In the given figure, DABC ~ DDCB, then AB × DB =
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P and Q are points on sides AB and AC respectively of ΔABC. If AP = 3 cm, PB = 6 cm, AQ = 5 cm and QC = 10 cm, then BC =
ABC is right triangle, right angled at C. If p is the length of the perpendicular from C to AB and a, b, c have the usual meaning, then
In the given figure, AD ⊥ BC, BE ⊥ AC, CF ⊥ AB, then AF2 + BD2 + CE2 =
ΔABC is such that AB = 3 cm, BC = 2 cm and CA = 2.5 cm. If ΔDEF ~ ΔABC and EF = 4 cm, then perimeter of ΔDEF is
In the given ΔABC, AD⊥BC and ∠A is right angled. Then AD2 =
In the given trapezium ABCD, AB || CD and AB = 2CD. If area of ΔAOB = 84 cm2, then the area of ΔCOD is
In the given figure, ∠ ABC = 90° and BD ⊥ AC . If BD = 8 cm, AD = 4 cm, then CD =
13 videos|44 docs|187 tests
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13 videos|44 docs|187 tests
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