L0Thediferencebetweenthetwoadjoiningsidescontainingtherightanigleofrig

1.	L0Thediferencebetweenthetwoadjoiningsidescontainingtherightanigleofrightangled triangle is 14 cm. The area of triangle is 120 cm . Verify this area by usingherons formula.(4)
Answer» Let the perpendicular be a cmBase = a+14 cm According to questionArea = 1201/2 b p =120a(a+14) = 240a^2 +14a-240 = 0a^2 +24a - 10a - 240=0a(a+24) - 10(a+24)= 0(a+24)(a-10)=0 a= - 24 and a=10Neglecting negative value, we geta=10 Perpendicular = 10 cmBase = 24 cmHypotenuse = 25 cm Now, semi perimeter = 59/2(s-a) = 39/2(s-b) = 11/2(s-c) = 9/2 Area of triangle = root( 5939119/16)= 3/4 * root 593911= 3/4 * 160 (approx.)= 3*40=120 cm sq. when perpendicular is 10cm and base is 24cmthen how can hypotenuse be 25cmaccording to my calculation hypotenuse is 26cm.

L0Thediferencebetweenthetwoadjoiningsidescontainingtherightanigleofrightangled triangle is 14 cm. The area of triangle is 120 cm . Verify this area by usingherons formula.(4)

Answer»

Let the perpendicular be a cmBase = a+14 cm

According to questionArea = 1201/2 b *p =120a(a+14) = 240a^2 +14a-240 = 0a^2 +24a - 10a - 240=0a(a+24) - 10(a+24)= 0(a+24)(a-10)=0

a= - 24 and a=10Neglecting negative value, we geta=10

Perpendicular = 10 cmBase = 24 cmHypotenuse = 25 cm

Now,

semi perimeter = 59/2(s-a) = 39/2(s-b) = 11/2(s-c) = 9/2

Area of triangle = root( 59*39*11*9/16)= 3/4 * root 59*39*11= 3/4 * 160 (approx.)= 3*40=120 cm sq.

when perpendicular is 10cm and base is 24cmthen how can hypotenuse be 25cmaccording to my calculation hypotenuse is 26cm.