Δ ABC ~ Δ PQR. A(Δ ABC) = 80 cm2, A(Δ PQR) = 125 cm2 and AB = 12 cm. Complete the following activity to find PQ.Solution :Δ ABC ~ Δ PQR .................... (given)\(\therefore \dfrac{A(\Delta ABC)}{A(\Delta PQR)} = \dfrac{\text{____}}{PQ^2}\) ......…
Q.3 (A)(1) Δ ABC ~ Δ PQR. A(Δ ABC) = 80 cm2, A(Δ PQR) = 125 cm2 and AB = 12 cm. Complete the following activity to find PQ.
Solution :
Δ ABC ~ Δ PQR .................... (given)
\(\therefore \dfrac{A(\Delta ABC)}{A(\Delta PQR)} = \dfrac{\text{____}}{PQ^2}\) .................... (theorem of areas of similar triangles)
\(\therefore \dfrac{80}{125} = \dfrac{12^2}{PQ^2}\)
\(\therefore \dfrac{16}{25} = \dfrac{12^2}{PQ^2}\)
\(\therefore \dfrac{\text{____}}{5} = \dfrac{12}{PQ}\) .................... (taking square roots)
\(\therefore PQ =\) ____ cm