Question

An isosceles trapezium ABCD has parallel sides 10 cm and 18 cm.
Its non parallel sides are 5 cm each. Find its height and area.
(Hint : Draw BE I DE and BF || AD
Then BF = 5 cm, EF = 5 FC = 4 cm. Find BE.]
1
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Answer 1

Answer:

Height=3cm

Hope this answer helps you!

Answer 2

Given:

• ABCD is an isosceles Trapezium
• Parallel sides are 10 cm and 18 cm
• Non-parallel side is 5cm

To Find:

• Height and area of the the figure

* See attached for the visual representation of the question.

Find the length DE:

$\text{DE} = \text{FC}$

$\text{DE} = (18 - 10) \div 2$

$\text{DE} = 4 \text { cm}$

Find the height AE:

Triangle ABE is a right angle triangle. So we can use Pythagoras Theorem to find the missing side.

$a^2 + b^2 = c^2$

$AE^2 + DB^2 = AD^2$

$AE^2 + 4^2 = 5^2$

$AE^2 = 5^2 - 4^2$

$AE^2 = 9$

$AE = 3 \text { cm}$

Answer: The height is 3 cm.

Find the area of the trapezium:

$\text{Area of a trapezium} = \dfrac{1}{2} (a + b) \times h$

$\text{Area of the trapezium} = \dfrac{1}{2} (10 + 18) \times (3)$

$\text{Area of the trapezium} = 42 \text { cm}^2$

Answer: The area is 42 cm²