Question

Consider two right-angled triangles inscribed in a square such that their hypotenuse intersect at a point O inside the square. What is the angle between the hypotenuse of the triangle and the base of the square?

• A. ${30}^{\circ }$
• B. ${90}^{\circ }$
• C. ${45}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is C

${45}^{\circ }$

In the square ABCD, we have right-angled triangles, $△$ ADC and $△$ BCD such that their hypotenuse intersect at the point O inside the square.

Now, since the diagonals of a square are equal in length, and every angle is a right angle, we have

$BD=AC$ and $\angle B={90}^{\circ }$.

Thus, $△BDC$ is a right-angled isosceles triangle.

$⟹\angle BDC={45}^{\circ }$ .....(Sum of angles of a triangle is ${180}^{\circ }$)