Question

The length of the sides of a triangle are 5 cm, 12 cm and 13 cm. Find the length

of perpendicular from the opposite vertex to the side measuring 13 cm.​

Answer 1

Answer:

30

Step-by-step explanation:

5 + 12 + 13 = 30

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Answer 2

Answer: The required length of perpendicular from the opposite vertex to the side measuring $13 \, \text{cm}$ is $\frac{60}{13} \,\text{cm}$.

Step-by-step explanation: Let the given sides of triangle $\triangle ABC$ (say) be denoted by $AB=\, 13 \text{cm},\, AC=\, 5\text{cm} \,\, \text{and} \, BC=\, 12 \text{cm}$.  From figure, $a= 13 \text{cm}, b= 12 \text{cm} \,\text{and} \, c= 5\text{cm}$. We need to find the length $h$ of perpendcular $CD$ from vertex $C$ on $AB.$

Then semi- perimeter (say $s$) of $\triangle ABC$ will be ,

$s= \, \frac{a+b+c}{2}$

   $=\frac{13+12+5}{2}$

  $= 15 \text{cm}$.

Using Heron's formula for area of a triangle,

area( $\triangle ABC$) $= \sqrt{s(s-a_(s-b)(s-c)$

                      $= \sqrt{15(15-13)(15-12)(15-3)$

                      $= \sqrt{ 15\times 2 \times 3 \times \times 10$

                      $=\, 30\, {cm}^2$.

Now, using general formula for the area of a triangle, we have

area ($\triangle ABC$) $= \frac{1}{2} \times h \times 13$

                 $30=\, \frac{13 h}{2}$

               $h =\, \frac{60}{13} \,\text{cm}$.

Or $h=\,4.62\, \text{cm}$.