Question

A triangle and a parallelogram have the same base and same area. If the sides of the
triangle are 15 cm, 14 cm and 13 cm and the parallelogram stands on the base 15 cm,
find the height of the parallelogram

Answer 1

AnswEr:-

Height of parallelogram = 5.6 cm

$\rule{200}{1}$

⋆ DIAGRAM OF TRIANGLE:-

$\setlength{\unitlength}{0.78 cm}\begin{picture}(12,4)\thicklines\put(5.4,5.8){ A }\put(11.2,5.8){ B }\put(8.4,10){ C }\put(6,6){\line(2,3){2.5}}\put(11,6){\line(-2,3){2.5}}\put(6,6){\line(1,0){5}}\put(5,7.9){ 15\:cm }\put(11,7.9){ 14\:cm }\put(8,5){ 13\:cm }\end{picture}$

Now using Heron's formula for finding area of triangle.

⋆ HERON'S FORMULA:-

$\star\:\large{\boxed{\rm{\blue{Area \: of \: triangle = \sqrt{s(s - a)(s - b)(s - c)} }}}}$

Now,s = semi-perimeter.

↠ s = (15 + 14 + 13)/2

↠ s = 42/2

↠ s = 21 cm

∴ Semi-perimeter of triangle = 21 cm

Now area:-

$\dashrightarrow\sf Area_{\triangle} = \sqrt{21(21 - 15)(21 - 14)(21 - 13)} \\\\\dashrightarrow\sf Area_{\triangle} = \sqrt{21 \times 6 \times 7 \times 8} \\\\\dashrightarrow\sf Area_{\triangle} = \sqrt{84 \times 84}\\\\\dashrightarrow\underline{\boxed{\rm{\red{Area_{\triangle} = 84 \: cm^2 }}}}$

∴ Area of triangle = 84 cm²

Now given condition:-

➩Area of triangle= Area of parallelogram

Given:-

• Base = 15 cm
• Area of parallelogram = 84 cm²

Now we know,

$\star\:\large{\boxed{\rm{\green{Area \: of \: parallelogram = Base \times Height }}}}$

According to question:-

$\twoheadrightarrow\sf 84 = 15 \times Height \\\\\twoheadrightarrow\sf Height = \dfrac{84}{15} \\\\\twoheadrightarrow{\underline{\boxed{\textsf{\textbf{Height = 5.6 cm }}}}}$

∴ Height of parallelogram = 5.6 cm

Answer 2

Answer:

⋆ TRAINGLE :

$\setlength{\unitlength}{1cm}\begin{picture}(6,8)\thicklines\put(1, .5){\line(2, 1){3}}\put(4, 2){\line(-2, 1){2}}\put(2, 3){\line(-2, -5){1}}\put(.7, .3){ A }\put(4.05, 1.9){ B }\put(1.7, 2.95){ C }\put(3.2, 2.5){ 13 cm }\put(0.6,1.7){ 14 cm }\put(2.7, 1.05){ 15 cm }\end{picture}$

$\underline{\bigstar\:\:\textsf{Semi Perimeter :}}$

$\longrightarrow\sf Semi\: Perimeter=\dfrac{Sum\:of\:Sides}{2}\\\\\\\longrightarrow\sf s=\dfrac{(15+14+13)}{2}\\\\\\\longrightarrow\sf s=\dfrac{42}{2}\\\\\\\longrightarrow\sf s=21$

$\rule{150}{1}$

$\underline{\bigstar\:\:\textsf{Area of Triangle :}}$

$:\implies\tt Area=\sqrt{s(s-a)(s-b)(s-c)}\\\\\\:\implies\tt Area = \sqrt{21 \times (21 - 15) \times (21 - 14) \times (21 - 13)}\\\\\\:\implies\tt Area = \sqrt{21 \times 6 \times 7 \times 8}\\\\\\:\implies\tt Area = \sqrt{7 \times 3 \times 3 \times 2 \times 7 \times 4 \times 2}\\\\\\:\implies\tt Area =7 \times 2 \times 3 \times 2\\\\\\:\implies\tt Area =84 \:{cm}^{2}$

$\rule{200}{2}$

⋆ PARALLELOGRAM :

$\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\thicklines\put(8.6,3){\large{A}}\put(7.7,0.9){\large{B}}\put(9.5,0.7){\sf{\large{15 cm}}}\put(11.1,0.9){\large{C}}\put(8,1){\line(1,0){3}}\put(11,1){\line(1,2){1}}\put(9,3){\line(3,0){3}}\put(9.1,2){\bf{\large{Area = 84 cm^2 }}}\put(9,1){\line(0,1){2}}\put(8,1){\line(1,2){1}}\put(12.1,3){\large{D}}\end{picture}$

$\underline{\bigstar\:\:\textsf{Area of Parallelogram :}}$

$\dashrightarrow\tt\:\:Area_{Parallelogram}=Area_{Triangle}\\\\\\\dashrightarrow\tt\:\:Base \times Height= 84\:cm^2\\\\\\\dashrightarrow\tt\:\:15\:cm \times Height=84\:cm^2\\\\\\\dashrightarrow\tt\:\: Height=\dfrac{84\:cm^2}{15\:cm}\\\\\\\dashrightarrow\:\:\underline{\boxed{\textsf{\textbf{Height = 5.6 cm}}}}$

⠀

$\therefore\:\underline{\textsf{Height of Parallelogram will be \textbf{5.6 cm}}}.$