Question

ABC and ADC are two equilateral triangles on a common base AC. Find the angles of the resulting quadrilateral. Show that it is a rhombus

• A. $\angle A={120}^o,\angle S={60}^o,\angle C={120}^o,\angle D={60}^o$
• B. $\angle A={110}^o,\angle S={60}^o,\angle C={120}^o,\angle D={60}^o$
• C. $\angle A={120}^o,\angle S={50}^o,\angle C={120}^o,\angle D={60}^o$
• D. $\angle A={120}^o,\angle S={60}^o,\angle C={110}^o,\angle D={70}^o$

Answer 1

The correct option is A $\angle A={120}^o,\angle S={60}^o,\angle C={120}^o,\angle D={60}^o$

$△ABCand△ADCaretwoequilateraltrianglesonacommonbaseAC.Sinceallthesidesofequilateraltrianglesareequal⟹AB=BC=AC=ADNowin△ABC\angle BAC=\angle ACB=\angle ABC\left(Equilateraltrianglehasallanglesequal\right)⟹\angle BAC=\angle ACB=\angle ABC=\mathrm{60...}\left(1\right)Nowin△ADC\angle ADC=\angle DCA=\angle CAD=\mathrm{60....}\left(2\right)In\square ABCD\angle BAD=\angle BAC+\angle CAD⟹\angle BAD=60+60=120\left[From\right(1\left)and\right(2\left)\right]Also,\angle BCD=\angle BCA+\angle ACD⟹\angle BCD=60+60=120\left[From\right(1\left)and\right(2\left)\right]Also\angle ABC=\angle ADC=60Since\angle BAD=\angle BCDand\angle ABC=\angle ADCandAB=BC=AD=CDHencethequadilateralisarhombusasithaveallthesidesandoppositeanglesequal$