Question

In the figure, CAKE and LAKS are rectangles. P,O,L and S are the midpoints of CE,AK,CA and KE.
Which of the following statements are true for the given figure?

• A. $\text{ar}△ACO=\text{ar}△KEO$
• B. $\text{ar}△PAK=\text{ar}△CEO$
• C. $\text{ar}△ACO=\text{ar}△CEO$
• D. $\text{ar}△CXP=\text{ar}△YOK$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $\text{ar}△ACO=\text{ar}△KEO$
B $\text{ar}△PAK=\text{ar}△CEO$
D $\text{ar}△CXP=\text{ar}△YOK$


From the figure:
CE = AK and CE $\parallel$ AK (CAKE is a rectangle)
PC = OK (P and O are the midpoints of CE and AK)
CL =AL ( L is the midpoint of CA)

$△ACO$ and $△KEO$ are having equal base and are between the same parallels.
$\therefore \text{ar}\left(△ACO\right)=\text{ar}\left(△KEO\right)$

$△PAK$ and $△CEO$ are having equal base and are between the same parallels.
$\therefore \text{ar}\left(△PAK\right)$ =$\text{ar}\left(△CEO\right)$

In $△CXP$ and $△YOK$

$\text{Area of triangle}=\frac{1}{2}base\times height$

$\text{ar}\left(△CXP\right)=\frac{1}{2}CP\times CL$

$\text{ar}\left(△YOK\right)=\frac{1}{2}OK\times AL$
but
PC = OK (P and O are the midpoints of CE and AK)
CL =AL ( L is the midpoint of CA)

$\therefore △CXP$ and $△YOK$ are having equal base and height.

$\therefore \text{ar}\left(△CXP\right)=\text{ar}\left(△YOK\right)$