trapezium ABCD into two triangles: △ABD and △BCD.

Area of the first triangle (△ABD):

The triangle has base a and height h.

Area of the second triangle (△BCD):

The triangle has base b and height h.

Area of the trapezium:

The trapezium is the sum of these two triangles.

Taking 1/2 h common:

Let the parallel sides of the trapezium be a and b, and its height be h.

Take two identical copies of the trapezium and arrange them by turning one copy around and joining the two non-parallel sides.

The two trapezia together form a parallelogram.

The base of the parallelogram is:

The height of the parallelogram is:

Since the parallelogram is made from two identical trapezia:

2 × Area of trapezium = (a + b)h

Let the diagonals of the kite be d₁ and d₂, and let them intersect at O.

For a kite, the diagonals are perpendicular and one diagonal bisects the other.

Suppose d₁ is the diagonal that is bisected.

Then,

Since the diagonals are perpendicular, the kite is divided into four right-angled triangles.

= d₁x / 2 + d₁y / 2

Since x + y = d₂:

Let the diagonals of the kite be AC and BD, intersecting at O.

Since the diagonals of a kite are perpendicular:

AC ⟂ BD

and one diagonal bisects the other.

Suppose AC bisects BD.

Then,

The diagonal AC divides the kite into two triangles:

△ABC and △ADC

Both have the same base AC.

Their heights are BO and DO, respectively.

Taking 1/2 AC common:

The area of rectangle ABCD is:

The area of rectangle PQRS is:

Now place two copies of ABCD along the side 2a and two copies along the side 2b.

This gives:

filling the larger rectangle completely.

Yes, 4 copies of rectangle ABCD fit exactly into rectangle PQRS.

All corresponding sides of △PQR are twice those of △ABC.

The corresponding height is also doubled.

Since both the base and height are multiplied by 2:

Now divide each side of △PQR into two equal parts and join the three midpoints.

This divides the larger triangle into four smaller triangles.

Each smaller triangle has side lengths a, b, and c.

Yes, 4 copies of △ABC fit exactly into △PQR.

All corresponding sides of △PQR are three times those of △ABC.

The corresponding height is also three times as large.

Now divide each side of △PQR into three equal parts and draw lines through the division points parallel to the sides.

This forms a triangular grid containing:

1 + 3 + 5 = 9 small triangles

Each small triangle has side lengths a, b, and c.

Yes, 9 copies of △ABC fit exactly into △PQR.

From the markings in the figure:

Let the whole triangle be △ABC.

The shaded region is the whole triangle minus the two unshaded corner triangles.

Upper unshaded triangle:

The upper triangle uses 1/2 of AB and 1/3 of AC.

of the whole triangle.

Lower-right unshaded triangle:

Its base is 1/3 AC and it has the same height as △ABC.

Shaded fraction:

Shaded fraction = 1 − 1/6 − 1/3

Let the vertices of the square ABCD be:

A(0, 0), B(a, 0), C(a, a), D(0, a)

Let the midpoints of BC, CD, DA, AB be E, F, G, H, respectively.

Their coordinates are:

E(a, a/2), F(a/2, a), G(0, a/2), H(a/2, 0)

The four joining lines are:

Their equations are:

CG: y = x / 2 + a / 2

The four intersections forming the shaded square are:

(4a/5, 2a/5), (3a/5, 4a/5)

(a/5, 3a/5), (2a/5, a/5)

The side length of the shaded square is:

The area of the outer square is:

Shaded area / Area of outer square = (a² / 5) / a²

Let the radius of each circle be r.

Then the diameter is:

The rectangle has:

Area of 3 circles:

Using π = 22/7:

For four circles:

The total area of four circles is:

Using π = 22/7:

From Question 17, both the 3-circle and 4-circle arrangements give:

The circles always occupy the fraction π/4 of the rectangle, regardless of the number of circles.

Let the radius of each circle be r.

Area of 10 circles:

Rectangle dimensions:

Area of 20 circles:

Rectangle area:

Area of 50 circles:

Rectangle area:

All three cases support the conjecture.

Suppose there are n identical circles, each of radius r, placed side by side in the same manner.

Each circle has diameter:

The total area of n circles is:

Cancelling nr²:

Using π = 22/7:

Fraction of rectangle occupied = π/4 ≈ 0.79

Let the length and breadth of each small rectangle be l cm and b cm, respectively (with l > b).

The large rectangle is composed of 9 identical small rectangles and has a total area of 72 cm².

Since Area = length × breadth:

From Fig. 6.47:

Since both rows span the identical width of the large rectangle:

Expressing l in terms of b:

Substitute equation (2) into equation (1):

⇒ b² = 32 / 5 = 6.4

Taking the square root:

b = √6.4 = √(32 / 5) = 4√10 / 5 cm ≈ 2.53 cm

Now find the length l using l = 5b / 4:

l = (5 / 4) × (4√10 / 5) = √10 cm ≈ 3.16 cm

The perimeter of a rectangle is given by:

Substitute l = √10 and b = 4√10 / 5:

= 18√10 / 5 cm = 3.6√10 cm

Using √10 ≈ 3.1623:

**Given:**

△ABC has base BC and opposite vertex A. Points D and E trisect the base BC, so:

BD = DE = EC = 1/3 BC

**To prove:**

**We know:**

**Proof:**

Both △ABD and △AEC share the same top vertex A. The perpendicular height from vertex A to the line BC (which contains the bases of both triangles) is the same for both triangles, say h.

Since points D and E trisect BC, we have BD = EC:

*(Note: In fact, the middle triangle △ADE also has Area = 1/2 × DE × h = Area(ABD), meaning the two trisection lines divide △ABC into three equal-area triangles, each having 1/3 the area of △ABC.)*

Because both triangles have the exact same base length (d = BD = EC) and the same height (h), we can dissect △ABD and reassemble its pieces to form △AEC through geometric dissection:

**Method 1 (Midline Cut and Rearrangement):**

1. **Cut the blue triangle:** Draw a line parallel to the base BD at half the height (h/2). This slices △ABD into two pieces:

2. **Form a parallelogram:** Rotate Piece 1 by 180° and join it along the upper edge of Piece 2. This forms a parallelogram of base d and height h/2.

3. **Reassemble to cover the red triangle:** An identical midline cut on the red triangle △AEC decomposes it into a matching top triangle and trapezium that assemble into the same base × half-height shape. Reversing the placement along the sloped edges of △AEC allows the pieces of the blue triangle to exactly cover the red triangle.

**Method 2 (Continuous Shear / Deck of Cards Perspective):**

Imagine slicing the blue triangle horizontally into very thin strips parallel to base BC (like a stack of playing cards). At any height y from the base, the width of the horizontal cross-section is:

Since △AEC has the exact same base d and height h, every horizontal strip at height y in △AEC has the identical width w(y). By sliding each strip horizontally to the right, the blue triangle smoothly shears into the exact shape of the red triangle without any change in area (Cavalieri's principle).

Let the side length of the square be 2r.

Before setting up the algebra, observe the total areas of the shapes:

Combined area of two semicircles = 1/2 πr² + 1/2 πr² = πr²

**Key Principle:**

The Quarter Circle and the two semicircles have the **EXACT SAME TOTAL AREA** (πr²).

Since both semicircles lie entirely within the quarter circle, the area lost by the two semicircles overlapping each other (Region A) must be **identically equal** to the area left uncovered inside the quarter circle (Region B)!

Refer to Fig. 6.49. We partition the area inside the quarter circle into four non-overlapping regions:

Now write the area equations:

1. Sum of the two semicircles:

⇒ C + D + 2A = πr² ... (1)

2. Area of the entire quarter circle:

The quarter circle is the union of all four distinct regions A, B, C, and D:

Equating (1) and (2) since both equal πr²:

C + D + 2A = A + B + C + D

By the Principle of Inclusion-Exclusion for two sets:

= (1/2 πr² + 1/2 πr²) − Area(A)

Region B is the portion of the quarter circle not covered by either semicircle:

The side of the given square is 2 units.

r = 2 / 2 = 1 unit

Each petal of the flower is bounded by two quarter-circle arcs, each of radius r = 1.

The length of one quarter-circle arc of radius r = 1 is:

= 1/4 × (2π × 1) = π / 2 units

Each of the 4 petals is bounded by two such arcs:

Since there are 4 identical petals in the flower:

Using π = 22/7:

Consider one petal of the flower. Each petal consists of two identical circular segments of a circle of radius r = 1 with a 90° central angle.

The area of one circular segment is:

Since each petal consists of two such circular segments:

There are 4 identical petals in the flower, so the total area of the flower is:

Using π = 22/7:

= 44 / 7 − 28 / 7

= 16 / 7 square units ≈ 2.29 square units

Let the radius of the larger circle be R and the radius of the smaller circle be r.

Since the chord BC touches the smaller circle at point A, the chord BC is tangent to the inner circle at A.

OA ⊥ BC

In the larger circle, OA is the perpendicular drawn from the centre O to the chord BC.

We know that the perpendicular from the centre of a circle to a chord bisects the chord.

Now consider the right-angled triangle △OAB (with ∠OAB = 90°):

Applying Pythagoras' theorem in △OAB:

⇒ R² = r² + l² / 4

Rearranging the terms:

The green region enclosed between the two concentric circles is an annulus. Its area is the difference between the areas of the two circles:

Let the two perpendicular legs of the right-angled triangle be a and b, and let the hypotenuse be c.

The area of a semicircle with diameter d (and radius d/2) is:

Summing the areas of the two smaller semicircles:

Since a² + b² = c², this gives:

which is exactly the area of the semicircle drawn on the hypotenuse.

Now examine the geometric regions in Fig. 6.52:

By Thales' theorem, the right-angled vertex of the triangle lies on the semicircle drawn on the hypotenuse c as diameter.

Let the two circular segments between the hypotenuse semicircle arc and the two legs be X and Y.

Then the larger semicircle (on the hypotenuse) consists of:

Similarly, the two smaller semicircles drawn on the legs a and b consist of:

From equation (1), the left-hand side of (2) equals the left-hand side of (3):

Cancelling (X + Y) from both sides:

Let the centres of the two circles be A and B, each with radius r.

Since each circle passes through the centre of the other circle:

Let the two circles intersect at points C and D.

Then the distance from each centre to the points of intersection equals the radius r:

In △ABC, all three sides are equal: AB = BC = CA = r.

Therefore, △ABC is an equilateral triangle, which means:

Similarly, in △ABD, AB = BD = DA = r, so △ABD is also an equilateral triangle:

The total central angle subtended by arc CBD at centre A is:

∠CAD = ∠CAB + ∠DAB = 60° + 60° = 120°

By symmetry, the central angle at centre B subtended by arc CAD is also:

Combined sector area = 2 × (1/3 πr²) = 2/3 πr²

Area of one equilateral triangle of side r:

The quadrilateral ACBD is a rhombus composed of two equilateral triangles △ABC and △ABD:

The required shaded lens-shaped common region is the union of the two circular segments on chord CD, which equals the sum of the two 120° sectors minus the rhombus ACBD (since the rhombus is counted twice in the two sectors):

Writing with a common denominator of 6:

Using π ≈ 3.14159 and √3 ≈ 1.73205:

4π − 3√3 ≈ 12.5664 − 5.1962 = 7.3702

Let the width and height of the rectangle be W and H respectively.

Let P be the common vertex of the three triangles A, B, and C.

Suppose the vertical line through P meets the top side of the rectangle, and the horizontal line through P meets the right side.

Triangle A has:

Triangle C has:

C = 1/2 (W − x)(H − y)

A + C = 1/2 x (H − y) + 1/2 (W − x)(H − y)

Taking 1/2 (H − y) as a common factor:

A + C = 1/2 (H − y)[x + W − x]

A + C = 1/2 W (H − y) ... (1)

Triangle B has:

B + C = 1/2 y (W − x) + 1/2 (W − x)(H − y)

Taking 1/2 (W − x) as a common factor:

B + C = 1/2 (W − x)[y + H − y]

B + C = 1/2 H (W − x) ... (2)

Also, from triangle C:

C = 1/2 (W − x)(H − y) ... (3)

Now consider the given expression:

[2(A + C)(B + C)] / C

Using equations (1), (2), and (3):

= { 2 × [1/2 W(H − y)] × [1/2 H(W − x)] } / [1/2 (W − x)(H − y)]

Multiplying the numerator factors:

= { 1/2 WH (W − x)(H − y) } / [1/2 (W − x)(H − y)]

Cancelling the common factor 1/2 (W − x)(H − y) from the numerator and denominator:

Since WH = Area of rectangle:

Let the radius of the quarter circle be R.

Then:

and

Therefore, △AOB is an isosceles right-angled triangle.

Let the two shaded regions be:

The quarter-circle sector has radius R and central angle 90°.

In right-angled triangle AOB, by Pythagoras' theorem:

The semicircle has AB as its diameter.

The area of a semicircle with diameter d is:

From equations (1) and (2):

Let S be the unshaded circular segment between chord AB and the quarter-circle arc AFB.

The quarter-circle sector consists of the triangle T and the circular segment S:

Similarly, the semicircle on AB consists of the lune L and the circular segment S:

From equation (3), the left-hand sides of (4) and (5) are equal:

Cancelling the common area Area(S) from both sides: