10th Grade Math Review Worksheet with area calculations and algebraic expressions.

Let's solve each problem step by step.

---

Problem 1: Find the area of the rectangle.

#### Given:
- Length of the rectangle = 5.2 m
- Width of the rectangle = 3 m

#### Formula for the area of a rectangle:
\[
\text{Area} = \text{Length} \times \text{Width}
\]

#### Calculation:
\[
\text{Area} = 5.2 \, \text{m} \times 3 \, \text{m} = 15.6 \, \text{m}^2
\]

#### Answer:
\[
\boxed{15.6}
\]

---

Problem 2: Find the area of the triangle.

#### Given:
- Base of the triangle = 6 cm
- Height of the triangle = 4.5 cm

#### Formula for the area of a triangle:
\[
\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}
\]

#### Calculation:
\[
\text{Area} = \frac{1}{2} \times 6 \, \text{cm} \times 4.5 \, \text{cm} = \frac{1}{2} \times 27 \, \text{cm}^2 = 13.5 \, \text{cm}^2
\]

#### Answer:
\[
\boxed{13.5}
\]

---

Problem 3: Simplify \( -2\sqrt{5} + 3\sqrt{5} \).

#### Given:
\[
-2\sqrt{5} + 3\sqrt{5}
\]

#### Explanation:
Both terms have the same radical part (\(\sqrt{5}\)), so we can combine them by adding their coefficients:
\[
-2\sqrt{5} + 3\sqrt{5} = (-2 + 3)\sqrt{5} = 1\sqrt{5} = \sqrt{5}
\]

#### Answer:
\[
\boxed{\sqrt{5}}
\]

---

Problem 4: Simplify \( 6\sqrt{3} - 2\sqrt{3} \).

#### Given:
\[
6\sqrt{3} - 2\sqrt{3}
\]

#### Explanation:
Both terms have the same radical part (\(\sqrt{3}\)), so we can combine them by subtracting their coefficients:
\[
6\sqrt{3} - 2\sqrt{3} = (6 - 2)\sqrt{3} = 4\sqrt{3}
\]

#### Answer:
\[
\boxed{4\sqrt{3}}
\]

---

Final Answers:

1. \(\boxed{15.6}\)
2. \(\boxed{13.5}\)
3. \(\boxed{\sqrt{5}}\)
4. \(\boxed{4\sqrt{3}}\)