Proof:
AB=AD (sides of a square are equal) AB||DC (opposite sides of a square are parallel) ABCD is parallelogram with consecutive sides equal. ABCD is a rhombus. (by definition) Since the diagonals of a rhombus are perpendicular to each other, AC ^ BD. AB..
AB=AD (sides of a square are equal) AB||DC (opposite sides of a square are parallel) ABCD is parallelogram with consecutive sides equal. ABCD is a rhombus. (by definition) Since the diagonals of a rhombus are perpendicular to each other, AC ^ BD. AB..Proof:
Let AC and BD intersect at right angles at O. In triangles AOD and COD, AO=OC (diagonals bisect each other) OD=OD (common side) AD = DC i.e., the adjacent sides are equal. By definition, ABCD is a rhombus. Hence the theorem is prov..
Let AC and BD intersect at right angles at O. In triangles AOD and COD, AO=OC (diagonals bisect each other) OD=OD (common side) AD = DC i.e., the adjacent sides are equal. By definition, ABCD is a rhombus. Hence the theorem is prov..Proof:
\ From equation(4), we get (Note that -x > 0, so multiplying this in equation by -x, the inequality remains same) Add 1 on both sides, Taking the reciprocal, we ha..
\ From equation(4), we get (Note that -x > 0, so multiplying this in equation by -x, the inequality remains same) Add 1 on both sides, Taking the reciprocal, we ha..Rhombus
A rhombus is a parallelogram having all sides equal. Its other properties are: (i) The diagonals bisect each other at right angles. (ii) The diagonals bisect the angles at each vertex...
A rhombus is a parallelogram having all sides equal. Its other properties are: (i) The diagonals bisect each other at right angles. (ii) The diagonals bisect the angles at each vertex...Proof:
Now in triangles, ABD and ABC, AB=AB (common side) (each angle is a right angle) AD=BC (opposite sides of parallelogram) BD=AC (corresponding parts of corresponding triangles) Hence the theorem is prov..
Now in triangles, ABD and ABC, AB=AB (common side) (each angle is a right angle) AD=BC (opposite sides of parallelogram) BD=AC (corresponding parts of corresponding triangles) Hence the theorem is prov..Proof: In triangles ABC and DBC, AB=DC (opposite sides of parallelogram) BC=BC (common side) AC=BD (given) (corresponding parts of corresponding triangles) But these angles are consecutive interior angles on the same side of transversal BC and AB||DC. By definition of rectangle, parall..
In triangles ABC and DBC, AB=DC (opposite sides of parallelogram) BC=BC (common side) AC=BD (given) (corresponding parts of corresponding triangles) But these angles are consecutive interior angles on the same side of transversal BC and AB||DC. By definition of rectangle, parall..Proof:
If Q coincides with O, then AO = BO. i.e., AQ=BQ If Q is distinct i.e., Q does not coincides with O, then compare triangles AOQ and BOQ. AO=BO ( O is the mid point of AB) (given) OQ=OQ ( common side) AQ=BQ ( CP..
If Q coincides with O, then AO = BO. i.e., AQ=BQ If Q is distinct i.e., Q does not coincides with O, then compare triangles AOQ and BOQ. AO=BO ( O is the mid point of AB) (given) OQ=OQ ( common side) AQ=BQ ( CP..Proof:
If r = s, there is nothing to prove. Now, If r < s, then n - r > n - s, then the above equation becomes Since both sides are products of (s-r), consecutive integers in Similarly it can be proved that n = r + s if r > s...
If r = s, there is nothing to prove. Now, If r < s, then n - r > n - s, then the above equation becomes Since both sides are products of (s-r), consecutive integers in Similarly it can be proved that n = r + s if r > s...Proof:
Draw AL, BM and CN perpendicular to x-axis. LM = x 2 -x 1 MN = x 3 -x 2 LN = x 3 -x 1 Area of D ABC = area of trap {ALMB + BMNC - ALNC} This can be expressed in the form of a determinant Another form which is convenient to use for the area of triangles but which is very much useful wh..
Draw AL, BM and CN perpendicular to x-axis. LM = x 2 -x 1 MN = x 3 -x 2 LN = x 3 -x 1 Area of D ABC = area of trap {ALMB + BMNC - ALNC} This can be expressed in the form of a determinant Another form which is convenient to use for the area of triangles but which is very much useful wh..Proof:
If P coincides with O then AO = OB or AP = PB. . If P does not coincides O, then compare triangles AOP and BOP. PA=PB ( given) AO=OB ( O is the mid point of AB) OP=OP ( common side) But ( Linear pair) ..
If P coincides with O then AO = OB or AP = PB. . If P does not coincides O, then compare triangles AOP and BOP. PA=PB ( given) AO=OB ( O is the mid point of AB) OP=OP ( common side) But ( Linear pair) ..See what our Users say :
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