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Geometry and Measure - Angle properties of triangles and quadrilaterals

Grade 7Cambridge (IGCSE)

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

🔑Concepts

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The sum of the interior angles of any triangle is always 180∘180^\circ. This property allows you to find a missing angle if two are known using the formula: Angle3=180∘−(Angle1+Angle2)Angle_3 = 180^\circ - (Angle_1 + Angle_2).

Triangle with interior angles labeled a, b, and c summing to 180 degrees.
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In an isosceles triangle, two sides are equal in length and the angles opposite those sides (base angles) are equal. In an equilateral triangle, all three sides are equal and each interior angle is 60∘60^\circ.

Isosceles triangle showing equal base angles marked x.
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The sum of the interior angles of any quadrilateral is 360∘360^\circ. This can be visualized by dividing the quadrilateral into two triangles, each contributing 180∘180^\circ.

Quadrilateral split into two triangles by a diagonal.
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Special quadrilaterals have unique properties: Parallelograms and Rhombuses have equal opposite angles; Trapeziums have at least one pair of parallel sides where co-interior angles between them sum to 180∘180^\circ.

📐Formulae

Sum of angles in a triangle: a+b+c=180∘a + b + c = 180^\circ

Sum of angles in a quadrilateral: a+b+c+d=360∘a + b + c + d = 360^\circ

Exterior angle of a triangle: ext=int1+int2ext = int_1 + int_2

Sum of interior angles of a polygon with nn sides: (n−2)×180∘(n - 2) \times 180^\circ

💡Examples

Problem 1:

In an isosceles triangle ABCABC, the angle at the vertex AA is 40∘40^\circ. Find the size of the two base angles BB and CC.

Solution:

70∘70^\circ

Explanation:

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

Problem 2:

A quadrilateral has three angles measuring 110∘110^\circ, 80∘80^\circ, and 75∘75^\circ. Calculate the size of the fourth angle.

Solution:

95∘95^\circ

Explanation:

The sum of interior angles in a quadrilateral is 360∘360^\circ. Add the known angles: 110∘+80∘+75∘=265∘110^\circ + 80^\circ + 75^\circ = 265^\circ. Subtract this sum from 360∘360^\circ: 360∘−265∘=95∘360^\circ - 265^\circ = 95^\circ.

Problem 3:

In triangle PQRPQR, the side QRQR is extended to a point SS. If angle PQR=50∘PQR = 50^\circ and angle QPR=60∘QPR = 60^\circ, find the exterior angle PRSPRS.

Solution:

110∘110^\circ

Explanation:

According to the exterior angle property, the exterior angle of a triangle is equal to the sum of the two interior opposite angles. Thus, angle PRS=anglePQR+angleQPR=50∘+60∘=110∘PRS = angle PQR + angle QPR = 50^\circ + 60^\circ = 110^\circ.

Problem 4:

In the triangle shown, one angle is a right angle and another is 35∘35^\circ. Find the value of the third angle xx.

Right-angled triangle with angles 90, 35, and x.

Solution:

x=180∘−(90∘+35∘)x = 180^\circ - (90^\circ + 35^\circ) x=180∘−125∘x = 180^\circ - 125^\circ x=55∘x = 55^\circ

Explanation:

Since the sum of angles in a triangle is 180∘180^\circ, we subtract the sum of the two known angles (the right angle 90∘90^\circ and the given 35∘35^\circ) from 180∘180^\circ.

Problem 5:

A kite has angles of 115∘115^\circ at its two opposite side vertices. If the top angle is 80∘80^\circ, find the size of the bottom angle yy.

Kite with angles 80, 115, 115, and y.

Solution:

y=360∘−(115∘+115∘+80∘)y = 360^\circ - (115^\circ + 115^\circ + 80^\circ) y=360∘−310∘y = 360^\circ - 310^\circ y=50∘y = 50^\circ

Explanation:

A kite has one pair of equal opposite angles. The interior sum of a quadrilateral is 360∘360^\circ. By subtracting the three known angles from 360∘360^\circ, we find the value of yy.