The area of an isosceles triangle is 240 2 cm and the length of each one of the equal side is 26 cm find its base
Answers
Step-by-step explanation:
(i)Height:
Consider an Isosceles triangle ABC,ABC, withAB=AC=26cm,BC=20cmAB=AC=26cm,BC=20cm
Draw a perpendicular ADAD on non-congruent side BCBC
In an Isosceles triangle a perpendicular bisects then non-congruent side
Hence, BD=\frac{1}{2} BC=\frac{1}{2}(20)=10cmBD=
2
1
BC=
2
1
(20)=10cm
In ΔABD,ΔABD,
⇒AB^{2}=AD^{2}+BD^{2}⇒AB
2
=AD
2
+BD
2
⇒26^{2}=AD^{2}+10^{2}⇒26
2
=AD
2
+10
2
⇒676=AD^{2}=100⇒676=AD
2
=100
⇒AD^{2}=100-676⇒AD
2
=100−676
⇒AD^{2}=576⇒AD
2
=576
⇒AD=\sqrt{576}⇒AD=
576
⇒AD=24cm⇒AD=24cm
(ii)Area:
First find out It's height by Pythagoras theorem,
⇒c^{2}=a^{2}+b^{2}⇒c
2
=a
2
+b
2
⇒26^{2}=10^{2}+b^{2}⇒26
2
=10
2
+b
2
⇒576=b^{2}⇒576=b
2
⇒b^{2}=576⇒b
2
=576
⇒b=\sqrt{576}⇒b=
576
⇒b=24⇒b=24
Now area=base×heightNow area=base×height
⇒20×24⇒20×24
⇒480cm^{2}⇒480cm
2