Question

the base of an isoseceles triangle measure 24 cm and it's area is 60cm^2 find it's perimeter

Answer 1

$\textbf{ Hello Mate }$

As we know that area of isosceles ∆

Area = ( x / 2 )√( y² + x² / 4 )

Here, x = 24 cm

$\frac{x}{2} \sqrt{( {y}^{2} - \frac{ {x}^{2} }{4} )} \\ \frac{x}{2} \sqrt{( {y}^{2} - {( \frac{x}{2} )}^{2} } \\ \frac{24}{2} \sqrt{ {y}^{2} - { (\frac{24}{2} )}^{2} } \\ 12 \sqrt{ {y}^{2} - 144} = 60 \\ \sqrt{ {y}^{2} - 144} = \frac{60}{12} \\ {y}^{2} - 144 = {5}^{2} \\ {y}^{2} = 25 + 144 \\ y = \sqrt{169} \: or \: 13$
Now, perimeter of isosceles ∆ = 2y + x
= 2(13) + 24
= 26 + 24 = 50 cm

$\boxed{ Answer : 50 cm }$

Have great day ahead

Answer 2

Let AD⊥BC be the altitude of the given triangle.
ΔADB ≅ ΔADC (By RHS Congruence Condition)
∴BD =DC=12 Cm
Area of triangle = 1/2 ×base × height
60= 1/2 ×24×height
height =60/12= 5 cm
height= AD= 5cm
In right triangle ADB
$AB^2= AD^2 +BD^2 \: (By \: Pythagoras \: theorem) \\ AB^2 = {5}^{2} + {12}^{2} \\ AB^2 = 25 + 144 \\ AB^2 = 169 \\ AB= 13 \: cm$
Hence, AB=13 cm
=> AC = 13 Cm (Triangle ABC is isosceles)
and BC( base of isosceles triangle = 24 cm (given)
Therefore, Perimeter of Triangle ABC = 24+13+13 = 50 cm