23) Find the area of a triangle two sides of which are 18 cm and 10 cm and the
perimeter is 42 cm
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2 thoughts on “23) Find the area of a triangle two sides of which are 18 cm and 10 cm and the<br />perimeter is 42 cm<br />​”

1. Answer:

   area of triangle = 1/2 × b × h

   1/2 × 18 × 10

   =90

   perimeter = sum of 3 side

   42 = 1st side + 2nd side + 3rd side

   42 = 18 + 10 + 3rd side

   42 = 28 + 3rd side

   42 -28 = 3rd side

   14 = 3rd side

   perimeter = sum of 3 sides

   perimeter = 14 + 18 + 10

   perimeter = 42

   so , area of triangle is 90 and perimeter of triangle is 42

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2. Answer:

   [tex]a = 18[/tex]

   [tex]b = 10[/tex]

   [tex]find \: c[/tex]

   [tex]perimeter = a + b + c[/tex]

   [tex]18 + 10 + c = 42[/tex]

   [tex]28 + c = 42[/tex]

   [tex]c = 42 – 28[/tex]

   [tex]c = 14[/tex]

   [tex]here \: [/tex]

   [tex]s \: = \frac{a + b + c}{2} \: = \frac{18 + 10 + 14}{2} [/tex]

   [tex] = \frac{42}{2 } = 21cm[/tex]

   [tex]area \: of \: triangle[/tex]

   [tex] \sqrt{s(s – a)(s – b)(s – c} [/tex]

   [tex] \sqrt{(21(21 – 18)(21 – 10)(21 – 14)} [/tex]

   [tex] \sqrt{21 \times 3 \times 11 \times 7} [/tex]

   [tex] \sqrt{7 \times 3 \times 3 \times 11} [/tex]

   [tex]3 \times 7 \times \sqrt{11} = 21 \sqrt{11} cm[/tex]

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