# Activities-Pythagoras theorem problems

Difficulty problems in Exercise 11.1 in pythagorus theorem chapter'

(4) In Right angled ∆ABC,∟BAC= 90°,∟B: ∟C = 1:2 andAC= 4cm.calculate thelenght of BC

'Solution

in some special right angled triangle

whose angle ratio 1:2:3 that is 30-60-90


has their sides ratio 1: ${\sqrt {3}}$ :2

in ▲ABC, BC = 2. AC

BC = 2.4

BC = 8 cm

(6) A door of width 6 mt has an archabove it having aheight of 2 mt , find the radius if the arch

Solution

In figure given AB=6 mt width of door CD=2 mt height of arch let OC is radius of arch OD= x mt jion OB, in ∆ODB ∟D= 90º

$OB^{2}=OD^{2}+DB^{2}$ $(x+2)^{2}=3^{2}+x^{2}$ $4+4x+9=9+x^{2}$ 4x=9-4

x=${\frac {5}{4}}$ x=1.25

But OC = 2+x

   OC= 2+1.25
OC= 3.25 mt


radius of arch is 3.25 mt

(7) The sides of a right angled triangle are in an AP. Show that sides are in ther ratio 3:4:5

Solution

IN right angled triangle ABC If ∟B=90º and sides are in AP

Let AB= a-d

   BC= a
AC= a+d


then $(a+d)^{2}=a^{2}+(a-d)^{2}$ $a^{2}+d^{2}+2ad=a^{2}+a^{2}+d^{2}-2ad$ $a^{2}=4ad$ a=4d

AB=a-d=4d-d=3d

BC= a=4d

AC= a+d+ 4d+d =5d

ratio of sides is 3d:4d:5d

if sides of the right angled triangle are in ratio 3:4:5 then their sides are in AP