Question

In a right angled triangle, the side opposite to the right angle is 10 cm .One of the perpendicular side is 6cm .The area of this triangle is ____sq.c ​

Answer 1

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

Given that,

• In a right angled triangle, the side opposite to the right angle is 10 cm.

• One of the perpendicular side is 6cm.

Let assume that

• The required triangle be ABC right angled at B such that, AC = 10 cm and BC = 6 cm.

Now, In right triangle ABC

Using Pythagoras Theorem, we have

$\rm :\longmapsto\: {AC}^{2} = {AB}^{2} + {BC}^{2}$

$\rm :\longmapsto\: {10}^{2} = {AB}^{2} + {6}^{2}$

$\rm :\longmapsto\: 100 = {AB}^{2} + 36$

$\rm :\longmapsto\: 100 - 36 = {AB}^{2}$

$\rm :\longmapsto\: 64 = {AB}^{2}$

$\rm :\longmapsto\: {8}^{2} = {AB}^{2}$

$\rm\implies \:AB = 8 \: cm$

Now,

$\rm :\longmapsto\:Area_{(\triangle ABC)} \: = \: \dfrac{1}{2} \times AB \times BC$

$\rm :\longmapsto\:Area_{(\triangle ABC)} \: = \: \dfrac{1}{2} \times 8 \times 6$

$\rm :\longmapsto\:Area_{(\triangle ABC)} \: = \: 4 \times 6$

$\rm :\longmapsto\: \boxed{\tt{ \: \: \: \: Area_{(\triangle ABC)} \: = \: 24 \: {cm}^{2} \: \: \: \: }}$

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