Given:$ABCDE$ is a pentagon. A line through \( \mathrm{B} \) parallel to \( \mathrm{AC} \) meets \( \mathrm{DC} \) produced at $F$.To do:We have to show that(i) \( \operatorname{ar}(\mathrm{ACB})=\operatorname{ar}(\mathrm{ACF}) \)(ii) \( \operatorname{ar}(\mathrm{AEDF})=\operatorname{ar}(\mathrm{ABCDE}) \)Solution:$ABCDE$ is a pentagon and $BF \| AC$.(i) $\triangle ACB$ and $\triangle ACF$ lie on the same base ... Read More

Given:Diagonals \( \mathrm{AC} \) and \( \mathrm{BD} \) of a trapezium \( \mathrm{ABCD} \) with \( \mathrm{AB} \| \mathrm{DC} \) intersect each other at \( \mathrm{O} \).To do:We have to prove that ar \( (\mathrm{AOD})=\operatorname{ar}(\mathrm{BOC}) \).Solution:$\triangle ABC$ and $\triangle ABD$ lie on the same base $AB$ and between the parallels ... Read More

Given:\( \mathrm{XY} \) is a line parallel to side \( \mathrm{BC} \) of a triangle \( \mathrm{ABC} \). \( \mathrm{BE} \| \mathrm{AC} \) and \( \mathrm{CF} \| \mathrm{AB} \) meet \( \mathrm{XY} \) at \( \mathrm{E} \) and \( F \) respectively.To do:We have to show that \( \operatorname{ar}(\mathrm{ABE})=\operatorname{ar}(\mathrm{ACF}) \).Solution:$BE ... Read More

Given:The side \( \mathrm{AB} \) of a parallelogram \( \mathrm{ABCD} \) is produced to any point \( \mathrm{P} \). A line through \( \mathrm{A} \) and parallel to \( \mathrm{CP} \) meets \( C B \) produced at \( Q \) and then parallelogram PBQR is completed.To do:We have to ... Read More

Given:\( \mathrm{D} \) and \( \mathrm{E} \) are points on sides \( \mathrm{AB} \) and \( \mathrm{AC} \) respectively of \( \triangle \mathrm{ABC} \) such that ar \( (\mathrm{DBC})=\operatorname{ar}(\mathrm{EBC}) \).To do:We have to prove that $DE\|BC$.Solution:ar \( (\mathrm{DBC})=\operatorname{ar}(\mathrm{EBC}) \)This implies, $\triangle DBC$ and $\triangle EBC$ are equal in area and ... Read More

Given:The sides of the wall are $15\ m, 11\ m$ and $6\ m$.To find:We have to find the area of the painted wall in colour.Solution:The sides of the wall are $15\ m, 11\ m$ and $6\ m$.By Heron's formula:$A=\sqrt{s(s-a)(s-b)(s-c)}$Since, $S=\frac{a+b+c}{2}$$S=\frac{15\ m+11\ m+6\ m}{2}$$S=\frac{32\ m}{2}$$S=16\ m$This implies, $A=\sqrt{16(16-15)(16-11)(16-6)}$$A=\sqrt{16(1)(5)(10)}$$A=\sqrt{800}\ m^2$$A=20\sqrt2\ m^2$Therefore, The ... Read More

Given:The sides of the triangle are $18\ cm$ and $10\ cm$ and the perimeter is $42\ cm$.To do:We have to find the area of the triangle.Solution:Let us assume the third side of the triangle as $x$This implies, The three sides of the triangle are $18\ cm, 10\ cm$ and $x\ ... Read More

Given:The sides of a triangle are in the ratio of $12: 17: 25$ and its perimeter is $540\ cm$.To find:We have to find the area of the triangle.Solution:Let the common ratio between the sides of the triangle be  $=\ a$This implies, The sides of the triangle as $12\ a, 17\ ... Read More

Given: An isosceles triangle has perimeter $30\ cm$ and each of the equal sides is $12\ cm$.To do:Let us assume the third side of the triangle as $x$We have, Two sides with equal length of $12\ cm$and perimeter as $30\ cm$We know that, Perimeter $P$ of a triangle with sides of ... Read More

Given:\( \mathrm{D}, \mathrm{E} \) and \( \mathrm{F} \) are respectively the mid-points of the sides \( \mathrm{BC}, \mathrm{CA} \) and \( \mathrm{AB} \) of a \( \triangle \mathrm{ABC} \).To do:We have to show that(i) BDEF is a parallelogram.(ii) \( \operatorname{ar}(\mathrm{DEF})=\frac{1}{4} \operatorname{ar}(\mathrm{ABC}) \)(iii) \( \operatorname{ar}(\mathrm{BDEF})=\frac{1}{2} \operatorname{ar}(\mathrm{ABC}) \)Solution:In $\triangle ABC$, By mid ... Read More