Answer:
Area of trapezium = 4.4132 R²
Step-by-step explanation:
Given, MNPK is a trapezoid
MN = PK and ∠NMK = 65°
OT = R.
⇒ ∠PKM = 65° and also ∠MNP = ∠KPN = x (say).
Now, sum of interior angles in a quadrilateral of 4 sides = 360°.
⇒ x + x + 65° + 65° = 360°
⇒ x = 115°.
Here, NS is a tangent to the circle and ∠NSO = 90°
consider triangle NOS;
line joining O and N bisects the angle ∠MNP
⇒ ∠ONS =
= 57.5°
Now, tan(57.5°) = 
⇒ 1.5697 = 
⇒ SN = 0.637 R
⇒ NP = 2×SN = 2× 0.637 R = 1.274 R
Now, draw a line parallel to ST from N to line MK
let the intersection point be Q.
⇒ NQ = 2R
Consider triangle NQM,
tan(∠NMQ) = 
⇒ tan65° =
⇒ QM =
QM = 0.9326 R .
⇒ MT = MQ + QT
= 0.9326 R + 0.637 R (as QT = SN)
⇒ MT = 1.5696 R
⇒ MK = 2×MT = 2×1.5696 R = 3.1392 R
Now, area of trapezium is (sum of parallel sides/ 2)×(distance between them).
⇒ A = (
) × (ST)
= (
) × 2 R
= 4.4132 R²
⇒ Area of trapezium = 4.4132 R²
1 - complementary = 90- 30 = 60
suplementary = 180- 30 = 150
2 - area = hb/2 = 360 = hb/2 = h = 72
3 - similar
4- see number 3
5 - asa, ssa, sas
6 - polygon that had all equal angle measures and sides (equiangular and equilateral)
7 - length x width = area so
240 / 24 = 10
8 - line
9 - triangle
10 - 180, as see in question 1
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Answer:
whatever the number of crocodiles is ___ multiply by 2 then add 4
____ x 2 = ____+4 =___
number of crocodiles____×2=___+4=__
( multiply symbol is x )
example : 5 crocodiles x 2 = 10 crocodiles + 4 = 14 turtles