Question

In PQR, DE || QR and DE = 1/4QR find ar pqr/
ar PDE​

Answer 1

Answer :

$\tt \dfrac{Area\;\triangle PQR}{Area\;\triangle PDE} = \dfrac{16}{1}$

Step-by-step explanation :

Given that :

• In PQR, DE || QR.
• DE =  1/4QR.

To Find :

Area of (ΔPQR)/(ΔPDE).

Solution :

In ΔPQR and ΔPDE,

⇒ ∠QPR =∠DPE

(Common)

⇒ ∠PQR = ∠PDE    

(Corresponding Angles)

⇒ ∠PRQ = ∠PED    

(Corresponding Angles)

Next step is also same,

⇒ ΔPQR ~ ΔPDES.

Values in Equation :

$\tt \implies \dfrac{Area\;\triangle PQR}{Area\;\triangle PDE} = \dfrac{QR^2}{DE^2}\\ \\ \\\implies \dfrac{Area\;\triangle PQR}{Area\;\triangle PDE} =\dfrac{QR^2}{\left(\frac{1}{4}QR\right)^2}\\ \\ \\\implies \dfrac{Area\;\triangle PQR}{Area\;\triangle PDE} = \dfrac{16}{1}\\ \\\\\bigstar \textbf{\underline{\underline{Refer the Attachment for the figure.}}}$

Answer 2

Answer:-

Refer the given attachment:-