Bài 3: Khái niệm về thể tích của khối đa diện
Hướng dẫn giải Bài 5 (Trang 26 SGK Toán Hình học 12)
<p>Cho tam gi&aacute;c ABC vu&ocirc;ng c&acirc;n ở A v&agrave; AB=a. Tr&ecirc;n đường thẳng qua C v&agrave; vu&ocirc;ng g&oacute;c với mặt phẳng (ABC) lấy</p> <p>điểm D sao cho CD=a. Mặt phẳng qua C vu&ocirc;ng g&oacute;c với BD, cắt BD tại F v&agrave; cắt AD tại E. T&iacute;nh thể t&iacute;ch khối tự</p> <p>diện CDEF theo a.&nbsp;</p> <p><strong>Giải</strong></p> <p><strong><img class="wscnph" src="https://static.colearn.vn:8413/v1.0/upload/library/21022022/hinh-27-7KvHcu.png" /></strong></p> <p>Ta c&oacute;: <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced open="" close="}"><mtable columnalign="right"><mtr><mtd><mi>B</mi><mi>A</mi><mo>&#8869;</mo><mi>C</mi><mi>D</mi></mtd></mtr><mtr><mtd><mi>B</mi><mi>A</mi><mo>&#8869;</mo><mi>C</mi><mi>A</mi></mtd></mtr></mtable></mfenced><mo>&#8658;</mo><mi>B</mi><mi>A</mi><mo>&#8869;</mo><mo>(</mo><mi>A</mi><mi>D</mi><mi>C</mi><mo>)</mo><mo>&#8658;</mo><mi>B</mi><mi>A</mi><mo>&#8869;</mo><mi>C</mi><mi>E</mi></math></p> <p>Mặt kh&aacute;c&nbsp;<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>B</mi><mi>D</mi><mo>&#8869;</mo><mo>(</mo><mi>C</mi><mi>E</mi><mi>F</mi><mo>)</mo><mo>&#8658;</mo><mi>B</mi><mi>D</mi><mo>&#8869;</mo><mi>C</mi><mi>E</mi></math></p> <p>Từ đ&oacute; suy ra&nbsp;<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>C</mi><mi>E</mi><mo>&#8869;</mo><mo>(</mo><mi>A</mi><mi>B</mi><mi>D</mi><mo>)</mo><mo>&#8658;</mo><mi>C</mi><mi>E</mi><mo>&#8869;</mo><mi>E</mi><mi>F</mi><mo>,</mo><mo>&#160;</mo><mi>C</mi><mi>E</mi><mo>&#8869;</mo><mi>A</mi><mi>D</mi></math></p> <p>V&igrave; tam gi&aacute;c ACD vu&ocirc;ng c&acirc;n&nbsp;</p> <p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>C</mi><mi>A</mi><mo>=</mo><mi>C</mi><mi>D</mi><mo>=</mo><mi>a</mi></math>, n&ecirc;n&nbsp;<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>C</mi><mi>E</mi><mo>=</mo><mfrac><mrow><mi>A</mi><mi>D</mi></mrow><mn>2</mn></mfrac><mo>=</mo><mfrac><mrow><mi>a</mi><msqrt><mn>2</mn></msqrt></mrow><mn>2</mn></mfrac></math></p> <p>Ta c&oacute;&nbsp;<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>B</mi><mi>C</mi><mo>=</mo><mi>a</mi><msqrt><mn>2</mn></msqrt><mo>,</mo><mo>&#160;</mo><mi>B</mi><mi>D</mi><mo>=</mo><msqrt><mn>2</mn><msup><mi>a</mi><mn>2</mn></msup><mo>+</mo><msup><mi>a</mi><mn>2</mn></msup></msqrt><mo>=</mo><mi>a</mi><msqrt><mn>3</mn></msqrt></math></p> <p>Trong tam gi&aacute;c vu&ocirc;ng BCD ta c&oacute; CF,BD=CB.CD(<math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mn>2</mn><msub><mi>S</mi><mrow><mi>B</mi><mi>C</mi><mi>D</mi></mrow></msub></math>)</p> <p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>&#8658;</mo><mi>C</mi><mi>F</mi><mo>=</mo><mfrac><mrow><mi>C</mi><mi>B</mi><mo>.</mo><mi>C</mi><mi>D</mi></mrow><mrow><mi>B</mi><mi>D</mi></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>a</mi><msqrt><mn>2</mn></msqrt><mo>.</mo><mi>a</mi></mrow><mrow><mn>2</mn><msqrt><mn>3</mn></msqrt></mrow></mfrac><mo>=</mo><mi>a</mi><msqrt><mfrac><mn>2</mn><mn>3</mn></mfrac></msqrt></math></p> <p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>&#8658;</mo><mi>E</mi><mi>F</mi><mo>=</mo><msqrt><mi>C</mi><msup><mi>F</mi><mn>2</mn></msup><mo>-</mo><mi>C</mi><msup><mi>E</mi><mn>2</mn></msup></msqrt><mo>=</mo><msqrt><mfrac><mn>2</mn><mn>3</mn></mfrac><msup><mi>a</mi><mn>2</mn></msup><mo>-</mo><mfrac><msup><mi>a</mi><mn>2</mn></msup><mn>2</mn></mfrac></msqrt><mo>=</mo><mfrac><mrow><mi>a</mi><msqrt><mn>6</mn></msqrt></mrow><mn>6</mn></mfrac></math></p> <p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>D</mi><mi>F</mi><mo>=</mo><msqrt><mi>D</mi><msup><mi>C</mi><mn>2</mn></msup><mo>-</mo><mi>C</mi><msup><mi>F</mi><mn>2</mn></msup></msqrt><mo>=</mo><msqrt><msup><mi>a</mi><mn>2</mn></msup><mo>-</mo><mfrac><mrow><mn>2</mn><msup><mi>a</mi><mn>2</mn></msup></mrow><mn>3</mn></mfrac></msqrt><mo>=</mo><mfrac><mrow><mi>a</mi><msqrt><mn>3</mn></msqrt></mrow><mn>3</mn></mfrac></math></p> <p>Diện t&iacute;ch tam gi&aacute;c CEF l&agrave;&nbsp;<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>S</mi><mrow><mi>C</mi><mi>E</mi><mi>F</mi></mrow></msub><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mi>E</mi><mi>C</mi><mo>.</mo><mi>E</mi><mi>F</mi><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mo>.</mo><mfrac><mrow><mi>a</mi><msqrt><mn>2</mn></msqrt></mrow><mn>2</mn></mfrac><mo>.</mo><mfrac><mrow><mi>a</mi><msqrt><mn>6</mn></msqrt></mrow><mn>6</mn></mfrac><mo>=</mo><mfrac><mrow><msup><mi>a</mi><mn>2</mn></msup><msqrt><mn>3</mn></msqrt></mrow><mn>12</mn></mfrac></math></p> <p>Thể t&iacute;ch khối tứ diện DCEF l&agrave;&nbsp;<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>V</mi><mrow><mi>D</mi><mi>C</mi><mi>E</mi><mi>F</mi></mrow></msub><mo>=</mo><mfrac><mn>1</mn><mn>3</mn></mfrac><mi>D</mi><mi>F</mi><mo>.</mo><msub><mi>S</mi><mrow><mi>C</mi><mi>E</mi><mi>F</mi></mrow></msub><mo>=</mo><mfrac><msup><mi>a</mi><mn>3</mn></msup><mn>36</mn></mfrac></math></p> <p>&nbsp;</p>
Hướng dẫn Giải Bài 5 (trang 26, SGK Toán 12, Hình học)
GV: GV colearn
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Hướng dẫn Giải Bài 5 (trang 26, SGK Toán 12, Hình học)
GV: GV colearn