Question

please solve this question

Answer 1

Continued….

∆ADP is right angled triangle

AD = 25 cm, DP = 15 cm, AP is altitude or perpendicula height.

AD^2 = DP^2 + AP^2 (Pythagoras theorem)

25^2= 15^2+ AP^2

25^2–15^2= AP^2

(25+15)(25–15) = AP^2

40*10= AP^2

400= AP^2

+or-20 = Ap ( taking sq.root on both sides)

AP = +20

Area of trapezium ABCD = 1/2(sum of the two parallel sides *height )

= (1/2)(40+10)*20

(1/2)*1000= 500  sq cm

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Answer 2

Draw a trapezium ABCD with :

AD = 10 cm

BC = 40 cm

AB = 25 cm

CD = 25 cm

Draw perpendiculars from A and D :

Let them meet at E and F respectively .

AE = DF =  h

Now :

BC = EF + BE + FC

But EF = AD = 10 cm

==> BC = 10 + BE + FC

==> 40 = 10 + BE + FC

==> BE + FC = 40 - 10

==> BE + FC = 30

Now BE = FC

So :

BE + BE = 30

==> 2 BE = 30

==> BE = 30/2

==> BE = 15

By Pythagoras theorem :

h² = 25² - 15²

==> h² = 625 - 225

==> h² = 400

==> h = 20

Area = 1 / 2 × [ sum of parallel sides ] × height

        = 1 / 2 × [ 10 + 40 ] × 20

        = 1 / 2 × 50 × 20

        = 50 × 10

        = 500 cm²

The area of trapezium is 500 cm²

Hope it helps :-)

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