Question

A triangle and a parallelogram have the same base and equal areas. If the altitude of the triangle is 12 cm, find the altitude of the parallelogram.​

Answer 1

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

Let assume that base of triangle and parallelogram be x cm.

Case :- 1

Base of triangle = x cm

Height of triangle, h = 12 cm

We know,

$\boxed{ \rm{ \:Area_{(triangle)} = \frac{1}{2} \times Base \times Height \: }}$

So, on substituting the values, we get

$\rm \: \:Area_{(triangle)} = \frac{1}{2} \times x \times 12 \:$

$\rm\implies \:\rm \: \:Area_{(triangle)} = 6x \: {cm}^{2} \:$

Case :- 2

Base of parallelogram = x cm

Let height of parallelogram = h cm

We know,

$\boxed{ \rm{ \:Area_{(parallelogram)} = Base \times Height \: }}$

So, on substituting the values, we get

$\rm\implies \:\rm \: Area_{(parallelogram)} \: = \: x \times h \: {cm}^{2}$

Now, it is given that,

$\rm \: Area_{(triangle)} \: = \: Area_{(parallelogram)}$

$\rm \: 6x = x \times h$

$\rm\implies \:\boxed{ \rm{ \:h \: = \: 6 \: cm \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

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

Answer 2

Let base is b

And height is h

Area of parallelogram = Area of triangle

$b \times h = \frac{1}{2} \times b \times 12$

$h = 6 \: cm$