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Geometry and Measure - Shapes and geometric reasoning

Grade 7IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

πŸ”‘Concepts

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Angles on a straight line sum to 180∘180^\circ and angles around a point sum to 360∘360^\circ.

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Vertically opposite angles are equal when two straight lines intersect.

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Parallel line rules: Alternate angles are equal (Z-shape), Corresponding angles are equal (F-shape), and Co-interior angles add up to 180∘180^\circ (U/C-shape).

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Types of Triangles: Scalene (no equal sides/angles), Isosceles (two equal sides/angles), and Equilateral (three equal sides/angles of 60∘60^\circ).

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Quadrilateral Properties: The sum of interior angles is always 360∘360^\circ. Key shapes include Parallelograms, Rhombuses, Trapezia, and Kites.

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Regular Polygons: All sides and all interior angles are equal.

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Symmetry: Line symmetry (folding) and Rotational symmetry (order of rotation).

πŸ“Formulae

Sum of interior angles of an nn-sided polygon = (nβˆ’2)Γ—180∘(n - 2) \times 180^\circ

Sum of exterior angles of any convex polygon = 360∘360^\circ

Individual exterior angle of a regular nn-sided polygon = 360∘n\frac{360^\circ}{n}

Individual interior angle of a regular nn-sided polygon = 180βˆ˜βˆ’exteriorΒ angle180^\circ - \text{exterior angle}

Area of a Triangle = 12Γ—baseΓ—height\frac{1}{2} \times \text{base} \times \text{height}

Area of a Parallelogram = baseΓ—height\text{base} \times \text{height}

πŸ’‘Examples

Problem 1:

In an isosceles triangle, the vertex angle is 40∘40^\circ. Find the size of the two base angles.

Solution:

70∘70^\circ each

Explanation:

The sum of angles in a triangle is 180∘180^\circ. Subtract the vertex angle: 180βˆ˜βˆ’40∘=140∘180^\circ - 40^\circ = 140^\circ. Since it is an isosceles triangle, the two base angles are equal. Therefore, 140∘÷2=70∘140^\circ \div 2 = 70^\circ.

Problem 2:

Calculate the size of one interior angle of a regular hexagon.

Solution:

120∘120^\circ

Explanation:

A hexagon has n=6n=6 sides. First, find the exterior angle: 360∘÷6=60∘360^\circ \div 6 = 60^\circ. Since the interior and exterior angles lie on a straight line, the interior angle is 180βˆ˜βˆ’60∘=120∘180^\circ - 60^\circ = 120^\circ.

Problem 3:

Two parallel lines are intersected by a transversal. If an alternate angle is 55∘55^\circ, what is the size of its corresponding co-interior angle?

Solution:

125∘125^\circ

Explanation:

If the alternate angle is 55∘55^\circ, the angle adjacent to it on the straight line is 180βˆ˜βˆ’55∘=125∘180^\circ - 55^\circ = 125^\circ. This adjacent angle is equal to the co-interior partner because co-interior angles must sum to 180∘180^\circ (180βˆ˜βˆ’55∘=125∘180^\circ - 55^\circ = 125^\circ).