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Shape and Space - Properties and Classification of Triangles

Grade 6IB

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

🔑Concepts

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Classification by Sides: Triangles can be categorized based on the lengths of their sides. An equilateral triangle has three equal sides (a=b=ca = b = c). An isosceles triangle has at least two equal sides. A scalene triangle has no equal sides.

Visual comparison of equilateral, isosceles, and scalene triangles.
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Classification by Angles: Triangles are also classified by their interior angles. An acute triangle has all angles less than 90∘90^\circ. A right-angled triangle has one angle exactly 90∘90^\circ. An obtuse triangle has one angle greater than 90∘90^\circ.

Diagram showing a right-angled triangle with a 90 degree square symbol and an obtuse triangle.
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The Angle Sum Property: The sum of the interior angles of any triangle is always 180∘180^\circ. This property allows us to find a missing angle if two are known using the formula Angle3=180∘−(Angle1+Angle2)Angle_3 = 180^\circ - (Angle_1 + Angle_2).

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The Triangle Inequality Theorem: For any triangle, the sum of the lengths of any two sides must be strictly greater than the length of the third side (a+b>ca + b > c). If this condition is not met, the three segments cannot form a closed triangle.

📐Formulae

Sum of interior angles: A+B+C=180∘A + B + C = 180^\circ

Area of a triangle: Area=12×base×heightArea = \frac{1}{2} \times \text{base} \times \text{height}

Perimeter of a triangle: P=a+b+cP = a + b + c

Triangle Inequality: a+b>ca + b > c, a+c>ba + c > b, and b+c>ab + c > a

💡Examples

Problem 1:

In an isosceles triangle, the vertex angle (the angle between the two equal sides) measures 70∘70^\circ. Calculate the size of the two remaining base angles.

Solution:

Step 1: Let the two equal base angles be represented by xx. Step 2: Since the sum of angles is 180∘180^\circ, the equation is x+x+70∘=180∘x + x + 70^\circ = 180^\circ. Step 3: Simplify to 2x+70∘=180∘2x + 70^\circ = 180^\circ. Step 4: Subtract 70∘70^\circ from both sides: 2x=110∘2x = 110^\circ. Step 5: Divide by 22: x=55∘x = 55^\circ.

Explanation:

This solution uses the property that an isosceles triangle has two equal angles and the fact that the total interior sum must be 180∘180^\circ.

Problem 2:

Find the area of a triangle where the base is 12 cm12\text{ cm} and the perpendicular height is 7 cm7\text{ cm}.

Solution:

Step 1: Identify the values: base(b)=12\text{base} (b) = 12 and height(h)=7\text{height} (h) = 7. Step 2: Use the area formula A=12×b×hA = \frac{1}{2} \times b \times h. Step 3: Substitute the values: A=12×12×7A = \frac{1}{2} \times 12 \times 7. Step 4: Calculate A=6×7=42A = 6 \times 7 = 42. Final Answer: 42 cm242\text{ cm}^2.

Explanation:

The area is calculated by taking half of the product of the base and the vertical height that is perpendicular to that base.

Problem 3:

In the triangle ABCABC shown, angle A=55∘A = 55^\circ and angle B=65∘B = 65^\circ. Calculate the size of angle CC.

Triangle ABC with angles at A and B labeled 55 and 65 degrees respectively.

Solution:

Angle C=180∘−(55∘+65∘)Angle\ C = 180^\circ - (55^\circ + 65^\circ) Angle C=180∘−120∘Angle\ C = 180^\circ - 120^\circ Angle C=60∘Angle\ C = 60^\circ

Explanation:

Since the sum of interior angles in a triangle is always 180∘180^\circ, we add the two known angles and subtract the result from 180∘180^\circ to find the third angle.

Problem 4:

A triangle has a base of 10 cm10\text{ cm} and a height of 8 cm8\text{ cm}. A second triangle is identical in shape but its base is 5 cm5\text{ cm}. Find the area of the first triangle.

Triangle with base 10cm and a dashed vertical line indicating a height of 8cm.

Solution:

Area=12×base×heightArea = \frac{1}{2} \times \text{base} \times \text{height} Area=12×10×8Area = \frac{1}{2} \times 10 \times 8 Area=5×8Area = 5 \times 8 Area=40 cm2Area = 40\text{ cm}^2

Explanation:

To find the area, we identify the base (10 cm10\text{ cm}) and the perpendicular height (8 cm8\text{ cm}) and apply the area formula for triangles.