Frank Brothers Solutions for ICSE Class 10 Mathematics Chapter 14 Symmetry

Exercise 14.1

Answer 1.
Answer: Construction of isosceles triangle with equal sides of 7 cm each:
Steps of construction:

1. Draw a line segment AB = 5 cm
2. With A as centre, cut an arc of 7 cm on one side of line segment AB.
3. With B as centre, cut an arc of 7 cm on same side of line segment AB. Let the point be C.
4. Join AC and BC. ABC is the required triangle.
5. Draw angle bisector of angle C meeting AB at D.
6. CD is perpendicular bisector of AB and AC=BC. Hence CD is the line of symmetry.

Isosceles triangle has only one line of symmetry.

[Diagram: This diagram shows an isosceles triangle ABC with vertex C at the top, base AB at the bottom (5 cm), and equal sides AC and BC both 7 cm long. A perpendicular line CD is drawn from vertex C to the midpoint D of base AB, showing the line of symmetry.]

In simple words: We make a triangle with two equal sides. Then we draw a line from the top corner to the middle of the base. This line divides the triangle into two identical halves.

📝 Teacher's Note: Show students how folding along the line of symmetry makes both halves match perfectly. An isosceles triangle looks the same on both sides of this line.

🎯 Exam Tip: Always mention that isosceles triangle has only one line of symmetry. Draw the perpendicular from the vertex angle to the base clearly.

Answer 2.
Answer: Construction of equilateral triangle with all sides 5.8 cm:
Steps of construction:

1. Draw a line segment BC = 5.8 cm.
2. With B and C as centre and radius = 5.8 cm, draw two arcs which intersect each other at A.
3. Join AB and AC. ABC is the required triangle.
4. Draw the bisectors AD, BE and CF of ∠A, ∠B and ∠C respectively.

Hence, AD, BE and CF are the lines of symmetry of triangle ABC.

[Diagram: This diagram shows an equilateral triangle ABC with all sides 5.8 cm. Three lines of symmetry are drawn: AD (from A to midpoint of BC), BE (from B to midpoint of AC), and CF (from C to midpoint of AB). All three lines meet at the center of the triangle.]

In simple words: We make a triangle with all three sides equal. This special triangle has three lines of symmetry - one from each corner to the middle of the opposite side.

📝 Teacher's Note: Cut an equilateral triangle from paper. Fold it along each line of symmetry to show how the triangle matches itself perfectly in three different ways.

🎯 Exam Tip: Write clearly that equilateral triangle has three lines of symmetry. Each line goes from a vertex to the midpoint of the opposite side.

Answer 3.
Answer: Construction of parallelogram:
Steps of construction:

1. Draw a line segment QR = 5.4 cm
2. At Q, draw a ray making an angle of 60 degrees with QR and cut QP = SR = 6 cm
3. P as centre draw an arc equal to 5.4 cm
4. R as centre draw an arc equal to 6 cm which intersects the first arc at S.
5. Join RS and PS. PQRS is the required parallelogram.
6. Join QS and PR which intersect each other at O.

There is no line of symmetry of parallelogram PQRS but it has one point symmetry which is O, the point of intersection of its diagonals.

[Diagram: This diagram shows parallelogram PQRS with sides QR and PS both 5.4 cm, and sides PQ and SR both 6 cm. The diagonals QS and PR intersect at point O in the center. Angle at Q is 60 degrees.]

In simple words: A parallelogram has no line of symmetry. But it has one special point in the center. If you rotate it 180 degrees around this point, it looks exactly the same.

📝 Teacher's Note: Explain that parallelograms have point symmetry but no line symmetry. The diagonals always meet at the center point which is the point of symmetry.

🎯 Exam Tip: Write "no line of symmetry" and "one point of symmetry at intersection of diagonals" for parallelogram. This is an important distinction.

Answer 4.
Answer: Construction of square with side 4.8 cm:
Steps of construction:

1. Draw a line segment AB = 4.8 cm.
2. At A and B, draw perpendiculars AX and BY
3. From AX and BY, cut off AD = BC = 4.8 cm
4. Join DC. ABCD is the required square.
5. Now draw perpendicular bisectors of AB and AD.
6. Also join the diagonals AC and BD.

The perpendicular bisectors and the diagonals are the lines of symmetry.

[Diagram: This diagram shows square ABCD with side length 4.8 cm. Four lines of symmetry are shown: two perpendicular bisectors of opposite sides (horizontal and vertical lines through the center), and two diagonals AC and BD. All four lines intersect at the center point.]

In simple words: A square has four lines of symmetry. Two lines divide it into equal halves horizontally and vertically. The other two lines are the diagonals that connect opposite corners.

📝 Teacher's Note: Use a paper square and fold it in four different ways to show the four lines of symmetry. Students can see how each fold creates perfect matching halves.

🎯 Exam Tip: Always write that square has 4 lines of symmetry: 2 perpendicular bisectors of opposite sides and 2 diagonals. Draw all four lines clearly.

Answer 5.
Answer: Construction of regular hexagon with side 3.8 cm:
Steps of construction:

1. Draw a line segment AB = 3.8 cm
2. At A and B, draw rays making an angle of 120° each and cut off AF = BC = 3.8 cm
3. Again at F and C, draw rays making an angle of 120° each and cut off CD = FE = 3.8 cm
4. Join DE. ABCDEF is the required hexagon.
5. Draw perpendicular bisectors of each of the opposite sides and also join AD, BE and CF. These six lines are the lines of symmetry.

[Diagram: This diagram shows regular hexagon ABCDEF with all sides 3.8 cm. Six lines of symmetry are drawn: three lines connecting opposite vertices (AD, BE, CF) and three lines connecting midpoints of opposite sides. All angles inside are 120 degrees.]

In simple words: A regular hexagon has six sides all equal. It has six lines of symmetry. Three lines connect opposite corners. Three lines connect midpoints of opposite sides.

📝 Teacher's Note: Show a hexagonal tile or honeycomb shape. Explain that 120° angles are used because 6 × 60° = 360° (full rotation). Each interior angle is 120°.

🎯 Exam Tip: Write that regular hexagon has 6 lines of symmetry. Mention both types: lines through opposite vertices and lines through midpoints of opposite sides.

Answer 6.
Answer: Construction of rhombus with one diagonal 8 cm:
Steps of construction:

1. Draw a line segment AB = 5 cm
2. With A as centre and radius 8 cm, and B as centre and radius 5 cm, draw arcs which intersect each other at C.
3. Join AC and BC.
4. Again with centre A and C and radius 5 cm, draw arcs which intersect each other at D
5. Join AD and CD. ABCD is the required rhombus.
6. Join BD.

Two diagonals AC and BD are the lines of symmetry.

[Diagram: This diagram shows rhombus ABCD with all sides 5 cm. One diagonal AC is 8 cm long. The two diagonals AC and BD intersect at point O in the center and are perpendicular to each other. Both diagonals are lines of symmetry.]

In simple words: A rhombus has four equal sides but different angles. It has two lines of symmetry - these are its two diagonals that cross each other at the center.

📝 Teacher's Note: Show students that rhombus diagonals are always perpendicular and bisect each other. Unlike a square, the diagonals of a rhombus are usually not equal in length.

🎯 Exam Tip: Write that rhombus has 2 lines of symmetry which are its diagonals. Remember that all sides are equal but angles are not 90 degrees.

Answer 7.
Answer:
Steps of construction:
(i) Draw a line segment BC = 8 cm
(ii) Draw its perpendicular bisector which intersects BC at D. With D as centre and BD or CD as radius, draw a semi-circle.
(iii) Produce the perpendicular bisector of BC which intersects the circle at A.
(iv) Join AB and AC. Triangle ABC is the required isosceles right-angled triangle.
The perpendicular bisector of hypotenuse BC is the line of symmetry.
In simple words: We make a triangle where two sides are equal and one angle is 90 degrees. The line that cuts the longest side in half is the symmetry line.

[Diagram: Shows an isosceles right triangle ABC with a semi-circle construction. Point D is the midpoint of BC (8 cm), and the perpendicular bisector passes through A and D, creating the line of symmetry.]

📝 Teacher's Note: Show students how to fold the triangle along the symmetry line. Both halves will match perfectly. This helps them see what line of symmetry means.

🎯 Exam Tip: Always draw the perpendicular bisector first. Mark the midpoint clearly. Write "isosceles right-angled triangle" in your answer to get full marks.

Answer 8.
Answer:
Steps of construction:
(i) Draw a line segment AC = 4.5 cm
(ii) With B and C as centres and 6 cm as radius, draw arcs which intersect each other at B.
(iii) Join AB and BC. △ABC is the required triangle.
(iv) Again with B and C as centres and 6 cm as radius, draw arcs which intersect each other at D.
(v) Join AD and DC. △ADC is the triangle which is the reflection of △ABC.
ABCD is the required quadrilateral and it is a rhombus.
In simple words: We make one triangle first. Then we make its mirror image on the other side. When we join all four points, we get a diamond shape called rhombus.

[Diagram: Shows rhombus ABCD with all sides equal to 6 cm and diagonal AC = 4.5 cm. Points B and D are mirror images across line AC.]

📝 Teacher's Note: Use a mirror and place it along AC. Students can see how triangle ABC becomes triangle ADC. This makes reflection very clear to understand.

🎯 Exam Tip: Always mention "reflection" in your answer. Draw both triangles with the same radius. Write "rhombus" as the final answer to score marks.

Answer 9.
Answer:
Steps of construction:
(i) Draw a line segment QR = 2.6 cm
(ii) At R draw a perpendicular to QR.
(iii) With Q as centre and radius 5.2 cm cut an arc on perpendicular to R at P.
(iv) Join PQ. △PQR is the required triangle.
(v) Produce QR to S such that RS = 2.6 cm
(vi) With S as centre and radius 5.2 cm cut an arc on perpendicular to R at P.
(vii) Join PS. △PSR is the triangle which is the reflection of △PQR.
△PQS is the required triangle and is an equilateral triangle.
In simple words: We make one triangle first. Then we make its copy on the other side. The big triangle we get has all three sides equal. This is called equilateral.

[Diagram: Shows triangle PQS where PQ = PS = 5.2 cm and QS = 5.2 cm total (QR + RS = 2.6 + 2.6 = 5.2 cm). Point R is the midpoint of QS.]

📝 Teacher's Note: Explain that QS = QR + RS = 2.6 + 2.6 = 5.2 cm. So all three sides PQ, PS, QS become 5.2 cm. That makes it equilateral.

🎯 Exam Tip: Always extend the base line to make S. Check that all three sides are equal in your final triangle. Write "equilateral triangle" clearly.

Answer 10.
Answer:
Steps of construction:
(i) Plot the points A, B and C as per given data.
(ii) Point D symmetrical about AB is a point with vertices x = -1 and y = 4 i.e. 3 units left of line AB.
(iii) Plot D(-1,4)
(iv) Join BC, AB, BD
The figure is an arrow.
In simple words: We plot three points first. Then we find the mirror image of one point. When we join all points, we get an arrow shape.

[Diagram: Shows coordinate plane with points A(2,0), B(2,8), C(5,4), and D(-1,4) plotted. Lines connect these points to form an arrow shape.]

📝 Teacher's Note: Show students that D is 3 units left of line AB, just like C is 3 units right of line AB. This is what symmetry means on a graph.

🎯 Exam Tip: Always plot points carefully on graph paper. Check that D is the same distance from line AB as point C. Write coordinates clearly.

Answer 11.
Answer:
Steps of construction:
(i) Plot the points P, Q and R as per given data.
(ii) Point S symmetrical about QR is a point with vertices x = 12 and y = 1 i.e. 5 unit right of line RQ.
(iii) Plot S(12,1)
(iv) Join PR, PS and RS
In simple words: We plot three points first. Then we find where the fourth point should be to make a symmetrical shape. We join all points to complete the figure.

[Diagram: Shows coordinate plane with points P(2,1), Q(7,1), R(7,5), and S(12,1) plotted. Lines connect these points to form a quadrilateral.]

📝 Teacher's Note: Point out that S is 5 units right of line QR, just like P is 5 units left of line QR. This creates a symmetric quadrilateral.

🎯 Exam Tip: Count squares carefully on graph paper. Make sure S is the same distance from QR as P is from QR. Join all four points to complete the shape.

Answer 12.
Answer:
Steps of construction:
(i) Plot the point A and B on the graph.
(ii) Plot point P whose vertices are x = 6 and y = 2. P is the point of symmetry.
(iii) Point symmetric to A(8,2) in the line x = 6 is C(4,2)
(iv) Point symmetric to B(6,4) in the line y = 2 is D(6,0)
(v) Join AP, PC, BP and PD.

Since BD = 4,
AD = \( \sqrt{(8-6)^2 + (2-0)^2} \)
= \( \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} \)

AB = \( \sqrt{(8-6)^2 + (2-4)^2} \)
= \( \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} \)

∴ BD² = AD² + AB²
4² = \( (\sqrt{8})^2 + (\sqrt{8})^2 \)
16 = 8 + 8
16 = 16
∴ ∠BAD = 90°

Clearly AB=BC=CD=DA, ∠BAD = 90° and AC and BD bisect each other at right angles, therefore ABCD is a square.
In simple words: We find points that are mirror images of A and B across the given lines. Then we connect all points to make a square shape.

[Diagram: This diagram shows a coordinate plane with points A(8,2), B(6,4), C(4,2), D(6,0), and P(6,2) marked. Lines connect these points to form a square ABCD with P as the center point where the diagonals meet.]

📝 Teacher's Note: Show students how to find mirror points by counting equal distances on both sides of the line of symmetry. Draw the line x = 6 and y = 2 clearly on the board.

🎯 Exam Tip: Always prove that all sides are equal and angles are 90° to show it is a square. Write the distance formula clearly and show all calculation steps.

Answer 13.
Answer:
Steps of construction:
(i) Plot point A and B on the graph.
(ii) Point symmetric to A(2,2) about x-axis is C(2,-2)
(iii) Point symmetric to B(5,5) about x-axis is D(5,-5)
(iv) Join AC, CD, BD.

The figure formed is a trapezium.
In simple words: We take the mirror image of points A and B across the x-axis (horizontal line). When we join all four points, we get a trapezium shape.

[Diagram: This diagram shows a coordinate plane with points A(2,2), B(5,5), C(2,-2), and D(5,-5) marked. Lines connect these points to form a trapezium ABCD.]

📝 Teacher's Note: Explain that reflection across x-axis means the y-coordinate changes sign but x-coordinate stays the same. Use a mirror placed on the x-axis to show this idea.

🎯 Exam Tip: When reflecting across x-axis, only change the sign of y-coordinate. Keep x-coordinate the same. Always state the final shape formed.

Answer 14.
Answer:
Steps of construction:
(i) Plot the point A, B and C on the graph.
(ii) Point symmetric to A(4,1) about x = 7 is D(10,1)
(iii) Point symmetric to B(2,3) about x = 7 is E(12,3)
(iv) Point symmetric to C(5,6) about x = 7 is F(9,6)
(v) Join AB, AC, BC, AD, DE, DF, EF and CF.

The figure formed is a trapezium ADCF with two equal scalene triangles (ABC and DEF) attached to it.
In simple words: We find mirror images of the three points across the vertical line x = 7. This creates a shape that looks like a house with two triangular roofs.

[Diagram: This diagram shows a coordinate plane with triangles ABC and DEF reflected across the line x = 7, forming a symmetric hexagon shape.]

📝 Teacher's Note: When reflecting across vertical line x = 7, the y-coordinate stays same but x-coordinate changes. Distance from line of symmetry remains equal on both sides.

🎯 Exam Tip: For reflection across line x = a, use formula: new x = 2a - old x, y stays same. Always describe the final shape clearly.

Answer 15.
Answer:
For the first figure:
Equal sides about line of symmetry: AB = BC = CD
Corresponding angles about line of symmetry: ∠ABC = ∠ADC

For the second figure:
Equal sides about line of symmetry: PS = PQ = QR
Corresponding angles about line of symmetry: ∠PSQ = ∠SRQ
In simple words: In symmetric shapes, the parts on both sides of the middle line are exactly the same. Equal sides and equal angles match up across the line.

[Diagram: Two geometric figures showing symmetrical quadrilaterals with dotted lines indicating the lines of symmetry.]

📝 Teacher's Note: Fold a paper shape along the line of symmetry to show students how both sides match perfectly. This makes the concept very clear.

🎯 Exam Tip: Always identify the line of symmetry first. Then find pairs of equal sides and equal angles on both sides of this line.

Answer 16.
Answer:
(i) ABC is an isosceles triangle. Draw AD ⊥ BC. AD bisects ∠A. Here AD is the line of symmetry. There is no point of symmetry.

(ii) Draw the angle bisector BM of ∠ABC. BM is the line of symmetry. There is no point of symmetry.
In simple words: For a triangle with two equal sides, the line from the top corner to the base middle is the line of symmetry. For a quarter circle, the line that cuts the right angle in half is the line of symmetry.

[Diagram: Two figures - (i) shows an isosceles triangle ABC with altitude AD as line of symmetry, (ii) shows a quarter circle with angle bisector BM as line of symmetry.]

📝 Teacher's Note: Cut out paper shapes and fold them to find the line of symmetry. Students can see that one half fits exactly on the other half.

🎯 Exam Tip: For isosceles triangle, line of symmetry is the altitude from vertex angle to base. For quarter circle, it is the angle bisector of the right angle.

Answer 17.
Answer: On completing the figure which is symmetrical about the given axis, it becomes an arrowhead.

[Diagram: Shows two triangular shapes. The first is a simple triangle with a dotted line (axis of symmetry) going through it vertically. The second shows the completed figure - when the triangle is reflected across the axis, it forms an arrowhead shape pointing up and down.]

On completing the figure which is symmetrical about the given axis, it becomes a rhombus.

[Diagram: Shows two quadrilateral shapes. The first is an incomplete quadrilateral with a dotted line (axis of symmetry). The second shows the completed figure - when reflected across the axis, it forms a rhombus (diamond shape).]

In simple words: When we draw the mirror image of a shape across a line, we get a symmetrical figure. The line acts like a mirror - what we see on one side must match the other side exactly.

📝 Teacher's Note: Use a real mirror to show students how reflection works. Place the mirror on the dotted line and show how the shape looks complete. This makes symmetry very clear to understand.

🎯 Exam Tip: Always draw the reflected part carefully. Make sure both sides of the symmetry line are exactly the same size and shape. Check by folding the paper along the line.

Answer 18.
Answer: On completing the figure about both axes it becomes a square.

[Diagram: Shows the process of completing a figure using two axes of symmetry (horizontal and vertical lines). The first image shows an incomplete triangular portion, and the second shows how reflecting it across both axes creates a complete square.]

On completing the figure about both axes it becomes a circle.

[Diagram: Shows the process of completing a quarter circle. The first image shows a quarter circle with two perpendicular axes. The second shows how reflecting it across both axes creates a complete circle.]

In simple words: When a shape has two lines of symmetry that cross each other, we can reflect the shape across both lines. This gives us four identical parts that make a complete figure.

📝 Teacher's Note: Show students how folding paper twice (once along each axis) creates four identical sections. When we unfold, we see the complete symmetrical shape. Use simple shapes like hearts or butterflies.

🎯 Exam Tip: When there are two axes of symmetry, reflect the given part across the first axis, then reflect the whole thing across the second axis. Draw all four parts to get the complete figure.