Question 14 Marks
Write the steps of construction for drawing a pair of tangents to a circle of radius $3\ cm$, which are inclined to each other at an angle of $60^\circ .$
Answer
Steps of Construction:
1. Taking $O$ on the plane of paper and draw a circle of radius $OA =3 cm$.
2. Produce $O A$ to $B$ such that $O A=A B=3 cm$.
3. Taking $A$ as the center draw a circle of radius $O A=A B=3 cm$. Supose it cuts the circle drawn in step 1 at $P$ and $Q .$
4. Join $B P$ and $B Q$ to get the desired tangents. View full question & answer→Question 24 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:
1. Draw a line segment $B C=9 cm$
2. Draw peependicular bisector $P Q$ of $B C$ Meeting it at $M$.
3. Fron QP cut-off a distance $M A=5 cm$.
4. Join $A B$ and $A C$.
Thus, isosceles $\triangle ABC$ is obtained.
5. Below $B C$, make an acute $\angle C B X$.
6. Along BX , mark off 4 points $B_1, B_2, B_3, B_4$, such that $B B_1,=B_1 B_2=B_1 B_3=B_3 B_4$.
7. Join $B_4 C$
8. From $B_3$, draw $B_3 C^{\prime} \| B_4 C$, meeting BC at $C ^{\prime}$.
9. From $C^{\prime}$, draw $C^{\prime} A^{\prime} \| C A$, meeting $A B$ at $A^{\prime}$.
Thus, $\triangle A ^{\prime} BC ^{\prime}$ is the required triangle similar to $\triangle ABC$ Such that each side of $\triangle A ^{\prime} BC ^{\prime}$ is $\frac{3}{4}$ times the corresponding side of $\triangle ABC$. View full question & answer→Question 34 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:
1. Drow a circle with the help of bangle.
2. Take a point $P$ outside the circle and take two chords $Q R$ and $S T$.
3. Draw perpendicular bisect of these chords.
4. Join $PO$ and bisect it, Let $U$ be the mid-point of $PO.$
5. Taking $U$ as centere, draw a circle of radius $OU$ , which will intersect the original circle at $V$ and $W .$
6. Join $PV$ and $PW$ are required taZngents.
View full question & answer→Question 44 Marks
Draw a circle of radius $3 \ cm$ . Draw a tangent to the circle making an angle of $30^{\circ}$ with a line passing through the centre.
Answer
Steps of Construction:
1. Draw a circle with center $O$ and radius $3 \ cm .$
2. Draw a radius $O A$ of this circle and product it to $B$.
3. Construct an $\angle AOP =60^{\circ}$ (Complement of $30^{\circ}$ )
4. Draw perpendicular to $OP$ at $P$ which intersects $OA$ produced at $Q $.
Thus, $PQ$ is the desired tangent such that $\angle OQP =30^{\circ}$. View full question & answer→Question 54 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 $A X$ mark 6 Point $A_1, A_2, A_3, A_4, A_5, A_6$, Such that $A A_1=A_1 A_2=A_2 A_3=A_3 A_4=A_4 A_5=A_5 A_6$.
- JJoin $\mathrm{A}_6 \mathrm{~B}$.
- 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 $A_6 B$.
Suppose these lines meet $A B$ at points $Q, R, S, T$ respectively. Thus, line segment $A B$ is divided into 6 equal parts such that $AT = TS = SR = RQ = QP = PB$. View full question & answer→Question 64 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 $AB = 7cm.$
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 $A A_1=A_1 A_2=A_2 A_3=A_3 A_4=A_4 A_5$
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 74 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 $B_1=B_1 B_2=B_2 B_3=B_3 B_4$
- 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 84 Marks
Construct a tangent to a circle of radius $4\ cm$ from a point on the concentric circle of radius $6\ cm$ and measure its length. Also, verify the measurement by actual calculation.
Answer
Steps of Construction:
1. Take a point $O$ on the plane of the paper and draw a circle of radius $O A=4 cm$. Also, draw a concentric circle of radius $O B=6 cm$.
2. Find the mid-point $C$ of $O B$ and draw a circle of radius $O C=B C$. Suppose this circle intersects the circle of radius $4 \ cm$ at $P$ and $Q$.
$3$. Join $B P$ and $B Q$ to get the desired tangents from a point Bon the circle of radius $6 \ cm .$
By actual measurment, we find that $B P=B Q=4.5 cm$
Verification:
In $\triangle\text{BOO},$ we have $OB = 6\ cm$ and $OP = 4\ cm$
$\therefore\text{OB}^2=\text{BP}^2+\text{OP}^2$
$\Rightarrow\text{BP}=\sqrt{\text{OB}^2-\text{OP}^2}$
$=\sqrt{36-16}=\sqrt{20}$
$=4.47\text{cm}=4.5\text{cm}$
Similarly, $\text{BQ} = 1.47\text{cm} = 4.5\text{cm}.$ View full question & answer→Question 94 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:
- Draw a line segment $AB = 6.5\ cm$
- Draw aray $AX$ making an acute $\angle\text{BAX}$ with $AB$
- Along $AX$ 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 $A A_1=A_1 A_2=A_2 A_3=A_3 A_4=A_4 A_5=A_5 A_6, \ldots A_{10} A_{11}$
- Join $A11B.$
- Through the point $A4$, draw a line parallel to $A_{11}B$ by making an angle equal to $\angle\text{AA}_{11}\text{B}$ at $A_4$. Suppose this line meets $AB$ at a point $C.$
Suppose this line meets $AB$ at a point $C.$
The Point $C$ so obtained is the required point, which divides $AB$ in the ratio $4 : 7.$ View full question & answer→Question 104 Marks
Draw two concentric circles of radii $4\ cm$ and $6\ cm$. Contruct a tangent to the smaller circle from a point on the larger circle. Measure the length of this tangent.
Answer
Steps of construction:
1. Take a point $O$ on the plane of the paper and draw a circle of radius $O A=4 cm$. Also, draw a concentric circle of radius $O B=6 cm$
2. Find the mid-point $C$ of $O B$ and draw a circle of radius $O C=B C$. Suppose this circle intrsects the circle of radius 4 cm at $P$ and $Q$.
3. Join $B P$ and $B Q$ to get the desired tangents from a point bon the circle of radius $6 \ cm .$
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 114 Marks
Construct a $\triangle\text{ABC},$ in which $BC = 5\ cm$, $\angle\text{C}=60^\circ$ and altitude from A is equal to $3\ cm$. 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 $BC = 5\ cm$
- Construct $\angle\text{BCP}=60^\circ$
- Draw a line $GH || BC$ at a distance of $3\ cm$, 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 124 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 $A X$ and five points $B_1, B_2, B_3, B_4, B_5$ on $B Y$ in such a way that $A A_1=A_1 A_2=A_2 A_3=A_3 A_4$.
Step 5. JJoin $A_4 B_5$
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 134 Marks
Draw two concentric circle of radii $3\ cm$ and $5\ cm$. Taking a point on the outer circle, construct the pair of tangents to the inner circle.
Answer
Steps of construction:
1. Draw a circle with radius $3 \ cm$ and centre $O .$
2. Draw another circle with radius $5 \ cm$ and same centre $O$.
3. Take a point $P$ on the circumference of larger circle and join $O$ to $p$.
4. Taking $O P$ as diameter draw another circle which intersects the smallest circle at $A$ and $B$.
5. Join $A$ to $P$ and $B$ to $P$.
Hence $A P$ and $B P$ are the required tangents. View full question & answer→Question 144 Marks
Draw a circle with centre $O$ and radius $4\ cm$. 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:
-
Draw a circle with centre $O$ and radius $4\ cm.$
-
Draw any diameter $AB.$
-
Draw line $\text{L}\perp\text{OA}$ such that $\angle\text{OAL}=90^\circ.$
-
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 154 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 $AB = 7.6\ cm$
Step 2. Draw a ray AX, making an acute angle $\angle{\text{BAX}}.$
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
$\mathrm{AA}_1=\mathrm{A}_1 \mathrm{~A}_2=\mathrm{A}_2 \mathrm{~A}_3=\mathrm{A}_3 \mathrm{~A}_4=\mathrm{A}_4 \mathrm{~A}_5=\mathrm{A}_5 \mathrm{~A}_6=\mathrm{A}_6 \mathrm{~A}_7=\mathrm{A}_7 \mathrm{~A}_8=\mathrm{A}_8 \mathrm{~A}_9=\mathrm{A}_9 \mathrm{~A}_{10}=\mathrm{A}_{10} \mathrm{~A}_{11}=\mathrm{A}_{11} \mathrm{~A}_{12}=\mathrm{A}_{12} \mathrm{~A}_{13}$
Step 4. Join $\mathrm{A}_{13} \mathrm{B}$.
Step 5. From $A_5$, draw $A_5 P$ parallel to $A_{13} B$ $\big($draw an angle equal to $\angle\text{AA}_{13}\text{B}\big),$ meeting AB 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 164 Marks
Draw a right triangle in which sides (other than hypotenuse) are of lengths $4\ cm$ and $3\ cm$. 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 $BC = 3\ cm.$
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 174 Marks
Draw a circle of radius $4\ cm$. Draw tangent to the circle making an angle of $60^\circ $ with a line passing through the centre.
Answer
Steps of construction:
1. Draw a circle with centre $O$ and radius $4 \ cm .$
2. Draw a radius $O A$ of the circle and produce it to $B$.
3. Construct an angle $\angle AOP$ equal to complement of $60^{\circ}$, i.e., equal to $30^{\circ}$.
4. Draw perpendicular to $O P$ at $P$ which intersects $O A$ produced at $Q$.
Thus, $PQ$ is the desired tangent such that $\angle OQP =60^{\circ}$. View full question & answer→Question 184 Marks
Draw two tangents to a circle of radius $3.5\ cm$ from a point $P$ at a distance of $6.2\ cm$ from its centre.
Answer
Steps of Construction:
1. Take a point $O$ on the plane of the paper and draw a circle of radius $3.5 cm .$
2. Mark a point $P$ at a distance of $6.2 cm$ from the center $O$ and Join $OP.$
3. Draw a right bisector of $O P$, intersecting $O P$ at $Q$.
4. Taking $Q$ as center and $OQ = PQ$ as radius, draw a circle to intersect the given circle at $T$ and $T ^{\prime}$.
5. Join $PT$ and $PT'$ to get the required tangents.
View full question & answer→MCQ 194 Marks
To construct a triangle similar to $\triangle\text{ABC}$ in which $BC = 4.5\ cm$, $\angle\text{B}=45^\circ$ and $\angle\text{C}=60^\circ,$ using a scale factor of $\frac35 , BC$ will be divided in the ratio.
AnswerTo construct a triangle similar to $\triangle\text{ABC}$ in which $BC = 4.5\ cm, \angle\text{B}=45^\circ$ 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 $D.$ View full question & answer→Question 204 Marks
Construct a $\triangle\text{ABC},$ with $BC = 7\ cm$, $\angle\text{B}=60^\circ$ and $AB = 6\ cm$. 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 $BC = 7\ cm.$
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 $B Y$, mark four points $B_1, B_2, B_3$ and $B_4$ such that $B B_1=B_1 B_2=B_2 B_3=B_3 B_4$.
Step 7. Join $\mathrm{CB}_4$.
Step 8. From $\mathrm{B}_3$, draw $\mathrm{B}_3 \mathrm{C}^{\prime} \| \mathrm{CB}_4$ meeting BC at $\mathrm{C}^{\prime}$.
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 214 Marks
Construct a $\triangle\text{ABC}$ in which $BC = 8\ cm$, $\angle\text{B}=45^\circ$ 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 $BC = 8\ cm.$
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 $B Z$, mark five points $Z_1, Z_2, Z_3, Z_4$ and $Z_5$ such that $B Z_1=Z_1 Z_2=Z_2 Z_3=Z_3 Z_4=Z_4 Z_5$.
Step 6. Join $\mathrm{CZ}_5$.
Step 7. From $Z_3$, draw $Z_3 C^{\prime} \| C Z_5$ meeting $B C$ at $C^{\prime}$.
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 224 Marks
Write the step of construct the tangents to a circle from an external point.
Answer
Steps of construction:
1. Take given circle and a point $P$ outside the circle. $O$ is centre of the circle.
2. Joint $OP.$
3. Bisect $OP$ and get its mid-point $M$.
4. Draw circle with centre $M$ and radius $=P M=M O$.
5. Circle drawn meets the given circle at $Q$ above $P O$ and at $Q^{\prime}$ below $P O$.
6. Join $P Q$ and $P Q^{\prime}$.
7. $P Q$ and $P Q^{\prime}$ are the required tangents drawn to the circle from the point $P$.
We observe that $PQ = PQ ^{\prime}$. View full question & answer→Question 234 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 $3\ cm$, 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 244 Marks
Draw a circle of radius $3.5\ cm$. Draw a pair of tangent to this circle which are inclined to each other at an angle of $60^\circ .$ Write the steps of construction.
Answer
Steps of construction:
1. Taking $O$ on the plane of paper and draw a circle of radius $O A=3.5 cm$.
2. Produce $O A$ to $B$ such that $O A=A B=3.5 cm$.
3. Taking $A$ as the center draw a circle of radius $A O=A B=3.5 cm$ Supose it cuts the circle draw in step 1 at $P$ and Q.
4. Join $B P$ and $B Q$ to get the desired tangents. View full question & answer→Question 254 Marks
Draw a circle of radius $3.5\ cm$. Take two point $A$ and $B$ on one of its extended diameter, each at a distance of $5\ cm$ from its centre. Draw tangents to the circle from each of these points $A$ and $B.$
Answer
Steps of Construction:
1. Draw a line segment $P Q$ of length $10 cm .$
2. Take the mid-point $O$ of $P Q$.
3. Draw the perpendicular bisectors of $P O$ and $O Q$ which intersect $P O$ at point $R$ and $O Q$ at point $S$.
4. With center $R$ and radius $RP$ draw a circle.
5. With center $S$ and radius $SQ$ draw a circle.
6. With center $O$ and radius $3.5 \ cm$ draw another circle which intersects the previous circles at the points $A , B , C$ and $D$ respectively.
7. Join $PA, PB, QC$ and $QD.$
Thus, $PA , PB , QC$ and $QD$ are the reduired tangents.
View full question & answer→Question 264 Marks
Construct a $\triangle\text{ABC}$ in which $AB = 6\ cm,$ $\angle\text{A}=30^\circ$ and $\angle\text{B}=60^\circ.$ Construct another $\triangle\text{AB}'\text{C}'$similar to $\triangle\text{ABC}$ with base $AB' = 8\ cm.$
Answer
Steps of Construction:
Step 1. Draw a line segment $A B=6 cm$.
Step 2. At A , draw $\angle XAB =30^{\circ}$
Step 3. At $B$, draw $\angle Y B A=60^{\circ}$. Suppose $A X$ and $B Y$ intersect at $C$.
Thus, $\triangle ABC$ is the required triangle.
Step 4. Produce $A B$ to $B^{\prime}$ such that $A B^{\prime}=8 cm$.
Step 5. From $B^{\prime}$, draw $B^{\prime} C^{\prime} \| B C$ meeting $A X$ at $C^{\prime}$.
Here, $\triangle AB ^{\prime} C ^{\prime}$ is the required triangle similar to $\triangle ABC$. View full question & answer→Question 274 Marks
Draw a line segment $AB$ of length $8\ cm$. Taking A as centre, draw a circle of radius $4\ cm$ and taking $B$ as centre, draw another circle of radius $3\ cm$. Construct tangents to each circle from the centre of the other circle.
Answer
Step of construction:
1. Draw a line segment $A B$ of length $8 \ cm .$
2. Draw the perpendicular bisector of $A B$ which intersect it at $C$.
3. With center $C$ and radius $C A$, draw a circle.
4. With centers $A$ and $B$ and radii $4 \ cm$ and $3 \ cm$ , draw two circle which intersect the previous circle at the points $P, Q, R$ and $S$. at the points $P, Q, R$ and $S$.
5. Join $A R, A S, B P$ and $B Q$.
Thus, $A R, A S, B P$ and $B Q$ are the required tangents. View full question & answer→Question 284 Marks
Draw a $\triangle\text{ABC},$ right-angled at $B$ such that $AB = 3\ cm$ and $BC = 4\ cm$. 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 $BC = 4\ cm$
- AT $B$, construct $\angle\text{MBC}=90^\circ.$
- Cut-off $BA = 3\ cm$ from $BM.$
- Join $AC.$
Thus, right-angled $\triangle\text{ABC}$ is obtained.
- Below BC, make an acute $\angle\text{CBX}.$
- Along $B X$, mark off 7 points $R_1, R_2, R_3, R_4, R_5, R_6, R_7$ such that $B R_1=R_1 R_2=R_2 R_3=R_3 R_4=\ldots=R_6 R_7$
- Join $\mathrm{R}_5 \mathrm{C}$.
- From $R_7$, draw $R_7 C_1 \| R_5 C$, meeting $B C$ produced at $C_1$.
- From $C_1$, draw $C_1 A_1 \| C A$, meeting $B A$ 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 294 Marks
Draw a circle of radius $3\ cm$. From a point P, $7\ cm$ away from the centre of the circle, draw two tangents to the circle. Also, measure the lengths of the tangents.
Answer
Steps of Construction:
1. Take a point $O$ on the plane of the paper and draw a circle of radius $3 cm .$
2. Mark a point $P$ at a distance od 7 cm from the center $O$ and Join $O P$.
3. Draw a right visector of $O P$, intersecting $O P$ at $Q$.
4. Taking $Q$ as center and $O Q-P Q$ as radius, draw a circle to intersect the given circle at $T$ and $T^{\prime}$.
5. Join $PT$ and $PT$' to get the required tangents.
By measurment,
$PT = PT' = 6.1cm$ View full question & answer→Question 304 Marks
Draw a circle of radius $4.2\ cm$. Draw a pair of tangents to this circle inclined to each other at an angle of $45^\circ .$
Answer
Steps of Construction:
1. Draw a ciecle with center $O$ and radius $4.5 cm$
2. Draw diameter $A B$
3. With $OB$ as base, draw $\angle BOC =45^{\circ}$
4. At A , draw a line perendicular to $O A$.
5. At $C$, draw a line perpendicular to $O C$.
These lines intersect each other at $P$.
Thus, $PA$ and $PC$ are the required tangents. View full question & answer→Question 314 Marks
Construct a $\triangle\text{PQR},$ in which $PQ = 6\ cm, QR = 7\ cm$ and $PR = 8\ cm$. 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 $Q X$, mark five points $R_1, R_2, R_3, R_4$ and $R_5$ such that $Q R_1=R_1 R_2=R_2 R_3=R_3 R_4=R_4 R_5$.
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 324 Marks
Construct an isosceles triangle whose base is $8\ cm$ and altitude $4\ cm$ 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 $4\ cm$, 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 334 Marks
Draw a circle of radius $4.8\ cm$. 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:
1. Taking $O$ as centre and radius equal to $4.8 \ cm$ draw acircle.
2. Take any point $P$ on the circle.
3. Draw any chord $P Q$ throut the given point $P$.
4. Take a point $R$ on the circle and $P$ and $Q$ to a point $R$.
5. Construct $\angle QPY =\angle PRQ$ on the opposite side of the chord PQ .
6. Produce $Y P$ to $X$ to get $Y P X$ as the required tangent. View full question & answer→