Question

Find the area of a right triangle whise sides are 14cm, (x+6) cm and (x+8)using herons formula

Answer 1

Answer:

$\begin{tabular}{|c |c | c|}\cline{1-3}\sf l/b &\sf b/l &\sf Hypotenuse \\\cline{1-3}\sf3 &\sf4 &\sf5 \\\sf5 &\sf12 &\sf13\\\sf7 &\sf24&\sf25 \\\sf8 &\sf15&\sf17\\\cline{1-3}\end{tabular}$

⠀

If it's a Right Angle Triangle, then it will must follow Pythagorean Triplet so by observing ; we can see that it will Follow Twice of [ 7, 24, 25 ] i.e. ( 14, 48, 50 )

⠀

• 14 cm
• x + 6 = 48 cm
• x + 8 = 50 cm

⠀

Value of x in the sides will be 42.

⠀

$\setlength{\unitlength}{1.5cm}\begin{picture}(6,2)\linethickness{0.4mm}\put(7.7,2.9){\large\sf{A}}\put(7.7,1){\large\sf{B}}\put(10.6,1){\large\sf{C}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\qbezier(10.5,1)(10,1.4)(8,2.9)\put(7.2,2){\sf{\large{48 cm}}}\put(8.8,0.7){\sf{\large{14 cm}}}\put(9.4,1.9){\sf{\large{50 cm}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\end{picture}$

⠀⠀⠀$\rule{160}{0.8}$

☢ $\underline{\textsf{Semi - Perimeter of the Triangle :}}$

$:\implies\sf Semi \:Perimeter=\dfrac{Sum\:of\:Sides}{2}\\\\\\:\implies\sf s = \dfrac{a + b + c}{2}\\\\\\:\implies\sf s = \dfrac{14 + 50 + 48}{2}\\\\\\:\implies\sf s = \dfrac{112}{2}\\\\\\:\implies\sf s =56$

$\rule{180}{1.5}$

☢ $\underline{\textsf{Area of the Right Angle Triangle :}}$

$\dashrightarrow\sf\:\:Area=\sqrt{s(s-a)(s-b)(s-c)}\\\\\\\dashrightarrow\sf\:\:Area=\sqrt{56(56-14)(56-50)(56-48)}\\\\\\\dashrightarrow\sf\:\:Area = \sqrt{56 \times 42 \times 6 \times 8}\\\\\\\dashrightarrow\sf\:\:Area = \sqrt{8 \times 7 \times 7 \times 6 \times 6 \times 8}\\\\\\\dashrightarrow\sf\:\:Area = 8 \times 7 \times 6\\\\\\\dashrightarrow\sf\:\:Area = 336 \:{cm}^{2}$

⠀

$\therefore\:\underline{\textsf{Hence, Area of the given Triangle is \textbf{336 cm ^\text2 }}}.$

Answer 2

$\sf{\huge{\mathfrak{\orange{\underline{Question}}}}}$

• Find the area of a right triangle whise sides are 14cm, (x+6) cm and (x+8)using herons formula

$\sf{\huge{\mathfrak{\orange{\underline{solution}}}}}$

$\setlength{\unitlength}{1.5cm}\begin{picture}(6,2)\linethickness{0.4mm}\put(7.7,2.9){\large\sf{A}}\put(7.7,1){\large\sf{B}}\put(10.6,1){\large\sf{C}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\qbezier(10.5,1)(10,1.4)(8,2.9)\put(7.2,2){\sf{\large{x + 6 cm}}}\put(8.8,0.7){\sf{\large{14 cm}}}\put(9.4,1.9){\sf{\large{x + 8 cm}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\end{picture}$

$\sf{\bold{\green{\underline{\underline{Given}}}}}$

• Triangle is a right angle triangle .
• sides are 14cm , x + 6 , x + 8 .

$\sf{\bold{\blue{\underline{\underline{To\:find}}}}}$

• area using herons formulae .

$\sf{\bold{\purple{\underline{\underline{Explanation}}}}}$

• triangle is a right angle triangle therefore verifies Pythagoras theorem .

$\sf{\bold{\red{\boxed{AC^2 = AB^2 + BC^2}}}}$

$\sf (x + 8)^2 = (x + 6)^2 + 14^2$

$\sf x^2 + 64 + 16x = x^2 + 36 + 12x + 196$

$\sf x^2 - x^2 + 16 x - 12x = 36 + 196 - 64$

$\sf 4x = 168$

$\sf x = \dfrac{168}{4}$

$\sf{\bold{\pink{\underline{\underline{x = 42cm}}}}}$

• putting value of x in sides of triangle

x + 6 = 42 + 6 = 48 cm

x + 8 = 42 + 8 = 50cm

• Herons formulae :-

$\sf{\bold{\red{\boxed{semi \:perimeter =\dfrac{ AB + BC + CA }{2}}}}}$

$\sf {semi\: perimeter = \dfrac{14 + 48 + 50 }{2}}$

$\sf{semi\: perimeter = \dfrac{112}{2}}$

$\sf{semi\: perimeter = 56cm}$

$\sf{\bold{\pink{\underline{\underline{semi\:perimeter = 56cm}}}}}$

$\sf{\bold{\red{\boxed{area = \sqrt{s(s-a)(s-b)(s-c)}}}}}$

$\sf\:\:Area=\sqrt{56(56-14)(56-50)(56-48)}$

$\sf\:\:Area = \sqrt{56 \times 42 \times 6 \times 8}$

$\sf\:\:Area = \sqrt{8 \times 7 \times 7 \times 6 \times 6 \times 8}.$

$\sf\:\:Area = 8 \times 7 \times 6$

$\sf\:\:Area = 336 \:{cm}^{2}$

$\sf{\bold{\pink{\underline{\underline{Area = 336cm^2}}}}}$

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