Simple Area Ratio

Geometry Level 4

$A$ , $B$ , $C$ and $D$ are the four vertices of a square. $P$ , $Q$ and $R$ are the mid-points of $AB$ , $BC$ and $AD$ respectively. $S$ is the mid-point of $CD$ and $T$ is the mid-point of $RS$ .

Find $\dfrac{\text{ar } \triangle QTS}{\text{ar } \triangle RDT}$

Answer to just one place of decimal.

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The answer is 2.0.

2 solutions

Chew-Seong Cheong
Mar 3, 2015

Since $RT=TS\quad \Rightarrow \dfrac {A_{\triangle QTS}} {A_{\triangle RDT}} = \dfrac {QS}{DT}$

Let $QC=CS=RD=1$ , and we note that $\triangle QCS$ and $\triangle RDT$ are isosceles right triangles, then we have:

$\dfrac {A_{\triangle QTS}} {A_{\triangle RDT}} = \dfrac {QS}{DT} = \dfrac {\sqrt{2}}{\frac{1}{\sqrt{2}}} = \boxed{2}$

Thanks for posting a solution sir , I had quite forgotten abt this question .

A Former Brilliant Member - 6 years, 3 months ago

Mariano PerezdelaCruz
Feb 18, 2015

Triangle Δ RTD equal to Δ DTS by symmetry. Then Δ DTS shares base TS with triangle Δ QTS and the height SQ is twice DT, so the area ratio is 2.

Simple Area Ratio

The answer is 2.0.

2 solutions

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