$$\triangle$$ABC $$\sim$$ $$\triangle$$PQR and PQ = 6 cm, QR = 8 cm and PR = 10 cm.If ar($$\triangle$$ABC) : ar($$\triangle$$PQR) = 1 : 4, then AB is equal to:

Given that,

$$\triangle$$ABC $$\sim$$ $$\triangle$$PQR

PQ = 6 cm, QR = 8 cm and PR = 10 cm

ar($$\triangle$$ABC) : ar($$\triangle$$PQR) = 1 : 4

Now as per the theorem of similarity of triangle,

$$\dfrac{ar(\triangle ABC)}{ar(\triangle PQR)}=\dfrac{AB^2}{PQ^2}=\dfrac{BC^2}{QR^2}=\dfrac{AC^2}{PR^2}$$
Now substituting the values,
$$\dfrac{AB^2}{PQ^2}=\dfrac{1}{4}-------(i)$$

taking square root of both side of equation (i) and substituting the values

$$\Rightarrow \dfrac{AB}{8}=\dfrac{1}{2}$$

$$\Rightarrow AB=\dfrac{8}{2}=4$$cm