This chapter focuses on the topic of viral geometry problems and the various approaches that can be used to tackle them. These problems can be quite complex and challenging, requiring a deep understanding of geometric concepts and creative problem-solving skills.
Rectangle Problems
Que 1: In the given figure ABCD is a rectangle then find the length of AE.

Solution: Extend the side CD to F
Theorem 1: Since ∠FDA=90°, we can draw a circumcircle of the △FDA. Here AF is the diameter of the Circle.

Theorem 2: Since FD is the tangent and EA is the secant of the circle.
So, ED2=EF×EA …….(i)
∠EDF=∠CDB (Vertically Opposite Angles)
∠CDB=∠ABD (Alternate angles)
△EDF~△EBA (Similar triangles)
\frac{EF}{AE}= \frac{ED}{BE}= \frac{8}{18}= \frac{4}{9}.
Let, EF=4k and AE=9k, Put these values in equation (i)
82=4k×9k
k=4/3
So, AE=9k= 12 cm
The Altitudes of a triangle
If the side lengths of △PQR are 13, 14, and 15 cm then find the value of \frac{1}{h_1}+\frac{1}{h_2}+\frac{1}{h_3}?

Solution: Let, the length of QR=a, PR=b, and PQ=c
Area of the triangle (△)= 1/2×Base×Height
\triangle=\frac{1}{2}\times a\times h_1.
\frac{1}{h_1}=\frac{a}{2\triangle} ………(i)
\frac{1}{h_1}+\frac{1}{h_2}+\frac{1}{h_3}=\frac{1}{2\triangle}(a+b+c)=\frac{s}{\triangle}=\frac{1}{r} ………(ii)
Here, s is the semi perimeter
2s=a+b+c
r is the inradius of the triangle
s=(13+14+15)/2=21
\triangle=\sqrt{s\times (s-a)(s-b)(s-c)}.
\triangle=\sqrt{21\times (21-13)(21-14)(21-15)}.
\triangle=\sqrt{21\times 8\times 7\times 6}.
\triangle=84 cm2
\frac{1}{h_1}+\frac{1}{h_2}+\frac{1}{h_3}=\frac{21}{84}=0.25