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The base and hypotenuse of a right triangle are respectively 6cm and 10cm long.find the area ??

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Answer 1

Question :-

the base and hypotenuse of a right triangle are respectively 6cm and 10cm long find the area

Given :-

• base = 6m
• hypotenuse = 10cm

To find :-

• area of the triangle

Solution :-

according to Pythagoras theorem

• $\mapsto \sf\boxed{\bold{\pink{ {h}^{2} \: = {b}^{2} + {a}^{2} }}} \:$

where,

• h = hypotenuse
• b = base
• a = altitude (height)

now, by putting values we get,

$\mapsto \rm \: {10}^{2} = {6}^{2} + {a}^{2}$

$\mapsto \rm {10}^{2} - {6}^{2} = {a}^{2}$

$\mapsto \rm100 - 36 = {a}^{2}$

$\mapsto \rm64 = {a}^{2}$

$\mapsto \rm \: a = 8cm$

here, we have all the sides of the triangle

• 1st side of triangle = 6cm
• 2nd side of triangle = 10cm
• 3rd side of triangle = 8cm

now,

• perimeter = sum of sides

now by putting values we get,

$\mapsto \rm \: perimeter = 6 + 10 + 8cm \\ \\ \mapsto \rm \: perimeter = 24cm$

now,

• $\mapsto \sf\boxed{\bold{\pink{ semi - perimeter \: = \frac{perimeter}{2} }}}$

$\rightarrow \tt \: s = \frac{24}{2} \\ \\ \rightarrow \tt \: s = 12cm$

now let's find area of triangle

by using heron's formula we get,

• $\mapsto \sf\boxed{\bold{\pink{ \sqrt{s(s - a)(s - b)(s - c)} }}}$

where,

• s = semi perimeter
• a,b,c are side of the triangle

now, by putting values we get,

$\rm :\longmapsto\: \sqrt{12(12 - 6)(12 - 10)(12 - 8)}$

$\rm :\longmapsto\: \sqrt{2 \times 6 \times 2 \times 4}$

$\rm :\longmapsto\: \sqrt{2 \times 2 \times 3 \times 2 \times 2 \times 2}$

$\rm :\longmapsto\: \sqrt{2 \times 2 \sqrt{3} \times 2 }$

$\rm :\longmapsto\:4 \sqrt{6} cm^{2}$

$\rm :\longmapsto\:4 \times 6cm^{2}$

$\rm :\longmapsto\:24cm$

so therefore the area of the triangle = 24cm