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# bisector-and-perpendicular-line-lesson-19.pptx

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# bisector-and-perpendicular-line-lesson-19.pptx

Bisector and Perpendicular line

Bisector and Perpendicular line

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### bisector-and-perpendicular-line-lesson-19.pptx

1. 1. 𝑫𝒆𝒇𝒊𝒏𝒊𝒕𝒊𝒐𝒏 𝒐𝒇 𝑫𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒇𝒓𝒐𝒎 𝒂 𝑷𝒐𝒊𝒏𝒕 𝒕𝒐 𝒂 𝑳𝒊𝒏𝒆 The distance from a point to a line is the length of the perpendicular segment from the point to the line. If the point lies on the line, the distance is zero. P C A B The distance from point P to AB is PC. Here PC ⊥ AB
2. 2. Theorem 35 A point on the bisector of an angle is equidistant from the sides of the angle. A D G F E C B Given: 𝑩𝑫 bisects ∠𝑨𝑩𝑪, G lies on 𝑩𝑫 Prove: 𝑮𝑬 = 𝑮𝑭
3. 3. A D G F E C B Given: 𝑩𝑫 bisects ∠𝑨𝑩𝑪, G lies on 𝑩𝑫 Prove: 𝑮𝑬 = 𝑮𝑭 𝐒𝐭𝐚𝐭𝐞𝐦𝐞𝐧𝐭𝐬 𝐑𝐞𝐚𝐬𝐨𝐧𝐬 1. 𝐁𝐃 bisects ∠𝐀𝐁𝐂, G lies on 𝐁𝐃 𝐆𝐢𝐯𝐞𝐧 2. ∠𝐀𝐁𝐃 ≅ ∠𝐂𝐁𝐃 𝐃𝐞𝐟𝐢𝐧𝐢𝐭𝐢𝐨𝐧 𝐨𝐟 𝐀𝐧𝐠𝐥𝐞 𝐁𝐢𝐬𝐞𝐜𝐭𝐨𝐫 3. 𝐆𝐄 ⊥ 𝐁𝐀 𝐆𝐅 ⊥ 𝐁𝐂 𝐃𝐞𝐟𝐢𝐧𝐢𝐭𝐢𝐨𝐧 𝐨𝐟 𝐭𝐡𝐞 𝐃𝐢𝐬𝐭𝐚𝐧𝐜𝐞 𝐟𝐫𝐨𝐦 𝐚 𝐏𝐨𝐢𝐧𝐭 𝐭𝐨 𝐚 𝐋𝐢𝐧𝐞 4. ∠𝐆𝐄𝐁 𝐚𝐧𝐝 ∠𝐆𝐅𝐁 𝐚𝐫𝐞 𝐫𝐢𝐠𝐡𝐭 𝐚𝐧𝐠𝐥𝐞𝐬. 𝐃𝐞𝐟𝐢𝐧𝐢𝐭𝐢𝐨𝐧 𝐨𝐟 𝐏𝐞𝐫𝐩𝐞𝐧𝐝𝐢𝐜𝐮𝐥𝐚𝐫 5. 𝐁𝐆 ≅ 𝐁𝐆 𝐑𝐞𝐟𝐥𝐞𝐱𝐢𝐯𝐞 𝐏𝐫𝐨𝐩𝐞𝐫𝐭𝐲 𝐨𝐟 𝐂𝐨𝐧𝐠𝐫𝐮𝐞𝐧𝐜𝐞 6. ∆𝐆𝐄𝐁 𝐚𝐧𝐝 ∆𝐆𝐅𝐁 𝐚𝐫𝐞 𝐫𝐢𝐠𝐡𝐭 𝐭𝐫𝐢𝐚𝐧𝐠𝐥𝐞𝐬. 𝐃𝐞𝐟𝐢𝐧𝐢𝐭𝐢𝐨𝐧 𝐨𝐟 𝐑𝐢𝐠𝐡𝐭 𝐓𝐫𝐢𝐚𝐧𝐠𝐥𝐞𝐬 7. ∆𝐆𝐄𝐁 ≅ ∆𝐆𝐅𝐁 𝐇𝐲𝐩𝐨𝐭𝐞𝐧𝐮𝐬𝐞 − 𝐀𝐜𝐮𝐭𝐞 𝐀𝐧𝐠𝐥𝐞 8. 𝐆𝐄 ≅ 𝐆𝐅 𝐂𝐏𝐂𝐓𝐂 9. 𝐆𝐄 = 𝐆𝐅 𝐃𝐞𝐟𝐢𝐧𝐢𝐭𝐢𝐨𝐧 𝐨𝐟 𝐂𝐨𝐧𝐠𝐫𝐮𝐞𝐧𝐭 𝐒𝐞𝐠𝐦𝐞𝐧𝐭𝐬 proof:
4. 4. Theorem 36 A point equidistant from the sides of an angle lies on the bisector of an angle A D G F E C B Given: 𝐺𝐸 ⊥ 𝐵𝐴; 𝐺𝐹 ⊥ 𝐵𝐶, 𝐺𝐸 = 𝐺𝐹 Prove: 𝐺 lies on the bisector of ∠𝐴𝐵𝐶. 𝑷𝒆𝒓𝒑𝒆𝒏𝒅𝒊𝒄𝒖𝒍𝒂𝒓 𝑩𝒊𝒔𝒆𝒄𝒕𝒐𝒓𝒔 𝒐𝒇 𝒂 𝑺𝒆𝒈𝒎𝒆𝒏𝒕 Recall that the perpendicular bisector of a segment is a line, ray, segment, or plane that is perpendicular to the segment at its midpoint.
5. 5. 𝑻𝒉𝒆𝒐𝒓𝒆𝒎 𝟑𝟕 A point on the perpendicular bisector of a segment is equidistant from the endpoint of a segment. C P D B A Given: 𝑪𝑫 ⊥ 𝒃𝒊𝒔𝒆𝒄𝒕𝒐𝒓 𝒐𝒇 𝑨𝑩, 𝑷 lies on 𝑪𝑫. Prove: 𝑷𝑨 = 𝑷𝑩
6. 6. C P D B A Given: 𝑪𝑫 ⊥ 𝒃𝒊𝒔𝒆𝒄𝒕𝒐𝒓 𝒐𝒇 𝑨𝑩, 𝑷 lies on 𝑪𝑫. Prove: 𝑷𝑨 = 𝑷𝑩 𝒑𝒓𝒐𝒐𝒇: 𝑆𝑡𝑎𝑡𝑒𝑚𝑒𝑛𝑡𝑠 𝑅𝑒𝑎𝑠𝑜𝑛𝑠 1. 𝐶𝐷 ⊥ bisector of 𝐴𝐵, 𝑃 lies on 𝐶𝐷. 𝐺𝑖𝑣𝑒𝑛 2. 𝐷𝑟𝑎𝑤 𝐴𝑃 𝑎𝑛𝑑 𝐵𝑃. 𝐿𝑖𝑛𝑒 𝑃𝑜𝑠𝑡𝑢𝑙𝑎𝑡𝑒 3. 𝐶𝐷 ⊥ 𝐴𝐵; 𝐷 𝑖𝑠 𝑡ℎ𝑒 𝑚𝑖𝑑𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝐴𝐵. 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝐵𝑖𝑠𝑒𝑐𝑡𝑜𝑟 4. ∠𝑃𝐷𝐴 𝑎𝑛𝑑 ∠𝑃𝐷𝐵 𝑎𝑟𝑒 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒. 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 5. ∆𝑃𝐷𝐴 𝑎𝑛𝑑 ∆𝑃𝐷𝐵 𝑎𝑟𝑒 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒. 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑅𝑖𝑔ℎ𝑡 𝑇𝑟𝑖𝑎𝑛𝑔𝑙𝑒𝑠 6. 𝐴𝐷 ≅ 𝐵𝐷 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑎 𝑀𝑖𝑑𝑝𝑜𝑖𝑛𝑡 7. 𝑃𝐷 ≅ 𝑃𝐷 𝑅𝑒𝑓𝑙𝑒𝑥𝑖𝑣𝑒 𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦 𝑜𝑓 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑐𝑒 8. ∆𝑃𝐷𝐴 ≅ ∆𝑃𝐷𝐵 𝐿𝑒𝑔 − 𝐿𝑒𝑔 𝑇ℎ𝑒𝑜𝑟𝑒𝑚 9. 𝑃𝐴 ≅ 𝑃𝐵 𝐶𝑃𝐶𝑇𝐶 10. 𝑃𝐴 = 𝑃𝐵 𝐷𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝐶𝑜𝑛𝑔𝑟𝑢𝑒𝑛𝑡 𝑆𝑒𝑔𝑚𝑒𝑛𝑡𝑠
7. 7. 𝑻𝒉𝒆𝒐𝒓𝒆𝒎 𝟑𝟖 A point equidistant from the endpoint of a segment lies on the perpendicular bisector of a segment. C P D B A Given: 𝐏𝐀 = 𝐏𝐁 Prove: P lies on the perpendicular bisector of AB.
8. 8. C P D B A Given: 𝐏𝐀 = 𝐏𝐁 Prove: P lies on the perpendicular bisector of AB. 𝒑𝒓𝒐𝒐𝒇: 𝑺𝒕𝒂𝒕𝒆𝒎𝒆𝒏𝒕𝒔 𝑹𝒆𝒂𝒔𝒐𝒏𝒔 1. 𝑳𝒆𝒕 𝑫 𝒃𝒆 𝒕𝒉𝒆 𝒎𝒊𝒅𝒑𝒐𝒊𝒏𝒕 𝒐𝒇 𝑨𝑩 𝑻𝒉𝒆𝒐𝒓𝒆𝒎 𝟔. 𝑬𝒗𝒆𝒓𝒚 𝒔𝒆𝒈𝒎𝒆𝒏𝒕 𝒉𝒂𝒔 𝒆𝒙𝒂𝒄𝒕𝒍𝒚 𝒐𝒏𝒆 𝒎𝒊𝒅𝒑𝒐𝒊𝒏𝒕 2. 𝑨𝑫 ≅ 𝑩𝑫 𝑫𝒆𝒇𝒊𝒏𝒊𝒕𝒊𝒐𝒏 𝒐𝒇 𝒂 𝑴𝒊𝒅𝒑𝒐𝒊𝒏𝒕 3. 𝑫𝒓𝒂𝒘 𝒍𝒊𝒏𝒆 𝑷𝑫 𝑳𝒊𝒏𝒆 𝑷𝒐𝒔𝒕𝒖𝒍𝒂𝒕𝒆 4. 𝑷𝑫 ≅ 𝑷𝑫 𝑹𝒆𝒇𝒍𝒆𝒙𝒊𝒗𝒆 𝑷𝒓𝒐𝒑𝒆𝒓𝒕𝒚 𝒐𝒇 𝑪𝒐𝒏𝒈𝒓𝒖𝒆𝒏𝒄𝒆 5. 𝑷𝑨 = 𝑷𝑩 𝑮𝒊𝒗𝒆𝒏 6. ∆𝑷𝑫𝑨 ≅ ∆𝑷𝑫𝑩 𝑺𝑺𝑺 𝑷𝒐𝒔𝒕𝒖𝒍𝒂𝒕𝒆 7. ∠𝑷𝑫𝑨 ≅ ∠𝑷𝑫𝑩 𝑪𝑷𝑪𝑻𝑪 8. ∠𝑷𝑫𝑨 𝒂𝒏𝒅 ∠𝑷𝑫𝑩 𝒇𝒐𝒓𝒎 𝒂 𝒍𝒊𝒏𝒆𝒂𝒓 𝒑𝒂𝒊𝒓. 𝑫𝒆𝒇𝒊𝒏𝒊𝒕𝒊𝒐𝒏 𝒐𝒇 𝑳𝒊𝒏𝒆𝒂𝒓 𝑷𝒂𝒊𝒓 9.∠𝑷𝑫𝑨 𝒂𝒏𝒅 ∠𝑷𝑫𝑩 𝒂𝒓𝒆 𝒔𝒖𝒑𝒑𝒍𝒆𝒎𝒆𝒏𝒕𝒂𝒓𝒚. 𝑳𝒊𝒏𝒆𝒂𝒓 𝑷𝒂𝒊𝒓 𝑷𝒐𝒔𝒕𝒖𝒍𝒂𝒕𝒆 10. ∠𝑷𝑫𝑨 𝒂𝒏𝒅 ∠𝑷𝑫𝑩 𝒂𝒓𝒆 𝒓𝒊𝒈𝒉𝒕 𝒂𝒏𝒈𝒍𝒆 𝑻𝒉𝒆𝒐𝒓𝒆𝒎 𝟏𝟔. 𝑰𝒇 𝒕𝒘𝒐 𝒂𝒏𝒈𝒍𝒆𝒔 𝒂𝒓𝒆 𝒃𝒐𝒕𝒉 𝒄𝒐𝒏𝒈𝒓𝒖𝒆𝒏𝒕 𝒂𝒏𝒅 𝒔𝒖𝒑𝒑𝒍𝒆𝒎𝒆𝒏𝒕𝒂𝒓𝒚, 𝒕𝒉𝒆𝒏 𝒆𝒂𝒄𝒉 𝒊𝒔 𝒂 𝒓𝒊𝒈𝒉𝒕 𝒂𝒏𝒈𝒍𝒆.
9. 9. 𝑻𝒉𝒆𝒐𝒓𝒆𝒎 𝟑𝟗 If a line contains two points each of which is equidistant from the endpoints of the segment, then the line is the perpendicular bisector of the segment. C P D B A Given: AP=BP, AC=BC Prove: PD is the ⊥ bisector of AB.
10. 10. C P D B A Given: AP=BP, AC=BC Prove: PD is the ⊥ bisector of AB. 𝐒𝐭𝐚𝐭𝐞𝐦𝐞𝐧𝐭𝐬 𝐑𝐞𝐚𝐬𝐨𝐧𝐬 1. 𝐀𝐏 ≅ 𝐁𝐏, 𝐀𝐂 ≅ 𝐁𝐂 𝐆𝐢𝐯𝐞𝐧 𝐚𝐧𝐝 𝐃𝐞𝐟𝐢𝐧𝐢𝐭𝐢𝐨𝐧 𝐨𝐟 𝐂𝐨𝐧𝐠𝐫𝐮𝐞𝐧𝐭 𝐒𝐞𝐠𝐦𝐞𝐧𝐭𝐬 2. 𝐏𝐂 ≅ 𝐏𝐂 𝐑𝐞𝐟𝐥𝐞𝐱𝐢𝐯𝐞 𝐏𝐫𝐨𝐩𝐞𝐫𝐭𝐲 𝐨𝐟 𝐂𝐨𝐧𝐠𝐫𝐮𝐞𝐧𝐜𝐞 3. ∆𝐏𝐂𝐀 ≅ ∆𝐏𝐂𝐁 𝐒𝐒𝐒 𝐏𝐨𝐬𝐭𝐮𝐥𝐚𝐭𝐞 4. ∠𝐀𝐏𝐃 ≅ ∠𝐁𝐏𝐃 𝐂𝐏𝐂𝐓𝐂 5. 𝐏𝐃 ≅ 𝐏𝐃 𝐑𝐞𝐟𝐥𝐞𝐱𝐢𝐯𝐞 𝐏𝐫𝐨𝐩𝐞𝐫𝐭𝐲 𝐨𝐟 𝐂𝐨𝐧𝐠𝐫𝐮𝐞𝐧𝐜𝐞 6. ∆𝐀𝐏𝐃 ≅ ∆𝐁𝐏𝐃 𝐒𝐀𝐒 𝐏𝐨𝐬𝐭𝐮𝐥𝐚𝐭𝐞 7. 𝐀𝐃 ≅ 𝐁𝐃 𝐂𝐏𝐂𝐓𝐂 8. 𝐃 𝐢𝐬 𝐭𝐡𝐞 𝐦𝐢𝐝𝐩𝐨𝐢𝐧𝐭 𝐨𝐟 𝐀𝐁 𝐃𝐞𝐟𝐢𝐧𝐢𝐭𝐢𝐨𝐧 𝐨𝐟 𝐌𝐢𝐝𝐩𝐨𝐢𝐧𝐭 9. ∠𝐏𝐃𝐀 ≅ ∠𝐏𝐃𝐁 𝐂𝐏𝐂𝐓𝐂 10.∠𝐏𝐃𝐀 𝐚𝐧𝐝 ∠𝐏𝐃𝐁 𝐟𝐨𝐫𝐦 𝐚 𝐥𝐢𝐧𝐞𝐚𝐫 𝐟𝐢𝐫𝐦 𝐃𝐞𝐟𝐢𝐧𝐢𝐭𝐢𝐨𝐧 𝐨𝐟 𝐋𝐢𝐧𝐞𝐚𝐫 𝐏𝐚𝐢𝐫 11. ∠𝐏𝐃𝐀 𝐚𝐧𝐝 ∠𝐏𝐃𝐁 𝐚𝐫𝐞 𝐬𝐮𝐩𝐩𝐥𝐞𝐦𝐞𝐧𝐭𝐚𝐫𝐲 𝐋𝐢𝐧𝐞𝐚𝐫 𝐏𝐚𝐢𝐫 𝐏𝐨𝐬𝐭𝐮𝐥𝐚𝐭𝐞 12. ∠𝐏𝐃𝐀 𝐚𝐧𝐝 ∠𝐏𝐃𝐁 𝐚𝐫𝐞 𝐫𝐢𝐠𝐡𝐭 𝐚𝐧𝐠𝐥𝐞𝐬 𝐓𝐡𝐞𝐨𝐫𝐞𝐦 𝟏𝟔. 𝐈𝐟 𝐭𝐰𝐨 𝐚𝐧𝐠𝐥𝐞𝐬 𝐚𝐫𝐞 𝐛𝐨𝐭𝐡 𝐜𝐨𝐧𝐠𝐫𝐮𝐞𝐧𝐭 𝐚𝐧𝐝 𝐬𝐮𝐩𝐩𝐥𝐞𝐦𝐞𝐧𝐭𝐚𝐫𝐲, 𝐭𝐡𝐞𝐧 𝐞𝐚𝐜𝐡 𝐢𝐬 𝐚 𝐫𝐢𝐠𝐡𝐭 𝐚𝐧𝐠𝐥𝐞 13. 𝐏𝐃 𝐢𝐬 𝐭𝐡𝐞 ⊥ 𝐛𝐢𝐬𝐞𝐜𝐭𝐨𝐫 𝐨𝐟 𝐀𝐁 𝐃𝐞𝐟𝐢𝐧𝐢𝐭𝐢𝐨𝐧 𝐨𝐟 𝐩𝐞𝐫𝐩𝐞𝐧𝐝𝐢𝐜𝐮𝐥𝐚𝐫 𝐁𝐢𝐬𝐞𝐜𝐭𝐨𝐫 proof:
11. 11. 𝑻𝒉𝒆𝒐𝒓𝒆𝒎 𝟒𝟎 In a plane, through a given points on a line, there is exactly one line perpendicular to the line. R S M P Q Given: MQ and point P on MQ Prove: a. There is at least on line PR ⊥ MQ b. There is only one line PR ⊥ MQ Part 1: Show that there is one line 𝑷𝑹 ⊥ 𝑴𝑸 through point P. Through the end point of 𝑷𝑸 , ray 𝑷𝑹 can be constructed in such a way that 𝒎∠𝑹𝑷𝑸 = 𝟗𝟎. ∠𝑹𝑷𝑸is therefore a right angle and hence, by the definition of perpendicular, 𝑷𝑹 ⊥ 𝑴𝑸 . Since rays 𝑷𝑹 ⊥ 𝑷𝑸 are contained in 𝑷𝑹 𝒂𝒏𝒅 𝑴𝑸 , respectively, therefore 𝑷𝑹 ⊥ 𝑴𝑸 .
12. 12. Part 2: Show that 𝑷𝑹 𝒊𝒔 𝒕𝒉𝒆 𝒐𝒏𝒍𝒚 𝒍𝒊𝒏𝒆 ⊥ 𝑴𝑸 through P. There is only one line 𝑷𝑹 ⊥ 𝑴𝑸 or other than 𝑷𝑹 there is another line 𝑷𝑹 ⊥ 𝑴𝑸 . Assume that there is another line 𝑷𝑺 ⊥ 𝑴𝑸 . Then by the definition of perpendicular, ∠𝑺𝑷𝑸 is a right angle and therefore 𝒎∠𝑹𝑷𝑸 = 𝟗𝟎. This contradicts the Angle Construction Postulate. Since the assumption leads to contradiction of postulate, it must be false. Therefore, there is only one line 𝑷𝑹 ⊥ 𝑴𝑸 . R S M P Q Given: MQ and point P on MQ Prove: a. There is at least on line PR ⊥ MQ b. There is only one line PR ⊥ MQ
13. 13. 𝑻𝒉𝒆𝒐𝒓𝒆𝒎 𝟒𝟏 There is one and only one line perpendicular to a given line through an external point. For Example: 𝐼𝑛 𝑡ℎ𝑒 𝑓𝑖𝑔𝑢𝑟𝑒, 𝑝𝑜𝑖𝑛𝑡 𝑀 𝑙𝑖𝑒𝑠 𝑜𝑛 𝐻𝐷, 𝑡ℎ𝑒 𝑏𝑖𝑠𝑒𝑐𝑡𝑜𝑟 𝑜𝑓 ∠𝐾𝐻𝑁. If 𝐾𝐻 = 4𝑥 − 15 and 𝑁𝐻 = 2𝑥 + 3. Find 𝑁𝐻. SOLUTION: 𝑲𝑯 = 𝑵𝑯 4𝑥 − 15 = 2𝑥 + 3 Substitute the value of 𝐾𝐻 and 𝑁𝐻. 4𝑥 − 2𝑥 = 3 + 15 Combine similar terms. 2𝑥 = 18 Simplify. 2𝑥 2 = 18 2 Divide both side by 2. 𝑥 = 9 𝑁𝐻 = 2𝑥 + 3 𝑁𝐻 = 2 9 + 3 Substitute the value of x 𝑁𝐻 = 18 + 3 Simplify. 𝑁𝐻 = 21 H D N M K
14. 14. EXAMPLE: 𝐼𝑛 𝑡ℎ𝑒 𝑓𝑖𝑔𝑢𝑟𝑒, 𝐷𝑇 𝑖𝑠 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝑏𝑖𝑠𝑒𝑐𝑡𝑜𝑟 𝑜𝑓 ∠𝑀𝐷𝑄. 𝑀𝐿 = 5𝑥 − 12, 𝑄𝐿 = 2𝑥 + 3, 𝐷𝑀 = 2𝑦 − 3, 𝑎𝑛𝑑 𝐷𝑄 = 𝑦 + 12. Find the perimeter of 𝑀𝐷𝑄𝐿. 𝑀𝐿 = 𝑄𝐿 𝐷𝑀 = 𝐷𝑄 5𝑥 − 12 = 2𝑥 + 3 2𝑦 − 3 = 𝑦 + 12 5𝑥 − 2𝑥 = 3 + 12 2𝑦 − 𝑦 = 12 + 3 3𝑥 = 15 𝑦 = 15 𝑥 = 5 𝑀𝐿 = 5𝑥 − 12 𝑄𝐿 = 2𝑥 + 3 𝐷𝑀 = 2𝑦 − 3 𝐷𝑄 = 𝑦 + 12 𝑀𝐿 = 5 5 − 12 𝑄𝐿 = 2(5) + 3 𝐷𝑀 = 2(15) − 3 𝐷𝑄 = 15 + 12 𝑀𝐿 = 25 − 12 𝑄𝐿 = 10 + 3 𝐷𝑀 = 30 − 3 𝐷𝑄 = 27 𝑀𝐿 = 13 𝑄𝐿 = 13 𝐷𝑀 = 27 Since, we are looking for the perimeter of 𝑀𝐷𝑄𝐿. Just add the length of the sides that we get. 𝑃 = 𝑀𝐿 + 𝑄𝐿 + 𝐷𝑀 + 𝐷𝑄 𝑃 = 13 + 13 + 27 + 27 𝑃 = 80 T L D Q M