Area of a Kite and Rhombus online exercise for - Free Printable

Problem Analysis

The task involves calculating the areas of rhombuses and kites using the formula for their respective areas. The general formulas are:

1. Area of a Rhombus:
\[
A = \frac{1}{2} \times d_1 \times d_2
\]
where \( d_1 \) and \( d_2 \) are the lengths of the diagonals.

2. Area of a Kite:
\[
A = \frac{1}{2} \times d_1 \times d_2
\]
where \( d_1 \) and \( d_2 \) are the lengths of the diagonals.

We will solve each part step by step.

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Problem 4: Find the area of each rhombus

#### Part (a)
- Diagonals: \( d_1 = 8 \, \text{cm} \), \( d_2 = 6 \, \text{cm} \)
- Formula:
\[
A = \frac{1}{2} \times d_1 \times d_2
\]
- Substituting the values:
\[
A = \frac{1}{2} \times 8 \times 6 = \frac{1}{2} \times 48 = 24 \, \text{cm}^2
\]

#### Part (b)
- Diagonals: \( d_1 = 3.6 \, \text{m} \), \( d_2 = 1.7 \, \text{m} \)
- Formula:
\[
A = \frac{1}{2} \times d_1 \times d_2
\]
- Substituting the values:
\[
A = \frac{1}{2} \times 3.6 \times 1.7 = \frac{1}{2} \times 6.12 = 3.06 \, \text{m}^2
\]

Answers for Problem 4:
- (a) \( 24 \, \text{cm}^2 \)
- (b) \( 3.06 \, \text{m}^2 \)

---

Problem 5: Find the area of each rhombus

#### Part (a)
- Diagonals: \( d_1 = 10.8 \, \text{cm} \), \( d_2 = 6 \, \text{cm} \)
- Formula:
\[
A = \frac{1}{2} \times d_1 \times d_2
\]
- Substituting the values:
\[
A = \frac{1}{2} \times 10.8 \times 6 = \frac{1}{2} \times 64.8 = 32.4 \, \text{cm}^2
\]

#### Part (b)
- Diagonals: \( d_1 = 13 \, \text{cm} \), \( d_2 = 7 \, \text{cm} \)
- Formula:
\[
A = \frac{1}{2} \times d_1 \times d_2
\]
- Substituting the values:
\[
A = \frac{1}{2} \times 13 \times 7 = \frac{1}{2} \times 91 = 45.5 \, \text{cm}^2
\]

Answers for Problem 5:
- (a) \( 32.4 \, \text{cm}^2 \)
- (b) \( 45.5 \, \text{cm}^2 \)

---

Problem 6: Complete to find the area of each kite

#### Part (a)
- Diagonals: \( d_1 = 10 \, \text{cm} \), \( d_2 = 8 \, \text{cm} \)
- Formula:
\[
A = \frac{1}{2} \times d_1 \times d_2
\]
- Substituting the values:
\[
A = \frac{1}{2} \times 10 \times 8 = \frac{1}{2} \times 80 = 40 \, \text{cm}^2
\]

#### Part (b)
- Diagonals: \( d_1 = 15 \, \text{cm} \), \( d_2 = 7 \, \text{cm} \)
- Formula:
\[
A = \frac{1}{2} \times d_1 \times d_2
\]
- Substituting the values:
\[
A = \frac{1}{2} \times 15 \times 7 = \frac{1}{2} \times 105 = 52.5 \, \text{cm}^2
\]

Answers for Problem 6:
- (a) \( 40 \, \text{cm}^2 \)
- (b) \( 52.5 \, \text{cm}^2 \)

---

Problem 7: Find the area of each kite

#### Part (a)
- Diagonals: \( d_1 = 19 \, \text{cm} \), \( d_2 = 6 \, \text{cm} \)
- Formula:
\[
A = \frac{1}{2} \times d_1 \times d_2
\]
- Substituting the values:
\[
A = \frac{1}{2} \times 19 \times 6 = \frac{1}{2} \times 114 = 57 \, \text{cm}^2
\]

#### Part (b)
- Diagonals: \( d_1 = 31 \, \text{cm} \), \( d_2 = 15 \, \text{cm} \)
- Formula:
\[
A = \frac{1}{2} \times d_1 \times d_2
\]
- Substituting the values:
\[
A = \frac{1}{2} \times 31 \times 15 = \frac{1}{2} \times 465 = 232.5 \, \text{cm}^2
\]

Answers for Problem 7:
- (a) \( 57 \, \text{cm}^2 \)
- (b) \( 232.5 \, \text{cm}^2 \)

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Final Answers:

\[
\boxed{
\begin{array}{ll}
\text{Problem 4:} & \text{(a) } 24 \, \text{cm}^2, \text{(b) } 3.06 \, \text{m}^2 \\
\text{Problem 5:} & \text{(a) } 32.4 \, \text{cm}^2, \text{(b) } 45.5 \, \text{cm}^2 \\
\text{Problem 6:} & \text{(a) } 40 \, \text{cm}^2, \text{(b) } 52.5 \, \text{cm}^2 \\
\text{Problem 7:} & \text{(a) } 57 \, \text{cm}^2, \text{(b) } 232.5 \, \text{cm}^2 \\
\end{array}
}
\]