# Frank Solutions for Chapter 14 Symmetry Class 10 ICSE Mathematics

**1. Construct an isosceles triangle whose equal sides are 7 cm each and the base side is 5 cm. Draw all its lines of symmetry.**

**Answer**

- Draw a line segment QR = 5 cm.
- With Q as a center and radius 7 cm, draw an arc.
- With R as a center and radius 7 cm, draw another arc, cutting the previous arc at P.
- Join PQ and PR. Then, Î”PQR is the required isosceles triangle.

- Now, draw an angle bisector with P as the center and meeting QR at S.
- PS is the perpendicular bisector of QR and PQ is equal to PR.

Therefore, PS is the line of symmetry. Isosceles triangle has only one line of symmetry.

**2. Construct a triangle ABC in which each side measures 5.8. Draw all the possible lines of symmetry.**

**Answer**

- Draw a line segment AC = 5.8 cm.
- With A as a center and radius 5.8 cm, draw an arc.
- With C as a center and radius 5.8 cm, draw another arc, cutting the previous arc at B.
- Join AB and CB. Then, Î”ABC is the required equilateral triangle.

- Now, draw angle bisectors with B as the center and meeting AC at P, with A as the center and meeting BC at Q, with C as the center and meeting AB at R.
- Therefore, BP, AQ, CR are the line of symmetry.

**3. Construct a parallelogram PQRS in which QR = 5.4 cm, SR = 6.0 cm and ****∠****Q = 60 ^{o}. Draw its lines of symmetry, if possible.**

**Answer**

- Draw a line segment QR = 5.4 cm.
- At Q, draw a ray making an angle of 60
^{o}with QR - Along a ray, set off QP = 6 cm.
- With P as a center and radius 5.4 cm, draw an arc.
- With R as a center and radius 6 cm, draw another arc, cutting the previous arc at S.
- Join PS and RS. Then, PQRS is the required parallelogram.

- So, QS and PR intersect each other at O.

Therefore, there is no line of symmetry in parallelogram PQRS.

**4. Construct a square of side 4.8 cm and draw all its lines of symmetry.**

**Answer**

- Draw a line segment PQ = 4.8 cm.
- At P and Q draw perpendiculars PM and QN.
- With P as center and radius equal to 4.8 cm, cut PM at S.
- With Q as center and radius equal to 4.8 cm, cut QN at R.
- Join RS, so PQRS is the required square.
- Now, join the diagonals of square PR and QS.
- Then, draw perpendicular bisectors of PQ and PS.

Therefore, the diagonals and perpendicular bisectors are the lines of symmetry of square PQRS.

**5. Construct a regular hexagon of side = 3.8 cm and draw all its lines of symmetry.**

**Answer**

- Draw a line segment LM = 3.8 cm.
- At L and M, draw a rays making an angle of 120
^{o}each, then cut off LQ and MN for 3.8 cm. - At N and F, draw a rays making an angle of 120
^{o}each, then cut off NO and QP for 3.8 cm. - Join OP, so LMNOPQ is the required regular polygon.
- Now, join the diagonals of regular hexagon LO, MP and NQ.
- Then, draw perpendicular bisectors of LM, NO and PQ.

Therefore, the diagonals and perpendicular bisectors are the lines of symmetry of regular hexagon LMNOPQ.

**6. Construct a rhombus ABCD with AB = 5 cm and AC = 8 cm. Draw it lines of symmetry.**

**Answer**

- Draw a line segment AB = 5 cm.
- With B as a center and radius 5 cm, draw an arc.
- With A as a center and radius 8 cm, draw another arc, cutting the previous arc at C.
- Join AC and BC, then we get Î”ABC the isosceles triangle.
- Again with A as a center and radius 5 cm, draw an arc.
- With C as a center and radius 5 cm, draw another arc, cutting the previous arc at D.
- Join AD and CD, then we get ABCD the required rhombus.
- Now, join the diagonal of rhombus BD.

Therefore, the diagonals are the lines of symmetry of rhombus ABCD.

**7. Construct an isosceles right-angled triangle, having hypotenuse = 8 cm. Draw its lines of symmetry.**

**Answer**

- Draw a line segment BC = 8 cm.
- Then draw its perpendicular bisector which intersects BC at D.
- With D as a center and BD or CD radius, draw a semi-circle.
- Now produce the perpendicular bisector of BC which intersects the circle at A
- Join AB and AC, so Î”ABC is the required isosceles right angled triangle.

Therefore, perpendicular bisector hypotenuse BC is the lines of symmetry of isosceles right angled triangle.

**8. Construct a Î”ABC in which BA = BC = 6 cm and AC = 4.5 cm. Taking AC as line of symmetry, obtain a point D to form a quadrilateral ABCD. Name the figure ABCD.**

**Answer**

- Draw a line segment AC = 4.5 cm.
- With A as a center and radius 6 cm, draw an arc.
- With C as a center and radius 6 cm, draw another arc cutting the previous arc at B.
- Join AB and BC, Then, Î”ABC is the isosceles triangle.

As per the condition given in the question,

- Taking AC as line of symmetry.
- With A as a center and radius 6 cm, draw an arc.
- With C as a center and radius 6 cm, draw another arc cutting the previous arc at D.
- Join AD and CD.

Therefore, ABCD is the required quadrilateral i.e. rhombus.

**9. Construct a Î”PQR in which ****∠****R = 90 ^{o}, PQ = 5.2 cm and QR = 2.6 cm. Complete the figure taking PR as the line of symmetry and name the figure.**

**Answer**

- Draw a line segment QS = 2.6 cm.
- With Q as a center and radius 5.2 cm, draw an arc.
- At R draw a perpendicular to QR to meet at P.
- Join PQ, so PQR is the required triangle.

As per the condition given in the question,

- Taking PR as the line of symmetry.
- Now, produce QR to S i.e. RS = 2.6 cm
- With Q as a center and radius 5.2 cm, draw an arc at p.
- Join PS, so PRS is the triangle.

Therefore, Î”PQS is the required triangle and also it is an equilateral triangle.

**10. Take a graph paper and mark the points A(2, 0), B(2, 8) and C(5, 4) on it. Taking AB as the line of symmetry, obtain and write the co-ordinates of point D. Complete the quadrilateral ABCD and give its geometrical name.**

**Answer**

- As per the given data plot the points A (2, 0), B(2, 8) and C(5, 4) on the graph.
- Join AB and BC.
- Condition given the question, taking AB as the line of symmetry.
- So, point D symmetrical about AB is a point with vertices x = -1 and y = 4 (because from point A to C in vertices x there are 3 units and in y there are 4 units)
- Now plot point D(-1, 4)
- Join BD.

Therefore, the obtained figure is an arrow.

**11.** **Take a graph paper and mark the points P(2, 1), Q(7, 1) and R(7, 5). Taking QR as the line of symmetry, obtain and write the co-ordinates of point S.**

**Answer**

- As per the given data plot the points P (2, 1), Q(7, 1) and R(7, 5) on the graph.
- Join PR and PQ.

Condition given the question, taking QR as the line of symmetry.

- So, point S symmetrical about QR is a point with vertices x = 12 and y = 1 (because from point Q to P in vertices x there are 5 units and in y there are 1 unit)
- Now plot point S(12, 1)
- Join SQ and SR.

Therefore, the obtained figure is an isosceles triangle.

**12. A(8, 2) and B(6, 4) are the vertices of a figure which is symmetrical about x = 6 and y = 2. Complete the figure and give the geometrical name of the figure.**

**Answer**

- As per the given data plot the points A (8, 2) and B(6, 8) on the graph.
- Then plot point M whose vertices are x = 6 and y = 2.
- Condition given the question, taking P as the point of symmetry.
- So, point symmetric to A(8, 2) in the line x = 6 is C(4, 2)
- Point symmetric to B(6, 4) in the line y = 2 is D(6, 0)
- Now join AP, PC, BP and PD

By using the distance formula, AD = √((8 – 6)^{2} + (2 – 0)^{2})

= √(2^{2} + 2^{2})

= √(4 + 4)

= √8

Then, AB = √((8 – 6)^{2} + (2 – 4)^{2})

= √(2^{2} + (-2^{2}))

= √(4 + 4)

= √8

So, from Pythagoras theorem BD^{2} = AD^{2} + AB^{2}

4^{2} = (√8)^{2} + (√8)^{2}

⇒ 16 = 8 + 8

⇒ 16 = 16

Therefore, ∠BAD = 90^{o}

Hence, it is clear that AB = BC = CD = DA, AC and BD bisect each other at right angles, so ABCD is a square.

**13. A(2, 2) and B(5, 5) are the vertices of a figure which is symmetrical about x – axis. Complete the figure and give its geometrical name.**

**Answer**

- As per the given data plot the points A (2, 2) and B(5, 8) on the graph.
- Condition given the question, a figure which is symmetrical about x – axis.
- So, point symmetric to A(2, 2) in the line x – axis is C(2, -2)
- Point symmetric to B(5, 5) in the line y = 2 is D(5, -5)
- Now join AB, AC, CD and BD

Therefore, the obtained figure is a trapezium.

**14. A(4, 1), B(2, 3) and C(5, 6) are the vertices of a figure which is symmetrical about x = 7. Complete the figure and give the geometrical name of the figure if any.**

**Answer**

- As per the given data plot the points A (4, 1), B(2, 3) and C(5, 6) on the graph.
- Condition given the question, a figure which is symmetrical about x = 7.
- So, point symmetric to A(4, 1) about x = 7 is D(10, 1)
- Point symmetric to B(2, 3) about x = 7 is E(12, 3)
- Point symmetric to C(5, 6) about x = 7 is F(9, 6)
- Now join AB, AC, BC, AD, DE, DF, EF and CF

Therefore, the obtained figure is a trapezium ADCF with two equal scalene triangles i.e. Î”ABC and Î”DEF are attached to it.

**15. In each of the following figures, the line of symmetry has been drawn with a dotted line. Identify the corresponding sides and the corresponding angles about the line of symmetry.**

**Answer**

**(i)**

In the given figure,

The corresponding sides about the line of symmetry is, PS = SR, PQ = QR

Then, corresponding angles bout line of symmetry is ∠SPQ = ∠SRQ

**(ii)**

In the given figure,

The corresponding sides about the line of symmetry is, AB = AD, BC = CD

Then, corresponding angles bout line of symmetry is ∠ABC = ∠ADC

**(iii)**

In the given figure,

The corresponding sides about the line of symmetry is, AB = BC, AD = DC

Then, corresponding angles bout line of symmetry is ∠DAB = ∠DCB

**(iv)**

In the given figure,

The corresponding sides about the line of symmetry is, PQ = PU, QR = UT

Then, corresponding angles bout line of symmetry is ∠PQR = ∠PUT, ∠QRT = ∠UTR

**16. Draw the line of symmetry of the following figures and also mark their points of symmetry, if any: **

**Answer**