In a quadrilateral ABCD, ?B = 90° and AD ² = AB ² + BC ² + CD ², then ?ACD is equal to:

Let ABCD is a rhombus in which AB = BC = CD = DA = diagonal CA.
In triangle ABC ,
AB = BC = CA , therefore ?B = 60°
?D = ?B = 60°
?A=180 - ?B = 180° - 60° =120°
?C = ?A=120°
Hence ,The ratio of diagonals = ?2 : 1

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 .

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

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 .