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Category: Geometry

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-

Question Number 199006 by adhigenz last updated on 26/Oct/23 $$\mathrm{Given}\:\mathrm{that}\:{ABCD}\:\mathrm{is}\:\mathrm{a}\:\mathrm{trapezium}\:\mathrm{such}\:\mathrm{that}\:{AD}//{BC}. \\ $$$$\mathrm{The}\:\mathrm{centroid}\:\mathrm{of}\:\bigtriangleup{ABD}\:\mathrm{lies}\:\mathrm{on}\:\mathrm{the}\:\mathrm{bisector}\:\mathrm{of}\:\angle{BCD}. \\ $$$$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{centroid}\:\mathrm{of}\:\bigtriangleup{ABC}\:\mathrm{lies}\:\mathrm{on}\:\mathrm{the}\:\mathrm{bisector}\:\mathrm{of}\:\angle{ADC}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Question-198933

Question Number 198933 by cherokeesay last updated on 25/Oct/23 Commented by mr W last updated on 26/Oct/23 Commented by Rasheed.Sindhi last updated on 26/Oct/23 $$\mathbb{T}\boldsymbol{\mathrm{han}}\Bbbk\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{help}}\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{diagram}}\:\boldsymbol{\mathrm{sir}}!…

Question-198763

Question Number 198763 by mr W last updated on 24/Oct/23 Commented by mr W last updated on 24/Oct/23 $$\mathrm{8}\:{small}\:{balls}\:{and}\:{a}\:{big}\:{ball}\:{on}\:{a} \\ $$$${table}\:{touch}\:{each}\:{other}. \\ $$$${if}\:{the}\:{radius}\:{of}\:{the}\:{small}\:{balls}\:{is}\:{r}, \\ $$$${find}\:{the}\:{radius}\:{of}\:{the}\:{big}\:{ball}.…

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-

Question Number 198643 by cortano12 last updated on 22/Oct/23 $$ \\ $$$$\mathrm{Given}\:\mathrm{an}\:\mathrm{isosceles}\:\mathrm{triangle}\:\mathrm{ABC} \\ $$$$\:\mathrm{which}\:\:\angle\mathrm{A}=\:\mathrm{30}°,\:\mathrm{AB}\:=\:\mathrm{AC}.\: \\ $$$$\mathrm{A}\:\mathrm{point}\:\mathrm{D}\:\mathrm{is}\:\mathrm{midpoint}\:\mathrm{of}\:\mathrm{BC}\:.\: \\ $$$$\mathrm{A}\:\mathrm{point}\:\mathrm{P}\:\mathrm{is}\:\mathrm{chosen}\:\mathrm{on}\:\mathrm{then} \\ $$$$\mathrm{segment}\:\mathrm{AD}\:\mathrm{and}\:\mathrm{a}\:\mathrm{point}\:\mathrm{Q}\:\mathrm{is} \\ $$$$\mathrm{chosen}\:\mathrm{on}\:\mathrm{the}\:\mathrm{side}\:\mathrm{AB}\:\mathrm{so}\:\mathrm{that} \\ $$$$\mathrm{BP}=\:\mathrm{PQ}. \\…

Question-198604

Question Number 198604 by ajfour last updated on 22/Oct/23 Commented by ajfour last updated on 22/Oct/23 $$\theta+\phi=\alpha\:\left({known}\right) \\ $$$${radius}\:{of}\:{arc}\:{is}\:{unity}. \\ $$$${If}\:{the}\:{two}\:{shaded}\:{parts}\:{are}\:{equal}, \\ $$$${find}\:\theta={f}\left(\alpha\right). \\ $$…