Question

In the figure, if $TD\parallel PO\parallel CH,$ then which of the following are true.

• A. $\text{ar}\left(△COP\right)=\text{ar}\left(△HOP\right)$
• B. $\text{ar}\left(△COP\right)=\text{ar}\left(△POD\right)$
• C. $\text{ar}\left(△POT\right)=\text{ar}\left(△POD\right)$
• D. $\text{ar}\left(△COT\right)=\text{ar}\left(△HPD\right)$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $\text{ar}\left(△COP\right)=\text{ar}\left(△HOP\right)$
C $\text{ar}\left(△POT\right)=\text{ar}\left(△POD\right)$
D $\text{ar}\left(△COT\right)=\text{ar}\left(△HPD\right)$


Theorem: Triangles on the same/equal base and between the same parallels are equal in area.

$\because △COP\text{and}△HOP$ are on the same base PO and between the same parallels PO and CH.
$\therefore \text{ar}\left(△COP\right)=\text{ar}\left(△HOP\right)$


$\because △POT\text{and}△POD$ are on the same base PO and between the same parallels PO and TD.
$\therefore \text{ar}\left(△POT\right)=\text{ar}\left(△POD\right)$

We have proved that
$\text{ar}\left(△COP\right)=\text{ar}\left(△HOP\right).......\left(1\right)$

$\text{ar}\left(△POT\right)=\text{ar}\left(△POD\right).......\left(2\right)$

Adding (1) and (2), we get

$\text{ar}\left(△COP\right)+\text{ar}\left(△POT\right)$ $=\text{ar}\left(△HOP\right)+\text{ar}\left(△POD\right)$

$\Rightarrow \text{ar}\left(△COT\right)=\text{ar}\left(△HPD\right)$