Question

rhe perimeter of an iscoscles triangleis 30 cm and each equal side be 12 cm ​

Answer 1

$\huge \sf \underline \red{Answer : }$

$\sf \underline \green{ \therefore \: Area \: of \: triangle = 9 \sqrt{15} {cm}^{2}}$

$\huge \sf \underline \orange{To \: Find \: }$

• Area of triangle

$\sf \huge \underline \blue{solution : }$

Given,

• perimeter of an iscoscles triangle is 30cm

• Each equal side be 12cm

$\sf \underline{we \: know \: that : }$

$\sf{ \boxed{ \underline{ \underline{ \red{ \tt{Area \: of \: triangle = \sqrt{s(s - a)(s - b)(s - c) \: }}}}}}}$

where,

• s is semi perimeter

• a,b,c is the sides of the triangle

Given:

• Triangle is iscoscles

• Iscoscles triangle two sides are equal so,

$\sf{a = b = 12cm}$

$\sf{Perimeter = 30cm}$

$\sf \underline{we \: know \: that : }$

$\sf{semi \: perimeter = \dfrac{ perimeter }{2}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ s= \dfrac{30}{2}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ s= 15cm}$

Now we have to find c we get,

$\sf{perimeter = 30cm}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{a + b + c = 30cm}$

where

• a is 12cm

• b is 12cm

• c is ?

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{12cm+ 12cm+ c = 30cm}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{24 + c = 30}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{c = 30 - 24}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{c = 6cm}$

Now we have to find the area of triangle:

$\sf{ \boxed{ \underline{ \underline{ \red{ \tt{Area \: of \: triangle = \sqrt{s(s - a)(s - b)(s - c) \: }}}}}}}$

where,

• a is 12cm

• b is 12 cm

• c is 6cm

• s is 15 cm

$\: \: \sf{ = \sqrt{15(15 - 12)(15 - 12)(15 - 6)}}$

$\: \: \sf{ = \sqrt{15(3)(3)(9)}}$

$\: \: \sf{ = \sqrt{15(9)(9)}}$

$\: \: \sf{ = \sqrt{9 \times 9} \times \sqrt{15}}$

$\: \: \sf{ = \sqrt{9 {}^{2} } \times \sqrt{15}}$

$\: \: \sf{ = (9) \times \sqrt{15}}$

$\: \: \sf{ = 9 \sqrt{15}}$

$\sf \underline \green{ \therefore \: Area \: of \: triangle = 9 \sqrt{15} {cm}^{2}}$

Answer 2

Answer:

answer

Explanation:

Perimeter of isosceles triangle=30cm

Length of equal sides=12cm

Let third side of triangle=xcm

According to problem,

x+12+12=30

x+24=30

x=30−24

x=6

∴ Third side of triangle=6cm

Using Heron's formula

Area of triangle=

s(s−a)(s−b)(s−c)

sq. units

where s=

2

a+b+c

s=

2

30

=15

Area of triangle=

15(15−12)(15−12)(15−6)

cm

2

=

15×3×3×9

cm

2

=3×3×

15

cm

2

=9

15

cm

2

∴ Area of triangle=9

15

cm

2

.