Question

the length of one side of right triangle exceeds the length of the other by 3cm if the hypotenuse is 15 cm then the length of the sides of the triangle​

Answer 1

Other Sides = 12 & 9 cm

Step-by-step explanation:

Given:

• Length of one side of right angled triangle exceeds other side by 3 cm.
• Measure of hypotenuse of triangle is 15 cm.

To Find:

• Measure of all sides of triangle ?

Solution: Let the length of one side be x cm. Therefore,

Length of another side = 3 more than x

Length of another side = (x + 3) cm

Since, it is right angled triangle so let's assume that

• Perpendicular = (x + 3) cm
• Base = x cm
• Hypotenuse = 15 cm

By using Pythagoras theorem in this ∆

H² = Base² + Perpendicular²

$\implies{\rm }$ 15² = x² + (x + 3)²

$\implies{\rm }$ 225 = x² + x² + 3² + 2•x•3

$\implies{\rm }$ 225 = 2x² + 9 + 6x

$\implies{\rm }$ 225 – 9 = 2x² + 6x

$\implies{\rm }$ 216 = 2(x² + 3x)

$\implies{\rm }$ 216/2 = x² + 3x

$\implies{\rm }$ 108 = x² + 3x

Now, solving the above equation by middle term splitting method.

x² + 3x – 108 = 0

x² + 12x – 9x – 108

x(x + 12) – 9 (x + 12)

(x + 12) or (x – 9)

x + 12 = 0 or x – 9 = 0

x = – 12 or x = 9

Since, length cannot be negative so we have to take positive value of x i.e 9.

• Base is x = 9 cm

• Perpendicular is x + 3 = 9 + 3 = 12 cm

Answer 2

Answer:

✡ Question ✡

➡ The length of one side of right triangle exceeds the length of the other by 3cm if the hypotenuse is 15 cm then the length of the sides of the triangle.

✡ Given ✡

➡Length of one side of right angled triangle exceeds other side by 3 cm.

➡Measure of hypotenuse of triangle is 15 cm.

✡ To Find ✡

➡Measure of all sides of triangle ?

✡ Solution ✡

➡ Let the length of one side be x cm.

▶ Therefore,

➡Length of another side = 3 more than x

➡Length of another side = (x + 3) cm

Hence, it is right angled triangle so let's assume that

✏Perpendicular = (x + 3) cm

✏Base = x cm

✏Hypotenuse = 15 cm

➡By using Pythagoras theorem in this ∆ ,

✡ H²=Base²+Perpendicular² ✡

=> 15² = x² + (x + 3)²

=> 225 = x² + x² + 3² + 2.x.3

=> 225 = 2x² + 9 + 6x

=> 225 – 9 = 2x² + 6x

=> 216 = 2(x² + 3x)

=> 216/2 = x² + 3x

=> 108 = x² + 3x

▶Now, we have to solve the above equation with the help of middle term splitting method.

=> x² + 3x – 108 = 0

=> x² + 12x – 9x – 108 = 0

=> x(x + 12) – 9 (x + 12) = 0

=> (x + 12) (x – 9) = 0

=> x + 12 = 0 | x – 9 = 0

=> x = – 12 | x = 9

x = 9Since, length cannot be negative so we have to take positive value of x i.e 9.

☞ Base is x = $\bold{\</em></strong><strong><em>l</em></strong><strong><em>a</em></strong><strong><em>r</em></strong><strong><em>g</em></strong><strong><em>e</em></strong><strong><em>{\fbox{\color{blue} {</em></strong><strong><em>9</em></strong><strong><em> </em></strong><strong><em>cm</em></strong><strong><em>}}}}$

☞Perpendicular is x+3=9+3 = $\bold{\</em><em>l</em><em>a</em><em>r</em><em>g</em><em>e</em><em>{\fbox{\color{</em><em>gree</em><em>n</em><em>} {</em><em>1</em><em>2</em><em> </em><em>cm</em><em>}}}}$

Step-by-step explanation:

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