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Lines & Triangles

Lines And Triangles

MODULES

Lines & Angles
Parallel Lines
Basics of Triangles
Types of Triangles
Triangle & its Segments
Area of a Triangle
Isosceles & Equilateral
Right-angled Triangles
Other Theorems
Congruency & Similarity
Past Questions

CONCEPTS & CHEATSHEET

Concept Revision Video

SPEED CONCEPTS

Lines & Triangles 1
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Lines & Triangles 2
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Lines & Triangles 3
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PRACTICE

Lines & Triangles : Level 1
Lines & Triangles : Level 2
Lines & Triangles : Level 3
ALL MODULES

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=21​bh

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)}=s(s−a)(s−b)(s−c)​

where
a,ba, ba,b and ccc are the lengths of the sides of the triangle and sss is the semi-perimeter, i.e.s=a+b+c2s = \dfrac{ a + b + c}{2}s=2a+b+c​

Area =12absinθ= \dfrac{1}{2} a b sin \theta=21​absinθ

where
θ\thetaθ is the angle subtended by sides aaa and bbb.

Area =r×s= r \times s=r×s

where
rrr is the inradius and sss is the semi-perimeter, i.e. s=a+b+c2s = \dfrac{a + b + c}{2}s=2a+b+c​

Area =abc4R= \dfrac{abc}{4\mathrm{R}}=4Rabc​ where a,ba, ba,b and ccc 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 \space17.05       (2) 27.85      27.85 \space \space \space\space \space \space27.85       (3) 22.45         22.45 \space\space \space\space \space \space\space \space \space22.45          (4) 32.25      32.25 \space\space \space\space \space \space32.25       (5) 26.25   26.25 \space \space \space26.25   

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}21​ ×\times× base ×\times× height =abc4R= \dfrac{abc}{4 \mathrm{R}}=4Rabc​ .

⇒ 12×a×3=a×9×17.54R\dfrac{1}{2} \times a \times 3 = \dfrac{a \times 9 \times 17.5}{4 \mathrm {R}}21​×a×3=4Ra×9×17.5​

aaa gets cancelled on both sides.

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


Answer: (
555) 26.2526.2526.25


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
∠\angle∠CAB, ∠\angle∠ABC and ∠\angle∠BCA, and a,ba, ba,b and ccc 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}}sinAa​=sinBb​=sinCc​

Cosine Rule: 

cos A
=b2+c2−a22bc= \dfrac{ b^2 + c^2 - a^2}{2bc}=2bcb2+c2−a2​

cos B
=c2+a2−b22ca= \dfrac{ c^2 + a^2 - b^2}{2ca}=2cac2+a2−b2​

cos C
=a2+b2−c22ab= \dfrac{ a^2 + b^2 - c^2}{2ab}=2aba2+b2−c2​


Example 14

In △\triangle△ ABC, AC === 8 cm, BC === 3 cm and ∠\angle∠ACB =60°= 60\degree=60°. 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}60°=2×8×382+32−x2​

⇒
12=73−x248\dfrac{1}{2} = \dfrac{73 - x^2}{48}21​=4873−x2​

⇒
x2=73−24x^2 = 73 -24x2=73−24
⇒
x=x =x= 7 cm


Answer: 777


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