Question 18

In a right angled triangle, the product of two sides is equal to half of the square of the third side i.e., hypotenuse. One of the acute angles must be

Solution

Let the sides of the triangle ABC(right angled at B) be 'a','b','c' and c is hypotenuse

Given that $$a\times b=$$ $$\frac{c^{2}}{2}$$ $$\Rightarrow$$ $$ c^{2}$$=2ab

We know that $$c^{2}$$=$$a^{2}+b^{2}$$

Substituting $$c^{2}$$ value in above equation

2ab$$=a^{2}+b^{2}$$

$$\Rightarrow$$ $$a^{2}-2ab+b^{2}=0$$

$$\Rightarrow$$ $$(a-b)^{2}=0$$

$$\Rightarrow a=b$$

In a triangle, if two sides are equal then the opposite angles must be equal

 We know that $$\angle A+\angle B+\angle C=180^\circ$$

Here $$\angle A$$=$$\angle C$$

$$90^\circ+2\angle A$$=$$180^\circ$$

$$\therefore \angle A$$=$$\angle C$$=$$45^\circ$$


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