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Find the base of an isosceles triangle whose area is 60m2 and length of equal sides is 13 cm |
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Answer» The AREA of isosceles Triangle=60m² The length of each sides=13cm Draw a perpendicular AD which is on BC with height H that's why BD=CD in isoceles traiangle ABD Now , USING pathagoras theorem AB²=AD²+BD² Here AB=60 AD=h BD=X 13²=h²+X²-------1 Similarly, find the area of traiangle ABC Area=1/2×base×height 60=1/2×2X×h 120=2Xh------(2) Adding equation 1 and 2 here 13²+120=h²+X²+2Xh here , using identity (a+b)²=a²+b²+2ab 169+120=(h+X)² 289=(h+X)² √289=h+X h+X=17 Solving it now h=17-x Now put the value of X in EQ 2 120=2X(17-X) 120=34X-2X² Dividing by 2 on both sides here X²-17X+60=0 Splitting the middle term here X²-12X-5X+60=0 X(X-12)-5(X-12)=0 (X-5)(X-12)=0 SO,X=5 , 12 Now, PUTTING the value of X=12 in EQ 2 120=2Xh 120=24h h=5m Here h=AD=5m BC=BD+CD=2X=2×12=24m Hence, The perpendicular of traiangle=5m The base of isoceles traiangle=24m |
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