III gimnazjum

Matematyka z plusem 3

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M834 80h400000v40h-400000z'/></svg>​ <section class="answer"><h4 class="answer-heading">Odpowiedź:</h4> D</section>

Rozwiązanie 1

a - długość boku równoległoboku h - długość wysokości równoległoboku i trapezu   Obliczamy pole trapezu (które jest równe polu równoległoboku): (7+11)⋅h⋅21​=a⋅h

r - promień koła 2πr=8   ∣:2π

Skoro promień trawnika wynosi 12 m, to jego średnica jest dwa razy większa, czyli droga przebyta przez psa to:  2⋅12 m=24 m   Właściciel przeszedł połowę obwodu okręgu, czyli:  21​⋅2⋅π⋅12 m=

P1​=π⋅2,52=6,25π   P2​=π⋅52=25π

<img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/28981/jioIThjyrATnPicqUyBGlQ%3D%3D_bedca7724fea10155c2397e60042a72ebd1ff61b1002f4126fb5fb1e05a099ca.jpg" width="609" height="870"> A. 1,3 cm+13,7 cm=15 cm&lt;20 cm   - okręgi rozłączne zewnętrznie B. 23,5 cm+2,2 cm=25,7 cm&gt;20 cm