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3 years ago

Triangle ABC is an isosceles triangle in which side

Mathematics

2 answers:

Maurinko [17]3 years ago

7 0

Answer:

10 + √10 units

Step-by-step explanation:

Shkiper50 [21]3 years ago

4 0

Answer:

Option B.

Step-by-step explanation:

Consider the below figure attached with this question.

It is given that triangle ABC is an isosceles triangle.

From the below figure it is clear that the vertices of triangle are A(-2,-4), B(2,-1) and C(3,-4).

Distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using this formula, we get

$AB=\sqrt{(2-(-2))^2+(-1-(-4))^2}=\sqrt{16+9}=5$

$BC=\sqrt{(3-2)^2+(-4-(-1))^2}=\sqrt{1+9}=\sqrt{10}$

$AC=\sqrt{(3-(-2))^2+(-4-(-4))^2}=\sqrt{25}=5$

Now,

Perimeter of triangle ABC = AB + BC + AC

$=5+\sqrt{10}+5$

$=10+\sqrt{10}$

So, perimeter of triangle ABC is $10+\sqrt{10}$ units.

Therefore, the correct option is B.

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Given LaTeX: f\left(x\right)=x^{^3}-3x+4f ( x ) = x 3 − 3 x + 4, determine the intervals where the function is increasing and wh

Vedmedyk [2.9K]

Answer:

Increasing: $x$ and $x>1$ .

Decreasing: $-1$

Step-by-step explanation:

We have been given a function $f(x)=x^3-3x+4$ . We are asked to determine the intervals, where the function is increasing and where it is decreasing.

First of all, we will find critical points of our given function by equating derivative of our given function to 0.

Let us find derivative of our given function.

$f'(x)=\frac{d}{dx}(x^3)-\frac{d}{dx}(3x)+\frac{d}{dx}(4)$

$f'(x)=3x^{3-1}-3+0$

$f'(x)=3x^{2}-3$

Let us equate derivative with 0 as find critical points as:

$0=3x^{2}-3$

$3x^{2}=3$

Divide both sides by 3:

$x^{2}=1$

Now we will take square-root of both sides as:

$\sqrt{x^{2}}=\pm\sqrt{1}$

$x=\pm 1$

$x=-1,1$

We know that these critical points will divide number line into three intervals. One from negative infinity to -1, 2nd -1 to 1 and 3rd 1 to positive infinity.

Now we will check one number from each interval. If derivative of the point is greater than 0, then function is increasing, if derivative of the point is less than 0, then function is decreasing.

We will check -2 from our 1st interval.

$f'(-2)=3(-2)^{2}-3=3(4)-3=12-3=9$