in a parallelogram SL and SM are the height one sides PQand QR respectively if area of a parallelogram is 21328m2 and pQ =182m and QR 142 m find the corresponding height to the bigger and smaller sides.
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Answer:
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Answer:
Area of parallelogram =120 cm
Area of parallelogram =120 cm 2
Area of parallelogram =120 cm 2 =15×PM
Area of parallelogram =120 cm 2 =15×PM⟹PM=8 cm
Area of parallelogram =120 cm 2 =15×PM⟹PM=8 cmNow in △PMS
Area of parallelogram =120 cm 2 =15×PM⟹PM=8 cmNow in △PMSPS
Area of parallelogram =120 cm 2 =15×PM⟹PM=8 cmNow in △PMSPS 2
Area of parallelogram =120 cm 2 =15×PM⟹PM=8 cmNow in △PMSPS 2 =SM
Area of parallelogram =120 cm 2 =15×PM⟹PM=8 cmNow in △PMSPS 2 =SM 2
Area of parallelogram =120 cm 2 =15×PM⟹PM=8 cmNow in △PMSPS 2 =SM 2 +PM
Area of parallelogram =120 cm 2 =15×PM⟹PM=8 cmNow in △PMSPS 2 =SM 2 +PM 2
Area of parallelogram =120 cm 2 =15×PM⟹PM=8 cmNow in △PMSPS 2 =SM 2 +PM 2
Area of parallelogram =120 cm 2 =15×PM⟹PM=8 cmNow in △PMSPS 2 =SM 2 +PM 2 10
Area of parallelogram =120 cm 2 =15×PM⟹PM=8 cmNow in △PMSPS 2 =SM 2 +PM 2 10 2
Area of parallelogram =120 cm 2 =15×PM⟹PM=8 cmNow in △PMSPS 2 =SM 2 +PM 2 10 2 =SM
Area of parallelogram =120 cm 2 =15×PM⟹PM=8 cmNow in △PMSPS 2 =SM 2 +PM 2 10 2 =SM 2
Area of parallelogram =120 cm 2 =15×PM⟹PM=8 cmNow in △PMSPS 2 =SM 2 +PM 2 10 2 =SM 2 +8
Area of parallelogram =120 cm 2 =15×PM⟹PM=8 cmNow in △PMSPS 2 =SM 2 +PM 2 10 2 =SM 2 +8 2
Area of parallelogram =120 cm 2 =15×PM⟹PM=8 cmNow in △PMSPS 2 =SM 2 +PM 2 10 2 =SM 2 +8 2
Area of parallelogram =120 cm 2 =15×PM⟹PM=8 cmNow in △PMSPS 2 =SM 2 +PM 2 10 2 =SM 2 +8 2 ⟹SM=10
Area of parallelogram =120 cm 2 =15×PM⟹PM=8 cmNow in △PMSPS 2 =SM 2 +PM 2 10 2 =SM 2 +8 2 ⟹SM=10 2
Area of parallelogram =120 cm 2 =15×PM⟹PM=8 cmNow in △PMSPS 2 =SM 2 +PM 2 10 2 =SM 2 +8 2 ⟹SM=10 2 −8
Area of parallelogram =120 cm 2 =15×PM⟹PM=8 cmNow in △PMSPS 2 =SM 2 +PM 2 10 2 =SM 2 +8 2 ⟹SM=10 2 −8 2
Area of parallelogram =120 cm 2 =15×PM⟹PM=8 cmNow in △PMSPS 2 =SM 2 +PM 2 10 2 =SM 2 +8 2 ⟹SM=10 2 −8 2 =6 cm