Question 1 :
Show that the bisectors of angles of a parallelogram form a rectangle .
Solution :
In triangle ADS,
<SAD = (1/2)<A ----(1)
<ADS = (1/2) <D ----(2)
(1) + (2)
<SAD + <ADS = (1/2) [<A + <D]
<SAD + <ADS = (1/2) (180)
<SAD + <ADS = 90
In triangle ADS,
<SAD + <ASD + <SDA = 180
<SAD + <SDA + <ASD = 180
90 + <ASD = 180
<ASD = 180 - 90
<ASD = 90, <PSR = 90 (Vertically opposite angle)
Hence PQRS is a rectangle.
Question 2 :
If a triangle and a parallelogram lie on the same base and between the same parallels, then prove that the area of the triangle is equal to half of the area of parallelogram.
Solution :
Let ΔABP and a parallelogram ABCD be on the same base AB and between the same parallels AB and PC.
To Prove : ar( ΔPAB ) = (1/2)ar( ABCD)
Draw BQ ||AP to obtain another parallelogram.ABQP and ABCD are on the same base AB and between the same parallels AB and PC.
Therefore, ar(ABQP) = ar(ABCD)
But ΔPAB ≅ ΔBQP( Diagonals PB divides parallelogram ABQP into two congruent triangles.
So ar (PAB) = ar(BQP) -----------(2)
ar (PAB) = (1/2)ar(ABQP) -----------------(3)
This gives ar (PAB) = (1/2)ar(ABCD)
Question 3 :
Iron rods a, b, c, d, e, and f are making a design in a bridge as shown in the figure. If a || b , c || d , e || f , find the marked angles between
(i) b and c
(ii) d and e
(iii d and f
(iv) c and f
Solution :
(i) Angle between b and c is 30.
(ii) Angle between d and e :
75 + angle between d and e = 180
Angle between d and e = 180 - 75
= 105
(iii) Angle between d and f :
In a parallelogram opposite angles will be equal.
Angle between d and is also 75.
(iv) Angle between c and f :
In a parallelogram opposite angles will be equal.
Angle between c and f is 105.
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