Math Problem Statement

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

Solution

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

Step 1: Area of Triangle ABDABD

Since A=90\angle A = 90^\circ, ABD\triangle ABD is a right-angled triangle. The area of a right-angled triangle can be calculated using the formula: Area of ABD=12×base×height\text{Area of } \triangle ABD = \frac{1}{2} \times \text{base} \times \text{height} Here, AB=32cmAB = 32 \, \text{cm} and AD=24cmAD = 24 \, \text{cm}, so: Area of ABD=12×32cm×24cm=12×768cm2=384cm2\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 BCDBCD

Now, BCD\triangle BCD is an isosceles triangle with BC=CD=25cmBC = 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=25cmBC = CD = 25 \, \text{cm} and BDBD can be calculated using the Pythagorean theorem from ABD\triangle ABD.

Step 2.1: Calculate BDBD

From ABD\triangle ABD, we use the Pythagorean theorem: BD2=AB2+AD2=322+242=1024+576=1600BD^2 = AB^2 + AD^2 = 32^2 + 24^2 = 1024 + 576 = 1600 BD=1600=40cmBD = \sqrt{1600} = 40 \, \text{cm}

Step 2.2: Use Heron's Formula for BCD\triangle BCD

The semi-perimeter ss of BCD\triangle BCD is: s=BC+CD+BD2=25+25+402=902=45cms = \frac{BC + CD + BD}{2} = \frac{25 + 25 + 40}{2} = \frac{90}{2} = 45 \, \text{cm} Now, apply Heron’s formula: Area of BCD=s(sa)(sb)(sc)\text{Area of } \triangle BCD = \sqrt{s(s-a)(s-b)(s-c)} where a=25cm,b=25cm,c=40cma = 25 \, \text{cm}, b = 25 \, \text{cm}, c = 40 \, \text{cm}: Area of BCD=45(4525)(4525)(4540)=45×20×20×5=90000=300cm2\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 ABCDABCD

The total area of quadrilateral ABCDABCD is the sum of the areas of ABD\triangle ABD and BCD\triangle BCD: Total Area=384cm2+300cm2=684cm2\text{Total Area} = 384 \, \text{cm}^2 + 300 \, \text{cm}^2 = 684 \, \text{cm}^2

Thus, the area of the quadrilateral ABCDABCD is 684 cm².

Would you like more details or have any questions?


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.

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Math Problem Analysis

Mathematical Concepts

Geometry
Area Calculation
Triangles
Quadrilaterals

Formulas

Area of a right-angled triangle = 1/2 * base * height
Pythagorean theorem: a^2 + b^2 = c^2
Heron's formula for triangle area: A = sqrt(s(s-a)(s-b)(s-c))

Theorems

Pythagorean theorem
Heron's formula

Suitable Grade Level

Grades 9-11