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, **ED**^{2}**=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)

8^{2}=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 (Viral Geometry Problems)

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 cm^{2}

**Carnot’s Theorem**

Carnot’s Theorem states that the sum of the signed lengths of the pedal lines from the circumcenter O to the sides of a triangle ABC is equal to the sum of the Inradius and Circumradius of the triangle.

**OX+OY+OZ=R+r**

Quadrilaterals AZOY, BZOX, and CXOY are Cyclic Quadrilaterals because pair of opposite angles is equal to 180°

In AZOY⇒ OZ.AY+OY.AZ=AO.ZY (Ptolemy’s Theorem)

OX, OY, and OZ are the perpendicular bisectors of the sides of the triangle because AB, BC, and CA are the chords of the circle and O is its center.

Let BC=a, CA=b, and AB=c

AZ=a/2 and AY=b/2

AO= R (length of circumradius of the triangle)

Since Z and Y are the mid-points of the sides so ZY=BC/2=a/2

OZ.AY+OY.AZ=AO.ZY

OZ.b/2+OY.c/2=R.a/2

OZ.b+OY.c=R.a ………….(i)

Similarly, in BZOX⇒ OZ.a+OX.c=R.b ……….(ii)

Similarly, in CXOY⇒ OX.b+OY.a=R.c ……….(iii)

Add all equations:

OX.(b+c)+OY.(a+c)+OZ.(a+b)=R.(a+b+c)

Put, 2s=a+b+c

Here ‘s’ is the semi-perimeter of the triangle

OX.(2s-a)+OY.(2s-b)+OZ.(2s-c)=R.2s

2s.(OX+OY+OZ)-(a.OX+b.OY+c.OZ)=R.2s ……..(iv)

Here, ar(△BOC)+ar(△COA)+ar(△AOB)=ar(△ABC)

½.BC.OX+½.AC.OY+½.AB.OZ=△

a.OX+b.OY+c.OZ=2△

Put this value in equation (iv):

2s.(OX+OY+OZ)-2△=R.2s

OX+OY+OZ=R+△/s

OX+OY+OZ=R+r

Here ‘r’ is the inradius of the triangle.

**Que: **If a=13 cm, b=14 cm, and c=15 cm then find the value of OX+OY+OZ.

**Solution: **s=21

△^{2}=s.(s-a)(s-b)(s-c)

△=84 cm^{2}

R=\frac{abc}{4\triangle}=\frac{13\times 14\times 15}{4\times 84}=8.125.

r=\frac{\triangle}{s}=\frac{84}{21}=4.

OX+OY+OZ=R+r= 12.125 cm

**Shaded Area inside a triangle**

**Que: **Find the ratio of the areas of △DEF and △ABC.

**Solution: **ar(△DAF)= ½×AD×AF×Sin(A)

ar(△ABC)=△=½×AB×AC×Sin(A)

ar(△DAF): ar(△ABC)= AD×AF: AB×AC= (AD/AB)×(AF/AC): 1

ar(△DAF)= (AD/AB)×(AF/AC)×△

ar(△DAF)= (5/7)×(2/6)×△=(5/7)×(1/3)×△

ar(△DBE)= (2/7)×(5/11)×△

ar(△ECF)= (6/11)×(4/6)×△=(6/11)×(2/3)×△

ar(△DEF)= △-△×[(5/7)×(1/3)+(2/7)×(5/11)+(6/11)×(2/3)]

= (62/231)×△

ar(△DEF): ar(△ABC)= 62: 231

**Conversion of quadrilateral to triangle**

**Que:** Find the area of quadrilateral ABCD, If ∠BAC=35°, ∠ACD=10°, ∠ADC=45°, and side AB=10 cm.

**Solution:** Flip the triangle CAD vertically and arrange it as shown in the picture

Now △ABD is an isosceles triangle because its two angles are 45°.

AB=BD=10 cm

Area of the △ABD=½×Base×Height

=½×10×10

=50 cm^{2}

**Miscellaneous**

**Que 6:** In the parallelogram ABCD, where IJ is parallel to AD, and ED and HF are perpendicular to DC, what is the ratio of the areas of quadrilateral AGJI to triangle HFC?

**Solution: **In △ADE, GJ∥AE

So, ar(△GJD): ar(△AED)=GD^{2}: AD^{2}=9:25

ar(AEJG)= 25-9= 16 unit …………..(i)

△IEJ ~△GJD (Similar Triangles)

ar(△IEJ): ar(△GJD)=IJ^{2}:GD^{2}=4:9

ar(△IEJ)=4 unit ………(ii)

ar(△HFC)=ar(△GJD)=9 unit

ar(AGJI): ar(△HFC)=16-4: 9=12:9= 4:3

**Que 7:** In the rectangle ABCD, the diagonal AC is extended to point E such that CE measures 5 cm. If the lengths of sides AB and BC are 8 cm and 6 cm respectively, what is the length of DE?

**Solution: **

Length of diagonal AC=10 cm

In △AED, AD∥FC

EF:FD=EC: CA=1:2

Let, EF=k and FD=2k ……..(i)

FC:AD=EC:EA=5:15

FC:6=1:3

FC=2 cm

In the right △DCF, DF^{2}=DC^{2}+FC^{2}

4k2=64+4=68

2k=2√17

k=√17

DE=3k= 3√17 cm