Solution: According to the given question, \[\Delta \text{ }ABC\text{ }\sim\text{ }\Delta \text{ }ADE\] So, we have \[AE/AC\text{ }=\text{ }DE/BC\] \[4/11\text{ }=\text{ }6.6/BC\] Or, \[BC=\left(...
(i) In \[\vartriangle \text{ }ADE\text{ }and\text{ }\vartriangle \text{ }ABC\] \[AE/EC\text{ }=\text{ }6/7.5\text{ }=\text{ }4/5\] \[AD/BD\text{ }=\text{ }4/5\] \[\left[ BD\text{ }=\text{ }AB\text{...
In \[\vartriangle \text{ }APQ\text{ }and\vartriangle \text{ }ABC\] \[\angle APQ\text{ }=\angle ABC\] [As PQ || BC, corresponding angles are equal.] \[\angle PAQ\text{ }=\angle BAC\] [Common angle]...
Solution: As, \[\vartriangle CPQ\text{ }\sim\text{ }\vartriangle CAB\text{ }by\text{ }AA\] criterion for similarity We have, \[CP/AC\text{ }=\text{ }CQ/CB\] \[CP/AC\text{ }=\text{ }4.8/8.4\text{...
Solution: (i) In \[\vartriangle CPQ\text{ }and\text{ }\vartriangle CAB\] \[\angle PCQ\text{ }=\angle APQ\] [As PQ || AB, corresponding angles are equal.] \[\angle C\text{ }=\angle C\] [Common angle]...
Solution: Because, \[\vartriangle ADE\text{ }\sim\text{ }\vartriangle ABC\text{ }by\text{ }AA\] criterion for similarity So, we have \[AD/AB\text{ }=\text{ }DE/BC\] \[3/8\text{ }=\text{ }DE/4.8\]...
Solution: (i) In \[\vartriangle ABC,\text{ }as\text{ }DE\text{ }||\text{ }BC\] Using BPT, \[AD/DB\text{ }=\text{ }AE/\text{ }EC\] So, \[AD/AB\text{ }=\text{ }AE/AC\] Now, \[AD/AB\text{ }=\text{...
Solution: (i) According to the given question, \[AD/DB\text{ }=\text{ }3/5\] And \[DE\text{ }||\text{ }BC\] Using Basic Proportionality theorem, \[AD/DB\text{ }=\text{ }AE/EC\] \[AE/EC\text{...