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GeometryQuestion and Answers: Page 14

Question Number 199023    Answers: 1   Comments: 1

Question Number 199006    Answers: 0   Comments: 0

Given that ABCD is a trapezium such that AD//BC. The centroid of △ABD lies on the bisector of ∠BCD. Show that the centroid of △ABC lies on the bisector of ∠ADC.

$$\mathrm{Given}\:\mathrm{that}\:{ABCD}\:\mathrm{is}\:\mathrm{a}\:\mathrm{trapezium}\:\mathrm{such}\:\mathrm{that}\:{AD}//{BC}. \\ $$$$\mathrm{The}\:\mathrm{centroid}\:\mathrm{of}\:\bigtriangleup{ABD}\:\mathrm{lies}\:\mathrm{on}\:\mathrm{the}\:\mathrm{bisector}\:\mathrm{of}\:\angle{BCD}. \\ $$$$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{centroid}\:\mathrm{of}\:\bigtriangleup{ABC}\:\mathrm{lies}\:\mathrm{on}\:\mathrm{the}\:\mathrm{bisector}\:\mathrm{of}\:\angle{ADC}. \\ $$

Question Number 198933    Answers: 1   Comments: 2

Question Number 198913    Answers: 3   Comments: 1

Question Number 199089    Answers: 2   Comments: 0

Question Number 198763    Answers: 3   Comments: 3

Question Number 198604    Answers: 1   Comments: 1

Question Number 198643    Answers: 0   Comments: 0

Given an isosceles triangle ABC which ∠A= 30°, AB = AC. A point D is midpoint of BC . A point P is chosen on then segment AD and a point Q is chosen on the side AB so that BP= PQ. Find the angle PQC

$$ \\ $$$$\mathrm{Given}\:\mathrm{an}\:\mathrm{isosceles}\:\mathrm{triangle}\:\mathrm{ABC} \\ $$$$\:\mathrm{which}\:\:\angle\mathrm{A}=\:\mathrm{30}°,\:\mathrm{AB}\:=\:\mathrm{AC}.\: \\ $$$$\mathrm{A}\:\mathrm{point}\:\mathrm{D}\:\mathrm{is}\:\mathrm{midpoint}\:\mathrm{of}\:\mathrm{BC}\:.\: \\ $$$$\mathrm{A}\:\mathrm{point}\:\mathrm{P}\:\mathrm{is}\:\mathrm{chosen}\:\mathrm{on}\:\mathrm{then} \\ $$$$\mathrm{segment}\:\mathrm{AD}\:\mathrm{and}\:\mathrm{a}\:\mathrm{point}\:\mathrm{Q}\:\mathrm{is} \\ $$$$\mathrm{chosen}\:\mathrm{on}\:\mathrm{the}\:\mathrm{side}\:\mathrm{AB}\:\mathrm{so}\:\mathrm{that} \\ $$$$\mathrm{BP}=\:\mathrm{PQ}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{PQC} \\ $$

Question Number 198479    Answers: 2   Comments: 2

Question Number 198421    Answers: 1   Comments: 3

Question Number 198391    Answers: 2   Comments: 1

Question Number 198235    Answers: 1   Comments: 0

Question Number 198158    Answers: 1   Comments: 0

Question Number 198054    Answers: 0   Comments: 3

In △MNO, MN=6units, MO=4 units and NO=12 units. If the bisector of the angle M meets NO at P, calculate NP.

$${In}\:\bigtriangleup{MNO},\:{MN}=\mathrm{6}{units},\:{MO}=\mathrm{4}\:{units} \\ $$$${and}\:{NO}=\mathrm{12}\:{units}.\:{If}\:{the}\:{bisector}\:{of}\:{the} \\ $$$${angle}\:{M}\:{meets}\:{NO}\:{at}\:{P},\:{calculate}\:{NP}. \\ $$

Question Number 198025    Answers: 1   Comments: 1

Question Number 197970    Answers: 2   Comments: 0

Question Number 197920    Answers: 2   Comments: 1

Question Number 197680    Answers: 1   Comments: 1

Question Number 197644    Answers: 1   Comments: 1

Question Number 197636    Answers: 1   Comments: 0

Question Number 197670    Answers: 1   Comments: 4

Question Number 197595    Answers: 2   Comments: 1

Question Number 197541    Answers: 2   Comments: 3

Montrer que x=((an+bm)/(m+n))

$$\boldsymbol{\mathrm{Montrer}}\:\boldsymbol{\mathrm{que}} \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{x}}=\frac{\boldsymbol{\mathrm{an}}+\boldsymbol{\mathrm{bm}}}{\boldsymbol{\mathrm{m}}+\boldsymbol{\mathrm{n}}} \\ $$

Question Number 197499    Answers: 1   Comments: 0

Question Number 197287    Answers: 0   Comments: 3

answer to the question number 197017 AF=FI & AG=GJ⇒FG=(1/2)IJ=(1/6)BC △FGH is squilatral ⇒ △FGH≈△ABC ⇒(S_(FGH) /S_(SBC) ) =(1/(36 )) ✓

$${answer}\:{to}\:{the}\:{question}\:{number} \\ $$$$\mathrm{197017} \\ $$$${AF}={FI}\:\&\:\:{AG}={GJ}\Rightarrow{FG}=\frac{\mathrm{1}}{\mathrm{2}}{IJ}=\frac{\mathrm{1}}{\mathrm{6}}{BC} \\ $$$$\bigtriangleup{FGH}\:\:{is}\:\:{squilatral}\:\Rightarrow\:\bigtriangleup{FGH}\approx\bigtriangleup{ABC} \\ $$$$\Rightarrow\frac{{S}_{{FGH}} }{{S}_{{SBC}} }\:=\frac{\mathrm{1}}{\mathrm{36}\:}\:\checkmark \\ $$$$ \\ $$

Question Number 197095    Answers: 1   Comments: 0

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