the difference between the two adjoining sides containing right angle of a right angled triangle is 14cm the area of triangle is120cm sq. verify this using herons formula
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Base = a+14 cm
According to question
Area = 120
1/2 b *p =120
a(a+14) = 240
a^2 +14a-240 = 0
a^2 +24a - 10a - 240=0
a(a+24) - 10(a+24)= 0
(a+24)(a-10)=0
a= - 24 and a=10
Neglecting negative value, we get
a=10
Perpendicular = 10 cm
Base = 24 cm
Hypotenuse = 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.
Draw a right triangle and name it ABC where angle B is 90°.
So, AB is perpendicular
BC is base
AC is hypotenus
Let AB be x
Then BC is x+14
According to area of triangle,
1/2×b×h=120cm²
1/2×(x+14)×x=120cm²
x²+14x=120×2
=240cm²
x²+14x-240=0
x(x+24)-10(x+24)=0
x+24=0. x-10=0
x=(-24). x=10
We'll not take x as( -24) as length of sides can never be negative
So x=10cm
So AB=x=10cm
BC=x+14=10+14=24cm
Then hypotenus AC=√10²+24²=√100+576
√676=26cm
So perimeter is 10+24+26=60cm
s=60/2=30cm
So (s-a)=30-10=20cm
(s-b)=30-24=6cm
(s-c)=40-26=4cm
So area by heron's formulae
=√s(s-a)(s-b)(s-c)
=√30×20×6×4cm²
=√14400cm²
=120cm²
.
. .Hence proved..