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For a right angled triangle, if the sum of the lengths of the hypotenuse and a side is kept constant, in order to have maximum area of the triangle, the angle between the hypotenuse and the side is

Correct Answer: 60°

Two right angled triangles are congruent if :
I. The hypotenuse of one triangle is equal to the hypotenuse of the other.
II. A side for one triangle is equal to the corresponding side of the other.
III. Sides of the triangles are equal.
IV. An angle of the triangle are equal.
Of these statements, the correct ones are combination of:

The hypotenuse of a right triangle is 2 centimeters more than linger side of the triangle. The shorter side of the triangle is 7 centimeters less than the longer side. Find the length of the hypotenuse.

The hypotenuse of a right-angled triangle 10 cm and its area is 24 cm^2. If the shorts side is halved and the longer side is double, the new hypotenuse becomes

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Consider the following statements :I. Every equilateral triangle is necessarily an isosceles triangle.II. Every right-angled triangle is necessarily an isosceles triangle.III. A triangle in which one of the median is perpendicular to the side it meets, is necessarily an isosceles triangle.The correct statements are:

PQR is a right angled triangle in which PQ = QR. If the hypotenuse of the triangle is 20 cm, then what is the area (in cm2) of the triangle PQR?

The area of a right angled triangle ABC, right angled at B, is 46 sq units. A median is drawn from A to BC which intersects at D. Find the area (in sq. units)of triangle ABD.

One side of a right-angled triangle is twice the other, and the hypotenuse is 10 cm. The area of the triangle is :

If a + b + c + d = 4, then find the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$     + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$     + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$     + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$     is?

If a + b + c + d = 4, then the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$     + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$     + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$     + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$     is?