A B is the diameter of the semicircle. C is a point on the circumference. O is the center if the semicircle.
If the radius of the semicircle is 5, ∠ C O A = 6 0 ∘ . Find the sum of the area of the region R and S . note region R and S is not the semicircle with the diameter AC and BC
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areas of R+S=area of the triangle ABC. It is 30-60-90 triangle. So its area =
2
3
∗
r
2
=
2
3
∗
5
2
=
2
1
.
6
5
0
6
Proof:areas of R+S=area of the triangle ABC.
A
C
2
+
B
C
2
=
A
B
2
,
∴
2
1
∗
π
∗
A
C
2
+
2
1
∗
π
∗
B
C
2
=
2
1
∗
π
∗
A
B
2
.
⟹
Sum of areas of semicircles on the legs =area of semicircle on the hypotenuse.
From both sides subtract area common to semicircles on the legs and hypotenuse.
That is the sum R+S=area of
Δ
A
B
C
.
This solution is not correct Ans 27.188
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file:///home/chronos/u-bb41696e22533f784022ff2f3bd1a43d45e55fef/MyFiles/Downloads/Untitled%20document.pdf - Thales' Theorem tells us that ∠ACB= 90° -∠COA=60∘ ,when there is a straight line, the other angle, ∠COB=180∘ - (60∘)= 120∘. -triangle COB is an isosceles, because OB=5 and OC=5 so the other angles in the triangle are each 30∘. -triangle ACO is an isosceles as well because AO=5 and CO=5 and since ∠COA=60∘ and because it is an isosceles, the other angles are also 60∘.(making a equilateral triangle) Let area enclosed by the line AC and arc AC be R1 Let area enclosed by line CB and arc CB be S1 Area of sector (AOC)=(60∘/360∘)pi(5)^2=R1+(25/4)(3^(0.5)) (where this is area of the equilateral triangle) Area of semicircle (pi/2)(5/2)^2=R+R1 (solve for R1 in the previous equation and plug into this one for R) same procedure for S, find area of the sector pi(r)^2(central angle/360∘) and subtract area of triangle you can use formula 0.5ab(sinC) where in this case, a=b=CO=OB and angle C=∠COB=120∘ to find S1 then, find the area of the semicircle(pi/2(r^2)) by finding CB by Pythagorean theorem because ∠ACB= 90°, AB=10 and AC=5 and dividing it by 2 because CB is the diameter finally, the area of the semicircle=S1+S , since you know S1 solve for S Then add S+R=21.651