From given figure , we have
AQ² = AC² + QC²
BP² = BC² + CP²
AQ² + BP² = (AC² + BC²) + (QC² + CP²) = AB² + PQ².

∴ AQ2 + BP2 = AB2 +		1	AB		×		1	AB			∵ PQ =	1	AB
		2					2					2

Correct Option: B

From given figure , we have
AQ² = AC² + QC²
BP² = BC² + CP²
AQ² + BP² = (AC² + BC²) + (QC² + CP²) = AB² + PQ².

∴ AQ2 + BP2 = AB2 +		1	AB		×		1	AB			∵ PQ =	1	AB
		2					2					2

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° ........... ( ⅰ )
4x – y = 10 ........... ( ⅱ )
From ( ⅰ ) and ( ⅱ ) , we get
⇒ y = 70 , x = 20
∴ The angles of the triangle are 20° , 70° , 90° .
Hence , the triangle will be Right-angle triangle .

Correct Option: A

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° ........... ( ⅰ )
4x – y = 10 ........... ( ⅱ )
From ( ⅰ ) and ( ⅱ ) , 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

Correct Option: C

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

Correct Option: D

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 .

ABCD is a parallelogram and AXCY is a quadrilateral .
Since X and Y are the mid-points of AB and DC respectively.

Correct Option: A

ABCD is a parallelogram and AXCY is a quadrilateral .
Since X and Y are the mid-points of AB and DC respectively.