In Fig. ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. Prove that ar (ABCDE) = ar (ΔAPQ). ____ and ____.

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Accepted Answer

I.Firstly consider the triangle (i) in the figure. As the side DE is parallel to the base of the triangle BC.
So, the side DE divides the sides of the triangle in the same ratio.
So, consider the ratio of the sides:

\frac{AD}{DB} = \frac{AE}{EC}

EC = AE \times \frac{DB}{AD}

Substitute the lengths of the sides AE, DB and AD as given in the diagram.

\Rightarrow EC = AE \times \frac{DB}{AD}

= 1 cm \times \frac{3 cm}{1.5 cm}

= 2 cm

So, the length of the side EC in the triangle (i) given in the figure is equal to 2cm.
So, the correct answer is “2cm”.
II.Firstly consider the triangle (ii) in the figure. As the side DE is parallel to the base of the triangle BC.
So, the side DE divides the sides of the triangle in the same ratio.
So, consider the ratio of the sides:

\frac{AD}{DB} = \frac{AE}{EC}

AD = DB \times \frac{AE}{EC}

Substitute the lengths of the sides AE , DB and AD as given in the diagram.

\Rightarrow AD = DB \times \frac{AE}{EC}

= 7.2 cm \times \frac{1.8 cm}{5.4 cm}

= 7.2 cm \times \frac{1}{3 cm} = 2.4 cm

So, the length of the side AD in the triangle (ii) given in the figure is equal to 2.4cm.
So, the length of the side EC in the triangle (i) given in the figure is equal to 2cm and the length of the side AD in the triangle (i) given in the figure is equal to 2.4cm.
So, the correct answer is “2.4cm”.

In Fig. ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. Prove that ar (ABCDE) = ar (ΔAPQ). and .

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