Solution: According to the given question, DE || AB By the basic proportionality theorem, $CD/AD\text{ }=\text{ }CE/BE$ Therefore, when a line is drawn parallel to one of the triangle's sides and...
Solution: According to the given question, $AB\text{ }=\text{ }4\text{ }cm$, $DE\text{ }=\text{ }6\text{ }cm$ $EF\text{ }=\text{ }9\text{ }cm$ $FD\text{ }=\text{ }12\text{ }cm$ Also, ∆ABC ∼ ∆DEF We...
Solution: Let an equilateral triangle of side 8 cm be ABC. $AB\text{ }=\text{ }BC\text{ }=\text{ }CA\text{ }=\text{ }8\text{ }cm$. (all sides of an equilateral triangle is equal) Construct an...
Solution: According to the given question, At point O, AC and PQ intersect each other and AB||DC. From triangles AOP and COQ, $\angle AOP\text{ }=~\angle COQ$[As they are vertically opposite angles]...
Solution: According to the given question, The given figure, PQRS is a trapezium in which PQ || RS and PQ = 3RS $PQ/RS\text{ }=\text{ }3/1\text{ }=\text{ }3$…(i) In triangles POQ and ROS, $\angle...
Solution: According to the given question, $\text{ }\Delta NSQ~\cong ~\Delta MTR$ $\angle 1\text{ }=~\angle 2$ Since, $\Delta NSQ\text{ }=\text{ }\Delta MTR$ As a result, $SQ\text{ }=\text{...
Solution: According to the given question, In triangle PQR, $P{{R}^{2}}~=\text{ }Q{{R}^{2}}$ and QM⊥PR Using the Pythagoras theorem, we obtain, $P{{R}^{2}}~=\text{ }P{{Q}^{2}}~+\text{ }Q{{R}^{2}}$...
Solution: As per the inquiry, Stomach muscle = 4 cm, DE = 6 cm EF = 9 cm FD = 12 cm Too, ∆ABC ∼ ∆DEF We have, By taking initial two terms, we have Also, by taking last two terms, we have,...
Solution: Leave ABC alone a symmetrical triangle of side 8 cm Stomach muscle = BC = CA = 8 cm. (all sides of a symmetrical triangle is equivalent) Draw height AD which is opposite to BC. Then, at...
Solution: As indicated by the inquiry, AC and PQ converge each other at the point O and AB||DC. From ∆AOP and ∆COQ, ∠AOP = ∠COQ [Since they are upward inverse angles] ∠APO = ∠CQO [since, AB||DC and...
Solution: As indicated by the inquiry, PQRS is a trapezium where PQ‖RS and PQ = 3RS PQ/RS = 3/1 = 3 … (I) In ∆POQ and ∆ROS, ∠SOR = ∠QOP [vertically inverse angles] ∠SRP = ∠RPQ [alternate angles] ∴...
SOLUTION: As indicated by the inquiry, \[\text{ }NSQ\cong MTR\] ∠1 = ∠2 Since, ∆NSQ = ∆MTR Along these lines, SQ = TR … .(I) Moreover, \[\angle 1\text{ }=\angle 2\Rightarrow PT\text{ }=\text{...
Solution: As per the inquiry, DE || AB Utilizing essential proportionality hypothesis, Compact disc/AD = CE/BE ∴ If a line is attracted corresponding aside of a triangle to such an extent that it...
Solution: As per the inquiry, In ∆PQR, PR2 = QR2 and QM⊥PR Utilizing Pythagoras hypothesis, we have, \[PR2\text{ }=\text{ }PQ2\text{ }+\text{ }QR2\] ∆PQR is correct calculated triangle at Q. From...