Astadium has central rectangular area 125m long by 80 m wide there are two semicircular ends.running tracks goes all way round

) Using formula for cicumference.

Circumference = Pi * diameter

Circumference = Pi * 80m

Circumference = 251.2m.

Total inside length of the curved ends.

2)Total distance round inside of track =

2(125) + 251.2m

= 250 + 251.2

= 501.2m

3) Area of circular ends =

Pi*r^2

= Pi*40^2

= 5024m^2

Area of central area (rectangle)

length * width

125m * 80m

= 10,000m^2

Total area inside track =

5,024m^2 + 10,000m^2

= 15,024m^2

Hope this helps.

:-)

A. AC = AC

Step-by-step explanation:

Given: In a quadrilateral ABCD, the diagonals intersect at point T.

Byron has used the Alternate Interior Angles Theorem to show ∠DAC≅∠BCA

and ∠BAC≅∠DCA

From the given options, we can only use option A i.e. AC=AC [reflexive property] such that

In ΔADC and ΔABC

∠DAC≅∠BCA  [ given ]

∠BAC≅∠DCA  [ given ]

AC=AC [reflexive property]

⇒ΔADC ≅ ΔABC [ASA congruency criteria]

Thus AD=BC [corresponding parts of congruent triangles are congruent]