CBSE Class 8 Maths Chapter 4: Practical Geometry NCERT Solutions

To construct a quadrilateral, we generally need five independent measurements. The process involves constructing triangles using the given sides and diagonals.

(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 cm.
2. With B as the center, draw an arc of radius 5.5 cm.
3. With A as the center, draw an arc of radius 7 cm. This arc intersects the previous arc at point C.
4. Join BC and AC. Now, triangle ABC is formed.
5. With A as the center, draw an arc of radius 6 cm.
6. With C as the center, draw an arc of radius 4 cm. This arc intersects the arc drawn in the previous step at point D.
7. Join AD and CD.

The resulting figure ABCD is the required quadrilateral.

(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 cm.
2. With J as the center, draw an arc of radius 4.5 cm.
3. With U as the center, draw an arc of radius 6.5 cm. This arc intersects the previous arc at point P.
4. Join PJ and PU. Now, triangle JUP is formed.
5. With P as the center, draw an arc of radius 5 cm.
6. With U as the center, draw an arc of radius 4 cm. This arc intersects the arc drawn in the previous step at point M.
7. Join MP and UM.

The resulting figure JUMP is the required quadrilateral.

(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. So, $MO\; =\; RE\; =\; 4.5\; \backslash text\{\; cm\}$ and $ME\; =\; OR\; =\; 6\; \backslash text\{\; cm\}$.

Steps of construction:

1. Draw a line segment $OR$ of length 6 cm.
2. With O as the center, draw an arc of radius 7.5 cm.
3. With R as the center, draw an arc of radius 4.5 cm. This arc intersects the previous arc at point E.
4. Join OE and RE. Now, triangle ORE is formed.
5. With E as the center, draw an arc of radius 6 cm (since $ME\; =\; OR$).
6. With O as the center, draw an arc of radius 4.5 cm (since $MO\; =\; RE$). This arc intersects the arc drawn in the previous step at point M.
7. Join OM and EM.

The resulting figure MORE is the required parallelogram.

(iv) Construction of Rhombus BEST

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

In a rhombus, all sides are equal. So, $BE\; =\; ES\; =\; ST\; =\; TB\; =\; 4.5\; \backslash text\{\; cm\}$. The diagonals of a rhombus bisect each other at right angles. The given $ET\; =\; 6\; \backslash text\{\; cm\}$ is a diagonal. Let's assume ET is the diagonal. The other diagonal is BS.

Steps of construction:

1. Draw a line segment $ET$ of length 6 cm.
2. Draw the perpendicular bisector of $ET$. Let it intersect $ET$ at O.
3. Since diagonals bisect each other, $EO\; =\; OT\; =\; 3\; \backslash text\{\; cm\}$.
4. The other diagonal BS will bisect ET at O. Also, $BO\; =\; OS$. Since all sides are 4.5 cm, we can find BO and OS. In right-angled triangle $BEO$, $BE^2\; =\; BO^2\; +\; EO^2$.
5. $(4.5)^2\; =\; BO^2\; +\; (3)^2$
6. $20.25\; =\; BO^2\; +\; 9$
7. $BO^2\; =\; 20.25\; -\; 9\; =\; 11.25$
8. $BO\; =\; \backslash sqrt\{11.25\}\; \backslash approx\; 3.35\; \backslash text\{\; cm\}$.
9. With E and T as centers, draw arcs of length $BO\; \backslash approx\; 3.35\; \backslash text\{\; cm\}$ on the perpendicular bisector. These points are B and S.
10. Join EB, SB, ST, and TB.

The resulting figure BEST is the required rhombus.