To do:We have to construct an equilateral triangle, given its side and justify the constructionSolution:Steps of construction:(i) Let us draw a line segment $BC$ of length $5\ cm$.(ii) Cut an arc of radius $5\ cm$ from point $B$ and an arc of $5\ cm$ from point $C$.(iii) Name the point ... Read More

To do:We have to construct the given angles and verify them by measuring them with a protractor.Solution:(i)Steps of construction:(a) Draw a ray $BC$.(b) With centre $B$ and a suitable radius, draw an arc meeting $BC$ at $E$.(c) Cut off arc $EF = FG$ from $E$.(d) Bisect the arc $FG$ at ... Read More

To do:We have to construct the given angles.Solution:(i)Steps of construction :(a) Draw a ray $AB$.(b) With centre $A$ and a suitable radius draw an arc meeting $AB$ at $C$.(c) With centre $C$ and the same radius as above draw another arc meeting the above arc at $D$.(d) Extend $AD$ to ... Read More

To do:We have to construct an angle of $45^o$ at the initial point of a given ray and justify the construction.Solution:Steps of construction:(a) Draw a ray $BC$.(b) With center $B$ and a suitable radius draw an arc meeting $BC$ at $E$.(c) With center $E$ and the same radius as above ... Read More

Given:$PQRS$ and $ABRS$ are parallelograms and \( \mathrm{X} \) is any point on side \( \mathrm{BR} \).To do:We have to show that(i) \( \operatorname{ar}(\mathrm{PQRS})=\operatorname{ar}(\mathrm{ABRS}) \)(ii) \( \operatorname{ar}(\mathrm{AXS})=\frac{1}{2} \mathrm{ar}(\mathrm{PQRS}) \)Solution:(i) Parallelograms $PQRS$ and $ABRS$ lie on the same base $SR$ and between the same parallels $SR$ and $PB$.This implies, $ar(PQRS) = ... Read More

To do:We have to construct an angle of $90^o$ at the initial point of a given ray and justify the construction.Solution:Steps of construction:(a) Draw a ray $AB$.(b) With centre $A$ and a suitable radius, draw an arc such that it cuts $AB$ at $C$.(c) With centre $C$ and the same ... Read More

Given:$P$ and \( Q \) are any two points lying on the sides \( D C \) and \( A D \) respectively of a parallelogram \( \mathrm{ABCD} \).To do:We have to show that ar \( (\mathrm{APB})=\operatorname{ar}(\mathrm{BQC}) \).Solution:$\triangle APB$ and parallelogram $ABCD$ lie on the same base $AB$ and between ... Read More

Given:\( \mathrm{P} \) is a point in the interior of a parallelogram \( \mathrm{ABCD} \). To do:We have to show that(i) \( \operatorname{ar}(\mathrm{APB})+\operatorname{ar}(\mathrm{PCD})=\frac{1}{2} \operatorname{ar}(\mathrm{ABCD}) \)(ii) \( \operatorname{ar}(\mathrm{APD})+\operatorname{ar}(\mathrm{PBC})=\operatorname{ar}(\mathrm{APB})+\operatorname{ar}(\mathrm{PCD}) \) Solution:Draw two lines $EF$ and $GH$ parallel to $AB$ and $BC$ respectively.(i) $\triangle APB$ and parallelogram $AEFB$ lie on the same base $AB$ and ... Read More

Given:$E, F, G$ and \( \mathrm{H} \) are respectively the mid-points of the sides of a parallelogram \( \mathrm{ABCD} \)To do:We have to show that \( \operatorname{ar}(\mathrm{EFGH})=\frac{1}{2} \operatorname{ar}(\mathrm{ABCD}) \).Solution:Join $EF, FG, GH, HE$ and $FH$.We know that, Opposite sides of a parallelogram are equal and parallel.This implies, $A D \| ... Read More

Given:A farmer was having a field in the form of a parallelogram PQRS. She took any point \( A \) on \( \mathrm{RS} \) and joined it to points \( \mathrm{P} \) and \( \mathrm{Q} \).The farmer wants to sow wheat and pulses in equal portions of the field separately.To ... Read More