Calculate the area of the shaded region between two concentric circles with radii 7 cm and 4 cm.

Let’s solve each problem one by one, step by step.

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Problem 1: Area of the shaded region (annulus)

We have two circles — a big outer circle and a small inner white circle. The shaded part is the ring between them.

- Radius of outer circle = 7 cm
- Radius of inner circle = 4 cm

Area of a circle = π × r²

So,

→ Area of outer circle = π × 7² = π × 49
→ Area of inner circle = π × 4² = π × 16

Shaded area = Outer area – Inner area
= 49π – 16π = 33π

If we use π ≈ 3.14:

33 × 3.14 = let’s calculate:

30 × 3.14 = 94.2
3 × 3.14 = 9.42
Total = 94.2 + 9.42 = 103.62 cm²

✔ So, shaded area = 103.62 cm²

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Problem 2: Two circles — bigger circle center O, radius 16 cm; smaller circle passes through O and touches the bigger circle

Let’s understand the setup:

- Big circle: center O, radius = 16 cm → so its diameter = 32 cm
- Small circle: passes through O and touches the big circle → that means the small circle is inside the big one, and it just barely touches the edge of the big circle.

Since the small circle passes through O (center of big circle) and also touches the outer edge of the big circle, the distance from O to the farthest point on the small circle must be equal to the radius of the big circle (16 cm).

That means: the diameter of the small circle = 16 cm → because from O to the touch point is 16 cm, and since O is on the small circle, the full diameter of the small circle is 16 cm.

Wait — let me think again.

Actually, if the small circle passes through O and touches the big circle externally? No — it says “touches the bigger circle” — and since it’s drawn inside, it’s internally tangent.

Better way: Let’s say the small circle has center C. Since it passes through O, then CO = radius of small circle (let’s call it r). Also, since it touches the big circle internally, the distance from O to C plus the radius of small circle equals radius of big circle.

Wait — no: If the small circle is inside the big one and touches it, then the distance between centers + small radius = big radius.

But here, the small circle passes through O — so distance from C to O is r (since O is on the small circle).

And since it touches the big circle, the distance from O to C plus r should equal 16? Wait — actually, if the small circle is inside and touches the big circle, then:

Distance between centers + small radius = big radius

But distance between centers is OC = r (because O is on the small circle)

So: r + r = 16 → 2r = 16 → r = 8 cm

Yes! That makes sense.

So small circle has radius 8 cm.

Now, shaded region = area of big circle minus area of small circle.

Big circle area = π × 16² = 256π
Small circle area = π × 8² = 64π
Shaded area = 256π – 64π = 192π

Using π ≈ 3.14:

192 × 3.14

Break it down:

200 × 3.14 = 628
Minus 8 × 3.14 = 25.12
So 628 – 25.12 = 602.88 cm²

✔ Shaded area = 602.88 cm²

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Problem 3: Isosceles triangle with base passing through center of circle. Diameter = 40 cm. Find shaded area.

First, diameter = 40 cm → radius = 20 cm

The figure shows a circle with an isosceles triangle inside it. The base of the triangle passes through the center O — meaning the base is a diameter? Wait, not necessarily — but in the diagram, it looks like the triangle is inscribed such that its base is vertical and goes through the center, and the apex is on the right side of the circle.

Actually, looking at the description: “the triangle is an isosceles triangle with its base passing through the centre of the circle.” And the shaded regions are the parts of the circle outside the triangle — specifically, left half-circle is fully shaded, and on the right, the two segments above and below the triangle are shaded.

Wait — actually, from the description and typical problems, this is likely: the triangle has its base as a chord passing through the center (so it's a diameter?), but wait — if the base passes through the center, and it’s isosceles, probably the base is horizontal or vertical.

But in the figure described, it seems the triangle is pointing to the right, with its base being a vertical line through the center — so the base is actually a diameter? Let me assume that.

Actually, re-reading: “its base passing through the centre of the circle” — doesn’t say the base is the diameter, but in many such problems, when they say that and show symmetry, the base is perpendicular to the axis of symmetry and passes through center.

But to make progress, let’s assume the triangle is formed by connecting three points on the circle: top, bottom, and rightmost point. Then the base would be from top to bottom — which is a vertical diameter. And the apex is at the right end of the horizontal diameter.

In that case, the triangle is made by points: (0,20), (0,-20), and (20,0) — assuming center at origin.

Then the base is from (0,20) to (0,-20) — length 40 cm (diameter), and height is from x=0 to x=20 — so height = 20 cm.

Area of triangle = (base × height)/2 = (40 × 20)/2 = 400 cm²

Area of circle = π × r² = π × 20² = 400π ≈ 400 × 3.14 = 1256 cm²

Shaded area = circle area – triangle area = 1256 – 400 = 856 cm²

But wait — is the entire triangle unshaded? In the description, it says “calculate the area of the shaded region”, and from the figure description, it seems the left half of the circle is shaded, and on the right, the two curved parts outside the triangle are shaded — so yes, total shaded = circle minus triangle.

But let me confirm: if the triangle is inside the circle, and shaded is everything else, then yes.

However, in some interpretations, if the base is not the full diameter, but just a chord through center — but if it passes through center and is straight, and the triangle is isosceles, likely the base is perpendicular to the line from center to apex.

But in standard problems like this, when diameter is given and triangle is isosceles with base through center, and apex on circumference, then the base is often the diameter.

Wait — another thought: if the base passes through the center, but is not necessarily the diameter — but for an isosceles triangle inscribed in a circle with base through center, the most symmetric case is when the base is perpendicular to the radius to the apex.

But to avoid confusion, let’s use coordinates.

Assume circle center at (0,0), radius 20.

Suppose the triangle has vertices at A(0,20), B(0,-20), and C(20,0). Then base AB is vertical, length 40, passes through center. Height from C to AB is horizontal distance = 20.

Area = (40 * 20)/2 = 400, as before.

Circle area = π*400 ≈ 1256

Shaded = 1256 - 400 = 856 cm²

But is this correct? Let me see if there’s another interpretation.

Perhaps the base is not the full diameter. Suppose the base is a chord through the center, but shorter. But the problem says "isosceles triangle with its base passing through the centre", and diameter is 40.

In many textbooks, this exact problem appears where the triangle is formed by two radii and a chord, but here it specifies base passing through center.

Another common version: the triangle has its apex at the top, and base horizontal through center — but same thing.

I think my calculation is correct.

But let me double-check with geometry.

The triangle described is actually a right-angled triangle? Points (0,20), (0,-20), (20,0). Distance from (0,20) to (20,0): sqrt((20)^2 + (-20)^2) = sqrt(800) = 20√2

Similarly, (0,-20) to (20,0): same.

Base from (0,20) to (0,-20): 40.

So it is isosceles, yes.

Area can also be calculated using shoelace formula:

Points: (0,20), (0,-20), (20,0)

Area = |(0*(-20 - 0) + 0*(0 - 20) + 20*(20 - (-20)))/2| = |0 + 0 + 20*40 / 2| = |800/2| = 400 — same.

So shaded area = circle - triangle = 400π - 400 = 400(π - 1)

With π≈3.14, 400*(2.14) = 856 cm²

✔ So shaded area = 856 cm²

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Final Answers:

1) 103.62 cm²
2) 602.88 cm²
3) 856 cm²

But let me write them neatly.

For problem 1, I used π=3.14, got 103.62
Problem 2: 602.88
Problem 3: 856

All seem correct.

Just to be thorough, for problem 3, is the shaded region really the whole circle minus the triangle? From the description: "the shaded region" — and in the figure, it shows the left semicircle shaded, and on the right, the two segments outside the triangle shaded — which together make up the entire circle except the triangle. Yes.

Sometimes people might think only the segments, but no — the left half is fully shaded, so total shaded is circle minus triangle.

Yes.

Final Answer:
1) 103.62 cm²
2) 602.88 cm²
3) 856 cm²