Home / iGCSE Mathematics (0580) :C4.6 Recognize rotational and line symmetry. iGCSE Style Questions Paper 3

iGCSE Mathematics (0580) :C4.6 Recognize rotational and line symmetry. iGCSE Style Questions Paper 3

Question

(a)Isosceles triangle diagram

The diagram shows an isosceles triangle and a straight line.
Work out the value of w.

(b)Rectangle diagram
ABCD is a rectangle.
AE is parallel to DBF.
Find the value of x and the value of y.

(c)Circle diagram
A, B and C are points on a circle.
AC is a diameter of the circle.
Find the value of a.

(d)Octagons and square diagram
Two regular octagons and a square meet at point P.
Show, by calculation, that the three interior angles at P add up to 360°.

▶️ Answer/Explanation
Answers:

(a) 56°
• Isosceles triangle → two base angles = 62° each
• Straight line angle = 180° → w = 180 – (62+62) = 56°

(b) x = 31°, y = 121°
• x = alternate angle to 31° (AE∥DBF)
• y = 180° – (90°-31°) = 121° (rectangle has 90° corners)

(c) 37°
• AC is diameter → angle ABC = 90° (semicircle theorem)
• a = 90° – 53° = 37°

(d) Proof
• Octagon interior angle = 135° (using (8-2)×180°/8)
• Square angle = 90° → Total = 135° + 135° + 90° = 360°

Question

(a)

Angle diagram

(i) Write down the mathematical name for this type of angle.

(ii) Measure this angle.

(b)

(i) Write down the mathematical name for an 8-sided polygon.

(ii) Work out the size of an interior angle of a regular 24-sided polygon.

(c)

Circle diagram

The diagram shows a circle, centre O, with diameter CE. A, B, C, D and E lie on the circumference of the circle.

(i) Find the value of x. Give a reason for your answer.

(ii) Find the value of y. Give a reason for your answer.

(iii) Draw a tangent to the circle at A.

▶️ Answer/Explanation
Answers:

(a)(i) Obtuse angle

Angles between 90° and 180° are called obtuse. The diagram shows such an angle.

(a)(ii) 134°

Using a protractor on the diagram, the angle measures approximately 134 degrees.

(b)(i) Octagon

An 8-sided polygon is called an octagon (from Greek “octa” meaning eight).

(b)(ii) 165°

Formula: (n-2)×180°/n = (24-2)×180°/24 = 22×7.5° = 165°.

(c)(i) x = 132°

Triangle AOB is isosceles (OA=OB as radii). So base angles are equal: (180°-96°)/2 = 42°. x = 180°-48° = 132° (angles on straight line).

(c)(ii) y = 17°

Angle ADE is 90° (angle in semicircle). So y = 90°-48°-25° = 17°.

(c)(iii)

Draw a line perpendicular to radius OA at point A – this is the tangent.
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