Difference between revisions of "Activities-Pythagoras theorem problems"

Revision as of 12:39, 11 July 2014

1 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

2 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

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

@@ Line 1: / Line 1: @@
-#In Right angled ∆ABC,∟BAC= 90°,∟B: ∟C = 1:2 andAC= 4cm.calculate thelenght of BC
+#1 In Right angled ∆ABC,∟BAC= 90°,∟B: ∟C = 1:2 andAC= 4cm.calculate thelenght of BC
 ''''''Solution'''''''''
 in some special right angled triangle
@@ Line 12: / Line 12: @@
 BC = 8 cm
-#A door of width 6 mt has an archabove it having aheight of 2 mt , find the radius if the arch
+#2 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>
+x=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
 # The sides of a right angled triangle are in an AP. Show that sides are in ther ratio 3:4:5