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CAT 2025 Lesson : Lines & Triangles - Area of a Triangle

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3.4 Area of a Triangle

Following are the five ways to compute the area of a triangle.

Area Figure
Area =12bh= \dfrac{1}{2} b h

where b is the base and h is the height or altitude of the triangle.
Area =s(s−a)(s−b)(s−c)= \sqrt{s(s - a)(s - b)(s - c)}

where a,ba, b and cc are the lengths of the sides of the triangle and ss is the semi-perimeter, i.e.s=a+b+c2s = \dfrac{ a + b + c}{2}
Area =12absinθ= \dfrac{1}{2} a b sin \theta

where θ\theta is the angle subtended by sides aa and bb.
Area =r×s= r \times s

where rr is the inradius and ss is the semi-perimeter, i.e. s=a+b+c2s = \dfrac{a + b + c}{2}
Area =abc4R= \dfrac{abc}{4\mathrm{R}} where a,ba, b and cc are the lengths of the sides of the triangle and R is the circumradius.

Example 13

In a triangle ABC, the lengths of the sides AB and AC equal 17.5 cm and 9 cm respectively. Let D be a point on the line segment BC such that AD is perpendicular to BC. If AD == 3 cm, then what is the radius (in cm) of the circle circumscribing the triangle ABC?
[CAT 2008]

(1) 17.05      17.05 \space \space \space\space \space \space (2) 27.85      27.85 \space \space \space\space \space \space (3) 22.45         22.45 \space\space \space\space \space \space\space \space \space (4) 32.25      32.25 \space\space \space\space \space \space (5) 26.25   26.25 \space \space \space

Solution

As the height and 2 sides of the triangle are given, and we are required to find the circumradius, we could equate two ways of computing the area of a triangle – 12\dfrac{1}{2} ×\times base ×\times height =abc4R= \dfrac{abc}{4 \mathrm{R}} .

⇒ 12×a×3=a×9×17.54R\dfrac{1}{2} \times a \times 3 = \dfrac{a \times 9 \times 17.5}{4 \mathrm {R}}

aa gets cancelled on both sides.

⇒ 6R=6 \mathrm {R}= 9 ×\times 17.5 ⇒ R=\mathrm {R}= 26.25

Answer: (55) 26.2526.25


3.5 Sine & Cosine Rules

Let us discuss some formulae linking triangles to trigonometry. In the below figure, A, B and C are the interior angles ∠\angleCAB, ∠\angleABC and ∠\angleBCA, and a,ba, b and cc are sides opposite these angles respectively.

Sine Rule: asinA=bsinB=csinC\dfrac{a}{\mathrm {sin A}} = \dfrac{b}{\mathrm {sin B}} = \dfrac{c}{\mathrm {sin C}}

Cosine Rule: 

cos A =b2+c2−a22bc= \dfrac{ b^2 + c^2 - a^2}{2bc}

cos B =c2+a2−b22ca= \dfrac{ c^2 + a^2 - b^2}{2ca}

cos C =a2+b2−c22ab= \dfrac{ a^2 + b^2 - c^2}{2ab}

Example 14

In △\triangle ABC, AC == 8 cm, BC == 3 cm and ∠\angleACB =60°= 60\degree. What is the length of AB (in cm)?

Solution

As two of the sides and one of the angles are known, we can apply the cosine rule.

Cos 60°=82+32−x22×8×360\degree = \dfrac{ 8^2 + 3^2 - x^2}{2 \times 8 \times 3}

⇒ 12=73−x248\dfrac{1}{2} = \dfrac{73 - x^2}{48}

⇒ x2=73−24x^2 = 73 -24
⇒ x=x = 7 cm

Answer: 77


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