Properties of Triangle

A triangle has a vertex at (1, 2) and the mid points of the two sides through it are (–1, 1) and (2, 3). Then the centroid of this triangle is :

In a $$\Delta PQR,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu}$$ If $$3{\mkern 1mu} \sin {\mkern 1mu} P + 4{\mkern 1mu} \cos {\mkern 1mu} Q = 6$$ and $$4\sin Q + 3\cos P = 1,$$ then the angle R is equal to :

A triangle ABC lying in the first quadrant has two vertices as A(1, 2) and B(3, 1). If $$\angle BAC = {90^o}$$ and area$$\left( {\Delta ABC} \right) = 5\sqrt 5$$ s units, then the abscissa of the vertex C is :

The triangle of maximum area that can be inscribed in a given circle of radius 'r' is :

The lengths of the sides of a triangle are 10 + x², 10 + x² and 20 $$-$$ 2x². If for x = k, the area of the triangle is maximum, then 3k² is equal to :

A straight line cuts off the intercepts $$\mathrm{OA}=\mathrm{a}$$ and $$\mathrm{OB}=\mathrm{b}$$ on the positive directions of $$x$$-axis and $$y$$ axis respectively. If the perpendicular from origin $$O$$ to this line makes an angle of $$\frac{\pi}{6}$$ with positive direction of $$y$$-axis and the area of $$\triangle \mathrm{OAB}$$ is $$\frac{98}{3} \sqrt{3}$$, then $$\mathrm{a}^{2}-\mathrm{b}^{2}$$ is equal to :

Let $$\left(5, \frac{a}{4}\right)$$ be the circumcenter of a triangle with vertices $$\mathrm{A}(a,-2), \mathrm{B}(a, 6)$$ and $$C\left(\frac{a}{4},-2\right)$$. Let $$\alpha$$ denote the circumradius, $$\beta$$ denote the area and $$\gamma$$ denote the perimeter of the triangle. Then $$\alpha+\beta+\gamma$$ is

Two vertices of a triangle $$\mathrm{ABC}$$ are $$\mathrm{A}(3,-1)$$ and $$\mathrm{B}(-2,3)$$, and its orthocentre is $$\mathrm{P}(1,1)$$. If the coordinates of the point $$\mathrm{C}$$ are $$(\alpha, \beta)$$ and the centre of the of the circle circumscribing the triangle $$\mathrm{PAB}$$ is $$(\mathrm{h}, \mathrm{k})$$, then the value of $$(\alpha+\beta)+2(\mathrm{~h}+\mathrm{k})$$ equals

In a triangle with sides $$a, b, c,$$ $${r_1} > {r_2} > {r_3}$$ (which are the ex-radii) then :

The sum of the radii of inscribed and circumscribed circles for an $$n$$ sided regular polygon of side $$a,$$ is :

If in a $$\Delta ABC$$ $$a\,{\cos ^2}\left( {{C \over 2}} \right) + c\,{\cos ^2}\left( {{A \over 2}} \right) = {{3b} \over 2},$$ then the sides $$a, b$$ and $$c$$ :

For a regular polygon, let $$r$$ and $$R$$ be the radii of the inscribed and the circumscribed circles. A $$false$$ statement among the following is :

In a triangle $$ABC$$, medians $$AD$$ and $$BE$$ are drawn. If $$AD=4$$, $$\angle DAB = {\pi \over 6}$$ and $$\angle ABE = {\pi \over 3}$$, then the area of the $$\angle \Delta ABC$$ is :

Let a, b, c be in arithmetic progression. Let the centroid of the triangle with vertices (a, c), (2, b) and (a, b) be $$\left( {{{10} \over 3},{7 \over 3}} \right)$$. If $$\alpha$$, $$\beta$$ are the roots of the equation $$a{x^2} + bx + 1 = 0$$, then the value of $${\alpha ^2} + {\beta ^2} - \alpha \beta$$ is :

In a triangle, the sum of lengths of two sides is x and the product of the lengths of the same two sides is y. If x² – c² = y, where c is the length of the third side of the triangle, then the circumradius of the triangle is :

The angles A, B and C of a triangle ABC are in A.P. and a : b = 1 : $$\sqrt 3$$. If c = 4 cm, then the area (in sq. cm) of this triangle is :

If in a triangle ABC, AB = 5 units, $$\angle B = {\cos ^{ - 1}}\left( {{3 \over 5}} \right)$$ and radius of circumcircle of $$\Delta$$ABC is 5 units, then the area (in sq. units) of $$\Delta$$ABC is :

Let $${{\sin A} \over {\sin B}} = {{\sin (A - C)} \over {\sin (C - B)}}$$, where A, B, C are angles of triangle ABC. If the lengths of the sides opposite these angles are a, b, c respectively, then :