Question

Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42cm.
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Answer 1

Answer:

$21 \sqrt{11}$

Step-by-step explanation:

PERIMETER OF TRIANGLE=42

PERIMETER OF TRIANGLE= SUM OF 3 SIDES

=42-(18+10)=14

SIDE OF TRIANGLES = 18,10&14

IT IS A SCALENE TRIANGLE SO,

SEMIPERETER = (a+b+c)/2

=42/2=21

AREA OF TRIANGLE =

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

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

$21 \sqrt{11}$

Answer 2

Answer:

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