Math, asked by harshita5574, 1 year ago

Two parallel sides DC and AB of a Trapezium are 12 cm and 36 cm respectively. If non parallel sides are each equal to 15 cm find the area of a Trapezium

Answers

Answered by ANGELNIVI
10
Let ABCD be the trapezium & AB = 36 cm, DC = 12 cm and AD = 12cm.

Now, EB = (AB-AE) = (AB-DC)

               = 36 - 12

               = 24cm.

In triangle EBC, CE = BC = 15cm.

CF is perpendicular to AB.

F is the midpoint of EB.

EF = 1/2 * AB

     = 1/2 * EB

    = 1/2 * 24

   = 12cm.

In right- angled triangle CFE, CF = 15cm, EF = 12cm.

By Pythagoras theorem, we have

CF = root CE^2 - EF^2

     = (15^2 - 12^2)

     = root 225 - 144

     = root 81

     = 9.

The distance between the parallel sides = 9cm.

Therefore, Area of the trapezium = 1/2 * (sum of parallel sides) * (distance)

                                                        = 1/2 * (12+36) * 9

                                                        = 432/2

                                                        = 216 cm^2.

Answered by PreetiGupta2006
9

➩Given:

(i) ABCD is a Trapezium

where DC||AB

(ii) AD=CB=15cm

(iii) DC=12cm

(iv) AB=36cm

➩To Find:

The Area of the Trapezium ABCD

➩Solution:

If DC||AB (given)

then DC||AE (parts of parallel lines)

Draw a line CE||AD

Now

AD||CE

DC||AE

Thus, AECD is a Parallelogram.

AD=CE (Opposite sides of parallelogram are always equal)

∴CE=15cm

Similarly,

DC=AE (Opposite sides of parallelogram are always equal)

∴AE=12cm

AE+EB=36cm

12+EB=36

EB=36-12

EB=24cm

Draw an Altitude CF⊥EB (Construction)

EF=FB (when an Altitude falls on a line ,it besects it or divide it into two equal parts)

EF+FB=24

FB+FB=24(as EF=FB)

2FB=24

FB = \frac{24}{2}

FB=12cm

Now in △CFB

we can use Pythagoras theorem as ∠CFB=90°

 {(CB)}^{2} = {(FB)}^{2} =  {(CF)}^{2}

 {(15)}^{2} = {(12)}^{2} =  {(CF)}^{2}

225 = 144 =  {(CF)}^{2}

225 - 144 = {(CF)}^{2}

81=  {(CF)}^{2}

 \sqrt{81}  = CF

CF=9cm

Also CF is the Height of the Trapezium ABCD

Area of Trapezium=

 \frac{1}{2} ×Height×(Sum \: of \: parallel \:sides)

= \frac{1}{2} ×9×(12+36)

 = \frac{1}{2} ×9×48

 =9×24

 = {(216)}cm^{2}

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