Answer:
6 cm (Option A)
Step-by-step explanation:
GIVEN :-
- ΔPQR whose area is 32 cm² & has base QR.
- ΔABC whose area is 8 cm² & has base BC = 3 cm
- ΔPQR is an enlarged figure of ΔABC.
TO FIND :-
FACTS TO KNOW BEFORE SOLVING :-
Lets say there are two triangles ΔABC & ΔXYZ , who are similar to each other.
[tex]=> \frac{Area \: of \: ABC}{Area of XYZ} = (\frac{AB}{XY})^ {2} = (\frac{BC}{YZ})^2 = (\frac{AC}{XZ} )^2[/tex]
SOLUTION :-
In the question it's given that ΔPQR is an enlarged figure of ΔABC.
⇒ ΔPQR is similar to ΔABC.
[tex]=> \frac{Area \: of \: PQR}{Area \: of \: ABC } = (\frac{QR}{BC})^2[/tex]
[tex]=>\frac{32}{8} = (\frac{QR}{3})^2[/tex]
Root squaring both the sides ,
[tex]=> \sqrt{(\frac{QR}{3})^2} = \sqrt{\frac{32}{8} }[/tex]
[tex]=> \frac{QR}{3} = \sqrt{4} =2[/tex]
Multiplying both the sides by 3 ,
[tex]=> \frac{QR}{3} \times 3 = 2 \times 3[/tex]
[tex]=> QR = 6 \:cm[/tex]
