Viral Geometry Problems (10 Brilliant Strategies for Effortless Problem-Solving)

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