Question

Find the area of triangle two sides of which are 11 cm and the perimeter is 42 cm

Answer 1

Two sides =11+11=22cm
Then third side=42-22=>20cm

As it's two side is equal then,it must be an isosceles triangle

Base=20
Each side=11cm

Height be 'x'
Half of vase=20/2=10

To find height we can use Pythagoras theorem

11²=x²+10²
121-100=x²
21=x²
√21=x
4.58cm


Area=1/2×base×height
=1/2×20×4 .58
=45.8cm²

Answer 2

Hii friend,

AB = BC = 11 cm

Perimeter = 42

AB + BC + AC = 42j

11 + 11 + AC = 42

AC = 42-22

AC = 20 cm

Since,

∆ABC have two sides equal and one side is different it means ∆ABC is isosceles triangle.

Therefore,

Semi Perimeter(S)= 1/2 × (AB+BC+AC) = 1/2×(11+11+20) = 1/2 × 42 = 21 cm

S-AB = 21-11 = 10

S-BC = 21-11 = 10 cm

S-AC = 21-20 = 1 cm

Therefore,

Area of∆ABC = ✓S(S-AB)(S-BC)(S-AC) = ✓21×10×10×1 = 2100 cm = 45.8 cm²

HOPE IT WILL HELP YOU...... :-)