Question

. The base of an isosceles triangle is 20cm and its perimeter is 42cm.
Find its area​

Answer 1

[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

$\boxed{\textbf{\large{Area of triangle ABC=45.83cm square }}}$

Answer 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