ABCD is a parallelogram and X, Y are the mid-points of sides AB and CD respectively. Then quadrilateral AXCY is a :

Let C(O, r) is a circle and let M is a point within it where O is the centre and r is the radius of the circle.
Let CD is another chord passes through point M.
We have to prove that AB < CD.
Now join OM and draw OL perpendicular to CD.
In right angle triangle OLM,
OM is the hypotenuse.
So OM > OL
? chord CD is nearer to O in comparison to AB.
? CD > AB
? AB < CD
So all chords of a circle of a circle at a given point within it, the smallest is one which is bisected at that point.
Hence required answer will be option D .

From given figure , we can see that
Let AB = 16 m , OM = 6 m , AM = BM = 8 m and OA = r
In triangle AMO ,
By pythagoras theorem ,
OA² = AM² + OM²
OA² = 8² + 6² = 64 + 36
OA² = 100 ? OA = ? 100 = 10
? OA = r = 10 m
The radius of the circular lawn = 10 m

Given :- ?A = x° , ?B = y° , ?C = ( y + 20 )° and 4x ? y = 10
We know that , ?A + ?B + ?C = 180°
? x + y + ( y + 20 ) = 180°
? x + 2y = 180 - 20 = 160° ........... ( ? )
4x ? y = 10 ........... ( ? )
From ( ? ) and ( ? ) , we get
? y = 70 , x = 20
? The angles of the triangle are 20° , 70° , 90° .
Hence , the triangle will be Right-angle triangle .

In a quadrilateral ABCD,
Given :- ?B = 90°
AB² + BC² + CD² = AC² + CD² = AD² { ? In triangle ABC , AB² + BC² = AC² }
So, in triangle ACD, angle opposite to AD = 90° ( by Converse of Pythagoras theorem)
? ?ACD = 90°