Square ABCD has sides of length 10 units. Isosceles triangle EBC has base BC, and the area common to triangle EBC and square ABCD is 80 square units. Find the length of the altitude to BC in ∆EBC

• AREA OF SQUARE ABCD = 100 SQUARE UNITS
  AREA IN COMMON IS 80 SQAURE UNITS
  SO AREA UNCOMMON = 20 SQAURE UNITS
  SO AD CUT'S AT 2 POINTS i.e B' and C'
  so two triangle formed ABB' AND CDC' with same area
  so area of ABB' and CDC'=5x+5x=10x
  10x=20
  x=2
  by triangle property
  B'C'/BC=h/h+10
  6/10=h/h+10
  3(h+10)=5h
  30=2h
  h=15
  so the answer is (h+10)=15+10=25
• 12 years agoHelpfull: Yes(2) No(0)