Math, asked by naikarekha511, 5 months ago

Find area of triangle where two sides are 9 and 12 cm and perimeter is 34 cm​

Answers

Answered by Anonymous
88

Answer :-

Perimeter of traingle = 34

Let the 3rd side be x

\sf x + 12 + 9 = 34

\sf x + 21 = 34

\sf x = 34 - 21

\sf x = 13

Sides of traingle = 13 cm , 12 cm and 9 cm

  • a = 13 cm
  • b = 12 cm
  • c = 9 cm
  • s = 34/2 = 17 cm

Substituting the value in Heron's formula -

\sf Area = \sqrt{(s)(s-a)(s-b)(s-c)}

\sf = \sqrt{17 \times ( 17 - 13 ) \times (17-12) \times (17-9)}

\sf = \sqrt{17 \times 4 \times 5 \times 8}

\sf = \sqrt{2720}

\sf = 52.15\: cm^2

➩ Area of traingle = 52.15 cm²

Answered by Anonymous
18

Answer:

Perimeter of traingle = 34

Let the 3rd side be x

➩ \sf x + 12 + 9 = 34x+12+9=34

➩ \sf x + 21 = 34x+21=34

➩ \sf x = 34 - 21x=34−21

➩ \sf x = 13x=13

Sides of traingle = 13 cm , 12 cm and 9 cm

  • a = 13 cm
  • b = 12 cm
  • c = 9 cm
  • s = 34/2 = 17 cm
  • Substituting the value in Heron's formula -

\sf Area = \sqrt{(s)(s-a)(s-b)(s-c)}Area= </p><p>(s)(s−a)(s−b)(s−c)

\sf = \sqrt{17 \times ( 17 - 13 ) \times (17-12) \times (17-9)}= </p><p>17×(17−13)×(17−12)×(17−9)

\sf = \sqrt{17 \times 4 \times 5 \times 8}= </p><p>17×4×5×8

\sf = \sqrt{2720}= </p><p>2720

\sf = 52.15\: cm^2=52.15cm </p><p>2

➩ Area of traingle = 52.15 cm²

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