I don't know the area of a parallelogram off the top of my head but you can use the area of the rectangle and the area of the triangle and add those. The area of the triangle is 1/2*base*height and use the pythagorean theorem to find the base. 3² + b² = 5²; 9 + b² = 25; b² = 16; b = 4. The area of the triangle is 1/2*4*3 or 6 in. Since there are two of them, the total area of the triangles are 12 in. The area of the rectangle is length*width or (12 - 4) * 3 which is 8 * 3 or 24in. Add these areas together to get 36in total
Answer:
49
Step-by-step explanation:
Positive 49 not -49
Answer:
Step-by-step explanation:
I think the only way you can solve this is to assume that <R means PRT in the given ratio. If I am wrong, I don't think the problem can be solved.
Find <T
Let <T = x
and <PRT = 3x
KLMN is a Parallelagram and therefore two adjacent angles are supplementary.
<PRT + <T = 180 degress
3x + x = 180 degrees
4x = 180
x = 45
So <T = 45
<PRT = 3*45 = 135
If RD is perpendicular to PS then <PDR = 90o
Here's the trick.
RD is also Perpendicular to RT
<MRD + <MRT = 90
<MRT = 180 - 90 - <T
<MRT = 180 - 90 - 45
<MRT = 45
Here comes your answer
=================
<MRD + MRT = 90
<MRD + 45 = 90
<MRD = 45
====================
Note: you must ignore everything to do with the diagram. It is not drawn to scale and the letters are not the same as in the question. The only thing you use is that the figure is a ||gm
Recall that
Your sums start at 
, so in order to apply these formulas directly, you need to compensate for the missing first two terms:
Answer:
Step-by-step explanation:
![\dfrac 12 \left[\sin(a+b)+\sin(a-b) \right]\\\\\\=\dfrac 12\left[ 2 \sin a \cos b\right]\\\\\\=\sin a \cos b](https://tex.z-dn.net/?f=%5Cdfrac%2012%20%5Cleft%5B%5Csin%28a%2Bb%29%2B%5Csin%28a-b%29%20%5Cright%5D%5C%5C%5C%5C%5C%5C%3D%5Cdfrac%2012%5Cleft%5B%202%20%5Csin%20a%20%5Ccos%20b%5Cright%5D%5C%5C%5C%5C%5C%5C%3D%5Csin%20a%20%5Ccos%20b)
