ABCD is a parallelogram P, Q, R and S are points on sides AB, BC, CD and DA respectively such that AP = DR. If the area of the parallelogram ABCD is 16 cm ², then the area of the quadrilateral PQRS is:

From given figure , we have
Let, (x + 1) be the hypotenuse and perpendicular = x - 1 , base = x.
By pythagoras theorem ,
? (x + 1)² = x² + (x + 1)² ? x = 4
? Hypotenuse = 4 + 1 = 5

From given figure , we have
In quadrilateral ABCD , we can see that
Diagonal AB = Diagonal CD ......... ( ? )
AO = OB ......... ( ? )
DO = CO ......... ( ? )
?DOB = 90° ......... ( ? )
From ( ? ) , ( ? ) , ( ? ) and ( ? ) , we get
From above , it is clear that In quadrilateral ABCD , diagonals are equal and they bisect each other at 90° .
? Quadrilateral ABCD is square.

From the given figure, In ?ABC
We know that , Exterior angle = Sum of two interior angles
?1 = ?A + ?5 ................. ( 1 )
and ?2 = ?A + ?6 ................. ( 2 )
Adding equations ( 1 ) and ( 2 ).
?1 + ?2 = 2?A + ?5 + ?6
= 2?A + (180° ? ?A) = ?A + 180°
The given question can be restated as the sum of two exterior angles exceeds ?A of the ?ABC by 2 right angles.

From above figure , we can see that
Join B and D and produce BD to E
Then, p + q = ? and s + t = x
We know that , Exterior angle = Sum of two interior angle
Now, s = p + ? (Exterior angle of a ? prop.)
Similarly, t = q + ? (Exterior angle of a ? prop.)
Adding, s + t = p + q + ? + ?
Putting s + t = x and p + q = ? ,
Hence x = ? + ? + ?

The circumcentre of a triangle is always the point of intersection of the Bisectors .Hence required answer will be bisectors .

In a triangle ABC,
Given :- AB = 3 cm , AC = 5 cm and BC = 6 cm
We know that BD : DC = AB : AC
BD : DC = 3 : 5
? Divided BC = 6 in the ratio 3 : 5
BC = 3x + 5x = 6 ? 8x = 6 ? x = 6/8 = 0.75
? BD = 2.25, CD = 3.75.

We know that when two circle neither touch nor intersect and one lies outside the other, then 4 common tangents can be drawn.
For 4 common tangents , we get
? r₁ + r₂ < c₁c₂
Hence ,The number of tangents = 4