Given: In a quadrilateral ABCD diagonals \( A C \) and \( B D \)  intersect at \( O \) such that \( O B=O D \).$AB = CD$.To do:We have to show that:(i) \( \operatorname{ar}(\mathrm{DOC})=\operatorname{ar}(\mathrm{AOB}) \)(ii) \( \operatorname{ar}(\mathrm{DCB})=\operatorname{ar}(\mathrm{ACB}) \)(iii) DA\| \( \mathrm{CB} \) or \( \mathrm{ABCD} \) is a ... Read More

Given:The sides of the triangular walls of the flyover are $122\ m, 22\ m$ and $120\ m$. The advertisements yield an earning of $\ RS. 5000\ per\ m^2$ per year. A company hired one of its walls for 3 months.To find:We have to find the rent paid by the company.Solution:The ... Read More

Given:Base $=12\ cm$ and the sum of the hypotenuse and another side $=18\ cm$.To do:We have to construct the right triangle with the given measurements.Solution:Steps of construction:(i) Let us draw the base $BC$ of length $12\ cm$.(ii) Now, construct an angle $XBC$ such that $\angle XBC=90^o$ from point B.(iii) Now, ... Read More

Given:$\angle Y=30^o, \angle Z=90^o$ and $XY+YZ+ZX=11\ cm$.To do:We have to construct the $\triangle XYZ$.Solution:Steps of construction:(i) Let us draw a line segment $AB$ of length $11\ cm$.(ii)  Now, construct an angle $LAB$ such that $\angle LAB=30^o$ from the point $A$(iii) Similarly, construct an angle $MBA$ such that $\angle MBA=90^o$ from ... Read More

Given:$QR=6\ cm, \angle Q=60^o$ and $PR-PQ=2\ cm$.To do:We have to construct a $\triangle PQR$.Solution:Steps of construction:(i) Let us draw a line segment $QR$ of length $6\ cm$.(ii) Now, construct an angle $RQX$ such that $\angle RQX=60^o$(iii) Now, by taking a measure of $PR-PQ=2\ cm$ with the compasses, let us draw ... Read More

Given:$BC=7\ cm, \angle B=75^o$ and $AB+AC=13\ cm$.To do:We have to construct a $\triangle ABC$.Solution:Steps of construction:(i) Let us draw a line segment $BC$ of length $7\ cm$.(ii) Now, construct an angle $CBX=75^o$ from point $B$.(iii) Now, by taking a measure of $AB+AC=13\ cm$ with the compasses, let us draw an ... Read More

Given:$ABC$ and $ABD$ are two triangles on the base $AB$.Line segment $CD$ is bisected by $AB$ at $O$.To do:We have to show that $ar(\triangle ABC) = ar(\triangle ABC)$.Solution:$CO = OD$Draw $CL \perp AB$ and $DM \perp AB$In $\Delta CLO$ and $\triangle D M O$, $\angle \mathrm{L}=\angle \mathrm{M}$$\mathrm{CO}=\mathrm{OD}$$\angle \mathrm{COL}=\angle \mathrm{DOM}$    ... Read More

Given:In a triangle \( \mathrm{ABC}, \mathrm{E} \) is the mid-point of median AD.To do:We have to show that \( \operatorname{ar}(\mathrm{BED})=\frac{1}{4} \operatorname{ar}(\mathrm{ABC}) \). Solution:We know that, A median divides a triangle into two triangles of equal areas.This implies, $\operatorname{ar}(\triangle \mathrm{ABD})=\operatorname{ar}(\triangle \mathrm{ADC})=\frac{1}{2} \operatorname{ar}(\triangle \mathrm{ABC})$............(i)In $\triangle \mathrm{ABD}$$\mathrm{BE}$ is the median.This implies, $\operatorname{ar}(\triangle \mathrm{BED})=\operatorname{ar}(\triangle \mathrm{BAE})=\frac{1}{2} ... Read More

To do:We have to show that the diagonals of a parallelogram divide it into four triangles of equal area. Solution:Let in a parallelogram $ABCD$ diagonals $AC$ and $BD$ intersect at $O$.We know that, The diagonals of a parallelogram bisect each other. This implies, $OA = OC$ and $OB = OD$.The median of a ... Read More

Given:$E$ is any point on median \( \mathrm{AD} \) of a \( \triangle \mathrm{ABC} \).To do:We have to show that ar \( (\mathrm{ABE})=\operatorname{ar}(\mathrm{ACE}) \). Solution:$AD$ is the median of $\triangle ABC$. This implies, $AD$ divides $\triangle ABC$ into two triangles of equal area.Therefore, $ar(\triangle ABD) = ar(\triangle ACD)$...........(i)$AD$ is the median ... Read More