Question

8. A triangle and a parallelogram have the same
area. If the sides of the triangles are 10 cm,
17 cm and 21 cm and the parallelogram stands
on the base 12 cm, find the height of the parallelogram

​

Answer 1

Answer:

$a = 10 \\ b = 17 \\ c = 21 \\ s = \frac{a + b+ c}{2} \\ \ = \frac{10 + 17 + 21}{2 } \\ = \frac{48}{2 } = 24 \\ \\ area \: of \: triagle = \sqrt{s( s- a)(s -b)(s -c } \\ = \sqrt{24(24 - 10)(24 - 17)(24 - 21)} \\ = \sqrt{24 \times 14 \times 7 \times 3} \\ = \sqrt{4 \times 3 \times 2 \times 2 \times 7 \times 7\times 3} \\ = 2 \times 2 \times 3 \times 7 \\ = 84$

$b = 21 \\ area \: of \: || gm \: = 84\\ b \times h = 84 \\ 21 \times h = 84\\ h = \frac{84}{21} \\ h= 4$

Answer 2

${\blue{\underline{\underline{\purple{\textsf{\textbf{Answer : }}}}}}}$

➙ Height of the ||gm is 14cm ${\boxed{\green{\checkmark{}}}}$

${\blue{\underline{\underline{\purple{\textsf{\textbf{Step-by-step explanation: }}}}}}}$

Question says that, Area of a triangle is equal to the area of a parallelogram.

Step 1 : Finding the semi-perimeter :

Given three sides of the triangle is :

➪ a = 10cm.

➪ b = 17cm.

➪ c = 21cm.

Here, to find the area we need to calculate the semi-perimeter first.

Semi-perimeter : $\sf \dfrac{a + b + c}{2}$

$\sf \to \: s = \frac{10 + 17 + 21}{2}$

$\sf \to \: s = \frac{48}{2}$

$\sf \to \: s = 24cm$

Step 2 : Finding the area of the triangle :

Area of a triangle is given by :

△ Herons formula :

$\sf \longrightarrow \sqrt{s(s - a)(s - b)(s - c)}$

Here, s is the semi-perimeter i.e., 24cm.

So,

$\sf \longrightarrow \sqrt{24(24 - 10)(24 - 17)(24 - 21)}$

$\sf \longrightarrow \sqrt{24(14)(7)(3)}$

$\sf \longrightarrow \sqrt{7,056}$

$\sf \longrightarrow 84cm^2$

${ \underline{ \sf {\therefore{A rea \: of \: the \triangle \: is { \textsf{ \textbf{84cm}}^{2} }}}}}$

Step 3 : Finding the height of the parallelogram :

Given that,

Area of the △ = Area of the ||gm

$\begin{gathered} { \textsf{ \textbf{here}}} \begin{cases} \sf{area \: of \: the \: \triangle \to {84cm}^{2} } \\ {\sf{area \: of \: the \: parallelogram \to \frac{1}{2}}} {\times b \times h } \\ b{ \sf{(base ) = 12cm(given)\:and }} \: h \: \sf{is \: the \: height}\\ \end{cases}\end{gathered}$

So,

$\implies 84 = \dfrac{1}{2} \times 12 \times h$

$\implies \dfrac{84 \times 2}{12} = h$

$\implies 14 = h$

${ \underline{ \sf{ \therefore{ the \: height \: of \: the \: paralellogram \: is { \textsf{ \textbf{14cm}}}}}}}$