Area of a Triangle

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We already know that \( Area \: of \: triangle = \frac{1}{2} \times base \times altitude. \)

In the \( \Delta ABC \) , side BC is the base and AD is its height,
Now,

\begin{align*} Area \: of \: \Delta ABC &= \frac{1}{2} \times base \times height \\ &= \frac{1}{2} \times BC \times AD\\ &= \frac{1}{2} bh \: (\because BC = b \: and \: AD = h )\end{align*}

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If three sides of a triangle a, b and c are known,
area of a triangle is obtained as,
$$ \Delta = \sqrt{s( s - a ) ( s - b) ( s - c )} $$

where, \(S = \frac{a + b + c}{2}\) (semi-perimeter of triangle)

Here, We can find the area of a triangle by another formula. If two sides and angle included between them is given.

\begin{align*} Area \: of \: \Delta ABC &= \frac{1}{2} \times base \times height \\ &= \frac{1}{2} \times BC \times AD\\ &= \frac{1}{2} \times a \times h \: \: \: ........ (1)\end{align*}

From triangle ADB,

\begin{align*} Sin \: B &= \frac{AD}{AB}\\ &= \frac{h}{c}\\ or, h &= c \: sin \: B \end{align*}

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Substituting the value of h in(1) we get,

\begin{align*} Area \: of \: \Delta ABC &= \frac{1}{2} \times a \times c \: Sin \: B \end{align*}

\(\boxed{\therefore Area \: of \Delta ABC = \frac{1}{2} \: ac \: sin \: B}\)

In the above triangle ABC \(\angle B \) is acute. The above formula is valid even if \(\angle B\) is obtuse.
similarly,


When two sides b,c and \(\angle A\)of \(\Delta ABC\) are given then,

\(\boxed{\therefore Area of \Delta ABC = \frac{1}{2} \: bc \: sin \: A}\)

and when two sides a, b and \(\angle C \) are given then,

\(\boxed{\therefore Area \: of \Delta ABC = \frac{1}{2} \: ab \: sin \: C}\)

Since \(\frac{1}{2}\) ab sin C, \(\frac{1}{2}\) bc sin A and \(\frac{1}{2}\) ac sin B represent the area of same triangle ABC.

So,

\begin{align*} \frac{1}{2} ab \: sin \: C &= \frac{1}{2} bc \: sin \: A = \frac{1}{2} ac \: sin \: B \\ \frac{\frac{1}{2} ab \: sin \: C}{\frac{1}{2} abc}&= \frac{\frac{1}{2}bc \: sin \: A}{\frac{1}{2}abc} = \frac{\frac{1}{2}ac \: sin \: B}{\frac{1}{2}abc}\\ or, \frac{sin \: C}{c} &= \frac{sin \: A}{a} = \frac{sin \: B}{b}\\ or, \frac{a}{sinA} &= \frac{b}{sin B} = \frac{c}{sinC}= constant\end{align*}

This is called the sine law of trigonometry.

The base and height of a triangle must be perpendicular to each other. In each of the examples below, the base is a side of the triangle. However, depending on the triangle, the height may or may not be a side of the triangle. For example, in the right triangle in Example 2, the height is a side of the triangle since it is perpendicular to the base. In the triangles in Examples 1 and 3, the lateral sides are not perpendicular to the base, so a dotted line is drawn to represent the height.

Solution:

\(\angle PRQ = \angle PQR = 75°\\ PR = 8 cm \\ [\therefore PQ = PR] \)

\begin{align*} \angle PQR + \angle PRQ + \angle QPR &= 180°\\ \angle QPR &= 180° - 75° - 75°\\ \angle QPR &= 30°\\ Area \:of\: \Delta PQR = ? \\ Area \: of \: \Delta PQR &= \frac{1}{2}PQ \times PR \:\: sin \angle QPR \\ &= \frac{1}{2} \times 8 \times 8 \times sin30° \\ &= 32 \times \frac{1}{2}\\ &= 16 \: cm^2 \:\: _{Ans} \end{align*}

Solution:

PQ = 6 cm
QR = 8 cm
\(\angle Q = x°\)
Area of \(\Delta PQR\) = ?

\begin{align*} x+x+75+45&=180°\\ or, 2x &= 180°-120°\\ or, x &= \frac{60}{2}\\ \therefore x &= 30° \end{align*}

\begin{align*} Area \: of\: \Delta PQR &= \frac{1}{2} \times PQ \times QR \: sin \angle PQR\\ &= \frac{1}{2} \times 6 \times 8 \times sin30°\\ &= 24 \times \frac{1}{2}cm^2 \\ &= 12 \: cm^2 \:\:\:\: _{Ans}\end{align*}

Solution:

\(\angle ACB=30°\\ AC= 8 cm \)
\(\text{Area of } \Delta ABC = 24cm^2\)
BC = ?

\begin{align*} \text{Area of } \Delta ABC &= \frac{1}{2}AC\times BC \: sin \angle ACB\\ 24&=\frac{1}{2} \times 8 \times BC \times sin30°\\ 24 &= 4 \times BC \times \frac{1}{2}\\ \frac{24}{2} &= BC\\ \therefore BC &= 12 \: cm \:\:_{ans} \end{align*}

Solution:

P = 12 cm
r = 15 cm
\(\angle PQR =65°\)
Area of\(\Delta PQR\) = ?
\begin{align*} Area \: of \: \Delta PQR &= \frac{1}{2}PQ \times QR \: sin \angle PQR\\ &= \frac{1}{2} \times 15 \times 12 \times sin65°\\ &= 90 \times 0.906 \\ &= 81.54cm^2 \:\:\: _{ans} \end{align*}

Solution:

AC = 4 cm
BC = 7 cm
\(\angle A = 110° \\ \angle B = 40°\\ \angle C = 180° - 110° - 40° = 30° \)

Now,

\begin{align*} Area \: of \: \Delta ABC &= \frac{1}{2}AC \times BC \times sin \angle ACB\\ &= \frac{1}{2}\times 4 \times 7 \times sin30°\\ &= 14 \times \frac{1}{2}\\ &= 7 \: cm^2 \: \: \: _{Ans} \end{align*}

Solution:

PR = 7 cm
QR = 8 cm
\(\angle PRQ\) = ?

\begin{align*} Area \: of \: \Delta PQR &= \frac{1}{2} PR \times QR sin \angle PRQ\\ &=\frac{1}{2} \times 7 \times 8 \times sin45° \\ &= 28 \times \frac{1}{\sqrt{2}}cm^2 \\ &= 14\sqrt{2} \end{align*}

Solution:

AB = 7 cm
BC = 16 cm
\(Area \: of \: \Delta ABC = 28 \sqrt{3} cm^2\)

\begin{align*} Area \: of \: \Delta ABC &= \frac{1}{2} \times AB \times BC \times sin \angle ABC \\ or, 28 \sqrt{3} &= \frac{1}{2} \times 7 \times 16 \times sin \angle ABC\\ or, \frac{28\sqrt{3}}{7 \times 8} &= sin \angle ABC \\ or, sin \angle ABC &= \frac{\sqrt{3}}{2}\\ or, sin \angle ABC &= sin 60°\\ \therefore \angle ABC &= 60° \:\:\:\: _{ans} \end{align*}

Solution:

Area of Parallelogram ABCD = 48 \(cm^2\)
AB = 8 cm
BC = 12 cm
\(\angle ABC\) = ?
Construction: Join A and C.
\begin{align*} Area \: of \: \Delta ABC &= \frac{1}{2} ABCD \:\:\: [\because \text{Diagonal bisect a parallelogram}] \\ &= \frac{1}{2} \times 48\\ &= 24 cm^2 \end{align*}

\begin{align*} Area \: of \: \Delta ABC &= \frac{1}{2} AB \times BC \times sin \angle ABC\\ 24 &= \frac{1}{2} \times 8 \times 12 \times sin \angle ABC \\ or, \frac{24}{4 \times 12} &= sin \angle ABC \\ or, sin \angle ABC &= \frac{1}{2} \\ or, sin \angle ABC &= Sin 30°\\ \therefore \angle ABC &= 30° \:\:\:\: _{Ans} \end{align*}

Solution:

AB = 10 cm
BC = 6 cm
\(\angle ABC = 60°\)
Area of parallelogram ABCD = ?
Construction: Join AC

\begin{align*} Area \: of \: \Delta ABC &= \frac{1}{2} AB \times BC \: Sin \angle ABC \\ &= \frac{1}{2} \times 10 \times 6 \times sin60°\\ &= 30 \times \frac{\sqrt{3}}{2} cm^2 \\ &= 15\sqrt{3} \: cm^2 \: \: \: _{Ans} \end{align*}

Solution:

Ab = 6 cm
BC = 8 cm
\(\angle ABC \) = 60°
Area f ABCD = ?
Construction : Join AC

\begin{align*} Area \: of \: \Delta ABC &= \frac{1}{2} AB \times BC \times sin \angle ABC \\ &= \frac{1}{2} \times 6 \times 8 \times sin60°\\ &= 24 \times \frac{\sqrt{3}}{2}\\ &= 12\sqrt{3} \:\:\:\: cm^2 \\ \\ Area \: of \: parallelogram\: ABCD &= 2 Area \: of\: \Delta ABC \\ &= 2 \times 12\sqrt{3} \\ &=24\sqrt{3} \:cm^2 \:\:\:_{ans} \end{align*}

Solution:

\(\angle CBP =60°\) \(\:\:\:\:\: \) [\(\because\) Angle of equilateral]
\(\angle BAD = \angle CBP = 60° \:\:\:\:\: [\because\) Corresponding angles ]
Construction : Join points B and D
AB = AD = CD = BC = BP = PC

\begin{align*} Area \: of\: \Delta ABD &= \frac{1}{2} AD \times AB \: Sin60°\\ &= \frac{1}{2} AD \times AB \times \frac{\sqrt{3}}{2} \\ &= \frac{\sqrt{3}}{4} AB^2\\ Area \: of \: \Delta BCD &= Area \: f \: \Delta ABD \\ &= \frac{\sqrt{3}}{4} AB^2\:\:\:\:\: [\because BD \: bisect \: the \: rhombus \: ABCD ]\\ Area \: of\: \Delta BCP &= \frac{1}{2} \times BC \times BP \times Sin60°\\ &= \frac{1}{2} \times AB \times AB \times \frac{\sqrt{3}}{4}\\ &= \frac{\sqrt{3}}{4} AB^2 \\ Area \: of \: \Delta APCD &= Area \: of\: \Delta ABD + Area \: of\: \Delta BCD + Area \: of\: \Delta BCP\\ 27\sqrt{3} &= \frac{\sqrt{3}}{4} AB^2 + \frac{\sqrt{3}}{4} AB^2 + \frac{\sqrt{3}}{4} AB^2 \\ or, 27\sqrt{3} &= \frac{\sqrt{3}AB^2 +\sqrt{3}AB^2+\sqrt{3}AB^2}{4} \\ or, 27\sqrt{3} &= \frac{3\sqrt{3}AB^2}{4}\\ or, AB^2 &= \frac{27\sqrt{3}\times 4}{3\sqrt{3}}\\ or, AB^2 &= 36\\ or, AB^2 &= 6^2\\ \therefore AB &= 6 \: cm \end{align*}

Solution:

XY = YZ = 12 cm
\(\angle XYZ = 30°\)
Area of \(\Delta XYZ\) = ?

\begin{align*} Area \: of \: \Delta XYZ &= \frac{1}{2} \times XY \times YZ \times sin \angle XYZ \\ &= \frac{1}{2} \times 12 \times 12 \times sin30° \\ &= 6 \times 12 \times \frac{1}{2} \\ &= 36 \: cm^2 \:\:\: _{Ans} \end{align*}

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  • What is the area of ( riangle)PQR whose p = 12 cm, r = 15 cm and  (measuredangle)PQR = 65° (sin 65° = 0.906)

    80.56 cm2
    82.56 cm2
    83.56 cm2
    81.56 cm2
  • In ( riangle)ABC, the sides a = 4 cm, b = 6 cm and (angle)C = 60°. Find the area of ( riangle)ABC.

    none
    6(sqrt{5})
    6(sqrt{4})
    6(sqrt{3})
  • What is the formulae of area of triangle, if base and height of a triangle is given?

    Area of ( riangle) = (frac{1}{2}) (p×h)
    Area of ( riangle) = (frac{1}{2}) (b×h)
    Area of ( riangle) = (frac{1}{3}) (b×h)
    Area of ( riangle) = (frac{1}{}) (p×h)
  • What is the formulae of area of triangle, if three sides of triangle is given?

    both
    none
    (sqrt{s(s-a) (s-b) (s-c)})
    (sqrt{(s-a) (s-b) (s-c)})
  • What is the formulae of the semiperimeter of the triangle?

    (frac{a+d+c}{2})
    (frac{a+b+d}{2})
    (frac{a+b+c}{2})
    none
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