Question

Find the area of a triangle whose sides are 5 cm ,12 cm and 13 cm . Also find the length of its altitude corresponding to the longest side. ​

Answer 1

• Given

• Sides of a triangle -

• Length = 5 cm
• Breadth = 12 cm
• Height = 13 cm

• To find

• Area of triangle.
• Length of its altitude corresponding to the longest side.

• Solution

To find area of triangle. We will use Heron's formula.

• Heron's formula -

$\red{ \bigstar} \boxed{\sf \green{Heron's \: formula \: = \sqrt{s(s - a)(s - b)(s - c)}}}$

• To find s -

$\red{ \bigstar} \boxed{\sf \green{s = \frac{a + b + c}{2}}}$

Substituting the given values -

$\sf{s = \dfrac{5 + 12 + 13}{2} }$

$\sf{s = \dfrac{30}{2} }$

$\sf{s = \cancel{\dfrac{30}{2}}}$

$\blue{\bigstar}$ s = 15

$: \implies\sf{Heron's \: formula \: = \sqrt{s(s - a)(s - b)(s - c)}}$

$: \implies\sf{ \sqrt{15(15 - 5)(15 - 12)(15 - 13)}}$

$: \implies\sf{ \sqrt{15(10)(3)(2)}}$

$: \implies\sf{ \sqrt{900}}$

$: \implies\sf{30}$

$\green{\bigstar}$ $\large\boxed{\sf{\red{Area~of~triangle~= 30~cm^2}}}$

━━━━━━━━━━━━━━━━━━━━━━

(II) Length of its altitude corresponding to the longest side.

• Longest side = 13 cm

Area = $\sf{\dfrac{1}{2}}$ × base × height

• $\implies$ $\sf{\dfrac{1}{2} \times 13 \times height}$

$\green{\bigstar}$ $\large\boxed{\sf{\red{Length~of~its~altitude~corresponding~to~the~longest~side= 4.615~cm}}}$

━━━━━━━━━━━

• Know more -

There three types of triangle -

• Equilateral triangle
• Isosceles triangle
• Scalene triangle 

Equilateral triangle -

• In this all three sides are equal.

• Area of equilateral triangle -

↬ $\sf{\dfrac{\sqrt{3}}{4}a^2}$

• Perimeter of equilateral triangle -

↬ 3a

where,

a = side of triangle

Isosceles triangle -

• In this two sides of triangle are equal.

• Area of isosceles triangle -

↬ $\sf{\dfrac{1}{2} \times base \times height}$

• Perimeter of isosceles triangle -

↬ 2a + b

Scalene triangle -

• In this no side is equal.

• Area of scalene triangle -

↬ Heron's formula = $\sf{\sqrt{s(s - a)(s - b)(s - c)}}$

• Perimeter of scalene triangle -

↬ Perimeter = Sum of all three sides

Answer 2

GivEn:

• Sides of ∆ = 5 cm, 12 cm and 13 cm

⠀⠀⠀⠀⠀⠀⠀

To find:

• Area of the triangle.
• Length of altitude corresponding to the largest side of the ∆.

⠀⠀⠀⠀⠀⠀⠀

Solution:

$\frak{Here} \begin{cases} & \sf{a = 5\;cm} \\ & \sf{b = 12\;cm} \\ & \sf{c = 13\;cm} \end{cases}$

✇ Finding semi - perimeter of ∆,

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf s = \dfrac{a + b + c}{2}$

$:\implies\sf s = \dfrac{5 + 12 + 13}{2}$

$:\implies\sf s = \cancel{\dfrac{30}{2}}$

$:\implies\bf s = 15\;cm$

✇ Now, Finding area of triangular field using Heron's Formula,

⠀⠀⠀⠀⠀⠀⠀

$\star\;{\boxed{\sf{\purple{Area = \sqrt{s(s - a)(s - b)(s - c)}}}}}$

$:\implies\sf \sqrt{15 (15 -5 )(15 - 12)(15 - 13)}$

$:\implies\sf \sqrt{15 \times 10 \times 3 \times 2}$

$:\implies\sf \sqrt{900}$

$:\implies{\boxed{\frak{\pink{30\;cm^2}}}}\;\bigstar$

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

✇ Now, Finding length of altitude corresponding to the largest side of the ∆.

⠀⠀⠀⠀⠀⠀⠀

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\linethickness{0.3mm}\qbezier(1, 0)(1,0)( 3,3)\qbezier(5,0)(5, 0)(3,3)\qbezier(5,0)(1,0)(1,0)\qbezier(1,0)(1,0)(4,1.5)\put(2.8,3.2){ \sf A }\put(2.5, - 0.3){5 cm}}\put(0.5,-0.2){\sf B}\put(4.5,1){12 cm}\put(5.1, - 0.3){\sf C}\put(1.1,1.8){13 cm}\put(4.2, 1.5){\sf D}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\boxed{ \bf @ValiantPrinceiss }}\end{picture}$

⠀

Here,

Largest side of triangular field is = 13 cm.

⠀⠀⠀⠀⠀⠀⠀

$\star\;{\boxed{\sf{\purple{Area = \dfrac{1}{2} \times (base) \times (height)}}}}$

$:\implies\sf 30 = \dfrac{1}{2} \times {13} \times (height)$

$:\implies{\boxed{\frak{\pink{Height = 4.615\;cm}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Hence,\;length\;of\; altitude\;is\; \bf{4.615\;cm}.}}}$