SilviaJendeukie

Since triangle PQR is the medial triangle of triangle MNO, QR is equidistant from NO and KL. Similarly, NO is equidistant from J and KL. Therefore, the altitude (J to QR) of triangle JQR is three times the altitude from P to QR. With the same base, the area of triangle JQR is three times triangle PQR which is 3 times 10, 30.

However, I have a question for you. Since triangle PQR is 10, and based on the information given in the question, I believe that MNO cannot be 20. If you draw out the diagram, it will be clearer that triangle MNO is four times triangle PQR. 

This answer happened to be wrong...:(

The numbers are different, and I am supposed to solve it using similar triangles.

THANK YOU

So...is it 20/(sqrt(2))?

I solved it.

Let x represent BX.

15^2 + x^2 (AB) + 6^2 + x^2 (BC) = 21^2 (AC)

Solving this equation gives us x = 3sqrt(10)

Therefore, BX = 3sqrt(10)

Hey, I just wanna know why are you using the same pfp and name as me...

They are not equal

I got it!

Let x represent the common difference.

\(a_1+a_3+a_5+a_7+a_9+a_{11} = \frac{6(a_1+a_1+10x)}{2} = 6a_1+30x = -2 \)

\(a_2+a_4+a_6+a_8+a_{10}+a_{12} = \frac{6(a_1+x+a_1+11x)}{2} = 6a_1+36x = 1\)

Solving the equation gives us \(x = \frac{1}{2}, a_1 = -\frac{17}{6}\)

From the given information, we know that there are 3 students in the math club, and there are twice as many students in the science club as in the math club which means there are \(3 \times 2 = 6\) students in the science club. However, we cannot simply add 6 to 3 because there are students who are in both clubs. There are 3, in this case, based on the question. Therefore, we need to subtract 3 from \(6 + 3 \) which will be 6.

It would take 1 hour for each member if there are 45 members to paint a wall, so it's 45 hours in total.

Let \(x\) represent how many hours 12 members need to paint the same wall.

\(12x=45 \times 1\)

\(x=\frac{45}{12}=\frac{15}{4}\)hrs

In minutes, it would be \(3 \times \frac{15}{4}=225 \) minutes.