In the following figure, ABCD to a trapezium with AB ‖ DC. If AB = 9 cm, DC = 18 cm, CF= 13.5 cm, AP = 6 cm and BE = 15 cm, Calculate: (i) EC (ii) AF - Noon Academy

In the following figure, ABCD to a trapezium with AB ‖ DC. If AB = 9 cm, DC = 18 cm, CF= 13.5 cm, AP = 6 cm and BE = 15 cm, Calculate: (i) EC (ii) AF

Class 10 | Exercise 15E | ICSE | Maths | Selina | Similarity (With Applications to Maps and Models)

In the following figure, ABCD to a trapezium with AB ‖ DC. If AB = 9 cm, DC = 18 cm, CF= 13.5 cm, AP = 6 cm and BE = 15 cm, Calculate: (i) EC (ii) AF

Selina Solutions Concise Class 10 Maths Chapter 15 ex. 15(E) - 2

Solution:

(i) In

$\Delta \text{ }AEB\text{ }and\text{ }\Delta \text{ }FEC,$

$\angle AEB\text{ }=\angle FEC$

[Vertically opposite angles]

$\angle BAE\text{ }=\angle CFE$

[Since, AB||DC]

Hence,

$\vartriangle AEB\text{ }\sim\text{ }\vartriangle FEC\text{ }by\text{ }AA$

criterion for similarity.

So, we have

$AE/FE\text{ }=\text{ }BE/EC\text{ }=\text{ }AB/FC$

$15/EC\text{ }=\text{ }9/13.5$

Or,

$EC\text{ }=\text{ }22.5\text{ }cm$

(ii) In

$\Delta \text{ }APB\text{ }and\text{ }\Delta \text{ }FPD$

$\angle APB\text{ }=\angle FPD$

[Vertically opposite angles]

$\angle BAP\text{ }=\angle DFP$

[Since, AB||DF]

Hence,

$\vartriangle APB\text{ }\sim\text{ }\vartriangle FPD\text{ }by\text{ }AA$

criterion for similarity.

So, we have

$AP/FP\text{ }=\text{ }AB/FD$

$6/FP\text{ }=\text{ }9/31.5$

Or,

$FP\text{ }=\text{ }21\text{ }cm$

So,

$AF\text{ }=\text{ }AP\text{ }+\text{ }PF$

$=\text{ }6\text{ }+\text{ }21\text{ }=\text{ }27\text{ }cm$

i

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