Math worksheet for calculating areas of shapes using algebraic expressions.

Let's solve each part of this worksheet step by step, forming expressions for the area of each shape in terms of $ x $ or in surd form.

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🔴 Section 1: RED

#### a. Rectangle
- Length = $ x + 3 $
- Width = $ x - 5 $

$$
\text{Area} = (x + 3)(x - 5) = x^2 - 5x + 3x - 15 = x^2 - 2x - 15
$$

✔ Answer: $ x^2 - 2x - 15 $

---

#### b. Rectangle
- Height = $ 2x + 3 $
- Width = $ 3x - 2 $

$$
\text{Area} = (2x + 3)(3x - 2) = 6x^2 - 4x + 9x - 6 = 6x^2 + 5x - 6
$$

✔ Answer: $ 6x^2 + 5x - 6 $

---

#### c. Right triangle
- Base = $ 2 - \sqrt{3} $
- Height = $ 5 + \sqrt{3} $

Area of triangle = $ \frac{1}{2} \times \text{base} \times \text{height} $

$$
= \frac{1}{2} (2 - \sqrt{3})(5 + \sqrt{3})
$$

Now expand:

$$
(2 - \sqrt{3})(5 + \sqrt{3}) = 2 \cdot 5 + 2 \cdot \sqrt{3} - 5\sqrt{3} - \sqrt{3} \cdot \sqrt{3}
= 10 + 2\sqrt{3} - 5\sqrt{3} - 3 = 7 - 3\sqrt{3}
$$

Then multiply by $ \frac{1}{2} $:

$$
\text{Area} = \frac{1}{2}(7 - 3\sqrt{3}) = \frac{7}{2} - \frac{3\sqrt{3}}{2}
$$

✔ Answer: $ \frac{7 - 3\sqrt{3}}{2} $

---

🟡 Section 2: AMBER

#### a. L-shaped figure

Break into two rectangles:

- Top rectangle:
- Width = $ 2x + 1 $
- Height = $ 6x - 2 - 2x = 4x - 2 $
- Area = $ (2x + 1)(4x - 2) $

- Bottom rectangle:
- Width = $ 3x + 4 - (2x + 1) = x + 3 $
- Height = $ 2x $
- Area = $ 2x(x + 3) $

Now compute both:

Top:
$$
(2x + 1)(4x - 2) = 8x^2 - 4x + 4x - 2 = 8x^2 - 2
$$

Bottom:
$$
2x(x + 3) = 2x^2 + 6x
$$

Total area:
$$
8x^2 - 2 + 2x^2 + 6x = 10x^2 + 6x - 2
$$

✔ Answer: $ 10x^2 + 6x - 2 $

---

#### b. Irregular quadrilateral (trapezoid-like)

We can split it into a rectangle and a right triangle.

From the diagram:
- Left side: height = $ 8x + 2 $
- Bottom base: $ 5x + 6 $
- Top base: $ 4x + 2 $
- Right side: height = $ 4x $

So, the vertical drop on the right is $ (8x + 2) - 4x = 4x + 2 $

The horizontal overhang on the bottom is $ (5x + 6) - (4x + 2) = x + 4 $

So we have:
- Rectangle: width $ 4x + 2 $, height $ 4x $
- Right triangle: base $ x + 4 $, height $ 4x + 2 $

But wait — actually, better to think of it as a rectangle with a triangle added.

Alternatively, use trapezoid formula, but since it's not a trapezoid with parallel sides, we'll break it.

Actually, it's easier to split into:
- A rectangle of width $ 4x + 2 $, height $ 4x $
- A right triangle on the right: base = $ (5x + 6) - (4x + 2) = x + 4 $, height = $ (8x + 2) - 4x = 4x + 2 $

Wait — no, the height difference is from top to bottom.

Better: The total height is $ 8x + 2 $, and the right side drops to $ 4x $, so the vertical leg of the triangle is $ (8x + 2) - 4x = 4x + 2 $

And the horizontal leg is $ (5x + 6) - (4x + 2) = x + 4 $

So:
- Rectangle area: $ (4x + 2) \times 4x = 16x^2 + 8x $
- Triangle area: $ \frac{1}{2}(x + 4)(4x + 2) $

Compute triangle:
$$
\frac{1}{2}(x + 4)(4x + 2) = \frac{1}{2}[4x^2 + 2x + 16x + 8] = \frac{1}{2}(4x^2 + 18x + 8) = 2x^2 + 9x + 4
$$

Total area:
$$
16x^2 + 8x + 2x^2 + 9x + 4 = 18x^2 + 17x + 4
$$

✔ Answer: $ 18x^2 + 17x + 4 $

---

#### c. Circle

Diameter = $ 4x - 2 $

So radius = $ \frac{4x - 2}{2} = 2x - 1 $

Area = $ \pi r^2 = \pi (2x - 1)^2 $

$$
= \pi (4x^2 - 4x + 1)
$$

✔ Answer: $ \pi(4x^2 - 4x + 1) $

---

🟢 Section 3: GREEN

#### a. L-shaped figure with surds

Dimensions:
- Total height: $ 10 + \sqrt{3} $
- Top width: $ 5 + \sqrt{6} $
- Bottom width: $ \sqrt{6} $
- Vertical segment: $ \sqrt{3} $

We can split into:
- Top rectangle: width $ 5 + \sqrt{6} $, height $ \sqrt{3} $
- Bottom rectangle: width $ \sqrt{6} $, height $ 10 + \sqrt{3} - \sqrt{3} = 10 $

So:
- Top area: $ (5 + \sqrt{6})\sqrt{3} = 5\sqrt{3} + \sqrt{18} = 5\sqrt{3} + 3\sqrt{2} $
- Bottom area: $ \sqrt{6} \times 10 = 10\sqrt{6} $

Total area:
$$
5\sqrt{3} + 3\sqrt{2} + 10\sqrt{6}
$$

✔ Answer: $ 5\sqrt{3} + 3\sqrt{2} + 10\sqrt{6} $

---

#### b. Irregular pentagon

Looks like a rectangle with a triangle cut off?

But actually, it's a trapezoid with a slanted side.

Wait — it has:
- Left side: $ 10 - \sqrt{5} $
- Right side: $ \sqrt{5} $
- Bottom: $ 3 + \sqrt{5} $
- Top: $ \sqrt{5} $
- But top is shorter than bottom? Wait — let’s analyze.

Actually, it appears that:
- The left side is vertical: $ 10 - \sqrt{5} $
- The right side is slanted down
- The top is $ \sqrt{5} $
- The bottom is $ 3 + \sqrt{5} $

But the top is only $ \sqrt{5} $ long, and the bottom is longer.

So likely, we can split it into:
- Rectangle: width $ \sqrt{5} $, height $ \sqrt{5} $
- Trapezoid or triangle?

Wait — better: Split into a rectangle and a right triangle.

From the left:
- The full height is $ 10 - \sqrt{5} $
- The right side is $ \sqrt{5} $, so the vertical drop is $ (10 - \sqrt{5}) - \sqrt{5} = 10 - 2\sqrt{5} $

Wait — perhaps it's a trapezoid with parallel sides being the top and bottom?

But top is $ \sqrt{5} $, bottom is $ 3 + \sqrt{5} $, and they are both horizontal.

Yes — this is a trapezoid with:
- Parallel sides: top = $ \sqrt{5} $, bottom = $ 3 + \sqrt{5} $
- Height = $ 10 - \sqrt{5} $ (vertical distance between them)

Area of trapezoid = $ \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height} $

$$
= \frac{1}{2} (\sqrt{5} + 3 + \sqrt{5}) (10 - \sqrt{5}) = \frac{1}{2}(3 + 2\sqrt{5})(10 - \sqrt{5})
$$

Now compute:

First: $ (3 + 2\sqrt{5})(10 - \sqrt{5}) $

$$
= 3 \cdot 10 + 3 \cdot (-\sqrt{5}) + 2\sqrt{5} \cdot 10 + 2\sqrt{5} \cdot (-\sqrt{5})
= 30 - 3\sqrt{5} + 20\sqrt{5} - 2 \cdot 5
= 30 - 3\sqrt{5} + 20\sqrt{5} - 10
= 20 + 17\sqrt{5}
$$

Then divide by 2:

$$
\text{Area} = \frac{1}{2}(20 + 17\sqrt{5}) = 10 + \frac{17\sqrt{5}}{2}
$$

✔ Answer: $ 10 + \frac{17\sqrt{5}}{2} $

---

#### c. Sector with a missing piece

This is a circle with a sector removed.

Given: Radius = $ 2 + \sqrt{8} $

Note: $ \sqrt{8} = 2\sqrt{2} $, so radius = $ 2 + 2\sqrt{2} $

But the angle is not given. However, it looks like a quarter circle with a smaller quarter circle removed?

Wait — actually, it's a semicircle with a sector removed?

Looking at the image: it appears to be a full circle with a sector missing — specifically, a quarter circle missing (90°), so remaining is $ \frac{3}{4} $ of a circle.

But the radius is $ 2 + \sqrt{8} = 2 + 2\sqrt{2} $

So area of full circle:
$$
\pi r^2 = \pi (2 + 2\sqrt{2})^2 = \pi [4 + 8\sqrt{2} + 8] = \pi (12 + 8\sqrt{2})
$$

Wait — let's compute:
$$
(2 + 2\sqrt{2})^2 = 4 + 2 \cdot 2 \cdot 2\sqrt{2} + (2\sqrt{2})^2 = 4 + 8\sqrt{2} + 8 = 12 + 8\sqrt{2}
$$

So full circle: $ \pi(12 + 8\sqrt{2}) $

If one-quarter is removed, then area = $ \frac{3}{4} \pi(12 + 8\sqrt{2}) $

Factor:
$$
= \frac{3}{4} \pi \cdot 4(3 + 2\sqrt{2}) = 3\pi(3 + 2\sqrt{2}) = 9\pi + 6\pi\sqrt{2}
$$

But is it really a quarter removed? The diagram shows a "bite" taken out — if it's a 90° sector, then yes.

Assuming it's a three-quarters circle, then:

✔ Answer: $ \frac{3}{4} \pi (2 + \sqrt{8})^2 = \frac{3}{4} \pi (12 + 8\sqrt{2}) = 9\pi + 6\pi\sqrt{2} $

Alternatively, leave as:
$$
\frac{3}{4} \pi (12 + 8\sqrt{2}) = 3\pi(3 + 2\sqrt{2})
$$

✔ Answer: $ 3\pi(3 + 2\sqrt{2}) $

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💀 KILLER QUESTION

Trapezium ABCD:
- AB = $ x $
- CD = $ x + 10 $ cm (10 cm more than AB)
- Distance between AB and CD = $ x + 5 $ cm (5 cm more than AB)

We need to find area of trapezium.

Formula for area of trapezium:
$$
\text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}
$$

Here:
- Parallel sides: AB and CD → $ x $ and $ x + 10 $
- Height = $ x + 5 $

So:
$$
\text{Area} = \frac{1}{2} (x + x + 10)(x + 5) = \frac{1}{2}(2x + 10)(x + 5)
$$

Simplify:
$$
= (x + 5)(x + 5) = (x + 5)^2
$$

✔ Answer: $ (x + 5)^2 $

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✔ Final Answers Summary

#### Section 1: RED
a. $ x^2 - 2x - 15 $
b. $ 6x^2 + 5x - 6 $
c. $ \frac{7 - 3\sqrt{3}}{2} $

#### Section 2: AMBER
a. $ 10x^2 + 6x - 2 $
b. $ 18x^2 + 17x + 4 $
c. $ \pi(4x^2 - 4x + 1) $

#### Section 3: GREEN
a. $ 5\sqrt{3} + 3\sqrt{2} + 10\sqrt{6} $
b. $ 10 + \frac{17\sqrt{5}}{2} $
c. $ 3\pi(3 + 2\sqrt{2}) $

#### KILLER QUESTION:
$ (x + 5)^2 $

Let me know if you'd like these simplified further or explained in steps!