Question

A semicircle with diameter PQ sits on an isosceles triangle PQR to form a region shaped like a two-dimensional ice- cream cone, as shown in the figure. If a(x) is the area of the semicircle and b(x) is the area of the triangle, find lim a(x)/b(x) as x→0

Answer 1

The limit of a(x)/b(x) as x approaches 0 is gotten as;

lim a(x)/b(x); x→0 = 0

• The image showing the semi circle and isosceles triangle is missing and so i have attached it.

• Formula for area of a semi circle is;

a(x) = A = ¹/₂πr²

• Area of triangle is;

b(x) = A =  ¹/₂ × base × height

• From the image, ∠PQR = θ

Thus; a(x) = ¹/₂π(10 sin (θ/2))²

a(x) =  ¹/₂π(100 sin² (θ/2))

Similarly;

b(x) =  ¹/₂(20 sin (θ/2)) × (10 cos (θ/2))

b(x) = 100 sin (θ/2) cos (θ/2)

• Thus;

a(x)/b(x) = ¹/₂π(100 sin² (θ/2)) ÷ 100 sin (θ/2) cos (θ/2)

100 will cancel out and the sin (θ/2) at the denominator will be eradicated

upon further simplification to give;

a(x)/b(x) = ¹/₂π(sin (θ/2)) ÷ cos (θ/2)

In trigonometry, we know that; tan θ = sin θ/cos θ

Thus;

a(x)/b(x) = ¹/₂π(sin (θ/2)) ÷ cos (θ/2) =  ¹/₂π(tan (θ/2))

We want to find the limit as θ approaches zero.

Thus;

lim θ → 0 =  ¹/₂π(tan (0/2))

⇒ ¹/₂π tan 0

⇒ 0

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