1,122
edits
Changes
Jump to navigation
Jump to search
mLine 1:
Line 1: − #In Right angled ∆ABC,∟BAC= 90°,∟B: ∟C = 1:2 andAC= 4cm.calculate thelenght of BC+ − + + + + + + + + Line 12:
Line 19: − #A door of width 6 mt has an archabove it having aheight of 2 mt , find the radius if the arch+ − # The sides of a right angled triangle are in an AP. Show that sides are in ther ratio 3:4:5+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
Activities-Pythagoras theorem problems (view source)
Revision as of 09:39, 30 October 2019
, 09:39, 30 October 2019added Category:Triangles using HotCat
''''''Solution'''''''''
'''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
in some special right angled triangle
whose angle ratio 1:2:3 that is 30-60-90
whose angle ratio 1:2:3 that is 30-60-90
BC = 8 cm
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
<math>OB^2=OD^2+DB^2</math>
<math>(x+2)^2=3^2+x^2</math>
<math>4+4x+9=9+x^2</math>
4x=9-4
x=<math>\frac{5}{4}</math>
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 <math>(a+d)^2=a^2+(a-d)^2</math>
<math>a^2+d^2+2ad=a^2+a^2+d^2-2ad</math>
<math>a^2=4ad</math>
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
[[Category:Triangles]]