Question

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

Answer 1

Answer:

Using the formulas

A=s(s﹣a)(s﹣b)(s﹣c)

P=a+b+c

s=a+b+c

2

Solving forA

A=1

4﹣P4+4P3a+4P3b﹣4(Pa)2﹣12P2ab﹣4(Pb)2+8Pa2b+8Pab2=1

4·﹣424+4·423·18+4·423·10﹣4·(42·18)2﹣12·422·18·10﹣4·(42·10)2+8·42·182·10+8·42·18·102≈69.64912cm²

Step-by-step explanation:

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Answer 2

Answer:

1st Method:-

→ We will use Herons Formula

Let the third side of triangle i.e. 'x'

perimeter of the given triangle = 42 cm

18 + 10 + x = 42

28 + x = 42

x = 42 - 28

x = 14

$\sf \: x = { \frac{perimeter}{2}} = \frac{42}{2} = 21cm$

By Herons Formula

$\sf{Area \: of \: the \: triangle}= \sqrt{s(s - a)(s - b)(s - c)}$

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

$\sf{ \sqrt{21(3)(11)(7) } = \sqrt[21]{11} } \: {cm}^{2}$

2nd Method:-

→ We will use area of triangle

$\sf{Area \: of \: Triangle = } \: A \frac{ h_{b}b }{2}$

Perimeter of triangle = 42 cm

perimeter of 2 sides = 18 cm and 10 cm.

$\sf{So, 18 + 10 = 28}$

$\sf{thus, perimeter \: of \: a \: triangle = 42}$

$\sf{42 - 28 } = 14$

$\sf{Hence, \: } \sqrt[21]{11 \: cm} \: or \: 14 \: is \: the \: exact \: answer.$