Question

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Answer 1

Answer:

sis

Step-by-step explanation:

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Answer 2

Answer:

1. Let ABCD be a trapezium with,

AB∥CD

AB=25m

CD=10m

BC=14m

AD=13m

Draw CE∥DA. So, ADCE is a parallelogram with,

CD=AE=10m

CE=AD=13m

BE=AB−AE=25−10=15m

In ΔBCE, the semi perimeter will be,

s= a+b+c

2

s= 14+13+15

2

s=21m

Area of ΔBCE,

A= √s(s−a)(s−b)(s−c)

= √21(21−14)(21−13)(21−15)

= √21(7)(8)(6)

= √7056

=84m²

Also, area of ΔBCE is,

A= 1/2 ×base×height

84= 1/2 ×15×CL

84×2

15

=CL

CL= ( 56 ) m

( 5 )

Now, the area of trapezium is,

A= 1/2(sumofparallelsides)(height)

A= ( 1 ) ×(25+10)(56/5)

( 2 )

A=196m²

Therefore, the area of the trapezium is 196m².

2.

Angle MON + Angle NOP + Angle MOQ + Angle QOP = 180 + 180 degrees = 360 degrees. Therefore all angles around O sum upto 360 degrees.

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