Two parallel sides of a trapezium are 29 cm and 21 cm. Non-parallel sides are equal and each is of length 8.5 cm. What is the area of the trapezium?
Finding the Area of an Isosceles Trapezium
The question asks us to find the area of a trapezium where we know the lengths of the two parallel sides and the lengths of the two equal non-parallel sides. This specific type of trapezium, where the non-parallel sides are equal, is called an isosceles trapezium.
The given information is:
- Length of one parallel side (\(a\)) = 29 cm
- Length of the other parallel side (\(b\)) = 21 cm
- Length of each non-parallel side (\(c\)) = 8.5 cm (since they are equal)
To find the area of any trapezium, we use the formula:
\(\text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}\)
Or, using the given variables:
\(\text{Area} = \frac{1}{2} \times (a + b) \times h\)
where \(h\) is the perpendicular distance between the parallel sides (the height of the trapezium).
We know \(a\) and \(b\), but we need to find the height \(h\). In an isosceles trapezium, we can find the height by drawing perpendicular lines from the endpoints of the shorter parallel side to the longer parallel side. This creates a rectangle in the middle and two congruent right-angled triangles on the sides.
Let the longer parallel side be \(a\) (29 cm) and the shorter parallel side be \(b\) (21 cm). The difference in length between the parallel sides is \(a - b = 29 - 21 = 8\) cm.
Since the two right-angled triangles formed are congruent (due to the trapezium being isosceles), the base of each triangle will be half of this difference:
\(\text{Base of triangle} (x) = \frac{a - b}{2} = \frac{29 - 21}{2} = \frac{8}{2} = 4 \text{ cm}\)
Now consider one of the right-angled triangles. The sides are:
- Base = 4 cm
- Height = \(h\) (the height of the trapezium)
- Hypotenuse = 8.5 cm (one of the non-parallel sides)
We can use the Pythagorean theorem (\( \text{hypotenuse}^2 = \text{base}^2 + \text{height}^2 \)) to find the height \(h\):
\(8.5^2 = 4^2 + h^2\)
Calculate the squares:
\(72.25 = 16 + h^2\)
Subtract 16 from both sides to find \(h^2\):
\(h^2 = 72.25 - 16\)
\(h^2 = 56.25\)
Take the square root to find \(h\):
\(h = \sqrt{56.25}\)
\(h = 7.5 \text{ cm}\)
Now that we have the height (\(h = 7.5\) cm) and the lengths of the parallel sides (\(a = 29\) cm, \(b = 21\) cm), we can calculate the area of the trapezium:
\(\text{Area} = \frac{1}{2} \times (a + b) \times h\)
\(\text{Area} = \frac{1}{2} \times (29 + 21) \times 7.5\)
\(\text{Area} = \frac{1}{2} \times (50) \times 7.5\)
\(\text{Area} = 25 \times 7.5\)
\(\text{Area} = 187.5 \text{ square cm}\)
So, the area of the isosceles trapezium is 187.5 square cm.
Revision Table: Trapezium Area Calculation
| Concept | Formula/Method | Value Used |
|---|---|---|
| Sum of parallel sides | \(a + b\) | \(29 + 21 = 50\) cm |
| Base of right triangle | \(\frac{a - b}{2}\) | \(\frac{29 - 21}{2} = 4\) cm |
| Pythagorean Theorem | \(c^2 = x^2 + h^2\) | \(8.5^2 = 4^2 + h^2\) |
| Height (\(h\)) | \(\sqrt{c^2 - x^2}\) | \(\sqrt{8.5^2 - 4^2} = \sqrt{72.25 - 16} = \sqrt{56.25} = 7.5\) cm |
| Area of Trapezium | \(\frac{1}{2} \times (a + b) \times h\) | \(\frac{1}{2} \times 50 \times 7.5 = 187.5\) cm\(^2\) |
Additional Information: Properties of Isosceles Trapezium
- An isosceles trapezium is a quadrilateral with one pair of parallel sides (bases) and the non-parallel sides being equal in length.
- The base angles of an isosceles trapezium are equal. That is, the two angles along one parallel base are equal, and the two angles along the other parallel base are also equal.
- The diagonals of an isosceles trapezium are equal in length.
- An isosceles trapezium has a line of symmetry that passes through the midpoints of the parallel sides.
Understanding these properties helps in solving geometry problems involving trapeziums, especially when calculating lengths, angles, or areas.
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