find the area of an isosceles triangle, each of whose equal sides measure 13 cm and whose base measure 20 cm.
Answers
Step-by-step explanation:
sides of triangle = 13,13,20
a=13 , b=13 , c=20
we find the area by heron's formula
semi perimeter(s) = (13+13+20)/2
46/2
23
s-a = 23-13 = 10
s-b = 23-13 = 10
s-c = 23-3 = 3
root of s×(s-a)×(s-b)×(s-c)
root of 23×10×10×3
10 root of 69 ans
OR
we can find the area by dividing it into two right angled triangle
base of right angle triangle = 10
hypotenuse =13
hight = ?
(height)^2 = (hypotenuse)^2 - (base)^2
(h)^2 = (13)^2 - (10)^2
(h)^2 = 169 - 100
(h)^2 = 69
(h) = root of 69
area = 1/2 × b × h
1/2 × 10 × root of 69
5 root of 69
area of both triangle
2 × 5 root of 69
10 root of 69 ans
Let the vertices of the given isosceles triangle be ABC respectively.
As it is an isosceles triangle, so two sides AB = AC = 13 cm
Length of the base BC = 20 cm
Construction: Draw perpendicular AD .
In triangle ADB,
BD = 10cm (As D is mid point of BC)
AB = 13 cm
AD² + BD² = AB² (Using Pythagoras theorem)
AD² + 10² = 13²
AD² = 13² - 10² = 8.3 cm
So, area of triangle ABC = 1/2 × 20 × 8.3 cm²
area of triangle ABC = 83 cm².