Question

The parallel sides of a trapezium are 40 cm and 70 cm. If the non-parallel
sides are equal, each being 25 cm, find the area of the trapezium.
​

Answer 1

Answer:Area of Trapezium is 1100 sq.cm.

Step-by-step explanation:

We have drawn the diagram for your reference.

Given:

The parallel sides of trapezium are 40 cm and 70 cm.

hence according to diagram:

AD = 40 cm

BC = 70 cm

Also Given:

Its non parallel sides are equal 25 cm each.

AB = 25 cm

CD =25 cm

So first we will draw imaginary lines AE and DF which are perpendicular BC such that;

EF = AD = 40 cm

so,

BE = FC = 15 cm each

Now In triangle Δ AEB.

m∠E = 90°

Hence By Pythagoras theorem we get;

Substituting the values we get;

Now Taking Square root on both side we get;

Now we need to find the Area of trapezium.

Area of trapezium is equal to half times sum of parallel side times distance between them.

framing in equation form we get;

Area of trapezium =

Hence Area of Trapezium is 1100 sq.cm.

Answer 2

ANSWER :-

Let ABCD be the given trapezium in which

⟹ AB || DC

• AB = 70 cm
• DC = 40 cm
• AD = BC = 25 cm

Draw CE | AB and CF || DA (see the fig.)

meeting AB at E and F respectively .

Clearly AFCD is a parallelogram

★ AF = CD = 40 cm &

★ CF = DA = 25 cm

[opposite sides of a || gm are equal]

FB =

⟹ AB - AF

⟹ ( 70 - 40 ) cm

⟹ 30 cm

In ∆ CFB , we have CF = CB = 25 cm

★ ∆ CFB is an isosceles triangle and CE I FB

⟹ E is the mid point of FB

⟹ FE = EB = ½ × FB

$⟹( \dfrac{1}{2} \times 30)cm$

⟹ 15 cm

From the right triangle CEF, we have

⟹ CE² = CF² - FE²

⟹ CE² = (25 cm)² - (15 cm)²

⟹ CE² = (625 - 225 cm)²

⟹ CE² = 400 cm²

⟹ CE = 20 cm

height of the trapezium = 20 cm

Area of trapezium ABCD

⟹ [ ½ ( 40 + 70 ) × 20 ] cm²

⟹ [ ½ ( 110 ) × 20 ] cm²

⟹ 1100 cm²