Pls help me out on this ill will give brainiest for the best answer

The area of a triangle is found by multiplying half of the measure of its base by its height. The area of the following right tr

GarryVolchara [31]

Answer:

$\sf Area\:of\:a\:triangle=\dfrac12 \times base \times height$

Given:

area = 18 cm²
base = 6 cm
height = (h + 1) cm

Substitute the given values into the equation and solve for h:

$\sf \implies 18 = \sf \dfrac12 \times 6 \times (h+1)$

$\sf \implies 18 = \sf \dfrac62 (h+1)$

$\sf \implies 18 =3(h+1)$

expand using distributive property of addition $a(b+c)=ab+ac$ :

$\sf \implies 18=3h+3$

subtract 3 from both sides:

$\sf \implies 3h=15$

divide both sides by 3:

$\sf \implies h=5$

<u>To verify</u>:

Half of the base is 3 cm.

If h=5, then the height of the triangle is 6 cm.

Multiplying 3 by 6 is 18.

This matches the given area, so we can verify that h = 5.

3 0

2 years ago

Read 2 more answers

A right △ABC is inscribed in circle k(O, r). Find the radius of this circle if:

natta225 [31]

We have been given that a right △ABC is inscribed in circle k(O, r).

m∠C = 90°, AC = 18 cm, m∠B = 30°. We are asked to find the radius of the circle.

First of all, we will draw a diagram that represent the given scenario.

We can see from the attached file that AB is diameter of circle O and it a hypotenuse of triangle ABC.

We will use sine to find side AB.

$\text{sin}=\frac{\text{Opposite}}{\text{Hypotenuse}}$

$\text{sin}(30^{\circ})=\frac{AC}{AB}$

$\text{sin}(30^{\circ})=\frac{18}{AB}$

$AB=\frac{18}{\text{sin}(30^{\circ})}$

$AB=\frac{18}{0.5}$

$AB=36$

Wee know that radius is half the diameter, so radius of given circle would be half of the 36 that is $\frac{36}{2}=18$ .

Therefore, the radius of given circle would be 18 cm.

8 0

3 years ago

HELP ME!!!!!!!!!! the problem is on the picture.

FrozenT [24]

(C) cause 17•3=51+39= 90 and that’s a 90 degree

8 0

3 years ago

3b+(-5)=(2b)³(-3)2+6=?

Flauer [41]

Answer:

b=-1

Step-by-step explanation:

given.3b+(-5)=(2b)³(-3)2+6=?

3b-5=8b-6+6

3b-5=8b

-5=8b-3b

-5=5b

-5/5=5b/5

b=-1

-3-5=-8-6+6=?

-8=-8=-8

7 0

2 years ago

There are three different cubes such that the first cube is 64 times the volume of the second, and the volume of the second cube

tankabanditka [31]

Answer:

4/3

Step-by-step explanation:

If the ratio between the volumes of the first and the second cube is 64, the ratio between the sides is the cubic root of the ratio between the volumes, so:

$V1 / V2 = 64$

$s1 / s2 = \sqrt[3]{64} = 4$

Doing the same for the second and third cubes, we have:

$V2 / V3 = 1/27$

$s2/ s3 = \sqrt[3]{1/27} = 1/3$

So the ratio of the first cube side and the third cube side is:

$s1 / s3 = (s1/s2) * (s2/s3) = 4 * (1/3) = 4/3$

5 0

3 years ago