Question

The hypotenuse of a right triangle is 6 1/2 centimetres and its area is
71/2 square centimetres. Calculate the lengths of its perpendicular sides​

Answer 1

Answer:

Let perpendicular sides of the triangle = 'a' and 'b'.

so (1/2)×a×b=area =7 1/2

or, ab/2=15/2 , or, ab=15

again, hypotenuse^2=a^2+b^2

or, a^2+b^2=(6 1/2)^2=(13/2)^2

we know, (a+b)^2=a^2+b^2+2ab

=(13/2)^2+2×15=169/4+30=289/4

a+b=17/2……..(1)

again, (a-b)^2=a^2+b^2–2ab

=(13/2)^2–2×15=169/4–30=49/4

a-b=7/2 ………(2)

(1)+(2), → 2a=17/2+7/2=24/2=12

a=12/2=6 cm

(1)-(2) , → 2b=17/2–7/2=10/2=5

b=5/2=2.5 cm

two sides are 2.5cm and 6 cm

check:-

√(6^2+2.5^2)=√(36+25/4)= √{(144+25)/4}=√(169/4)=13/2 =6 1/2

Step-by-step explanation:

$Hope \: it \: helps.$

Answer 2

Let perpendicular sides of the triangle = 'a' and 'b'.

so (1/2)×a×b=area =7 1/2

or, ab/2=15/2 , or, ab=15

again, hypotenuse^2=a^2+b^2

or, a^2+b^2=(6 1/2)^2=(13/2)^2

we know, (a+b)^2=a^2+b^2+2ab

=(13/2)^2+2×15=169/4+30=289/4

a+b=17/2……..(1)

again, (a-b)^2=a^2+b^2–2ab

=(13/2)^2–2×15=169/4–30=49/4

a-b=7/2 ………(2)

(1)+(2), → 2a=17/2+7/2=24/2=12

a=12/2=6 cm

(1)-(2) , → 2b=17/2–7/2=10/2=5

b=5/2=2.5 cm

two sides are 2.5cm and 6 cm

check:-

√(6^2+2.5^2)=√(36+25/4)= √{(144+25)/4}=√(169/4)=13/2 =6 1/2

Step-by-step explanation:

hope it's helpful,,

:-)

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