Question

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

Answer 1

Answer:21√11cm²

Step-by-step explanation:

PERIMETER=42cm

THIRD SIDE=42-18-10=14

AREA OF TRIANGLE=√S(S-A)(S-B)(S-C)

S=14+10+28/2=21

AREA=√21(21-18)(21-10)(21-14)

=√21*3*11*7

=21√11cm²    

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Answer 2

$\large\sf{Perimeter=a+b+c}$

$\large\sf{42=18+10+c}$

$\large\sf{42=28+c}$

$\large\sf{42-28=c}$

$\large\sf{14=c}$

$\large\sf{s=\frac{42}{2}}$

$\large\sf{s=21}$

$\large\sf{By\:Heron's\:Formula,}$

$\large\sf{\sqrt{s(s - a)(s - b)(s - c)}}$

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

$\large\sf{\sqrt{21 \times 3 \times 11 \times 7}}$

$\large\sf{\sqrt{21 \times 21 \times 11}}$

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