Question

A wire of length 96m is bent to form a right angled triangle with hypotenuse 40m find the area of triangle

Answer 1

Answer:

Perimeter = 96 m

Hypotenuse = 40 m

Let base be X

B+h+hypo = 96 m

X + h + 40 = 96

h = ( 96 - 40 - X ) m

h = ( 56 - X ) m

b^2+h^2 = ( hypo . )^2 ( Acc. to Pythagoras Theorem )

=> X^2 + h^2 = ( 40 )^2

=> X^2 + ( 56 - X )^2 = 1600

=> X^2 + 3136 - 112X + X^2 = 1600

=> 2X^2 -112X + 3136-1600 = 0

=> 2X^2-112X+1536=0

=> 2 ( X^2 - 56 + 768 ) = 0

=> X^2 - 56X + 768 = 0

=> X^2 - ( 32X + 24X ) + 768 = 0

=> X^2 - 32X + 24X + 768 = 0

=> X(X-32) - 24 ( X - 32 ) = 0

=> ( X - 24 ) ( X - 32 ) = 0

X = 24. X = 32

B= 24 cm. b = 32 cm

therefore, h = ( 56- 24 ) therefore , h=( 56 - 32 )

= 32 cm. =24 cm

therefore , area = 1/2 × b ×h

= ( 1/2 × 32 × 24 ) cm2

= 384 cm 2

Answer 2

The area of triangle is 384 m².

Given - Wire length and hypotenuse

Find - Area of triangle

Solution - The length of wire will be equal to the perimeter of the right angled triangle. Perimeter of triangle refers to the sum of all the sides of triangle which are hypotenuse, base and perpendicular.

Let us represent hypotenuse, base and perpendicular be H, B and P.

So, forming the equation now -

96 = 40 + b + P

B + P = 96 - 40

B + P = 56

B = 56 - P

Now, using Pythagoras theorem to find the dimensions of triangle.

H² = B² + P²

Keep the value of Base in formula-

40² = (56 - P)² + P²

Expanding the expression -

1600 = 3136 + P² - 112P + P²

1600 = 3136 - 112P + 2P²

Rewriting the equation to form quadratic equation -

2P² - 112P + 3136 - 1600 = 0

2P² - 112P + 1536 = 0

P² - 56P + 768 = 0

Taking factors -

P² - 32P - 24P + 768 = 0

P (P - 32) - 24 (P - 32) = 0

(P - 32) (P - 24) = 0

P = 32, 24

If perpendicular is 32 -

B = 56 - 32

B = 24 m

If perpendicular is 24 -

B = 56 - 24

B = 32 m

So, Base and perpendicular will be 32 and 24 m.

Area = $\frac{1}{2}$ × B × P

Area = $\frac{1}{2} \times 32 \times 24$

Area = 16 × 24

Area = 384 m².

Thus, the area of triangle is 384 m².

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