. The base of an isosceles triangle is 20cm and its perimeter is 42cm.
Find its area
Answers
[Note :refer the figure in the attachment ]
Step-by-step explanation:
we know,
◾As if we consider the isosceles triangle ,there two sides are equal in length but other remaining side is distinct from those two sides, I. e unequal in length
◾let us consider the isosceles triangle is [ triangle ABC], were the sides are
side a, side b , side c
◾the base is side AB, I. e side b ,the base is given base = 20cm
Side b = base = 20cm
As we know,
◾Perimeter of triangle = sum of all sides of triangle
42 = side a + side b + side c
In the isosceles triangle ABC ,the equal sides are side a and side c Therefor side a = side c
42 = side c + side c + side b (base)
42 = 2 side c + 20
42 - 20 = 2 side c
22 / 2 = side c
side c = 11cm = side a
◾Now, draw a perpendicular through vertex A to the side b
so consider the point M that perpendicular joining to the side b
◾To find the height of triangleABC ,let us assume the triangle AMB
In triangle AMB
side AB = 11 cm, side BM = 1/2 side BC (base) = 1/2 ( 20) = 10cm
side BM = 10cm
/_ AMB = 90°
So, by Pythagoras theorem
(side AB )^2 = (side AM) ^2 + (side BM)^2
(11)^2 =(side AM) ^2 + ( 10 )^2
121 - 100 = (side AM) ^2
√ 21 cm = side AM = height of triangle ABC
◾In triangle ABC,
side AM is the height of a triangle
Therefor, we know the formula to find the area of triangle
Area of triangle
= 1/2 (base) x (height)
Area of triangle ABC
= 1/2 ( side BC (base) ) x ( height)
= 1/2 ( 20 ) x ( √ 21 )
= 10 √21 = 10 x ( 4.583 )
=45.83 cm^2
Answer:
45.83 square
Step-by-step explanation:
we know,
◾As if we consider the isosceles triangle ,there two sides are equal in length but other remaining side is distinct from those two sides, I. e unequal in length
◾let us consider the isosceles triangle is [ triangle ABC], were the sides are
side a, side b , side c
◾the base is side AB, I. e side b ,the base is given base = 20cm
Side b = base = 20cm
As we know,
◾Perimeter of triangle = sum of all sides of triangle
42 = side a + side b + side c
In the isosceles triangle ABC ,the equal sides are side a and side c Therefor side a = side c
42 = side c + side c + side b (base)
42 = 2 side c + 20
42 - 20 = 2 side c
22 / 2 = side c
side c = 11cm = side a
◾Now, draw a perpendicular through vertex A to the side b
so consider the point M that perpendicular joining to the side b
◾To find the height of triangle ABC ,let us assume the triangle AMB
In triangle AMB
side AB = 11 cm, side BM = 1/2 side BC (base) = 1/2 ( 20) = 10cm
side BM = 10cm
/_ AMB = 90°
So, by Pythagoras theorem
(side AB )^2 = (side AM) ^2 + (side BM)^2
(11)^2 =(side AM) ^2 + ( 10 )^2
121 - 100 = (side AM) ^2
√ 21 cm = side AM = height of triangle ABC
◾In triangle ABC,
side AM is the height of a triangle
Therefor, we know the formula to find the area of triangle
Area of triangle
= 1/2 (base) x (height)
Area of triangle ABC
= 1/2 ( side BC (base) ) x ( height)
= 1/2 ( 20 ) x ( √ 21 )
= 10 √21 = 10 x ( 4.583 )
=45.83 cm^2