Question

The diagonals of a cyclic quadrilateral \[ABCD\] intersect at \[P\] and the area of the triangle \[APB\] is \[24\] square \[cm\]. if \[AB{\text{ }} = {\text{ }}8cm{\text{ }} and {\text{ }}CD{\text{ }} = {\text{ }}5cm,\] then what is the area of the triangle \[CPD?\]
A. $ 24c{m^2} $
(B) $ 15c{m^2} $
(C) $ 12.5c{m^2} $
(D) $ 9.375c{m^2} $

Answer 1

Hint:This type of question demands us to solve it with the help of construction.
Draw the demands and try to understand what is given data and what is required.
Students are advised to make the use of construction to establish the relationship between the properties of the shape and the diagram.
Students are advised to refer to the properties of a cyclic quadrilateral before attempting the question.
To answer this question we need to know, \[ABCD\] is a cyclic quadrilateral, then angle found by chord \[AD\] in same segment \[i.e.,\] $ \angle DCA = \angle ABD $ , similar is the case for other chords also.
Also, ratio of area of similar triangles \[ = \] $ {\left( {ratio{\text{ }}of{\text{ }}sides} \right)^2} $ .

Complete step-by-step answer:
let us understand what the question provides and what it demands.
The question says,the diagonals of a cyclic quadrilateral \[ABCD\] intersect at \[P\]and the area of the triangle \[APB\]is \[24\]square \[cm\]. if \[AB{\text{ }} = {\text{ }}8cm{\text{ }}and{\text{ }}CD{\text{ }} = {\text{ }}5cm,\]
Thus, we have to find the area of the triangle \[CPD\].
Diagram for assistance:-

From the figure, it is clear that by the property of the cyclic quadrilateral,
 $
 \Rightarrow \angle DCA = \angle ABD \\
 \Rightarrow \angle DCP = \angle ABP{\text{ }}\left\{ {as{\text{ }}P{\text{ }}is{\text{ intersection }}of{\text{ }}AC{\text{ }}and{\text{ }}BD} \right\} \\
  $
Similarly, angle formed by chord \[BC\] in same segment $ \angle CDB = \angle BAC $
 $
\Rightarrow \angle CDP = \angle BAP{\text{ }}\left\{ {as{\text{ }}P{\text{ }}is{\text{ intersection }}of{\text{ }}AC{\text{ }}and{\text{ }}BD} \right\} \\
\Rightarrow \angle APB = \angle DPC{\text{ }}\left( {Opposite{\text{ }}angles} \right) \\
  $
In $ \Delta APB{\text{ }}and{\text{ }}\Delta CPD, $
 $
  \angle ABP = \angle DCP \\
  \angle BAP = \angle CDP \\
  \angle APB = \angle DPC \\
  \Delta APB \approx \Delta CPD \\
  $
Ratio of similar triangle \[ = \] $ {\left( {ratio{\text{ }}of{\text{ }}sides} \right)^2} $
 $ \dfrac{{Area{\text{ }}of\Delta APB}}{{Area{\text{ }}of\Delta CPD}} = {\left( {\dfrac{{AB}}{{CD}}} \right)^2} $
 $
\Rightarrow \dfrac{{24}}{{Area{\text{ }}of{\text{ }}\Delta CPD}} = {\left( {\dfrac{8}{5}} \right)^2} \\
 \Rightarrow Area{\text{ }}of{\text{ }}\Delta CPD = \dfrac{{24 \times {5^2}}}{{{8^2}}} \\
 \Rightarrow Area{\text{ }}of{\text{ }}\Delta CPD = \dfrac{{75}}{8}c{m^2} = 9.375c{m^2} \\
  $

So, the correct answer is “Option D”.

Note: If an angle of one triangle is congruent to the corresponding angle of another triangle and the lengths of the sides including these angles are in proportion, the triangles are similar. The corresponding sides of similar triangles are in proportion.
Students can make mistakes in the writing down the ratios and the angles as a matter of confusion. Thus, I am advised to approach every step by referring to the diagram.