UPSC CDS Math Quadrilateral and Polygon Questions

Let h1 units and h2 units be the heights of the poles at C and D respectively.
Now,∠ABC = ∠BCD

(6−2)×180°

6

=

720°

6

=120°
In ΔABC,
AB = BC
⇒∠ACB = ∠BAC
Remember: Equal sides have equal angles opposite to them.
Using angle sum property, we have
∠ACB = ∠BAC = 30°
Draw BP ⊥ AC.
Therefore, P is the mid-point of AC.
AC = 2AP
In right ΔABP,
cos‌30°=

√3

2

=

AP

AB

⇒AP=

√3a

2

∴AC=2×

√3a

2

=√3a
∠ACD =∠BCD - ∠ACB
=120°−30°=90°
In right ΔACD,
=√4a2=2a
In right ΔACG,
tan‌30°=

1

√3

=

CG

AC

⇒

h1

√3a

=

1

√3

⇒h1=a
In right ΔADH,
tan60°=√3=

DH

AD

⇒

h2

2a

=√3
⇒h2=2√3a
∴h1:h2=a:2√3a=1:2√3