Question

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

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

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

Let assume that the required triangle be ABE and sides of triangle be denoted by a, b, c respectively.

So, a = 13 cm, b = 14 cm and c = 15 cm

Now, Semi - perimeter (s) of triangle is given by

$\boxed{\sf{  \:\rm \: s \:= \: \frac{a + b + c)}{2} \: \: }}$

So, on substituting the values of a, b, c, we get

$\rm \: s \: = \: \dfrac{13 + 14 + 15}{2}$

$\rm \: s \: = \: \dfrac{42}{2}$

$\rm\implies \:s \: = \: 21 \: cm$

Now, area of triangle using Heron's Formula is given by

$\boxed{\sf{  \:\rm \: Area_{\triangle} \: = \: \sqrt{s(s - a)(s - b)(s - c)} \: }}$

So, on substituting the values, we get

$\rm \: Area_{\triangle}$

$\rm \: =  \: \sqrt{21(21 - 13)(21 - 14)(21 - 15)}$

$\rm \: =  \: \sqrt{21(8)(7)(6)}$

$\rm \: =  \: \sqrt{7 \times 3 \times 2 \times 2 \times 2 \times 7 \times 3 \times 2}$

$\rm \: =  \: 7 \times 3 \times 2 \times 2$

$\rm \: =  \:84 \: {cm}^{2}$

Now, it is given that A triangle and a parallelogram have the same base and the same area, the parallelogram stands on the base 14cm. Let assume that Height of parallelogram to the corresponding base is h cm.

We know,

$\rm \: Area_{ { \parallel}^{gm} } \: = \: base \: \times \: height$

$\rm \: Area_{ { \parallel}^{gm} } \: = \: 14 \: \times \: h \: {cm}^{2}$

So, as we have

$\rm \: Area_{\triangle} \: = \: Area_{ { \parallel}^{gm} }$

$\rm \:  \: 84 = 14 \times h$

$\rm \: h \: = \: \dfrac{84}{14}$

$\rm\implies \:h \: = \: 6 \: cm$

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$\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$

Answer 2

Step-by-step explanation:

Answer:

$\boxed{\huge{\mathbb\pink{REFERR \:TO\: THE\:\: ATTACHMENT }}}$

Step-by-step explanation:

hope it help you.

thanks