Question

In a trapezium ABCD, seg AB || seg DC seg BD⊥seg AD,seg AC⊥seg BC,If AD=15, BC=15 and AB=25. Find A(꯳ABCD)

Answer 1

Answer:

A(꯳ABCD) 192 sq units

Step-by-step explanation:

in Δ ABD

AB² = AD² + BD²

=> 25² = 15² + BD²

=> BD = 20

Similalrly AC = 20

Area of ΔABD = (1/2) AD * BD

= (1/2) * 15 * 20

= 150 sq unit

if DM ⊥ AB

then 150 = (1/2) AB * DM

=> 300 = 25 * DM

=> DM = 12

similarly CN ⊥ AB

=> CN = 12

in Δ ADM

AM² = AD² - DM²

=> AM² = 15² - 12²

=> AM = 9

& CN = 9

MN = AB - AM - CN = 25 - 9 - 9 = 7

=> CD = MN = 7

Area of Trapezium = (1/2)( AB + CD) * DM

= (1/2) (25 + 7 ) * 12

= 192 sq units

Answer 2

we can find out the area of triangle ADB

so height for this triangle wud be equal to the height of trapezium.

so we get height = 12cm.

now we can easily get to CD=7cm

and we know area of trapezium =1/2*(AB+CD)*HEIGHT

NOW,1/2*(25+7)*12 = 192 cm.

how to get CD=7 cm

when we will drop two perpendicular lines from D to AB as point E and from C to AB as point F.

So,we can see that EF = CD (as the line drawns were perpendicular)

Now In triangleADE,

We know AD=15 and DE=12

So we can easily find AE = 9

Similarly from triangleBCF,

we will get BF = 9cm

therefore,EF = AB -(AE+BF) = 25 - 18 =7cm.

Hope it Helps.:)