Question

if the base of an isosceles triangle whose area is12 cm² and one if the equal side is 5 cm,is:​

Answer 1

Answer:

area=1/2xBxH

12=1/2xBxH

24=BxH

B=24÷5

B=4.8

Step-by-step explanation:

Mark me brainliest

Answer 2

Answer:

Let equal sides be (a)=5cm and base (b)=?

Area of an isosceles triangle=12sq.cm

Area of an isosceles triangle= $\rm{ \dfrac{b}{4} \sqrt{ 4a^2 - b^2}}$

12 = $\rm { \dfrac{b}{4} \sqrt{ 4 \times (5)^2 - b^2}}$

48 = $\rm { b\sqrt{100-b^2}}$

Squaring both sides  

$\rm{ (b^2  −64)=0 ⇒ b^2  = 64 ⇒ b = ±8}$

or

[tex]\rm{(b^2 −36)=0⇒b^2 =36⇒b=±6}[/tex]

Hence, the base is ±8,±6.