Question

Two parallel sides of an isosceles trapezium are 10 cm and 20 cm and its non-parallel sides are each equal to
13 cm. Find the area of the trapezium.
​

Answer 1

Given :

DC = 10 cm , AB = 20 , DA - CB = 13 cm.

Through C, draw CM parallel to DA meeting AB at M

Now, AM- DC= 10 em, MB= 20 cm - 10 cm = 10 cm

Draw CN ⊥ MB cm

Since, ΔCMB is isosceles so CN bisects MB

MN = NB = 5 cm

Now, from rt. ΔCMN,

CN = √CM² - MN²

= √13² -5²

= √169 - 25

= √144

= 12

Area of trapezium ABCD = ½ x height x (sum of the parallel sides)

= ½ x 12 x (10 + 20) cm²

= 6 x 30

= 180 cm²

Answer 2

Step-by-step explanation:

Given :-

• Two parallel sides of an isosceles trapezium are 10 cm and 20 cm .

• its non-parallel sides are each equal to 13 cm.

To Find : -

• Find the area of the trapezium.

Solution : -

CF ⊥ AB

So, F is the midpoint of EB.

Therefore, EF = ¹/₂ × EB = 1/2× 10= 5cm.

Thus, in right-angled ∆CFE, we have CE = 13 cm, EF = 5 cm.

By Pythagoras’ theorem : -

CF = [√CE² - EF²]

CF = √(13² - 5²)

CF= √169-25= √144 = √12×12

CF= 12cm

Thus, the distance between the parallel sides is 12cm.

Area of trapezium ABCD = ¹/₂ × (sum of parallel sides) × (distance between them)

Area of trapezium ABCD = ¹/₂ × (20 + 10) × 12 cm²

Area of trapezium ABCD =

= 1/2×(30)×12

= 30 × 6

= 180 cm²

Hence, Area of trapezium ABCD= 180 cm²

more information : -

Trapezoid : -

• A trapezoid, also known as a trapezium, is a flat closed shape having 4 straight sides, with one pair of parallel sides.

• The parallel sides of a trapezium are known as the bases, and its non-parallel sides are called legs. .

• The parallel sides can be horizontal, vertical or slanting.