Ratio of areas of similar figures

**AeroScizor** · January 20th 2011, 03:05 AM

Hi all,
Here goes the question:

Ratio of areas of similar figures-figure1.jpg

In the diagram, AC,RQ and BP are parallel lines. AR=9cm,RQ=6cm,BP=8cm, and RB=3cm. Calculate the ratio of:
(i)area of triangle ARQ:area of trapezium RBPQ'
(ii)area of triangle BQR:area of triangle ARQ
(iii)area of triangle ABC:area of triangle BQR

Another question:

Ratio of areas of similar figures-figure2.jpg

In the diagram, AB=5cm,BE=4cm,EC=2cm,CD=3cm and DA=1cm. If DE=xcm, using a geometrical argument, deduce the value of x.

Thanks in advance

**Prove It** · January 20th 2011, 03:36 AM

In Q2 are the two triangles similar?

**AeroScizor** · January 20th 2011, 03:51 AM

nope it isn't stated in the question but anyway it doesn't look similar

**AeroScizor** · January 20th 2011, 04:36 AM

please can someone help me because i need it for my assisgnment. thanks again.

**Prove It** · January 20th 2011, 04:38 AM

We're not supposed to do work that contributes to a student's final grade, so you'll need to show us what you've tried...

**AeroScizor** · January 20th 2011, 04:45 AM

you got me wrong, by saying assignment i am referring to normal homework. I mean i have no idea how to start it off so can anyone provide me with hints?

**Soroban** · January 20th 2011, 05:42 AM

Hello, AeroScizor!

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$\text{In the diagram: }\:AC \parallel RQ \parallel BP.$

$AR=9,\; RQ=6,\;BP=8,\;RB=3.$

$\text{Calculate the ratios of the areas of:}$

$(i)\;\Delta ARQ : \text{quad }RBPQ$

We have: . $\Delta ARQ \sim \Delta ABP$

Hence: . $AR:AB \:=\:9:12 \;=\;3:4$

Then: . $\text{area }\Delta ARQ:\text{area }\Delta ABP \;=\;9:16$

Therefore: . $\Delta ARQ:\text{quad }RBPQ \;=\;9:7$

$(ii)\;\Delta BQR : \Delta ARQ$

$\Delta BQR$ has a base of 3.
$\Delta ARQ$ has a base of 9.
. . Their bases are in the ratio $3:9 \:=\:1:3$

They have the same height.

Therefore, their areas are in the ratio $1:3.$

$(iii)\; \Delta ABC : \Delta BQR$

We have: . $\Delta ABC \sim \Delta BQR$

Their bases are in the ratio $12:3 \:=\:4:1$
Their heights are also in the ratio $4:1$

Therefore, their areas are in the rato $16:1$

**AeroScizor** · January 20th 2011, 05:43 AM

thanks a million. i really wanted to know the thought process

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