Question 15 Marks
Write the steps of construction for drawing a pair of tangents to a circle of radius 3cm, which are inclined to each other at an angle of 60º.
Answer
Steps of Construction:
- Taking O on the plane of paper and draw a circle of radius OA = 3cm.
- Produce OA to B such that OA = AB = 3cm.
- Taking A as the center draw a circle of radius OA = AB = 3cm. Supose it cuts the circle drawn in step 1 at P and Q.
- Join BP and BQ to get the desired tangents.
View full question & answer→Question 25 Marks
Construct an isosceles triangle whose base is $9\ cm$ and altitude $5\ cm$. Construct another triangle whose side are $\frac34$ of the corresponding sides of the first isosceles triangle.
Answer
Steps of construction:
- Draw a line segment $BC = 9\ cm$
- Draw peependicular bisector $PQ$ of $BC$ Meeting it at $M.$
- Fron $QP$ cut-off a distance $MA = 5\ cm.$
- Join $AB$ and $AC.$
Thus, isosceles $\triangle\text{ABC}$ is obtained.
- Below $BC$, make an acute $\angle\text{CBX}.$
- Along $BX, $ mark off $4$ points $B_1, B_2, B_3, B_4, $ such that $BB_1, = B_1B_2 = B_1B_3 = B_3B_4.$
- Join $B_4C$
- From $B_3, $ draw $B_3C' || B_4C,$ meeting $BC $ at $C'.$
- From $C',$ draw $C' A' || CA$, meeting $AB$ at $A'.$
Thus, $\triangle\text{A}'\text{BC}'$ is the required triangle similar to $\triangle\text{ABC}$ Such that each side of $\triangle\text{A}'\text{BC}'$ is $\frac34\text{times}$ the corresponding side of $\triangle\text{ABC}.$ View full question & answer→Question 35 Marks
Draw a circle with the help of a bangle. Take any point P outside the circle. Construct the pair of tangents from the point P to the circle.
Answer
Steps of construction:
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Drow a circle with the help of bangle.
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Take a point P outside the circle and take two chords QR and ST.
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Draw perpendicular bisect of these chords.
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Join PO and bisect it, Let U be the mid-point of PO.
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Taking U as centere, draw a circle of radius OU, which will intersect the original circle at V and W.
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Join PV and PW are required tangents.
View full question & answer→Question 45 Marks
Draw a circle of radius 3cm. Draw a tangent to the circle making an angle of 30° with a line passing through the centre.
Answer
Steps of Construction:
- Draw a circle with center O and radius 3cm.
- Draw a radius OA of this circle and product it to B.
- Construct an $\angle\text{AOP}=60^\circ$ (Complement of 30°)
- Draw perpendicular to OP at P which intersects OA produced at Q.
Thus, PQ is the desired tangent such that $\angle\text{OQP}=30^\circ.$ View full question & answer→Question 55 Marks
Draw a line segment $AB$ of length $5.4\ cm$. Divide it into six equal parts. Write the steps of construction.
Answer
Steps of Construction:
- Draw a line segment $AB = 5.4\ cm$
- Draw aray $AX$ making an acute $\angle\text{BAX}$ with $AB$
- Along $AX$ mark $6$ Point $A_1, A_2, A_3, A_4, A_5, A_6, $ Such that $AA_1 = A_1A_2 = A_2A_3 = A_3A_4 = A_4A_5 = A_5A_6.$
- Join $A_6B.$
- Through the point $A_5, $ draw a line parallel to $A_6B$ by making an angle equal to $\angle\text{AA}_6\text{B}$ at $A_5. $ Suppose this line meets $AB $ at a point $P.$
- Similarly, though points $A_4, A_3, A_2, A_1,$ draw lines parallel to $A_6B.$
Suppose these lines meet $AB $ at points $Q, R, S, T$ respectively.
Thus, line segment $AB$ is divided into 6 equal parts such that
$AT = TS = SR = RQ = QP = PB.$ View full question & answer→Question 65 Marks
Draw a line segment $AB$ of length $7\ cm$. Using ruler and compasses, find a point $P$ on $AB$ such that $\frac{\text{AP}}{\text{AB}}=\frac{3}{5}.$
Answer
Steps of Construction:
Step 1. Draw a line segment $AB = 7\ cm.$
Step 2. Draw a ray $AX$, making an acute angle $\angle\text{BAX}.$
Step 3. Along $AX,$ mark $5$ points $($greater of $3$ and $5) A_1, A_2, A_3, A_4$ and $A_5$ such that $AA_1= A_1A_2 = A_2A_3= A_3A_4 = A_4A_5$
Step 4. Join $A_5B.$
Step 5. From $A_3,$ draw $A_3P$ parallel to $A_5B$ $\big($draw an angle equal to $\angle\text{AA}_5\text{B}\big),$ meeting $AB$ in $P.$
Here, $P$ is the point on $AB$ such that $\frac{\text{AP}}{\text{PB}}=\frac32$ or $\frac{\text{AP}}{\text{AB}}=\frac35.$ View full question & answer→Question 75 Marks
Construct a $\triangle\text{ABC},$ in which $BC = 6.5\ cm$, $AB = 4.5\ cm$ and $\angle\text{ABC}=60^\circ.$
Construct a triangle similar to this triangle whose sides are $\frac{3}{4}$ the corresponding sides of $\triangle\text{ABC}.$
Answer
Steps of Construction:
- Draw a line segment $BC = 6.5\ cm$
- At B, construct $\angle\text{CBX}=60^\circ$
- With B as center and radius $4.5\ cm$, draw an arc intersecting BX at A
- Join AC to obtain $\triangle\text{ABC}$
- Below BC, make an acute $\angle\text{CBY}$
- Along BY, mark off $4$ points $\Big($ greater of $3$ and in $\frac34\Big)$ $B_1, B_2, B_3, B_4$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4$
- Join $B_4C$
- From point $B_3$, draw a line parallel to $B_4C$ intersecting $BC$ at $C'$
- From Point $C'$, draw a line parallel to $AC$ intersecting $AB$ at $A'$
Thus, $\triangle\text{A}'\text{BC}'$ is the required triangle. View full question & answer→Question 85 Marks
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.
Answer
Steps of Construction:
- Take a point O on the plane of the paper and draw a circle of radius OA = 4cm. Also, draw a concentric circle of radius OB = 6cm.
- Find the mid-point C of OB and draw a circle of radius OC = BC. Suppose this circle intersects the circle of radius 4cm at P and Q.
- Join BP and BQ to get the desired tangents from a point Bon the circle of radius 6cm.
By actual measurment, we find that BP = BQ = 4.5cmVerification:
In $\triangle\text{BOO},$ we have OB = 6cm and OP = 4cm $\therefore\text{OB}^2=\text{BP}^2+\text{OP}^2$ $\Rightarrow\text{BP}=\sqrt{\text{OB}^2-\text{OP}^2}$ $=\sqrt{36-16}=\sqrt{20}$ $=4.47\text{cm}=4.5\text{cm}$ Similarly, $\text{BQ} = 1.47\text{cm} = 4.5\text{cm}.$ View full question & answer→Question 95 Marks
Draw a line segment $AB$ of length $6.5\ cm$ and divide it in the ratio $4 : 7$. Measure each of the two parts.
Answer
Steps of Construction:
1. Draw a line segment $A B=6.5 cm$
2. Draw aray $A X$ making an acute $\angle B A X$ with $A B$
3. Along $A X$ mark $(4+7) 11$ Point $A_1, A_2, A_3, A_4, A_5, A_6, A_7, A_8, A_9 A_{10} A_{11}$
Such that $A_1=A_1 A_2=A_2 A_3=A_3 A_4=$ $A _4 A_5= A _5 A_6 \ldots A_{10} A_{11}$
4. Join A11B.
5. Through the point $A 4$, draw a line parallel to $A_{11} B$ by making an angle equal to $\angle A A_{11} B$ at $A_4$.Suppose this line meets $A B$ at a point $C$.
Suppose this line meets $A B$ at a point $C$.
The Point $C$ so obtained is the required point, which divides $A B$ in the ratio $4: 7$. View full question & answer→Question 105 Marks
Draw two concentric circles of radii 4cm and 6cm. Contruct a tangent to the smaller circle from a point on the larger circle. Measure the length of this tangent.
Answer
Steps of construction:
- Take a point O on the plane of the paper and draw a circle of radius OA = 4cm. Also, draw a concentric circle of radius OB = 6cm
- Find the mid-point C of OB and draw a circle of radius OC = BC. Suppose this circle intrsects the circle of radius 4cm at P and Q.
- Join BP and BQ to get the desired tangents from a point bon the circle of radius 6cm.
By actual mesurement, we find that BP = BQ = 4.5cm
Verification:
In $\triangle\text{BOO},$ we have OB = 6cm and OP = 4cm
$\therefore\text{OB}^2=\text{BP}^2+\text{OP}^2$
$\Rightarrow\text{BP}\sqrt{\text{OB}^2-\text{OP}^2}$
$=\sqrt{36-16}=\sqrt{20}$
$=4.47\text{cm}=4.5\text{cm}$
Similarly, BQ = 4.47cm = 4.5cm.
View full question & answer→Question 115 Marks
Construct a $\triangle\text{ABC},$ in which BC = 5cm, $\angle\text{C}=60^\circ$ and altitude from A is equal to 3cm. Construct a $\triangle\text{ADE}$ similar to $\triangle\text{ABC},$ such that each side of $\triangle\text{ADE}$ is $\frac32$ times the corresponding side of $\triangle\text{ABC}$ Write the steps of construction.
Answer
Step of constuction:
- Draw a line segment BC = 5cm
- Construct $\angle\text{BCP}=60^\circ$
- Draw a line GH || BC at a distance of 3cm, intersecting CP at A.
- Join AB and draw altitude $\text{AM}\perp\text{BC}.$ Thus, $\triangle\text{ABC}$ is obtained.
- Extend AB to D such that $\text{AD}=\frac32\text{AB}.$
- Draw DE || BC, cutting AC produced at E.
Then, $\triangle\text{ADE}$ is the required triangle similar to $\triangle\text{ABC}$ such that each side of $\triangle\text{ADE}$ is $\frac32\text{times}$ the corresponding side of $\triangle\text{ABC}$ View full question & answer→Question 125 Marks
Draw a line segment of length $8\ cm$ and divide it internally in the ratio $4 : 5.$
Answer
Step Of Construction:
Step 1. Draw a line segment $AB = 8cm.$
Step 2. Draw a ray $AX$ making an acute angle $\angle\text{BAX}=60^\circ$ with $AB.$
Step 3. Draw a ray $BY$ parallel to $AX$ by making an acute angle $\angle\text
{ABY}=\angle\text{BAX}.$
Step 4. Mark four points $A_1, A_2, A_3, A_4$ on $AX$ and five points $B_1, B_2, B_3, B_4, B_5$ on $BY$ in such a way that $AA_1= A_1A_2= A_2A_3= A_3A_4.$
Step 5. Join $A_4B_5$
Step 6.Let this line intersect $AB$ at a point $P$. Thus, P is the point dividing the line segment$ AB$ internally in the ratio of $4 : 5.$ View full question & answer→Question 135 Marks
Draw two concentric circle of radii 3cm and 5cm. Taking a point on the outer circle, construct the pair of tangents to the inner circle.
Answer
Steps of construction:
- Draw a circle with radius 3cm and centre O.
- Draw another circle with radius 5cm and same centre O.
- Take a point P on the circumference of larger circle and join O to p.
- Taking OP as diameter draw another circle which intersects the smallest circle at A and B.
- Join A to P and B to P.
Hence AP and BP are the required tangents.
View full question & answer→Question 145 Marks
Draw a circle with centre O and radius 4cm. Draw any diameter AB of this circle. Construct tangents to the circle at each of the two end points of the diameter AB.
Answer
Steps of Construction:
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Draw a circle with centre O and radius 4cm.
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Draw any diameter AB.
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Draw line $\text{L}\perp\text{OA}$ such that $\angle\text{OAL}=90^\circ.$
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Draw line $\text{M}\perp\text{OB}$ such that $\angle\text{OBM}=90^\circ.$
Thus, LA and LB are the required tangent.
View full question & answer→Question 155 Marks
Draw a line segment of length $7.6\ cm$ and divide it in the ratio $5 : 8$. Measure the two parts.
Answer
Steps of Construction:
Step 1. Draw a line segment $A B=7.6 cm$
Step 2. Draw a ray $A X$, making an acute angle $\angle B A X$.
Step 3. Along $A X$, mark $(5+8=) 13$ points $A_1, A_2, A_3, A_4, A_5, A_6, A_7, A_8, A_9, A_{10}, A_{11}, A_{12}$ and $A_{13}$ such that
$AA_1=A_1 A_2=A_2 A_3=A_3 A_4=A_4 A_5=A_5 A_6=A_6 A_7=A_7 A_8=A_8 A_9=A_9 A_{10}=A_{10} A_{11}=A_{11} A_{12}=A_{12} A_{13}$
Step 4. Join $A_{13} B$.
Step 5. From $A_5$, draw $A_5 P$ parallel to $A_{13} B$ (draw an angle equal to $\angle A A_{13} B$ ), meeting $A B$ in $P$.
Here, P is the point on AB which divides it in the ratio $5: 8$.
$\therefore$ Length of $A P=2.9 cm$ (Approx)
Length of $B P=4.7 cm$ (Approx) View full question & answer→Question 165 Marks
Draw a right triangle in which sides (other than hypotenuse) are of lengths 4cm and 3cm. Then, construct another triangle whose sides are $\frac53\text{times}$ the corresponding sides of the given triangle.
Answer
Steps of Construction:
Step 1. Draw a line segment BC = 3cm.
Step 2. At B, draw $\angle\text{XBC}=90^\circ.$
Step 3. With B as centre and radius 4cm, draw an arc cutting BX at A.
Step 4. Join AC. Thus, a right $\triangle\text{ABC}$ is obtained.
Step 5. Extend BC to D such that $\text{BD}=\frac{5}{3}\text{BC}=\frac53\times3\text{cm}=5\text{cm}.$
Step 6. Draw DE || CA, cutting BX in E.
Here, $\triangle\text{BDE}$ is the required triangle similar to $\triangle\text{ABC}$ such that each side of $\triangle\text{BDE}$ is $\frac53\text{times}$ the corresponding side of $\triangle\text{ABC}.$ View full question & answer→Question 175 Marks
Draw a circle of radius 4cm. Draw tangent to the circle making an angle of 60° with a line passing through the centre.
Answer
Steps of construction:
- Draw a circle with centre O and radius 4cm.
- Draw a radius OA of the circle and produce it to B.
- Construct an angle $\angle\text{AOP}$ equal to complement of 60°, i.e., equal to 30°.
- Draw perpendicular to OP at P which intersects OA produced at Q.
Thus, PQ is the desired tangent such that $\angle\text{OQP}=60^\circ.$ View full question & answer→Question 185 Marks
Draw two tangents to a circle of radius 3.5cm from a point P at a distance of 6.2cm from its centre.
Answer
Steps of Construction:
-
Take a point O on the plane of the paper and draw a circle of radius 3.5cm.
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Mark a point P at a distance of 6.2cm from the center O and Join OP.
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Draw a right bisector of OP, intersecting OP at Q.
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Taking Q as center and OQ = PQ as radius, draw a circle to intersect the given circle at T and T'.
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Join PT and PT' to get the required tangents.
View full question & answer→Question 195 Marks
To construct a triangle similar to $\triangle\text{ABC}$ in which BC = 4.5cm, $\angle\text{B}=45^\circ$ and $\angle\text{C}=60^\circ,$ using a scale factor of $\frac35$ BC will be divided in the ratio.
-
3 : 4
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4 : 7
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3 : 10
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3 : 7
AnswerTo construct a triangle similar to $\triangle\text{ABC}$ in which BC = 4.5cm, $\angle\text{B}=45^\circ$ and $\angle\text{C}=60^\circ,$ using a scale factor of $\frac35$ BC will be divided in the ratio 3 : 4.
Here, $\triangle\text{ABC}\sim\triangle\text{AB}'\text{C}'$
BC' : C'C = 3 : 4
or BC' : BC = 3 : 7
Hence, the correct answer is option A. View full question & answer→Question 205 Marks
Construct a $\triangle\text{ABC},$ with $BC = 7cm,$ $\angle\text{B}=60^\circ$ and AB = 6cm. Construct another triangle whose sides are $\frac34\text{times}$ the corresponding sides of $\triangle\text{ABC}.$
Answer
Steps of Construction:
Step 1. Draw a line segment $BC = 7cm.$
Step 2. At $B,$ draw $\angle\text{XBC}=60^\circ.$
Step 3. With $B$ as centre and radius 6cm, draw an arc cutting the ray $BX$ at $A.$
Step 4. Join $AC.$ Thus, $\triangle\text{ABC}$ is the required triangle.
Step 5. Below BC, draw an acute angle $\angle\text{YBC}.$
Step 6. Along BY, mark four points $B_1, B_2, B_3$ and B_4such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4.$
Step 7. Join $CB_4.$
Step 8. From $B_3,$ draw $B_3C' || CB_4$ meeting $BC$ at $C'.$
Step 9. From $C',$ draw $A'C' || AC$ meeting $AB$ in $A'.$
Here, $\triangle\text{ABC}$ is the required triangle whose sides are $\frac34\text{times}$ the corresponding sides of $\triangle\text{ABC}.$ View full question & answer→Question 215 Marks
Construct a $\triangle\text{ABC}$ in which $BC = 8cm, $ $\angle\text{B}=45^\circ$ and $\angle\text{C}=60^\circ.$ Construct another triangle similar to $\triangle\text{ABC}$ such that its sides are $\frac35$ of the corresponding sides of $\triangle\text{ABC}.$
Answer
Steps of Construction:
Step 1. Draw a line segment $BC = 8cm.$
Step 2. At $B$, draw $\angle\text{XBC}=45^\circ.$
Step 3. At $C,$ draw $\angle\text{YCB}=60^\circ.$ Suppose $BX$ and $CY$ intersect at $ A.$
Thus, $\triangle\text{ABC}$ is the required triangle.
Step 4. Below $BC,$ draw an acute angle $\angle\text{ZBC}.$
Step 5. Along $BZ, $ mark five points $Z_1, Z_2, Z_3, Z_4$ and $Z_5$ such that $BZ_1 = Z_1Z_2 = Z_2Z_3 = Z_3Z_4 = Z_4Z_5.$
Step 6. Join $CZ_5$.
Step 7. From $Z_3,$ draw $Z_3C' || CZ_5$ meeting $BC$ at $C'.$
Step 8. From $C',$ draw $A'C' || AC$ meeting $AB$ in $A'.$
Here, $\triangle\text{AB}'\text{C}'$ is the required triangle whose sides are $\frac{3}{5}$ of the corresponding sides of $\triangle\text{ABC}.$ View full question & answer→Question 225 Marks
Write the step of construct the tangents to a circle from an external point.
Answer
Steps of construction:
- Take given circle and a point P outside the circle. O is centre of the circle.
- Joint OP.
- Bisect OP and get its mid-point M.
- Draw circle with centre M and radius = PM = MO.
- Circle drawn meets the given circle at Q above PO and at Q’ below PO.
- Join PQ and PQ’.
- PQ and PQ’ are the required tangents drawn to the circle from the point P.
We observe that PQ = PQ’.
View full question & answer→Question 235 Marks
Construct a triangle with sides $5\ cm, 6\ cm$ and $7\ cm$ and then another triangle whose sides are $\frac75$ of the corresponding sides of first triangle.
Answer
Steps of Construction:
Step 1. Draw a line segment $BC = 4\ cm.$
Step 2. With $B$ as centre, draw an angle of $90^\circ.$
Step 3. With $B$ as centre and radius equal to $3cm$, cut an arc at the right angle and name it $A.$
Step 4. Join $AB$ and $AC.$
Thus, $\triangle\text{ABC}$ is obtained .
Step 5. Extend $BC$ to $D$, such that $\text{BD}=\frac75\text{BC}=\frac75(4)\text{cm}=5.6\text{cm}.$
Step 6. Draw $DE || CA,$ cutting AB produced to $E.$ View full question & answer→Question 245 Marks
Draw a circle of radius 3.5cm. Draw a pair of tangent to this circle which are inclined to each other at an angle of 60°. Write the steps of construction.
Answer
Steps of construction:
- Taking O on the plane of paper and draw a circle of radius OA = 3.5cm.
- Produce OA to B such that OA = AB = 3.5cm.
- Taking A as the center draw a circle of radius AO = AB = 3.5cm Supose it cuts the circle draw in step 1 at P and Q.
- Join BP and BQ to get the desired tangents.
View full question & answer→Question 255 Marks
Draw a circle of radius 3.5cm. Take two point A and B on one of its extended diameter, each at a distance of 5cm from its centre. Draw tangents to the circle from each of these points A and B.
Answer
Steps of Construction:
-
Draw a line segment PQ of length 10cm.
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Take the mid-point O of PQ.
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Draw the perpendicular bisectors of PO and OQ which intersect PO at point R and OQ at point S.
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With center R and radius RP draw a circle.
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With center S and radius SQ draw a circle.
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With center O and radius 3.5cm draw another circle which intersects the previous circles at the points A, B, C and D respectively.
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Join PA, PB, QC and QD.
Thus, PA, PB, QC and QD are the reduired tangents.
View full question & answer→Question 265 Marks
Construct a $\triangle\text{ABC}$ in which AB = 6cm, $\angle\text{A}=30^\circ$ and $\angle\text{B}=60^\circ.$ Construct another $\triangle\text{AB}'\text{C}'$similar to $\triangle\text{ABC}$ with base AB' = 8cm.
Answer
Steps of Construction:
Step 1. Draw a line segment AB = 6cm.
Step 2. At A, draw $\angle\text{XAB}=30^\circ$
Step 3. At B, draw $\angle\text{YBA}=60^\circ.$ Suppose AX and BY intersect at C.
Thus, $\triangle\text{ABC}$ is the required triangle.
Step 4. Produce AB to B' such that AB' = 8cm.
Step 5. From B', draw B'C' || BC meeting AX at C'.
Here, $\triangle\text{AB}'\text{C}'$ is the required triangle similar to $\triangle\text{ABC}.$ View full question & answer→Question 275 Marks
Draw a line segment AB of length 8cm. Taking A as centre, draw a circle of radius 4cm and taking B as centre, draw another circle of radius 3cm. Construct tangents to each circle from the centre of the other circle.
Answer
Step of construction:
-
Draw a line segment AB of length 8cm.
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Draw the perpendicular bisector of AB which intersect it at C.
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With center C and radius CA, draw a circle.
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With centers A and B and radii 4cm and 3cm, draw two circle which intersect the previous circle at the points P, Q, R and S. at the points P, Q, R and S.
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Join AR, AS, BP and BQ.
Thus, AR, AS, BP and BQ are the required tangents.
View full question & answer→Question 285 Marks
Draw a $\triangle\text{ABC},$ right-angled at $B$ such that $AB = 3cm$ and $BC = 4cm.$ Now, Construct a triangle a triangle similar to $\triangle\text{ABC},$ each of whose sides is $\frac75\text{times}$ the correponding side of $\triangle\text{ABC}.$
Answer
Steps of construction:
- Draw a line segment $BC = 4cm$
- AT $B$, construct $\angle\text{MBC}=90^\circ.$
- Cut-off $BA = 3cm$ from $BM.$
- Join $AC.$
Thus, right-angled $\triangle\text{ABC}$ is obtained.
- Below $BC$, make an acute $\angle\text{CBX}.$
- Along $BX$, mark off 7 points $R_1, R_2, R_3, R_4, R_5, R_6, R_7$ such that $BR_1 = R_1R_2 = R_2R_3 = R_3R_4 = ... = R_6R_7$
- Join $R_5C.$
- From $R_7$, draw $R_7C_1 || R_5C$, meeting BC produced at $C_1.$
- From $C_1$, draw $C_1A_1 || CA$, meeting BA produced at $A_1.$
Then, $\triangle\text{A}_1\text{BC}_1$ is the required triangle similar to $\triangle\text{ABC}$ such that each side of $\triangle\text{A}_1\text{BC}_1$ is $\frac75\text{times}$ the corresponding side of $\triangle\text{ABC}.$ View full question & answer→Question 295 Marks
Draw a circle of radius 3cm. From a point P, 7cm away from the centre of the circle, draw two tangents to the circle. Also, measure the lengths of the tangents.
Answer
Steps of Construction:
- Take a point O on the plane of the paper and draw a circle of radius 3cm.
- Mark a point P at a distance od 7cm from the center O and Join OP.
- Draw a right visector of OP, intersecting OP at Q,
- Taking Q as center and OQ - PQ as radius, draw a circle to intersect the given circle at T and T'.
- Join PT and PT' to get the required tangents.
By measurment,
PT = PT' = 6.1cm
View full question & answer→Question 305 Marks
Draw a circle of radius 4.2cm. Draw a pair of tangents to this circle inclined to each other at an angle of 45°.
Answer
Steps of Construction:
- Draw a ciecle with center O and radius 4.5cm
- Draw diameter AB
- With OB as base, draw $\angle\text{BOC}=45^\circ$
- At A, draw a line perendicular to OA.
- At C, draw a line perpendicular to OC.
These lines intersect each other at P. Thus, PA and PC are the required tangents. View full question & answer→Question 315 Marks
Construct a $\triangle\text{PQR},$ in which $PQ = 6cm, QR = 7cm$ and $PR = 8cm.$ Then, construct another triangle whose sides are $\frac{4}{5}\text{ times}$ the corresponding sides of $\triangle\text{PQR}.$
Answer
Steps of construction:
Step 1. Draw a line segment $QR = 7cm.$
Step 2. With $Q$ as centre and radius 6cm, draw an arc.
Step 3. With $R$ as centre and radius 8cm, draw an arc cutting the previous arc at $P$.
Step 4. Join $PQ$ and $PR$. Thus, $\triangle\text{PQR}$ is the required triangle.
Step 5. Below $QR$, draw an acute angle $\angle\text{RQX}.$
Step 6. Along $QX$, mark five points $R_1, R_2, R_3, R_4$ and $R_5$ such that $QR_1 = R_1R_2 = R_2R_3 = R_3R_4= R_4R_5.$
Step 7. Join $RR_5.$
Step 8. From $R_4$, draw $R_4R' || RR_5$ meeting $QR$ at $R'.$
Step 9. From $R'$, draw $P'R' || PR$ meeting $PQ$ in $P'$. Here, $\triangle\text{P}'\text{QR}'$ is the required triangle, each of whose sides are $\frac{4}{5}\text{ times}$ the corresponding sides of $\triangle\text{PQR}.$ View full question & answer→Question 325 Marks
Construct an isosceles triangle whose base is 8cm and altitude 4cm and then another triangle whose sides are $1\frac12\text{times}$ the corresponding sides of the isosceles triangle.
Answer
Steps of Construction:
Step 1. Draw a line segment BC = 8cm.
Step 2. Draw the perpendicular bisector XY of BC, cutting BC at D.
Step 3. With D as centre and radius 4cm, draw an arc cutting XY at A.
Step 4. Join AB and AC. Thus, an isosceles $\triangle\text{ABC}$ whose base is 8cm and altitude 4cm is obtained.
Step 5. Extend BC to E such that $\text{BE}=\frac{3}{2}\text{BC}=\frac32\times8\text{cm}=12\text{cm}.$
Step 6. Draw EF || CA, cutting BA produced in F.
Here, $\triangle\text{BEF}$ is the required triangle similar to $\triangle\text{ABC}$ such that each side of $\triangle\text{BEF}$ is $1\frac12$ or $\Big(\frac32\Big)\text{times}$ the corresponding side of $\triangle\text{ABC}.$ View full question & answer→Question 335 Marks
Draw a circle of radius 4.8cm. Taking a point P on it. Without using the center of the circle, construct a tangent at the point P. Write the steps of construction.
Answer
Steps of construction:
- Taking O as centre and radius equal to 4.8cm draw acircle.
- Take any point P on the circle.
- Draw any chord PQ throut the given point P.
- Take a point R on the circle and P and Q to a point R.
- Construct $\angle\text{QPY}=\angle\text{PRQ}$ on the opposite side of the chord PQ.
- Produce YP to X to get YPX as the required tangent.
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