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In the figure, given below, straight lines AB and CD intersect at P; and AC || BD. Prove that: (i) ∆APC and ∆BPD are similar. (ii) If BD = 2.4 cm, AC = 3.6 cm, PD = 4.0 cm and PB = 3.2 cm; find the lengths of PA and PC.
In the figure, given below, straight lines AB and CD intersect at P; and AC || BD. Prove that: (i) ∆APC and ∆BPD are similar. (ii) If BD = 2.4 cm, AC = 3.6 cm, PD = 4.0 cm and PB = 3.2 cm; find the lengths of PA and PC.
Class 10 | Exercise 15A | ICSE | Maths | Selina | Similarity (With Applications to Maps and Models)
In the figure, given below, straight lines AB and CD intersect at P; and AC || BD. Prove that: (i) ∆APC and ∆BPD are similar. (ii) If BD = 2.4 cm, AC = 3.6 cm, PD = 4.0 cm and PB = 3.2 cm; find the lengths of PA and PC.

Selina Solutions Concise Class 10 Maths Chapter 15 ex. 15(A) - 1

Answer:

(I) In \[\vartriangle \mathbf{APC}\text{ }\mathbf{and}\text{ }\vartriangle \mathbf{BPD}\]we have

\[\angle APC\text{ }=\angle BPD\] [Vertically inverse angles]

\[\angle ACP\text{ }=\angle BDP\] [Alternate points as, AC || BD]

Accordingly, \[\vartriangle APC\text{ }\sim\text{ }\vartriangle BPD\]by \[AA\] similarity criterion

 

(ii) So, by comparing portions of comparative triangles, we have

\[PA/PB\text{ }=\text{ }PC/PD\text{ }=\text{ }AC/BD\]

Given, \[BD\text{ }=\text{ }2.4\text{ }cm,\text{ }AC\text{ }=\text{ }3.6\text{ }cm,\text{ }PD\text{ }=\text{ }4.0\text{ }cm\text{ }and\text{ }PB\text{ }=\text{ }3.2\text{ }cm\]

\[PA/\left( 3.2 \right)\text{ }=\text{ }PC/4\text{ }=\text{ }3.6/2.4\]

\[PA/3.2\text{ }=\text{ }3.6/2.4\text{ }and\text{ }PC/4\text{ }=\text{ }3.6/2.4\]

Consequently,

\[PA\text{ }=\text{ }\left( 3.6\text{ }x\text{ }3.2 \right)/2.4\text{ }=\text{ }4.8\text{ }cm\] and

\[PC\text{ }=\text{ }\left( 3.6\text{ }x\text{ }4 \right)/2.4\text{ }=\text{ }6\text{ }cm\]

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