Given:
square with sides measuring 7 cm.
3 triangles attached to three sides of the square. A line bisecting one triangle is measured at 4 cm.
Area of a square = s² = (7cm)² = 49 cm²
Area of a triangle = hb/2 = (4cm*7cm)/2 = 14 cm²
Area of the 3 triangles = 14 cm² x 3 = 42 cm²
Total area of the logo = 49 cm² + 42 cm² = 91 cm²
23 +30 (15 feet each way) = 53
28 + 30 = 58
58 x 53 = 3074ft
Answer:
m∠QPM=43°
Step-by-step explanation:
see the attached figure to better understand the problem
we know that
m∠NPQ=m∠MPN+m∠MPQ
we have
m∠NPQ=(9x-25)°
m∠MPN=(4x+12)°
m∠MPQ=(3x-5)°
substitute the given values and solve for x
(9x-25)°=(4x+12)°+(3x-5)°
(9x-25)°=(7x+7)°
9x-7x=25+7
2x=32
x=16
Find the measure of angle QPM
Remember that
m∠QPM=m∠MPQ
m∠MPQ=(3x-5)°
substitute the value of x
m∠MPQ=(3(16)-5)=43°
therefore
m∠QPM=43°
Let us say that the intersection point of lines
AB and CD is called point E. The lines AB and CD are perpendicular to each
other which also means that the triangle CEB is a right triangle.
Where the line CB is the radius of the circle
while the side lengths are half of the whole line segment:
EB = 0.5 AB = 0.5 (8 ft) = 4 ft
CE = 0.5 CD = 0.5 (6 ft) = 3 ft
Now using the hypotenuse formula since the
triangle is right triangle, we can find for the radius or line CB:
CB^2 = EB^2 + CE^2
CB^2 = (4 ft)^2 + (3 ft)^2
CB^2 = 16 ft^2 + 9 ft^2
CB^2 = 25 ft^2
<span>CB = 5 ft = radius</span>