The DAV Class 7 Maths Book Solutions and DAV Class 7 Maths Chapter 13 Worksheet 2 Solutions of Symmetry offer comprehensive answers to textbook questions.
DAV Class 7 Maths Ch 13 WS 2 Solutions
Question 1.
Find the line of symmetry of any 3 line segments on different sheets of paper.
Answers:
Draw the perpendicular bisectors of the line segment which is the line of symmetry of the line segment. In above figures;
l1 is the line of symmetry of \(\overline{\mathrm{AB}}\)
l2 is the line of symmetry of \(\overline{\mathrm{PQ}}\)
and l3 is the line of symmetry of \(\overline{\mathrm{MN}}\).
Question 2.
Check if the dotted lines are the line of symmetry of the given line segments.
Answer:
(i) The dotted line is not the perpendicular bisector of PQ, so it is not a line of symmetry.
(ii) The dotted line is the perpendicular bisector of XY, so it is a line of symmetry.
Question 3.
Lines l and m are the lines of symmetry of the line segments XY and YZ respectively. If XA = 4 cm and YZ = 6 cm, find AY, YB, XZ.
Answer:
l is the line of symmetry of XY.
∴ XA = AY
⇒ 4 = AY
⇒ AY = 4 cm
m is the line of symmetry of YZ, YZ = 6 cm
∴ YB = BZ = \(\frac{6}{2}\) = 3 cm
XZ = 2XA + YZ = 2(4) + 6 = 8 + 6 = 14 cm
Hence, AY = 4 cm, YB = 3 cm and XZ = 14 cm
Question 4.
Find the line of symmetry of the following angles on different sheets of paper.
(a) 60°
(b) 150°
(c) 45°
Answer:
(a) OP is the bisector of ∠AOB, so it is the line of symmetry of ∠AOB where ∠AOB = 60°
(b) SQ is the bisector of ∠RST, so it is the line of symmetry of ∠RST, where ∠RST = 150°
(c) NL is the bisector of ∠MNO, so it is the line of symmetry of ∠MNO, where ∠MNO = 45°.
Question 5.
If dotted lines represent the lines of symmetry of the given angles, find x in each case.
Answer:
In fig. (z) The dotted line is the bisector of the angle
x = \(\frac{80^{\circ}}{2}\) = 40°
In figure (ii) the dotted line is the bisector of the angle
x = 2 × 55° = 110°
Question 6.
The dotted line represents the line of symmetry of ΔABC. If ∠ABC = 40° and OB = 4.5 cm, find.
(a) ∠BAO
(b) Measure of OC. Give reasons.
Answer:
(a) In ΔABC, OA is the line of symmetry 8
∴ ΔABC is an isosceles triangle ∠B = ∠C
∠A + ∠B + ∠C = 180°
∠A + 40° + 40° = 180°
∴ ∠A = 180° – 80° = 100°
AO is the line of symmetry
AO is the bisector of ∠A
∴ ∠BAO = \(\frac{100^{\circ}}{2}\) = 50°
(b) OA is the line of symmetry of BC
OA is the perpendicular bisector of BC BO = OC = 4.5 cm
Hence, ∠BAO = 50° and OC = 4.5 cm
Question 7.
Construct an equilateral ΔABC of side 4 cm. Find all its line of symmetry.
Answer:
Draw ΔABC in which AB = BC = AC = 4 cm.
Draw the bisectors of ∠A, ∠B and ∠C
Al, Bm and Cn are the line of symmetry of an equilateral ΔABC.
Question 8.
Will a scalene triangle have any line of symmetry?
Answer:
No, scalene triangle has no line of symmetry.
Question 9.
Under what circumstances will a right angled triangle have a line of symmetry? Give reasons.
Answer:
A right angled triangle will have a line of symmetry if it is an isosceles right triangle. Because the triangle cannot be an equilateral triangle and scalene triangle has no line of symmetry.
Question 10.
Can a right angled triangle have more than one line of symmetry under any circumstances?
Answer:
No, right angled triangle will have only one line of symmetry only if it is isosceles triangle.