Question

the semi-perimeter of ∆ABC whose sides are 10cm,8cm and 12cm is​

Answer 1

Answer:

142 g/mol

Step-by-step explanation:

Na. -2 moles × 23 = 46

S - 1 moles × 32 = 32

O - 4 moles × 16 = 64

46+32+64 = 142g/mol

Answer 2

Answer:

Given :-

Sides :-

a = 10cm

b = 8cm

c = 12 cm

Solution :-

Diagram

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(1, 0)(1,0)(3,3)\qbezier(5,0)(5,0)(3,3)\qbezier(5,0)(1,0)(1,0)\put(2.85,3.2){ \bf A }\put(0.5,-0.3){ \bf C }\put(5.2,-0.3){ \bf B }\end{picture}$

As we all know ,

The perimeter of Triangle ↬ sum of all sides

Perimeter ↬ ( a + b + c )

From given

➤ a = 24 cm

➤ b = 7 cm

➤ c = 25 cm

Now,

Applying the Heron's Formula

$\bold{According \: to \: heron's \: formula} \\ \tt: \implies s = \frac{a + b + c}{2} \\ \\ \tt: \implies s = \frac{10 + 8 + 12}{2} \\ \\ \tt: \implies s = \frac{30}{2} \\ \\ \bf{\tt: \implies s = 15}$

Here s is the half perimeter ( semi perimeter )

You can even find area of triangle now ,

$\bold{For \: Area \: of \: triangle} \\ \tt: \implies Area \: of \: triangle = \sqrt{s(s - a)(s - b)(s - c)} \\ \\ \tt: \implies Area \: of \: triangle = \sqrt{15(15 - 10)(15 - 8)(15 - 12)} \\ \\ \tt: \implies Area \: of \: triangle = \sqrt{15 \times 5 \times 7\times 3 } \\ \\ \tt: \implies Area \: of \: triangle = \sqrt{1575} \\ \\ \bf{\tt: \implies Area \: of \: triangle = 39.68 \: {cm}^{2} }$

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