Contents

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

$\frac{1}{h_1}+\frac{1}{h_2}+\frac{1}{h_3}=\frac{21}{84}=0.25$