Answer:
Area of the shaded region= ![500inch^2](https://tex.z-dn.net/?f=500inch%5E2)
Step-by-step explanation:
Area of the shaded region= Area of the triangle- Area of the rectangle
Dimensions of the triangle:
Base= 40 inch
Height= 40 inch
Area of a triangle=![\frac{1}{2}*Base*Height](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%2ABase%2AHeight)
![=\frac{1}{2}*40*40\\\\ =\frac{1}{2} *1600\\\\=800 inch^2](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B2%7D%2A40%2A40%5C%5C%5C%5C%20%3D%5Cfrac%7B1%7D%7B2%7D%20%2A1600%5C%5C%5C%5C%3D800%20inch%5E2)
Area of the rectangle= Length*Width
Length=30 inch
Width=10 inch
Area:
![= 30*10\\\\=300 inch^2](https://tex.z-dn.net/?f=%3D%2030%2A10%5C%5C%5C%5C%3D300%20inch%5E2)
Area of the shaded region= ![800-300= 500 inch^2](https://tex.z-dn.net/?f=800-300%3D%20500%20inch%5E2)
Answer:
yea ur correct
Step-by-step explanation:
use PhotoMath or something like that to check next time, it'll save a lot of time
Answer:
![KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08\\ \\KN=45\sin 50^{\circ}\approx 34.47](https://tex.z-dn.net/?f=KL%3D45%5Ctan%2050%5E%7B%5Ccirc%7D%5Csin%2050%5E%7B%5Ccirc%7D%5Capprox%2041.08%5C%5C%20%5C%5CKN%3D45%5Csin%2050%5E%7B%5Ccirc%7D%5Capprox%2034.47)
Step-by-step explanation:
Given:
KL ║ NM ,
LM = 45
m∠M = 50°
KN ⊥ NM
NL ⊥ LM
Find: KN and KL
1. Consider triangle NLM. This is a right triangle, because NL ⊥ LM. In this triangle,
LM = 45
m∠M = 50°
So,
![\tan \angle M=\dfrac{\text{opposite leg}}{\text{adjacent leg}}=\dfrac{NL}{LM}=\dfrac{NL}{45}\\ \\NL=45\tan 50^{\circ}](https://tex.z-dn.net/?f=%5Ctan%20%5Cangle%20M%3D%5Cdfrac%7B%5Ctext%7Bopposite%20leg%7D%7D%7B%5Ctext%7Badjacent%20leg%7D%7D%3D%5Cdfrac%7BNL%7D%7BLM%7D%3D%5Cdfrac%7BNL%7D%7B45%7D%5C%5C%20%5C%5CNL%3D45%5Ctan%2050%5E%7B%5Ccirc%7D)
Also
(angles LNM and M are complementary).
2. Consider triangle NKL. This is a right triangle, because KN ⊥ NM . In this triangle,
(alternate interior angles)
(angles KNL and KLN are complementary).
So,
![\sin \angle KNL=\dfrac{\text{opposite leg}}{\text{hypotenuse}}=\dfrac{KL}{LN}=\dfrac{KL}{45\tan 50^{\circ}}\\ \\KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08](https://tex.z-dn.net/?f=%5Csin%20%5Cangle%20KNL%3D%5Cdfrac%7B%5Ctext%7Bopposite%20leg%7D%7D%7B%5Ctext%7Bhypotenuse%7D%7D%3D%5Cdfrac%7BKL%7D%7BLN%7D%3D%5Cdfrac%7BKL%7D%7B45%5Ctan%2050%5E%7B%5Ccirc%7D%7D%5C%5C%20%5C%5CKL%3D45%5Ctan%2050%5E%7B%5Ccirc%7D%5Csin%2050%5E%7B%5Ccirc%7D%5Capprox%2041.08)
and
![\cos \angle KNL=\dfrac{\text{adjacent leg}}{\text{hypotenuse}}=\dfrac{KN}{LN}=\dfrac{KN}{45\tan 50^{\circ}}\\ \\KN=45\tan 50^{\circ}\cos 50^{\circ}=45\sin 50^{\circ}\approx 34.47](https://tex.z-dn.net/?f=%5Ccos%20%5Cangle%20KNL%3D%5Cdfrac%7B%5Ctext%7Badjacent%20leg%7D%7D%7B%5Ctext%7Bhypotenuse%7D%7D%3D%5Cdfrac%7BKN%7D%7BLN%7D%3D%5Cdfrac%7BKN%7D%7B45%5Ctan%2050%5E%7B%5Ccirc%7D%7D%5C%5C%20%5C%5CKN%3D45%5Ctan%2050%5E%7B%5Ccirc%7D%5Ccos%2050%5E%7B%5Ccirc%7D%3D45%5Csin%2050%5E%7B%5Ccirc%7D%5Capprox%2034.47)