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Minchanka [31]
3 years ago
12

Area of a triangle two sides of which are 18 cm and 12 cm and the perimeter is 44 cm.​

Mathematics
2 answers:
cestrela7 [59]3 years ago
7 0

Answer:

≈ 83.9 cm²

Step-by-step explanation:

<u>Given:</u>

  • P = a+b+c = 44 cm
  • a = 18 cm
  • b = 12 cm

<u>Then:</u>

  • c= P -(a+b) = 44 -(18+12) = 14 cm

<u>Area of  the triangle is found by using Herons formula:</u>

  • A = √s(s-a)(s-b)(s-c),

where s = P/2 = 44/2 = 22 cm

  • A = √22(22-18)(22-12)(22-14) = √22*4*10*8= √7040 ≈ 83.9 cm²
atroni [7]3 years ago
7 0

Answer:

\huge \boxed{\mathrm{83.9  \ cm^2}}

Step-by-step explanation:

Two sides of the triangle are given.

The perimeter is given.

We need to solve for the third side.

P=a+b+c

P= \sf perimeter

a,b,c= \sf side \ lengths

44=18+12+c

c=14

The measure of the third side is 14 cm.

When three sides of the triangle are given, we can solve for the area using Heron’s formula.

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

s=\sf semi \ perimeter

\displaystyle s=\frac{P}{2} =\frac{44}{2} =22

Plugging in the values and evaluating.

A=\sqrt{22(22-18)(22-12)(22-14)}

A = 83.904708...

The area of the triangle is approximately 83.9 cm².

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Hello!

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Given:

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\sin \theta

To find:

The quadrant of the terminal side of \theta and find the value of \sin\theta.

Solution:

We know that,

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In Quadrant II: Only sin and cosec are positive.

In Quadrant III: Only tan and cot are positive.

In Quadrant IV: Only cos and sec are positive.

It is given that,

\cos \theta =\dfrac{3}{5}

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Here cos is positive and sine is negative. So, \theta must be lies in Quadrant IV.

We know that,

\sin^2\theta +\cos^2\theta =1

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It is only negative because \theta lies in Quadrant IV. So,

\sin \theta=-\sqrt{1-\cos^2\theta}

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\sin \theta=-\sqrt{1-(\dfrac{3}{5})^2}

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Step-by-step explanation:

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