CBSE Class 8 Mathematics Chapter 4: Practical Geometry NCERT Solutions

To construct a quadrilateral, we need sufficient information about its sides and angles. The following steps outline the construction for each given quadrilateral:

(i) Construction of Quadrilateral ABCD

Given: $AB\; =\; 4.5\; \backslash text\{\; cm\},\; BC\; =\; 5.5\; \backslash text\{\; cm\},\; CD\; =\; 4\; \backslash text\{\; cm\},\; AD\; =\; 6\; \backslash text\{\; cm\},\; AC\; =\; 7\; \backslash text\{\; cm\}$

Steps of construction:

1. Draw a line segment $AB$ of length $4.5\; \backslash text\{\; cm\}$.
2. With $B$ as the center, draw an arc of radius $5.5\; \backslash text\{\; cm\}$.
3. With $A$ as the center, draw an arc of radius $7\; \backslash text\{\; cm\}$. This arc intersects the previous arc at point $C$.
4. Join $BC$ and $AC$.
5. With $A$ as the center, draw an arc of radius $6\; \backslash text\{\; cm\}$.
6. With $C$ as the center, draw an arc of radius $4\; \backslash text\{\; cm\}$. This arc intersects the arc drawn in the previous step at point $D$.
7. Join $AD$ and $CD$.

Thus, the quadrilateral $ABCD$ is constructed.

(ii) Construction of Quadrilateral JUMP

Given: $JU\; =\; 3.5\; \backslash text\{\; cm\},\; UM\; =\; 4\; \backslash text\{\; cm\},\; MP\; =\; 5\; \backslash text\{\; cm\},\; PJ\; =\; 4.5\; \backslash text\{\; cm\},\; PU\; =\; 6.5\; \backslash text\{\; cm\}$

Steps of construction:

1. Draw a line segment $JU$ of length $3.5\; \backslash text\{\; cm\}$.
2. With $J$ as the center, draw an arc of radius $4.5\; \backslash text\{\; cm\}$.
3. With $U$ as the center, draw an arc of radius $6.5\; \backslash text\{\; cm\}$. These two arcs intersect at point $P$.
4. Join $PJ$ and $PU$.
5. With $P$ as the center, draw an arc of radius $5\; \backslash text\{\; cm\}$.
6. With $U$ as the center, draw an arc of radius $4\; \backslash text\{\; cm\}$. These two arcs intersect at point $M$.
7. Join $MP$ and $UM$.

Thus, the quadrilateral $JUMP$ is constructed.

(iii) Construction of Parallelogram MORE

Given: $OR\; =\; 6\; \backslash text\{\; cm\},\; RE\; =\; 4.5\; \backslash text\{\; cm\},\; EO\; =\; 7.5\; \backslash text\{\; cm\}$

In a parallelogram, opposite sides are equal. Therefore, $MO\; =\; RE\; =\; 4.5\; \backslash text\{\; cm\}$ and $ME\; =\; OR\; =\; 6\; \backslash text\{\; cm\}$.

Steps of construction:

1. Draw a line segment $OR$ of length $6\; \backslash text\{\; cm\}$.
2. With $O$ as the center, draw an arc of radius $7.5\; \backslash text\{\; cm\}$.
3. With $R$ as the center, draw an arc of radius $4.5\; \backslash text\{\; cm\}$. These two arcs intersect at point $E$.
4. Join $OE$ and $RE$.
5. With $E$ as the center, draw an arc of radius $6\; \backslash text\{\; cm\}$ (since $ME\; =\; OR$).
6. With $O$ as the center, draw an arc of radius $4.5\; \backslash text\{\; cm\}$ (since $MO\; =\; RE$). These two arcs intersect at point $M$.
7. Join $OM$ and $EM$.

Thus, the parallelogram $MORE$ is constructed.

(iv) Construction of Rhombus BEST

Given: $BE\; =\; 4.5\; \backslash text\{\; cm\},\; ET\; =\; 6\; \backslash text\{\; cm\}$

A rhombus has all sides equal. So, $BE\; =\; ES\; =\; ST\; =\; TB\; =\; 4.5\; \backslash text\{\; cm\}$. The diagonal $ET$ is given as $6\; \backslash text\{\; cm\}$.

Steps of construction:

1. Draw a line segment $ET$ of length $6\; \backslash text\{\; cm\}$.
2. Draw the perpendicular bisector of $ET$. Let it intersect $ET$ at $O$.
3. Since the diagonals of a rhombus bisect each other at right angles, $EO\; =\; OT\; =\; 6/2\; =\; 3\; \backslash text\{\; cm\}$.
4. With $E$ as the center, draw an arc of radius $4.5\; \backslash text\{\; cm\}$ on the perpendicular bisector. Mark this point as $B$.
5. With $T$ as the center, draw an arc of radius $4.5\; \backslash text\{\; cm\}$ on the perpendicular bisector. Mark this point as $S$.
6. Join $EB,\; TB,\; ES,\; TS$.

Thus, the rhombus $BEST$ is constructed.