How i solve the question of quadratic equations of exercise 4.2 of question no.5 by easy method. It is very important of board exam. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
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The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides. Q.5
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Let us say, the base of the right triangle be x cm.
Given, the altitude of right triangle = (x – 7) cm
From Pythagoras theorem, we know,
Base2 + Altitude2 = Hypotenuse2
∴ x2 + (x – 7)2 = 132
⇒ x2 + x2 + 49 – 14x = 169
⇒ 2x2 – 14x – 120 = 0
⇒ x2 – 7x – 60 = 0
⇒ x2 – 12x + 5x – 60 = 0
⇒ x(x – 12) + 5(x – 12) = 0
⇒ (x – 12)(x + 5) = 0
Thus, either x – 12 = 0 or x + 5 = 0,
⇒ x = 12 or x = – 5
Since sides cannot be negative, x can only be 12.
Therefore, the base of the given triangle is 12 cm and the altitude of this triangle will be (12 – 7) cm = 5 cm.