Constructions - Class 10 Maths - Extra Questions

Draw a line segment of length 10 cm. Divide it into four equal parts. Measure each of these parts.

Steps of construction:
Step I - draw a line segment AB = 10 cm with the help of a ruler.
Step II - With vertex A, by using compass take more than half of AB and draw arcs, one each side of sides of AB
Step III - With vertex B, by using a compass and with the same radius cut the previous arcs on both sides of AB at R and S respectively. Join RS.
Mark the point O on AB where RS intersects it.
Step IV - With vertex A, by using compass take more than half of AO and draw arcs, one each side of sides of AO
Step V - With vertex O, by using a compass and with same radius cut the previous arcs on both sides of AO at P and Q respectively. Join PQ.
Mark the point M on AB where PQ intersects it.
Step VI - With vertex O, by using compass take more than half of OB and draw arcs, one each side of sides of OB
Step VII - With vertex B, by using a compass and with the same radius cut the previous arcs on both sides of OB at T and U respectively. Join TU.
Mark the point N on AB where TU intersects it.

Each part measures $$+2.5\ cm$$

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Draw a circle of radius $$3\ cm$$. Take two points $$P$$ and $$Q$$ on one of its extended on both sides of its centre, each at a distance of $$7\ cm$$ on opposites of its centre. Draw tangents to the circle from each of these two points $$P$$ and $$Q$$

Step 1: Draw a circle with centre $$O$$ and radius $$3\ cm$$ using a compass.
Step 2: Draw a secant passing through the centre. Mark points $$P$$ and $$Q$$ on opposite sides of the centre at a distance of $$7\ cm$$ from $$O$$.
Step 3: Place the compass on $$P$$ and draw two arcs on opposite sides of $$OP$$. Now place the compass on $$O$$ and draw two arcs intersecting the arcs drawn from point $$P$$.
Step 4: Join the intersection points of the arcs to obtain the perpendicular bisector of $$OP$$. Mark the mid point of $$OP$$ as $$M_1$$
Step 5: From $$M_1$$ draw a circle with radius $$=M_1P= M_1O$$
Step 6: Mark the intersection points of the circle drawn from $$M_1$$ with the circle drawn from $$O$$ as $$A$$ and $$B$$.
Step 7: Join $$P-A$$ and $$P-B$$
Step 8: Repeat steps $$3$$ to $$7$$ for point $$Q$$ and obtain tangents $$QC$$ and $$QD$$

$$PA,PB,QC$$ and $$QD$$ are the required tangents.

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Construct a tangent to a circle of radius $$4cm$$ from a point on the concentric circle of radius $$6cm$$ and measure its length. Also verify the measurement by actual calculation.

Construction the two conuntric circle with to $$r_1 = 4 cm$$  and $$r_2 =6cm$$  with centre be $$C$$
locate a point $$P$$ on  outer circle such that $$CP= 6cm$$ and lets assume $$M$$ be the mid point of CP such that $$CM=MP=3cm$$
Now draw a circle with centre $$M$$ and radius $$= 2 cm $$ and lets say its intersect the inner circle at point $$A$$ and $$B$$.
Now join the points $$A,P$$ and $$B,P$$ so we get our tangents $$AP$$ and $$BP$$
Since $$\triangle CAP$$ is a right angled triangle with $$\angle A = 90^o$$
therefore using pythogorus therom
$$(CA)^2+(AP)^2=(CP)^2$$
$$(PA)^2=36-9$$
$$AP=3\sqrt{3} cm$$


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Draw a circle with the help of a bangle. Take a point $$P$$ outside the circle. Construct the pair of tangents from this point $$P$$ to the circle.

Step 1: Draw a circle by tracing the surface of a bangle.
Step 2: Take a point $$P$$ outside the circle. Draw secant $$P-R-Q$$
Step 3: Place the compass on $$P$$ and mark two arcs on opposite side of $$PQ$$. Now, place the compass on $$Q$$ and draw two arcs intersecting the arcs drawn from $$P$$.
Step 4: Join the intersection points of the arcs to obtain the perpendicular bisector of $$PQ$$. Mark the mid point of $$PQ$$ as $$M$$.
Step 5: Place the compass on $$M$$ and taking $$PM$$ as radius draw a semicircle passing through $$P$$ and $$Q$$.
Step 6: From point $$R$$, draw a perpendicular to $$PQ$$ and let it intersect the semicircle drawn in step $$5$$ at $$C$$.
Step 7: Place the compass at $$P$$ and with $$PC$$ as radius, mark two arcs on the original circle and mark them as $$A$$ and $$B$$.
Step 8: Join $$P-A$$ and $$P-B$$

$$PA$$ and $$PB$$ are the required tangents.

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Draw a triangle $$ABC$$ with side $$BC = 6$$ cm, $$\angle C = 30^o$$ and$$\angle A = 105^o$$. Then construct another triangle whose sides are $$\dfrac{2}{3}$$ times the corresponding sides of $$\Delta ABC$$.

Given:
$$BC = 6$$ cm, $$\angle C= 30^o$$ and $$\angle A = 105^o$$
We know that in $$\Delta ABC, \angle A + \angle B + \angle C = 180^o$$
Therefore, $$\angle B = 180^o - (\angle A + \angle C)$$
$$= 180^o − (105^o + 30^o)\Rightarrow  45^o$$
Steps of construction:
1. Draw a line $$BC = 6$$ cm.
2. Draw a ray CN making an angle of $$30^o$$ at $$C.$$
3. Draw a ray BM making an angle of $$45^o$$ at $$B.$$
4. Locate the point of intersection of rays $$CN$$ and $$BM$$ and name it as $$A.$$
5. $$ABC$$ is the triangle whose similar triangle is to be drawn.
6. Draw any ray $$BX$$ making an acute angle with $$BC$$ on the side opposite to the vertex $$A.$$
7. Locate 3 (Greater of $$2$$ and $$3$$ in $$\dfrac{2}{3}$$) points $$B_1, B_2$$ and $$B_3$$ on $$BX$$ so that $$BB_1 = B_1B_2 = B_2B_3$$.
8. Join $$B_3C$$ and draw a line through $$B_2$$ (Smaller of $$2$$ and $$3$$ in $$\dfrac{2}{3}$$) parallel to $$B_3C$$ to intersect $$BC$$ at $$C'.$$
9. Draw a line through $$C'$$ parallel to the line $$CA$$ to intersect $$BA$$ at $$A'.$$
10. $$A'BC'$$ is the required similar triangle whose sides are $$\dfrac{2}{3}$$ times the corresponding sides of $$\Delta ABC$$.

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Class 10 Maths Extra Questions