Question

the perimeter of an isosceles triangle is 14cm and the ratio of the lateral side to the base is 5:4. Find the area of the triangle.

Answer 1

 Let sides are 5x, 5x and 4x. 
Therefore, 5x+5x+4x = 14 
14x= 14 
x=1 
Therefore sides are 5cm, 5cm and 4cm. 
S= a+b+c/2 = 5+5+4/2 = 7 
Area= √ s(s-a)(s-b)(s-c) 
=√ 7(7-5)(7-5)(7-4) 
=√ 7x 2x 2x 3 
2√21
hope it helps u.............

Answer 2

Area of triangle is $2\sqrt{21}\ cm^2$.

Step-by-step explanation:

Given:

The perimeter of an isosceles triangle is 14 cm and the ratio of the lateral side to the base is 5:4.

Now, let the equal sides be $5a$ and the base be $4a$   [given]

Perimeter of the triangle = Sum of all sides = 14

⇒ $5a + 5a + 4a = 14$

⇒ $14a = 14$

⇒ $a=1$

Now, base, $b=4a=4(1)=4\ cm$

Lateral sides, $l =5a=5(1)=5\ cm$

For an isosceles triangle, the altitude from the vertex opposite the base divides the triangle into two equal right angled triangles bisecting the base of the original triangle.

So, the height of the right angled triangle is the altitude, the base is half of the base of the original triangle and hypotenuse is the lateral side.

Using Pythagoras Theorem, we have

$l^2=h^2+(\frac{b}{2} )^2$

$5^2=h^2+2^2$

$25=h^2+4$

$25-4=h^2$

$h^2=21$

$h=\sqrt{21}\ cm$

Area of triangle = $\frac{1}{2}\times base\times height$

                         = $\frac{1}{2}\times 4\times \sqrt{21}$

∴ Area of triangle = $2\sqrt{21}\ cm^2$