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CAT 2025 Lesson : Lines & Triangles - Isosceles & Equilateral

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3.6 Types of Triangles – Sides

Definition & Features Figure
Scalene: A triangle where length of each side and each of the angles are different.
Isosceles: A triangle where two sides and the two angles opposite to them are equal. In △\triangle ABC, AD is the median, drawn from the vertex bound by the equal sides. AD is also the altitude, angle bisector and perpendicular bisector. This does not apply for the other two medians.

Area = b44a2−b2= \dfrac{b}{4}\sqrt{4a^2 - b^2}
where a is the length of the equal sides

Equilateral : A triangle where all sides and angles are equal. The angles are equal to 60°60\degree A median drawn from any vertex is also its altitude, angle bisector and perpendicular bisector, as shown. In ∆ABC, AD, BE, CF are the medians, altitudes, angle bisectors and perpendicular bisectors.

For a given perimeter, the area of a triangle is maximised when all its sides are equal, i.e. an equilateral triangle. Likewise, for a given area, perimeter is minimised when the triangle is equilateral.
In an equilateral triangle where a is the length of a side,

Area = 34a2 =  \dfrac{ \sqrt{3}}{4}a^2; Height == h =32a= \dfrac{ \sqrt{3}}{2}a

Inradius = 13×=  \dfrac{1}{3} \times h =a23= \dfrac{a}{2 \sqrt{3}}; Circumradius:  23×\dfrac{2}{3} \times h =a3= \dfrac{a}{ \sqrt{3}}

Example 15

In △\triangle ABC, AB == AC, BC =103= 10 \sqrt{3} cm and ∠\angleBAC =120°= 120\degree. What is the area of △\triangle ABC (in cm2^2)?

Solution

AD is the median of △\triangle ABC. As AB = AC, AD is also the angle bisector and altitude.

In right-angled △\triangle ADC,

sin 60°60\degree =DCAC = \mathrm {\dfrac {DC} {AC}}⇒ 32=53AC\dfrac{\sqrt{3}}{2} = \dfrac{5\sqrt{3}}{\mathrm{AC}}⇒ AC == 10

Using Pythagoras theorem,
AD =102−(53)2=25=5= \sqrt{ 10^2 - ( 5 \sqrt{3})^2}= \sqrt{25} = 5

Where AB == BC =a= a, AC =b= b and the height AD =h= h, area can be computed in following 3 ways.
Area =12×b×h=12×103×5=253= \dfrac{1}{2} \times b \times h = \dfrac{1}{2} \times 10\sqrt{3} \times 5 = 25 \sqrt{3}

Area of isosceles triangle = b44a2−b2= \dfrac{b}{4}\sqrt{4a^2 - b^2} = 10344×102−(103)2= \dfrac{10\sqrt{3}}{4}\sqrt{4\times 10^2 - (10\sqrt{3})^2} =253= 25\sqrt{3}

Area =12×a×a×sinθ=12×10×10×sin120°=253= \dfrac{1}{2} \times a \times a \times sin\theta= \dfrac{1}{2} \times 10 \times 10 \times sin120\degree = 25\sqrt{3}

Answer: 25325\sqrt{3}

Note: Trigonometry is covered in the last lesson under Geometry.


Example 16

In an equilateral triangle ABC, whose length of each side is 3 cm, D is the point on BC such that BD =12= \dfrac{1}{2} CD. What is the length of AD?

(1) 5 cm     \sqrt{5} \space \mathrm{ cm} \space \space \space\space \space (2) 6 cm     \sqrt{6} \space \mathrm{ cm} \space \space \space\space \space (3) 7 cm         \sqrt{7} \space \mathrm{ cm} \space\space \space\space \space \space\space \space \space (4) 8 cm      \sqrt{8} \space \mathrm{ cm} \space\space \space\space \space \space (5) None of the above    \space \space \space

Solution

In equilateral △\triangle ABC, AE is the ⊥\bot bisector.
DE == BE −- BD == 1.5 −- 1 == 0.5
The height AE =3a2=32= \dfrac{\sqrt{3}a}{2} = \dfrac{\sqrt{3}}{2}.

In right-angled △\triangleAED, AD =AE2+DE2= \mathrm{\sqrt{AE^2 + DE^2}}

AD =274+14=7= \sqrt{\dfrac{27}{4} + \dfrac{1}{4}} = \sqrt{7}

Answer: (33) 7 cm\sqrt{7} \space \mathrm{ cm}


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