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rjkz [21]
3 years ago
8

Which is the correct formula to find the area of ABC?

Mathematics
2 answers:
Alex777 [14]3 years ago
5 0

Answer:

none of them

Step-by-step explanation:

The area (A) of a triangle is calculated as

A = 0.5 × base × perpendicular height

Here the base is BC and perpendicular height is AD, hence

Area of ΔABC = 0.5 × BC × AD

Neporo4naja [7]3 years ago
4 0

Answer:  C. \text{area of }\triangle ABC=\dfrac{1}{2}\times BC\times AD

Step-by-step explanation:

The common formula is used to find the area of a triangle is

\dfrac{1}{2}\times\text{base}\times\text{altitude}

In the given picture , we have a triangle ABC , in which Altitude (making right angle with base) of triangle = AD with corresponding base = BC

Thus , the area of the triangle will be :

\text{area of }\triangle ABC=\dfrac{1}{2}\times BC\times AD

Therefore , the correct options is

C. \text{area of }\triangle ABC=\dfrac{1}{2}\times BC\times AD

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\frac{-3+ 2\sqrt{3}}{2}\textrm{ and }\frac{-3- 2\sqrt{3}}{2}

Step-by-step explanation:

3(2b+3)^{2}=36\\\frac{3(2b+3)^{2}}{3}=\frac{36}{3}\\(2b+3)^{2}=12

Taking square root both sides, we get

\sqrt{(2b+3)^{2}}=\pm \sqrt{12}\\2b+3=\pm \sqrt{4\times 3}\\2b+3=\pm 2\sqrt{3}\\2b+3-3=\pm 2\sqrt{3}-3\\2b=-3\pm 2\sqrt{3}\\b=\frac{-3\pm 2\sqrt{3}}{2}\\b=\frac{-3+ 2\sqrt{3}}{2}\textrm{ or }b=\frac{-3- 2\sqrt{3}}{2}

Therefore, the values of b are:

\frac{-3+ 2\sqrt{3}}{2}\textrm{ and }\frac{-3- 2\sqrt{3}}{2}

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