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: $${\sqrt3}$$ :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