Let ∠RTS=∠RST = a (say)
∠QUA=∠QSU= b(say)
then we know , at point S, a+40+b=180. so, a+b=140 we'll use this later.
consider trianglePQR, ∠P+∠Q+∠R=180
i.e.P+(180-2b)+(180-2a)=180
P+180+180-2(a+b)=180 ⇒P+180-2(a+b)=0 ⇒P=2(140)- 180=280-180=100
hence,answer is E
C. 0.4 times 6 is 2.4. Hope that helps.
Diagonal of the parallelogram divides the parallelogram in to two equal areas.
So area of parallelogram = 2(area of triangle)
According to the given diagram,
AB= 8, AD = 5 and BD = 11
So according to the Heron's formula,
Area of triangle = 
and a, b and c are the three sides of the triangle
Area of triangle ABD =
So, area of parallelogram ABCD = 2(area of triangle ABD)
area of parallelogram ABCD = 2 (18.33)
area of parallelogram ABCD = 36.66
area of parallelogram ABCD = 36.7 sq. units
Answer:
45.1 km^2
hope this helps
have a good day :)
Step-by-step explanation: