Let AD be x.
tana = 4/x --- (1).
tana = x/9 --- (2).
Equating (1) and (2).
4/x = x/9
Cross Multiply.
x² = 36
x = 6 units.
Again, x is AD, the height of the ascribed trapezoid AB...
a = ½(24)
a = 12 units.
a is the radius of the half by circle.
sinb = 8/24
b = asin(⅓)°
tan(asin(⅓)) = c/12
c = 4.2426406871 units.
tand = 12/4.2426406871
d = 70.5287793656°
e = ½(d)
e = 35.2643...
a = (r-7) units.
b = (r-14) units.
Therefore, r, radius of the inscribed circle is;
r² = a²+b²
r² = (r-7)²+(r-14)²
r² = r²-14r+49+r²-28r+196
r²-42r+245 = 0
Resolving the above quadratic equation...
a = 6 units.
a is the side length of the square.
Therefore;
Area square is;
6*6
= 36 square units.
Area inscribed shaded triangle is;
½(6*6)
= 18 square units.
It implies;
T...
a² = 24²+7²
a = √(625)
a = 25 units.
b = 7+a
b = 32 units.
c² = 32²+24²
c = √(1600)
c = 40 units.
d = ½(c)
d = 20 units.
It implies, the required length, x is;
25² = x²+20²
x²...
Calculating r, radius of the inscribed half circle.
a = (r-6) units.
b = (r-3) units.
Therefore;
r² = a²+b²
r² = (r-6)²+(r-3)²
r² = r²-12r+36+r²-6r+9
r²-18r+45 = 0
Resolving the a...
Let the side length of the regular hexagon be 1 unit.
Therefore, area A (area inscribed square) is;
1²
= 1 square unit.
Calculating area B.
a² = 1²+(½)²
a² = 1+¼
a = √(5/4)
a = ½√(5...
Let the single side length of the square be (x+10) unit.
Calculating x.
10²=(5+x)²+5²
100-25-25=10x+x²
Therefore;
x²+10x-50=0
Resolving the above quadratic equation via completing the squ...
Calculating x, length AC = length AB.
a = 14+4
a = 18 units.
b = ½(18)
b = 9 units.
c²+9² = x²
c = √(x²-81) units.
d = 9-4
d = 5 units.
It implies;
13² = √(x²-81)²+5²
169 =...
Let CD be b units.
Let BC be a units.
c = 2+4
c = 6 units.
Observing similar plane shape (right-angled) side length ratios.
a - 6
6 - b
Cross Multiply.
ab = 36 --- (1).
Let alph...
