Question

Find the area of trapeizium ABCD;AB||DC i which AB =9cm, DC=4cm, AD=5cm and BC=5cm.​

Answer 1

Step-by-step explanation:

ABCD is an isosceles trapezium with AB= 9 cm,DC = 4 cm, AD = 5 cm and BC = 5 cm.

The distance between AB and CD = [5^2–{(9–4)/2}^2]^0.5

= [25–6.25]^0.5

= 4.33 cm

So the area of the trapezium, ABCD = (9+4)*4.33/2

= 28.15 sq cm. Answer.

Answer 2

• GIVEN:-

AB = 9 cm (1st parallel side)

BC = 5 cm (2nd parallel side)

AD = 5 cm ( distance between them)

(refer to the attachment)

• To Find:-

Area of trapezium.

• SOLUTION:-

Formula of area of trapezium,

$\large\boxed{\sf{Area=\dfrac{1}{2}\times(sum\:of\:parallel\:sides)\times(distance\:between\:them)}}$

According to the question,

$\large\Longrightarrow{\sf{Area=\dfrac{1}{2}\times(9+4)\times5}}$

$\large\Longrightarrow{\sf{Area=\dfrac{1}{2}\times13\times5}}$

$\large\Longrightarrow{\sf{Area=\dfrac{1}{2}\times65}}$

$\large\Longrightarrow{\sf{Area=\dfrac{65}{2}}}$

$\large\therefore{\sf{Area=32.5\:{cm}^{2}}}$

$\huge\orange\therefore\boxed{\sf{\orange{Area=32.5\:{cm}^{2}}}}$