Question

Find the height of the trapezium whose parallel sides are 77 cm and 60 cm and other sides are 25 cm and 26 cm

Answer 1

Answer:

Draw AE || BC, through A.

Draw AF perpendicular to  CD

For triangle ADE,

s = 21 cm,

Area of triangle ADE=    = root s (s-a) (s-b) (s-c)

= root 21 * 7 * 8 *6

= 84 cm^2

1/2*base* height = 84 cm ^2

Height =   84*2/15 = 56/5

= 11.2 cm

Answer 2

Answer:

24 cm

Step-by-step explanation:

Figure in attachment :

Let ABCD be our trapezium .

CD = 77 cm

AB = 60 cm

AD = 26 cm

BC = 25 cm

Let FC = x

In BFC,

BF² = BC² - FC²

⇒ h² = 25² - x² ........(1)

Also in ADE

AD² = DE² + AE²

⇒ 25² = h² + ( 77 - 60 - x )²

⇒ h² = 26² - ( 17 - x )² .........(2)

From 1 and 2 :

25² - x² = 26² - ( 17 - x )²

⇒ 25² - x² = 26² - ( 17² + x² - 34 x )

⇒ 625 - x² = 676 - 289 - x² + 34 x

⇒ 34 x = 238

⇒ x = 238/34 = 7

Now in BFC :

h² = 25² - x²

⇒ h² = 625 - 7²

⇒ h² = 625 - 49

⇒ h² = 576

⇒ h = 24

Hence height is 24 cm .