{"id":"informator-maturalny-matematyka-2010/zad/68","paper_id":"informator-maturalny-matematyka-2010","number":"68","points":2,"ptype":"open","subject":"matematyka","category":"informator-maturalny","year":2010,"month":null,"level":null,"text":"Zadanie 68. (2 pkt)\nOblicz pole trójkąta równoramiennego ABC, w którym\n24\nAB\ni\n13\n= BC\nAC","answer":null,"answer_text":null,"solution":"Trójkąt równoramienny: AB=24 (podstawa), AC=BC=13. Wysokość h opuszczona na AB: h=sqrt(13^2-12^2)=sqrt(169-144)=sqrt(25)=5.\nPole = (1/2)*24*5=60.\nOdpowiedź: 60.","image":"img/informator-maturalny-matematyka-2010/zad-68.webp","solution_image":null,"topics":null,"page_from":85,"source":"ai","answer_source":null,"answer_text_source":null,"solution_source":"ai","text_source":"ocr","source_label":"Matematyka · Informator maturalny · 2010","subject_label":"Matematyka","category_label":"Informator maturalny","text_html":"<p>Zadanie 68. (2 pkt)<br>Oblicz pole trójkąta równoramiennego ABC, w którym<br>24<br>AB<br>i<br>13<br>= BC<br>AC</p>","solutions":[{"source":"ai","label":"AI","kind":"text","html":"<p>Trójkąt równoramienny: AB=24 (podstawa), AC=BC=13. Wysokość h opuszczona na AB: h=sqrt(13^2-12^2)=sqrt(169-144)=sqrt(25)=5.<br>Pole = (1/2)<em>24</em>5=60.<br>Odpowiedź: 60.</p>"}]}