Q2. Calculate the area of a quadrilateral ABCD in which angle A = 900, AB = 32cm, AD = 24cm and BC = CD = 25cm. 

The given quadrilateral \( ABCD \) has one right angle at \( A \), i.e., \( \angle A = 90^\circ \). We can divide the quadrilateral into two triangles, \( \triangle ABD \) and \( \triangle BCD \), and calculate the area of each triangle separately.

### Step 1: Area of Triangle \( ABD \)
Since \( \angle A = 90^\circ \), \( \triangle ABD \) is a right-angled triangle. The area of a right-angled triangle can be calculated using the formula:
\[
\text{Area of } \triangle ABD = \frac{1}{2} \times \text{base} \times \text{height}
\]
Here, \( AB = 32 \, \text{cm} \) and \( AD = 24 \, \text{cm} \), so:
\[
\text{Area of } \triangle ABD = \frac{1}{2} \times 32 \, \text{cm} \times 24 \, \text{cm} = \frac{1}{2} \times 768 \, \text{cm}^2 = 384 \, \text{cm}^2
\]

### Step 2: Area of Triangle \( BCD \)
Now, \( \triangle BCD \) is an isosceles triangle with \( BC = CD = 25 \, \text{cm} \). To find the area, we first need to determine the height of this triangle. 

Using Heron's formula:
- The sides of the triangle are \( BC = CD = 25 \, \text{cm} \) and \( BD \) can be calculated using the Pythagorean theorem from \( \triangle ABD \).

### Step 2.1: Calculate \( BD \)
From \( \triangle ABD \), we use the Pythagorean theorem:
\[
BD^2 = AB^2 + AD^2 = 32^2 + 24^2 = 1024 + 576 = 1600
\]
\[
BD = \sqrt{1600} = 40 \, \text{cm}
\]

### Step 2.2: Use Heron's Formula for \( \triangle BCD \)
The semi-perimeter \( s \) of \( \triangle BCD \) is:
\[
s = \frac{BC + CD + BD}{2} = \frac{25 + 25 + 40}{2} = \frac{90}{2} = 45 \, \text{cm}
\]
Now, apply Heron’s formula:
\[
\text{Area of } \triangle BCD = \sqrt{s(s-a)(s-b)(s-c)}
\]
where \( a = 25 \, \text{cm}, b = 25 \, \text{cm}, c = 40 \, \text{cm} \):
\[
\text{Area of } \triangle BCD = \sqrt{45(45-25)(45-25)(45-40)} = \sqrt{45 \times 20 \times 20 \times 5} = \sqrt{90000} = 300 \, \text{cm}^2
\]

### Step 3: Total Area of Quadrilateral \( ABCD \)
The total area of quadrilateral \( ABCD \) is the sum of the areas of \( \triangle ABD \) and \( \triangle BCD \):
\[
\text{Total Area} = 384 \, \text{cm}^2 + 300 \, \text{cm}^2 = 684 \, \text{cm}^2
\]

Thus, the area of the quadrilateral \( ABCD \) is **684 cm²**.

Would you like more details or have any questions?

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### Related questions:
1. How is Heron's formula applied to find the area of a triangle?
2. How can the Pythagorean theorem be used in different triangle problems?
3. What are the properties of an isosceles triangle?
4. How would the problem change if one of the sides was different?
5. Can you use trigonometry to calculate the area of a quadrilateral?

**Tip:** Always check if a quadrilateral can be split into simpler geometric shapes to make calculations easier.

Q2. Calculate the area of a quadrilateral ABCD in which angle A = 90°, AB = 32cm, AD = 24cm and BC = CD = 25cm.

Find the area of a quadrilateral ABCD where angle A = 90°, AB = 32 cm, AD = 24 cm, and BC = CD = 25 cm. This step-by-step solution covers triangle area calculations using the Pythagorean theorem and Heron's formula. Ideal for students in Grades 9-11.

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