5 Marks Questions

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.

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Question 25 Marks

Construct an isosceles triangle whose base is 9cm and altitude 5cm. 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 = 9cm
Draw peependicular bisector PQ of BC Meeting it at M.
Fron QP cut-off a distance MA = 5cm.
Join AB and AC.

Thus, isosceles $\triangle\text{ABC}$ is obtained.

Below BC, make an acute $\angle\text{CBX}.$
Along $B X$, mark off 4 points $B_1, B_2, B_3, B_4$, such that $B_1, B_1 B_2=B_1 B_3=B_3 B_4$.
Join $\mathrm{B}_4 \mathrm{C}$
From $B_3$, draw $B_3 C^{\prime} \| B_4 C$, meeting $B C$ at $C^{\prime}$.
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 $\triangle\text{ABC}.$

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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:

Drow a circle with the help of bangle.
Take a point P outside the circle and take two chords QR and ST.
Draw perpendicular bisect of these chords.
Join PO and bisect it, Let U be the mid-point of PO.
Taking U as centere, draw a circle of radius OU, which will intersect the original circle at V and W.
Join PV and PW are required tangents.

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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 $\angle\text{OQP}=30^\circ.$

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Question 55 Marks

Draw a line segment AB of length 5.4cm. Divide it into six equal parts. Write the steps of construction.

Answer

Steps of Construction:

Draw a line segment AB = 5.4cm
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 $A A_1=A_1 A_2=A_2 A_3=A_3 A_4=A_4 A_5=A_5 A_6$.
Join $\mathrm{A}_6 \mathrm{~B}$.
. Through the point $\mathrm{A}_5$, draw a line parallel to $\mathrm{A}_6 \mathrm{~B}$ by making an angle equal to $\angle \mathrm{AA}_6 \mathrm{~B}$ at $\mathrm{A}_5$. Suppose this line meets $A B$ 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 AB at points Q, R, S, T respectively.
Thus, line segment AB is divided into 6 equal parts such that
$AT = TS = SR = RQ = QP = PB.$

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Question 65 Marks

Draw a line segment AB of length 7cm. 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 $A X$, mark 5 points (greater of 3 and 5 ) $A_1, A_2, A_3, A_4$ and $A_5$ such that $A A_1=A_1 A_2=A_2 A_3=A_3 A_4=A_4$ $\mathrm{A}_5$
Step 4. Join $\mathrm{A}_5 \mathrm{~B}$.
Step 5. From $A_3$, draw $A_3 P$ parallel to $A_5 B$ (draw an angle equal to $\angle A_5 B$ ), meeting $A B$ in $P$.
Here, P is the point on AB such that $\frac{\text{AP}}{\text{PB}}=\frac32$ or $\frac{\text{AP}}{\text{AB}}=\frac35.$

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Question 75 Marks

Construct a $\triangle\text{ABC},$ in which BC = 6.5cm, AB = 4.5cm 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 = 6.5cm
At B, construct $\angle\text{CBX}=60^\circ$
With B as center and radius 4.5cm, draw an arc intersecting BX at A
Join AC to obtain $\triangle\text{ABC}$
Below BC, make an acute $\angle\text{CBY}$
Along $B Y$, mark off 4 points (greater of 3 and in $\frac{3}{4}$ ) $B_1, B_2, B_3, B_4$ such that $B B_1=B_1 B_2=B_2 B_3=B_3 B_4$
Join $B_4 C$
From point $B_3$, draw a line parallel to $B_4 C$ intersecting $B C$ at $C^{\prime}$
From Point C', draw a line parallel to AC intersecting AB at A'

Thus, $\triangle\text{A}'\text{BC}'$ is the required triangle.

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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, $\text{BQ} = 1.47\text{cm} = 4.5\text{cm}.$

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Question 95 Marks

Draw a line segment AB of length 6.5cm and divide it in the ratio 4 : 7. Measure each of the two parts.

Answer

Steps of Construction:

Draw a line segment AB = 6.5cm
Draw aray AX making an acute $\angle\text{BAX}$ with AB
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 $A A_1=A_1 A_2=A_2 A_3=A_3 A_4=A_4 A_5=A_5 A_6 \ldots A_{10} A_{11}$
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.

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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.

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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 $\triangle\text{ABC}$

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Question 125 Marks

Draw a line segment of length 8cm and divide it internally in the ratio 4 : 5.

Answer

Step Of Construction:
Step 1. Draw a line segment $A B=8 cm$.
Step 2. Draw a ray $A X$ making an acute angle $\angle B A X=60^{\circ}$ with $A B$.
Step 3. Draw a ray $B Y$ parallel to $A X$ by making an acute angle $\angle A B Y=\angle B A X$.
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. Join $A_4 B_5$
Step 6. Let this line intersect $A B$ at a point $P$. Thus, $P$ is the point dividing the line segment $A B$ internally in the ratio of $4: 5$.

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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.

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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:

Draw a circle with centre O and radius 4cm.
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.

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Question 155 Marks

Draw a line segment of length 7.6cm and divide it in the ratio 5 : 8. Measure the two parts.

Answer

Steps of Construction:
Step 1. Draw a line segment AB = 7.6cm
Step 2. Draw a ray AX, making an acute angle $\angle{\text{BAX}}.$
Step 3. Along $A X$, mark $\left(5+8=13\right.$ 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 $A_1=A_1 A_2=A_2 A_3=A_3 A_4=A_4 A_5=A_5 A_6=A_6 A_7=A_7 A_8=A_8 A_9=A_9 A_{10}=A_{10} A_{11}=A_{11} A_{12}=A_{12} A_{13}$
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 AP = 2.9cm (Approx)
Length of BP = 4.7cm (Approx)

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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 $\triangle\text{ABC}.$

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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 $\angle\text{OQP}=60^\circ.$

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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.
Mark a point P at a distance of 6.2cm from the center O and Join OP.
Draw a right bisector 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.

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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
4 : 7
3 : 10
3 : 7

Answer

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.

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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 $B_4$ such that $B_1=B_1 B_2=B_2 B_3=B_3 B_4$.
Step 7. Join $\mathrm{CB}_4$.
Step 8. From $B_3$, draw $B_3 C^{\prime} \| C B_4$ meeting $B C$ at $C^{\prime}$.
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 $\triangle\text{ABC}.$

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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.$\text { Along } B Z \text {, mark five points } Z_1, Z_2, Z_3, Z_4 \text { and } Z_5 \text { such that } B Z_1=Z_1 Z_2=Z_2 Z_3=Z_3 Z_4=Z_4 Z_5 \text {. }$
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 $\triangle\text{ABC}.$

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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’.

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Question 235 Marks

Construct a triangle with sides 5cm, 6cm and 7cm 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 = 4cm.
Step 2. With B as centre, draw an angle of $90^o.$
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.

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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.

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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.
Take the mid-point O of PQ.
Draw the perpendicular bisectors of PO and OQ which intersect PO at point R and OQ at point S.
With center R and radius RP draw a circle.
With center S and radius SQ draw a circle.
With center O and radius 3.5cm draw another circle which intersects the previous circles at the points A, B, C and D respectively.
Join PA, PB, QC and QD.

Thus, PA, PB, QC and QD are the reduired tangents.

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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 $\triangle\text{ABC}.$

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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.
Draw the perpendicular bisector of AB which intersect it at C.
With center C and radius CA, draw a circle.
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.
Join AR, AS, BP and BQ.

Thus, AR, AS, BP and BQ are the required tangents.

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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 $\angle\text{CBX}.$
6. Along $B X$, mark off 7 points $R_1, R_2, R_3, R_4, R_5, R_6, R_7$ such that $B R_1=R_1 R_2=R_2 R_3=R_3 R_4=\ldots=R_6 R_7$
7. Join $R _5 C$.
8. From $R_7$, draw $R_7 C_1 \| R_5 C$, meeting $B C$ produced at $C_1$.
9. 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 $\triangle\text{ABC}.$

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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

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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.

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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 $Q R=7 cm$.
Step 2. With $Q$ as centre and radius 6 cm , draw an arc.
Step 3. With $R$ as centre and radius 8 cm , draw an arc cutting the previous arc at $P$.
Step 4. Join $P Q$ and $P R$. Thus, $\triangle P Q R$ is the required triangle.
Step 5. Below $Q R$, draw an acute angle $\angle R Q X$.
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 RR5.
Step 8. From $R_4$, draw $R_4 R^{\prime} \| R R_5$ meeting $Q R$ at $R^{\prime}$.
Step 9. From $R^{\prime}$, draw $P^{\prime} R^{\prime} \| P R$ meeting $P Q$ in $P^{\prime}$. Here, $\triangle P^{\prime} Q R^{\prime}$ is the required triangle, each of whose sides are $\frac{4}{5}$ times the corresponding sides of $\triangle PQR$.

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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 $\triangle\text{ABC}.$

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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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