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

Grade 7IB

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

🔑Concepts

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Triangles can be classified by their sides: Scalene (all sides different), Isosceles (two sides equal), and Equilateral (all three sides equal). They can also be classified by their angles: Acute, Obtuse, and Right-angled. The sum of interior angles in any triangle is always 180∘180^\circ.

A triangle labeled ABC illustrating the interior angle sum property.
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Quadrilaterals are four-sided polygons. The sum of their interior angles is 360∘360^\circ. Specific types include Parallelograms (opposite sides parallel and equal), Rectangles (parallelogram with four 90∘90^\circ angles), Rhombuses (parallelogram with four equal sides), and Squares (regular quadrilateral).

A parallelogram showing opposite sides are parallel.
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The area of a triangle is calculated using the perpendicular height (hh) and the base (bb). Even if the triangle is obtuse, the height is the vertical distance from the vertex to the line containing the base.

Right-angled triangle showing base and perpendicular height.
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A trapezium (or trapezoid) is a quadrilateral with at least one pair of parallel sides. The area is the average of the parallel sides multiplied by the height.

📐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 40∘40^\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 180∘180^\circ, so: 40∘+x+x=180∘40^\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=70∘x = 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.

Problem 3:

Calculate the area of the following parallelogram which has a base of 12 cm12\text{ cm} and a perpendicular height of 7 cm7\text{ cm}.

Parallelogram with base 12 cm and height 7 cm.

Solution:

  1. Identify the formula for the area of a parallelogram: A=b×hA = b \times h
  2. Substitute the given values: b=12b = 12 and h=7h = 7.
  3. Calculate: A=12×7=84A = 12 \times 7 = 84
  4. The area is 84 cm284\text{ cm}^2.

Explanation:

The area of a parallelogram is simply the base multiplied by the vertical height, not the slant height of the sides.

Problem 4:

Find the value of the missing angle xx in a quadrilateral where the other three angles are 110∘110^\circ, 85∘85^\circ, and 75∘75^\circ.

Quadrilateral with angles 110, 85, 75 and x.

Solution:

  1. The sum of angles in a quadrilateral is 360∘360^\circ.
  2. Set up the equation: x+110∘+85∘+75∘=360∘x + 110^\circ + 85^\circ + 75^\circ = 360^\circ
  3. Add the known angles: 110+85+75=270110 + 85 + 75 = 270.
  4. Subtract from 360360: x=360∘−270∘x = 360^\circ - 270^\circ
  5. x=90∘x = 90^\circ

Explanation:

By applying the angle sum property of quadrilaterals, we can find any unknown angle if the other three are provided.