Question

find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is42cm​

Answer 1

Answer:

Question ⤵

Fine the area of a triangle two sides of which are 18cm and 10cm and the perimeter is 42 cm.

Answer ⬇

Given :

Perimeter = 42 cm.

Sides = 18 cm , 10cm .

Let third side be a .

Sum of all sides = perimeter .

a + 18 + 10 = 42

a = 42 - 28

a = 14 cm.

Now, Area = 1/2 × 18 × 12

Area = 108 cm² .

#Ayush.

Thanks..

Answer 2

Answer:

${\tt{{\underline{\bold{\huge{Question}}}}}}$

☞︎︎︎find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is 42cm.

${\tt{{\underline{\bold{\huge{Required \: Answer:}}}}}}$

➪Find The Area Of A Triangle:-

➪A,B,C are the sides of the triangle:-

➪And Perimeter Is 42 cm.

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⠀⠀

☯︎Fᴏʀᴍᴜʟᴀ Oғ Aʀᴇᴀ Oғ Tʀɪᴀɴɢʟᴇ :-

☞︎︎︎

$\sqrt{s(s - a)(9 - b)(s - 0)}$

☝︎Here, s is the semi- perimeter,

☯︎Fᴏʀᴍᴜʟᴀ Oғ Sᴇᴍɪ-Pᴇʀɪᴍᴇᴛᴇʀ :-

$semi \: - \: perimeter = \frac{perimeter}{2}$

➪Semi - Perimeter = 42/2

➪S = 21

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⠀

☯︎Nᴏᴡ, Tʜᴇ Aʀᴇᴀ Oғ Tʀɪᴀɴɢʟᴇ :-

∴c=42−(18+10)cm =14 cm

∴ᴀʀᴇᴀ ᴏғ ᴛʀɪᴀɴɢʟᴇ:-

$\sqrt{21.(21 - 18)(21 - 10)(21 - 14)}$

=

$\sqrt{21.3.7.11 \: {cm}^{2} }$

=

$\sqrt{3 \times 7 \times 3 \times 7 \times 11 \: {cm}^{2} }$

=

$\sqrt{ {(3 \times 7)}^{2} \times 11 \: {cm}^{2} }$

=

$21 \sqrt{11 {cm}^{2} }$

=Thus The Area Required Of Angle Is

$21 \sqrt{11} {cm}^{2}$

➪☯︎Aɴsᴡᴇʀ=

$21 \sqrt{11} {cm}^{2}$