Question

the measure of one side of right angle triangle is 42 cm. if difference in length and it's hypotenuse and other side is 14 cm. find measure of 2 unknown sides​

Answer 1

Answer:

Note : refer the figure in the attachment ]

let us suppose,

the triangle is triangle ABC

AB = 42m, hypotenuse = AC = ?

BC = ?

◾we have given the relation between, AC (hypotenuse) and BC ( other side)

i.e difference between lengths of hypotenuse and other side is 14m.

AC - BC = 14m

AC ( hypotenuse) = 14 + BC ....(1)

◾As we have given triangle is right angle triangle, /_ ABC = 90°

We know,

★Pythagoras theorem , ( hypotenuse) ^2 = ( side 1 )^2 + ( side 2 )^2

Therefor, by Pythagoras theorem,

(AC) ^2 = ( AB) ^2 + ( BC )^2

from equation (1) substitute the value of AC

(14 + BC)^2 = (42 ) ^2 + ( BC) ^2

⟹ ( (14 )^2 + 2 x (14 ) x ( BC) + (BC) ^2)= 1,764 - ( BC) ^2

⟹ 196+28(BC)+(BC)^2-(BC)^2 = 1,764

⟹ 28(BC) = 1764 - 196

⟹ ( BC) = 1,568 / 28

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◾Now, To find AC( hypotenuse) ,use Pythagoras theorem

(AC)^2 = ( AB)^2 + (BC)^2

( AC) ^2 = ( 42 )^2 + ( 56 )^2

= 1,764 + 3,136

= 4,900

AC = √ ( 4,900)

= 70 m

therefor,

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Now, we know the formula to find the area of triangle,

★Area of triangle = (1/2 )x base x height

here, base = side BC and height = side AB

Therefor, Area of triangle ABC

= (1/ 2 )x ( BC) x ( AB)

= ( 1/2 ) x ( 56 ) x ( 42 )

= 28 x 42

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Verification by heron's formula:

★heron's formula

Area of triangle

= √[ s (s - a) (s - b) (s - c) ]

first, we have to find s ( semiperimeter)

s =( a + b + c ) /2 [ since, a, b, c are the sides of triangle ]

S =[ ( AB) + (BC) +( AC) ] / 2

= ( 42 + 56 + 70 ) / 2

= ( 98 + 70 ) / 2

= 168 / 2

= 84

Therefor apply the heron's for area of triangle,

1,176

= √ [ s ( s - (AB)) ( s - (BC)) ( s - (AC))]

1,176 =√ [ 84 ( 84 - 42 ) ( 84 - 56 ) ( 84 - 70)]

1,176 = √[ 84 x ( 42 )x ( 28 ) x ( 14) ]

1,176 = √ [ 84 x 14 x 42 x 28 ]

1,176 = √ [ 1, 176 x 1, 176 ]

1,176 = 1, 176

hence verified

Answer 2

afnan has given in a very complex way .....

refer to his answer.............