All categories

3 years ago

A triangle has an area of 30 square centimeters.

Mathematics

1 answer:

Daniel [21]3 years ago

6 0

Answer: 10.9 cm

Step-by-step explanation:

The equation to find the area of a triangle is

Base x height x 1/2 = area

Since you are given the area, rearrange or “do the opposite” to find the base

(Area x 2 ) / height = base

30 x 2 = 60

60/ 5.5 = 10.9

Hope this helped :)

You might be interested in

Calculate the area of triangle ABC with altitude, CD, given

defon

Answer:

37.5 square units

Step-by-step explanation:

Given:

A(-6,-4), B(6,5), C(-1,6), and D(2, 2), where CD is the altitude if ∆ABC

Required:

Area of ∆ABC

SOLUTION:

Area of a ∆ABC =½*AB*CD

✍️ $AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{6 -(-6))^2 + (5 -(-4))^2$

$AB = \sqrt{12^2 + 9^2}$

$AB = \sqrt{144 + 81}$

$AB = \sqrt{225} = 15$

✍️ $CD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{2 -(-1))^2 + (2 - 6)^2$

$CD = \sqrt{(3)^2 + (-4)^2}$

$CD = \sqrt{9 + 16}$

$CD = \sqrt{25} = 5$

✍️Area of a ∆ABC =½*AB*CD

= ½*15*5

✅ Area = 37.5 square units

5 0

3 years ago

Can someone please help with these 2 word problems ASAP.

PolarNik [594]

Answer:

4. 4.66hours

5. time_trip = 2.777 h

Step-by-step explanation:

4. A jet plane travels 2 times the speed of a commercial airplane. The distance between Vancouver and Regina is 1730 km. If the flight from Vancouver to Regina on a commercial airplane takes 140 minutes longer than a jet plane, what is the time of a commercial plane ride of this route?

We know that

Speed_commercial = distance/ time_comm

Speed_commercial = 1730 km/ time_comm

Then,

Speed_jet = distance/ time_jet = 2*Speed_commercial

Speed_jet = 1730 km/ time_jet = 2*Speed_commercial

time_comm = 2.333 h + time_jet

1730 km/ time_jet = 2*1730 km/ time_comm

time_comm = 2*time_jet

time_comm = 2*(time_comm - 2.333h)

time_comm = (2*time_comm - 4.666h)

time_comm = 4.666 h

5. A man goes fishing in a river and wants to know how long it will take him to get 10km upstream to his favourite fishing location. The speed of the current is 3 km/hr and it takes his boat twice as long to go 3km upstream as is does to go 4km downstream. How long will it take his boat to get to his fishing spot?

Distance = 10 km upstream

Speed_current = 3 km/h

Upstream

(Speed_boat - Speed_current) = 3 km / (2*time_downstream)

Downstream

(Speed_boat + Speed_current) = 4 km / (time_downstream)

We subtract both equations

(Speed_boat + Speed_current) - (Speed_boat - Speed_current) = 4 km / (time_downstream) - 3 km / (2*time_downstream)

2*Speed_current = (5/2 km)/ time_downstream

2*(3 km/h)= (5/2 km)/ time_downstream

time_downstream = 0.416 h

(Speed_boat + 3km/h) = 4 km / (0.416 h)

Speed_boat = 6.6 km/h

Trip upstream

(Speed_boat - Speed_current) = 10 km / (time_trip)

time_trip = 10 km/(3.6 km/h) = 2.777 h

time_trip = 2.777 h

3 0

3 years ago

WILL MARK YOU BRAINLIEST JUST GIVE THE ANSWER!!

Igoryamba

Your answer would be the second option

7 0

3 years ago

Read 2 more answers

Which expression is equivalent to 3(m − 3) + 4? 3m + 1 3m − 5 3m + 13 3m − 3

Volgvan

Answer:

3m - 5

Step-by-step explanation:

3(m - 3) + 4 Distribute the 3

3m - 9 + 4 Simplify

3m - 5

6 0

3 years ago

Read 2 more answers

In ΔLMN, l = 870 cm, ∠N=117° and ∠L=17°. Find the length of n, to the nearest 10th of a centimeter.

MAXImum [283]

We have been given that in ΔLMN, l = 870 cm, ∠N=117° and ∠L=17°. We are asked to find the length of n to the nearest 10th of a centimeter.

We will use law of sines to find the length of n.

Law of sine states the relation between the angles of a triangle and their corresponding side.

$\frac{a}{\text{sin(A)}}=\frac{b}{\text{sin(B)}}=\frac{c}{\text{sin(C)}}$ , where a, b and c are corresponding sides to angles A, B and C respectively.