Given: $AM$ is a median of a triangle $ABC$.To do: To find whether $AB + BC + CA > 2 AM?$Solution:Let us consider $\Delta ABM$ and $\Delta AMC$It is a known fact that the sum of the triangle of any two sides in a triangle should be greater than the length of ... Read More

Given: Sides:$(i).\ 2 cm, \ 3 cm, \ 5 cm$$(ii).\ 3 cm, \ 6 cm, \ 7 cm$$(iii).\ 6 cm, \ 3 cm, \ 2 cm$To do: To check whether it is possible to have a triangle with the given sides.Solution:In a triangle, the sum of its two sides is always ... Read More

Given: A triangle PQR.To do: To take any point O in the interior of a triangle PQR. And to find whether :$(i).\ OP + OQ > PQ?$$(ii).\ OQ + OR > QR?$$(iii).\ OR + OP > RP?$Solution:$(i)$. Join $OR$, $OQ$ and $OP$In $\triangle OPQ$, $OP+OR>PQ$yes, the $POQ$ form is a triangle.$(ii)$ ... Read More

Given: Triangles in the above-given diagram with unknown angles $x$ and $y$.To do: To find the values of unknown angles $x$ and $y$ in each case.Solution:For convenience, we name all the triangles given in the diagram as $\triangle ABC$. $(i).\ \angle y+\angle ACD=180^{\circ}$       [linear pair]$\Rightarrow \angle y\ +\ 120^{\circ}=180^{\circ}$$\Rightarrow ... Read More

Given: Triangles given in the above diagram with unknown angle $x$.To do: To find unknown angle $x$ in each case.Solution:$(i)$. By using the angle sum property of the triangle$x+50^{\circ}+60^{\circ}=180^{\circ}$$\Rightarrow x=180^{\circ}-110^{\circ}$$\Rightarrow x=70^{\circ}$$(ii)$. By using the angle sum property of the triangle$90^{\circ}+30^{\circ}+x=180^{\circ}$$\Rightarrow x=180^{\circ}-120^{\circ}$$\Rightarrow x=60^{\circ}$$(iii)$. By using the angle sum property of the ... Read More

To do: We have to find the value of unknown interior angle $x$ in each case.Solution: For convenience, we shall name all the triangles given in the diagram as $\triangle ABC$.We know that, An exterior angle of a triangle is equal to the sum of its interior opposite angles.Therefore, (i) $\angle ... Read More

To do: We have to find the unknown exterior angle $x$ in each case.Solution:For convenience, we shall name all the triangles given in the diagram as $\triangle ABC$.We know that, An exterior angle of a triangle is equal to the sum of its interior opposite angles.Therefore, (i) $\angle BAC+\angle ABC=x$$\Rightarrow ... Read More

To do:We have to verify whether the median and altitude of an isosceles triangle can be the same.Solution:Follow the steps:Draw a $\triangle PQR$  with $PQ = PR$.Let us draw a line segment $PS$ perpendicular to $QR$.$PS$ is the altitude of the triangle.It can be observed that the length of $QS$ ... Read More

To do:We have to draw rough sketches for the following:(a) In $\triangle ABC, BE$ is a median.(b) In $\triangle PQR, PQ$ and $PR$ are altitudes of the triangle.(c) In $\triangle XYZ, YL$ is an altitude in the exterior of the triangle.Solution :(a)A median of a triangle is a line segment joining a ... Read More

Given: In $\triangle PQR, D$ is the mid-point of $\overline{QR}$.To do: We have to name $\overline{PM}, PD$ in $\triangle PQR$ and find whether $QM = MR$.Solution:An altitude of a triangle is the perpendicular drawn from the vertex of the triangle to the opposite side.In the given figure $\overline{PM}$ is perpendicular ... Read More