Math, asked by StarTbia, 1 year ago

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)

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Answers

Answered by amitnrw
72

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

Answered by wasifthegreat786
15

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.:)

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