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The area of a right-angled triangle is 600 cm^2. If the base of the triangle exceeds the altitude by 10 cm, find the dimensions of the triangle.

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Question

The area of a right-angled triangle is 600 cm2. If the base of the triangle exceeds the altitude by 10 cm, find the dimensions of the triangle.

Sum
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Solution 1

Area of a triangle = `("Height" xx "Base")/2`

Here, Height and base are x and (x + 10) and the area is 600 

Hence, `x xx (x + 10) xx 1/2 = 600`

⇒ x2 + 10x = 1200 

⇒ x2 + 10 x – 1200 = 0 

⇒ x2 + 40x – 30x – 1200 = 0 

⇒ x(x + 40} – 30(x + 40) = 0 

⇒ (x + 40)(x – 30) = 0, hence x = 30.

Base = 30 + 10 = 40 cms.

Hence h2= 402 + 302 = 2500  

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Solution 2

Let the altitude of the triangle be x cm 

Therefore, the base of the triangle will be (x + 10) cm

Area of triangle = `1/2x (x + 10) = 600`

⇒ (x + 10) = 1200

⇒ x2 + 10x – 1200 = 0

⇒ x2 + (40 – 30)x – 1200 = 0

⇒ x2 + 40x – 30x – 1200 = 0

⇒ x(x + 40) – 30(x + 40) = 0

⇒ (x + 40) (x – 30) = 0

⇒ x = –40 or x = 30

⇒ x = 30   ...[∵ Altitude cannot be negative]

Thus, the altitude and base of the triangle are 30 cm and (30 + 10 = 40) cm, respectively.

(Hypotenuse)2 = (Altitude)2 + (Base)2

⇒ (Hypotenuse)2 = (30)2 (40)2

⇒ (Hypotenuse)2 = 900 + 1600 = 2500

⇒ (Hypotenuse)2 = (50)2

⇒ (Hypotenuse) = 50

Thus, the dimensions of the triangle are: 

Hypotenuse = 50 cm

Altitude = 30 cm

Base = 40 cm

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Chapter 6: Problems Based On Quadratic Equations - Exercise 7.1

APPEARS IN

Frank Mathematics Part 2 [English] Class 10 ICSE
Chapter 6 Problems Based On Quadratic Equations
Exercise 7.1 | Q 25
R.S. Aggarwal Mathematics [English] Class 10
Chapter 4 Quadratic Equations
EXERCISE 4D | Q 66. | Page 229
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