Question : Let $\triangle ABC \sim \triangle RPQ$ and $\frac{{area}(\triangle {ABC})}{{area}(\triangle {PQR})}=\frac{4}{9}$. If AB = 3 cm, BC = 4 cm and AC = 5 cm, then RP (in cm) is equal to:
Option 1: 6
Option 2: 5
Option 3: 4.5
Option 4: 12
Correct Answer: 4.5
Solution : When $Δ ABC ∼ Δ RPQ$ $\frac{AB}{RP}=\frac{BC}{PQ}=\frac{AC}{QR}=\frac{\sqrt{area(ABC)}}{\sqrt{area(RPQ)}}$ $\therefore \frac{AB}{RP} = \sqrt\frac{4}{9}$ ⇒ $\frac{AB}{RP} = \frac{2}{3}$ ⇒ $RP = \frac{3}{2} × AB=\frac{3}{2} × 3=4.5$ cm Hence, the correct answer is 4.5.
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Question : Let $\triangle {ABC} \sim \triangle {RPQ}$ and $\frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle P Q R)}=\frac{4}{9}$. If ${AB}=3 {~cm}, {BC}=4 {~cm}$ and ${AC}=5 {~cm}$, then ${PQ}$ (in ${cm}$ ) is equal to:
Question : In $\triangle$ABC, D is the median from A to BC. AB = 6 cm, AC = 8 cm, and BC = 10 cm.The length of median AD (in cm) is:
Question : For a triangle ABC, D and E are two points on AB and AC such that $\mathrm{AD}=\frac{1}{6} \mathrm{AB}$, $\mathrm{AE}=\frac{1}{6} \mathrm{AC}$. If BC = 22 cm, then DE is _______. (Consider up to two decimals)
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