Given:Diagonals \( \mathrm{AC} \) and \( \mathrm{BD} \) of a quadrilateral \( \mathrm{ABCD} \) intersect each other at \( \mathrm{P} \). To do:We have to show that ar \( (\mathrm{APB}) \times \operatorname{ar}(\mathrm{CPD})=\operatorname{ar}(\mathrm{APD}) \times \operatorname{ar}(\mathrm{BPC}) \).Solution:Draw $AM$ perpendicular to $BD$ and $CN$ perpendicular to $BD$$ar(\triangle ABP) = \frac{1}{2}\times BP \times AM$…………..(i)$ar(\triangle APD) ... Read More

Given:\( P \) and \( Q \) are respectively the mid-points of sides \( \mathrm{AB} \) and \( \mathrm{BC} \) of a triangle \( \mathrm{ABC} \) and \( \mathrm{K} \) is the mid-point of \( \mathrm{AP} \)To do:We have to show that(i) \( \operatorname{ar}(\mathrm{PRQ})=\frac{1}{2} \operatorname{ar}(\mathrm{ARC}) \)(ii) ar \( (\mathrm{RQC})=\frac{3}{8} \) ... Read More

Given:\( \mathrm{ABC} \) is a right triangle right angled at \( \mathrm{A} . \mathrm{BCED}, \mathrm{ACFG} \) and \( \mathrm{ABMN} \) are squares on the sides \( \mathrm{BC}, \mathrm{CA} \) and \( \mathrm{AB} \) respectively. Line segment \( \mathrm{AX} \perp \mathrm{DE} \) meets \( \mathrm{BC} \) at Y. To do:We have to ... Read More

Given :The curved surface area of a right circular cylinder of height $14\ cm$ is $88\ cm^2$.To find:we have to find the diameter of the base of the cylinder.Solution: Let the radius of the base of the cylinder be $r$ cm.The curved surface area of a cylinder of radius $r$ and height ... Read More

Given:It is required to make a closed cylindrical tank of height $1\ m$ and base diameter $140\ cm$ from a metal sheet.To do:We have to find the area of the sheet required. Solution:Height of the cylinder $(h) = 1\ m$$= 100\ cm$Diameter of the cylinder $= 140\ cm$This implies, Radius $(r)=\frac{140}{2}$$=70 ... Read More

Given:A metal pipe is \( 77 \mathrm{~cm} \) long. The inner diameter of a cross section is \( 4 \mathrm{~cm} \), the outer diameter being \( 4.4 \mathrm{~cm} \)To do:We have to find its(i) inner curved surface area, (ii) outer curved surface area, (iii) total surface area.Solution:Length of the metal ... Read More

Given:$D$ and \( \mathrm{E} \) are two points on \( \mathrm{BC} \) such that \( \mathrm{BD}=\mathrm{DE}=\mathrm{EC} \).To do:We have to show that \( \operatorname{ar}(\mathrm{ABD})=\operatorname{ar}(\mathrm{ADE})=\operatorname{ar}(\mathrm{AEC}) \).Solution:In $\triangle ABE$$BD=DE$This implies, $AD$ is the median.We know that, The median of a triangle divides it into two parts of equal areas.This implies, $ar(\triangle ABD) ... Read More

Given:$ABCD, DCFE$ and $ABFE$ are parallelograms.To do:We have to prove that $ar(\triangle ADE) = ar(\triangle BCF)$.Solution:$ABCD$ is a parallelogram.This implies, $AD = BC$Similarly, In parallelogram $ABEF$, $AE = BF$In parallelogram $CDEF$, $DE = CF$In $\triangle ADE$ and $\triangle BCF$, $AD = BC$$DE = CF$$AE = BF$Therefore, by SSS axiom, $\triangle ... Read More

Given:$D$ and \( \mathrm{E} \) are two points on \( \mathrm{BC} \) such that \( \mathrm{BD}=\mathrm{DE}=\mathrm{EC} \).To do:We have to show that \( \operatorname{ar}(\mathrm{ABD})=\operatorname{ar}(\mathrm{ADE})=\operatorname{ar}(\mathrm{AEC}) \).Solution:In $\triangle ADP$ and $\triangle QCP$, $\angle APD = \angle QPC$            (Vertically opposite angles)$\angle ADP = \angle QCP$        ... Read More

Given:$ar(\mathrm{DRC})=ar(\mathrm{DPC})$ and $ar(\mathrm{BDP})=\operatorname{ar}(\mathrm{ARC})$To do:We have to show that both the quadrilaterals \( \mathrm{ABCD} \) and \( \mathrm{DCPR} \) are trapeziums.Solution:$ar (\triangle DPC) = ar(\triangle DRC)$.....……(i)$ar(\triangle BDP) = ar(\triangle ARC)$……...(ii)Subtracting (i) from (ii), we get, $ar(\triangle BDP) - ar(\triangle DPC) = ar(\triangle ARC) - ar(\triangle DRC)$$ar(\triangle BDC) = ar(\triangle ADC)$Triangles $BDC$ ... Read More