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The length of the hypotenuse of a right-angled triangle exceeds the length of the base by 2 cm and exceeds twice the length of the altitude by 1 cm. Find the length of each side of the triangle.

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Questions

The length of the hypotenuse of a right-angled triangle exceeds the length of the base by 2 cm and exceeds twice the length of the altitude by 1 cm. Find the length of each side of the triangle. 

The length of the hypotenuse of a right-angled triangle exceeds the length of the base by 2 cm and exceeds twice the length of the altitude by 1 cm. Find the length of all sides. 

Sum
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Solution

Let the base and altitude of the right-angled triangle be x and y cm, respectively.

Therefore, the hypotenuse will be (x + 2) cm. 

∴ (x + 2)2 = y2 + x2   ...(i) 

Again, the hypotenuse exceeds twice the length of the altitude by 1 cm. 

∴ h = (2y + 1) 

⇒ x + 2 = 2y + 1 

⇒ x = 2y – 1 

Putting the value of x in (i), we get: 

(2y – 1 + 2)2 = y2 + (2y – 1)2 

⇒ (2y + 1)2 = y2 + 4y2 – 4y + 1 

⇒ 4y2 + 4y + 1 = 5y2 – 4y + 1 

⇒ –y2 + 8y = 0

⇒ y2 – 8y = 0 

⇒ y(y – 8) = 0 

⇒ y = 8 cm 

∴ x = 16 – 1

= 15 cm 

∴ h = 16 + 1

= 17 cm 

Thus, the base, altitude and hypotenuse of the triangle are 15 cm, 8 cm and 17 cm, respectively. 

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Chapter 5: Quadratic equations - Exercise 5E [Page 92]

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Nootan Mathematics [English] Class 10 ICSE
Chapter 5 Quadratic equations
Exercise 5E | Q 23. | Page 92
R.S. Aggarwal Mathematics [English] Class 10
Chapter 4 Quadratic Equations
EXERCISE 4D | Q 70. | Page 229
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