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Geometric Twins - Angles of Isosceles and Equilateral Triangles

Grade 7CBSE

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

🔑Concepts

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An Isosceles triangle is a triangle with at least two equal sides. The angles opposite to these equal sides are also equal. These are known as the base angles, while the third angle is the vertex angle.

Isosceles triangle ABC where sides AB and AC are equal and base angles B and C are marked as x.
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An Equilateral triangle has all three sides equal. Consequently, all three interior angles are equal. Since the sum of angles in a triangle is 180∘180^\circ, each angle must be 60∘60^\circ.

Equilateral triangle with all three interior angles labeled as 60 degrees.
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In an isosceles triangle, if the vertex angle is known, the base angles can be found using the formula: Base Angle=180∘−Vertex Angle2\text{Base Angle} = \frac{180^\circ - \text{Vertex Angle}}{2}.

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The Angle Sum Property states that the sum of all internal angles of any triangle (including isosceles and equilateral) is always 180∘180^\circ.

📐Formulae

∠A+∠B+∠C=180∘\angle A + \angle B + \angle C = 180^\circ

If AB=AC in △ABC, then ∠B=∠C\text{If } AB = AC \text{ in } \triangle ABC, \text{ then } \angle B = \angle C

Each angle of an equilateral triangle=60∘\text{Each angle of an equilateral triangle} = 60^\circ

Base angle of Isosceles Triangle=180∘−Vertex Angle2\text{Base angle of Isosceles Triangle} = \frac{180^\circ - \text{Vertex Angle}}{2}

💡Examples

Problem 1:

In an isosceles triangle, the vertex angle is 70∘70^\circ. Find the measure of the base angles.

Solution:

Let the triangle be △ABC\triangle ABC with vertex angle ∠A=70∘\angle A = 70^\circ. Let the base angles be ∠B\angle B and ∠C\angle C. Since it is an isosceles triangle, ∠B=∠C=x\angle B = \angle C = x. Using the angle sum property: 70∘+x+x=180∘70^\circ + x + x = 180^\circ 70∘+2x=180∘70^\circ + 2x = 180^\circ 2x=180∘−70∘2x = 180^\circ - 70^\circ 180−70110\begin{array}{r} 180 \\ - 70 \\ \hline 110 \end{array} 2x=110∘2x = 110^\circ x=110∘2=55∘x = \frac{110^\circ}{2} = 55^\circ. So, the base angles are 55∘55^\circ each.

Explanation:

We use the property that base angles of an isosceles triangle are equal and then apply the Angle Sum Property (180∘180^\circ) to solve for the unknown.

Problem 2:

One of the base angles of an isosceles triangle is 45∘45^\circ. Determine the measure of the vertex angle.

Solution:

Let the base angles be ∠B\angle B and ∠C\angle C. Given ∠B=45∘\angle B = 45^\circ. Since base angles are equal, ∠C=45∘\angle C = 45^\circ. Let the vertex angle be VV. V+45∘+45∘=180∘V + 45^\circ + 45^\circ = 180^\circ V+90∘=180∘V + 90^\circ = 180^\circ V=180∘−90∘V = 180^\circ - 90^\circ 180−9090\begin{array}{r} 180 \\ - 90 \\ \hline 90 \end{array} The vertex angle is 90∘90^\circ.

Explanation:

Since two angles are 45∘45^\circ each, their sum is 90∘90^\circ. Subtracting this from 180∘180^\circ gives the vertical angle.

Problem 3:

If a triangle is equilateral and one of its angles is given as (x+20)∘(x + 20)^\circ, find the value of xx.

Solution:

We know that every angle in an equilateral triangle is 60∘60^\circ. Therefore, we can set up the equation: x+20=60x + 20 = 60 x=60−20x = 60 - 20 60−2040\begin{array}{r} 60 \\ - 20 \\ \hline 40 \end{array} Thus, x=40x = 40.

Explanation:

Because all angles in an equilateral triangle are equal to 60∘60^\circ, any expression representing an angle must equal 6060.

Problem 4:

In △PQR\triangle PQR, PQ=PRPQ = PR and ∠Q=55∘\angle Q = 55^\circ. Find the measure of ∠P\angle P.

Isosceles triangle PQR with PQ=PR and angle Q marked as 55 degrees.

Solution:

  1. Since PQ=PRPQ = PR, △PQR\triangle PQR is an isosceles triangle.
  2. In an isosceles triangle, angles opposite to equal sides are equal. Therefore, ∠R=∠Q=55∘\angle R = \angle Q = 55^\circ.
  3. Using the Angle Sum Property: ∠P+∠Q+∠R=180∘\angle P + \angle Q + \angle R = 180^\circ.
  4. ∠P+55∘+55∘=180∘\angle P + 55^\circ + 55^\circ = 180^\circ.
  5. ∠P+110∘=180∘\angle P + 110^\circ = 180^\circ.
  6. ∠P=180∘−110∘=70∘\angle P = 180^\circ - 110^\circ = 70^\circ.

Explanation:

We identify the triangle as isosceles because two sides are equal. We equate the base angles and then subtract their sum from 180∘180^\circ to find the vertex angle.

Problem 5:

Find the value of xx in the given equilateral triangle LMNLMN where one angle is represented as (2x−10)∘(2x - 10)^\circ.

Equilateral triangle LMN with one angle at the base labeled as 2x - 10.

Solution:

  1. In an equilateral triangle, every interior angle is equal to 60∘60^\circ.
  2. Therefore, we can set up the equation: 2x−10=602x - 10 = 60.
  3. 2x=60+102x = 60 + 10.
  4. 2x=702x = 70.
  5. x=702=35x = \frac{70}{2} = 35.

Explanation:

Since all angles in an equilateral triangle are 60∘60^\circ, any algebraic expression representing an angle can be equated to 6060 to solve for the variable.