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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.
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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
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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