krit.club logo

Geometry and Measurement - Properties of 2D Shapes (Triangles and Quadrilaterals)

Grade 7IB

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

🔑Concepts

Sum of Interior Angles in a Triangle: The sum of the three interior angles of any triangle is always 180180^\circ. Visually, if you imagine cutting the three corners of a triangle and placing them side-by-side, they would form a perfect straight line.

Classification of Triangles by Sides: Triangles are categorized as Equilateral (three equal sides and three 6060^\circ angles), Isosceles (two equal sides and two equal base angles), and Scalene (no equal sides or angles). Visually, an equilateral triangle looks perfectly symmetrical from all sides, while an isosceles triangle resembles a tall, balanced tent.

Classification of Triangles by Angles: Triangles are also defined by their largest angle. A Right-Angled triangle contains one 9090^\circ angle (indicated by a small square in the corner), an Obtuse triangle has one angle greater than 9090^\circ (looking 'wide'), and an Acute triangle has all angles less than 9090^\circ.

Sum of Interior Angles in a Quadrilateral: The interior angles of any four-sided shape (quadrilateral) always sum to 360360^\circ. This is visually logical because any quadrilateral can be split into two triangles by drawing a single diagonal line, and 2×180=3602 \times 180^\circ = 360^\circ.

Properties of Parallelograms and Rectangles: A parallelogram has two pairs of parallel and equal opposite sides. A rectangle is a special type of parallelogram where all four interior angles are exactly 9090^\circ. Visually, a parallelogram looks like a rectangle that has been pushed or tilted to one side.

Rhombus and Square Properties: A rhombus is a quadrilateral with four equal sides; its diagonals bisect each other at 9090^\circ. A square is a regular quadrilateral, meaning it is both a rectangle (four 9090^\circ angles) and a rhombus (four equal sides). Visually, a rhombus looks like a 'diamond' shape.

Trapeziums and Kites: A trapezium (or trapezoid) has at least one pair of parallel sides. A kite has two pairs of equal-length sides that are adjacent to each other. Visually, a kite looks like a traditional flying kite with two shorter sides at the top and two longer sides at the bottom, while its diagonals cross at a right angle.

The Exterior Angle Theorem: The exterior angle of a triangle is equal to the sum of the two opposite interior angles. Visually, if you extend one side of a triangle outward, the angle formed outside is equal to the sum of the two angles 'inside' that aren't touching it.

📐Formulae

Sum of angles in a triangle: A+B+C=180\angle A + \angle B + \angle C = 180^\circ

Sum of angles in a quadrilateral: A+B+C+D=360\angle A + \angle B + \angle C + \angle D = 360^\circ

Area of a Triangle: A=12×base×height=12bhA = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}bh

Area of a Rectangle: A=length×width=lwA = \text{length} \times \text{width} = lw

Area of a Parallelogram: A=base×perpendicular height=bhA = \text{base} \times \text{perpendicular height} = bh

Area of a Trapezium: A=12(a+b)hA = \frac{1}{2}(a + b)h, where aa and bb are parallel sides.

Perimeter of any 2D shape: P=side lengthsP = \sum \text{side lengths}

💡Examples

Problem 1:

In an isosceles triangle ABCABC, the vertex angle A\angle A is 4040^\circ. Find the size of the two base angles, B\angle B and C\angle C.

Solution:

  1. Let the base angles be xx. Since it is an isosceles triangle, B=C=x\angle B = \angle C = x.
  2. The sum of angles is 180180^\circ, so: 40+x+x=18040^\circ + x + x = 180^\circ.
  3. Simplify: 40+2x=18040 + 2x = 180.
  4. Subtract 4040 from both sides: 2x=1402x = 140.
  5. Divide by 22: x=70x = 70^\circ. Final Answer: B=70\angle B = 70^\circ and C=70\angle C = 70^\circ.

Explanation:

This solution uses the property that base angles of an isosceles triangle are equal and the triangle angle sum theorem.

Problem 2:

Calculate the area of a trapezium where the parallel sides are 8 cm8\text{ cm} and 12 cm12\text{ cm}, and the perpendicular height is 5 cm5\text{ cm}.

Solution:

  1. Identify the values: a=8a = 8, b=12b = 12, and h=5h = 5.
  2. Use the formula: A=12(a+b)hA = \frac{1}{2}(a + b)h.
  3. Substitute the values: A=12(8+12)×5A = \frac{1}{2}(8 + 12) \times 5.
  4. Calculate inside the brackets: A=12(20)×5A = \frac{1}{2}(20) \times 5.
  5. Multiply: A=10×5=50 cm2A = 10 \times 5 = 50\text{ cm}^2. Final Answer: The area is 50 cm250\text{ cm}^2.

Explanation:

The area is found by taking the average of the parallel bases and multiplying by the vertical height between them.