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#### Page No 522:

#### Answer:

**Steps of Construction:**

**Step 1**. Draw a line segment AB = 7 cm.

**Step 2**. Draw a ray AX, making an acute angle $\angle $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_{1}A_{2} = A_{2}A_{3 }= A_{3}A_{4} = A_{4}A_{5}

**Step 4**. Join A_{5}B.

**Step 5**. From A_{3}, draw A_{3}P parallel to A_{5}B (draw an angle equal to $\angle $AA_{5}B), meeting AB in P.

Here, P is the point on AB such that $\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{3}{2}$ or $\frac{\mathrm{AP}}{\mathrm{AB}}=\frac{3}{5}$.

#### Page No 522:

#### Answer:

(i) **Steps of Construction:**

**Step 1**. Draw a line segment AB = 8 cm

**Step 2**. Draw a ray AX, making an acute angle $\angle $BAX.

**Step 3**. Along AX, mark (4 + 5 =) 9 points A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, A_{7}, A_{8} and A9 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}

**Step 4**. Join A_{9}B.

**Step 5**. From A_{4}, draw A_{4}D parallel to A_{9}B (draw an angle equal to $\angle $AA_{9}B), meeting AB in D.

Here, D is the point on AB which divides it in the ratio 4 : 5.

(ii) **Steps of Construction:**

**Step 1**. Draw a line segment AB = 7.6 cm

**Step 2**. Draw a ray AX, making an acute angle $\angle $BAX.

**Step 3**. Along AX, 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 $AA_{13}B), meeting AB in P.

Here, P is the point on AB which divides it in the ratio 5 : 8.

∴ Length of AP = 2.9 cm (Approx)

Length of BP = 4.7 cm (Approx)

#### Page No 522:

#### Answer:

**Steps of Construction**

**Step 1**. Draw a line segment QR = 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 PQ and PR. Thus, ∆PQR is the required triangle.

**Step 5**. Below QR, draw an acute angle $\angle $RQX.

**Step 6**. Along QX, mark five points R_{1}, R_{2}, R_{3}, R_{4 }and R_{5 }such that QR_{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_{4}R' || RR_{5} meeting QR at R'.

**Step 9**. From R', draw P'R' || PR meeting PQ in P'.

Here, ∆P'QR' is the required triangle, each of whose sides are $\frac{4}{5}$ times the corresponding sides of ∆PQR.

#### Page No 522:

#### Answer:

Steps of Construction :

Step 1. Draw a line segment BC = 4 cm.

Step 2. With B as centre, draw an angle of 90^{o}.

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, △ ABC is obtained .

Step 5. Extend BC to D, such that BD = $\frac{7}{5}$ BC = $\frac{7}{5}$ (4) cm = 5.6 cm.

Step 6. Draw DE ∥ CA, cutting AB produced to E.

Thus, △EBD is the required triangle, each of whose sides is $\frac{7}{5}$ the corresponding sides of ∆*ABC*.

#### Page No 523:

#### Answer:

Given: In Δ*ABC*,

*BC *= 6 cm,

∠*B* = 60°

*AB *= 5 cm

Steps of construction:

(1) Draw a line segment *AB* = 5 cm.

(2) From the point *B*, draw an ∠*ABY* = 60°

(3) Taking *B* as center, 6 cm radius, draw an arc on the ray *BY*,

Let the point where the arc intersects the ray named as *C*.

(4) Join *AC*.

Hence, Δ*ABC* is the required triangle.

Now, we construct another triangle whose sides are $\frac{3}{4}$ times the corresponding sides of Δ*ABC*.

Steps of construction:

(1) Draw any ray *BX* making an acute angle with *BA* on the side opposite to the vertex *C*.

(2) Mark four points *B*_{1}, *B*_{2}, *B*_{3 }and *B*_{4} on *BX*, so that *BB*_{1} = *B*_{1}*B*_{2} = *B*_{2}*B*_{3} = *B*_{3}*B*_{4}.

(3) Join *B*_{4}*A* and draw a line through *B*_{3 }parallel to *B*_{4}*A *to intersect *AB* at* A'*.

(4) Draw a line through *A'* parallel to *AC *to intersect *BC* at *C'*.

Hence, Δ*A'BC'* is the required triangle.

#### Page No 523:

#### Answer:

**Steps of Construction**

**Step 1**. Draw a line segment AB = 6 cm.

**Step 2**. At A, draw $\angle $XAB = 30º.

**Step 3**. At B, draw $\angle $YBA = 60º. Suppose AX and BY intersect at C.

Thus, ∆ABC is the required triangle.

**Step 4**. Produce AB to B' such that AB' = 8 cm.

**Step 5**. From B', draw B'C' || BC meeting AX at C'.

Here, ∆AB'C' is the required triangle similar to ∆ABC.

#### Page No 523:

#### Answer:

**Steps of Construction**

**Step 1**. Draw a line segment BC = 8 cm.

**Step 2**. At B, draw $\angle $XBC = 45º.

**Step 3**. At C, draw $\angle $YCB = 60º. Suppose BX and CY intersect at A.

Thus, ∆ABC is the required triangle.

**Step 4**. Below BC, draw an acute angle $\angle $ZBC.

**Step 5**. Along BZ, mark five points Z_{1}, Z_{2}, Z_{3}, Z_{4 }and Z_{5 }such that BZ_{1} = Z_{1}Z_{2} = Z_{2}Z_{3} = Z_{3}Z_{4} = Z_{4}Z_{5}.

**Step 6**. Join CZ_{5}.

**Step 7**. From Z_{3}, draw Z_{3}C' || CZ_{5} meeting BC at C'.

**Step 8**. From C', draw A'C' || AC meeting AB in A'.

Here, ∆A'BC' is the required triangle whose sides are $\frac{3}{5}$ of the corresponding sides of ∆ABC.

#### Page No 523:

#### Answer:

To construct a triangle similar to ∆ABC in which BC = 4.5 cm, $\angle $B = 45º and $\angle $C = 60º, using a scale factor of $\frac{3}{7}$, BC will be divided in the ratio 3 : 4.

Here, ∆ABC ∼ ∆A'BC'

BC' : C'C = 3 : 4

or BC' : BC = 3 : 7

Hence, the correct answer is option A.

#### Page No 523:

#### Answer:

**Steps of Construction**

**Step 1**. Draw a line segment BC = 8 cm.

**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 ∆ABC whose base is 8 cm and altitude 4 cm is obtained.

**Step 5**. Extend BC to E such that BE = $\frac{3}{2}$BC = $\frac{3}{2}\times 8$ cm = 12 cm.

**Step 6**. Draw EF || CA, cutting BA produced in F.

Here, ∆BEF is the required triangle similar to ∆ABC such that each side of ∆BEF is $1\frac{1}{2}\left(\mathrm{or}\frac{3}{2}\right)$ times the corresponding side of ∆ABC.

#### Page No 523:

#### Answer:

**Steps of Construction**

**Step 1**. Draw a line segment BC = 3 cm.

**Step 2**. At B, draw $\angle $XBC = 90º.

**Step 3**. With B as centre and radius 4 cm, draw an arc cutting BX at A.

**Step 4**. Join AC. Thus, a right ∆ABC is obtained.

**Step 5**. Extend BC to D such that BD = $\frac{5}{3}$BC = $\frac{5}{3}\times 3$ cm = 5 cm.

**Step 6**. Draw DE || CA, cutting BX in E.

Here, ∆BDE is the required triangle similar to ∆ABC such that each side of ∆BDE is $\frac{5}{3}$ times the corresponding side of ∆ABC.

#### Page No 529:

#### Answer:

**Steps of Construction**

**Step 1**. Draw a circle with O as centre and radius 3 cm.

**Step 2**. Mark a point P outside the circle such that OP = 7 cm.

**Step 3**. Join OP. Draw the perpendicular bisector XY of OP, cutting OP at Q.

**Step 4**. Draw a circle with Q as centre and radius PQ (or OQ), to intersect the given circle at the points T and T'.

**Step 5**. Join PT and PT'.

Here, PT and PT' are the required tangents.

PT = PT' = 6.3 cm (Approx)

#### Page No 529:

#### Answer:

**Steps of Construction**

**Step 1**. Draw a circle with O as centre and radius 3.5 cm.

**Step 2**. Mark a point P outside the circle such that OP = 6.2 cm.

**Step 3**. Join OP. Draw the perpendicular bisector XY of OP, cutting OP at Q.

**Step 4**. Draw a circle with Q as centre and radius PQ (or OQ), to intersect the given circle at the points T and T'.

**Step 5**. Join PT and PT'.

Here, PT and PT' are the required tangents.

#### Page No 529:

#### Answer:

**Steps of Construction**

**Step 1**. Draw a circle with O as centre and radius 3 cm.

**Step 2**. Mark a point P and Q on one of its diameters extended on both sides outside the circle such that OP = OQ = 7 cm.

**Step 3**. Join OP and OQ. Draw the perpendicular bisector XY of OP and X'Y' of OQ, cutting OP at L and OQ at M.

**Step 4**. Draw a circle with L as centre and radius PL (or OL), to intersect the given circle at the points A and B. Draw another circle with M as centre and radius MQ (or OM), to intersect the given circle at the points C and D.

**Step 5**. Join PA and PB. Join QC and QD.

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

#### Page No 529:

#### Answer:

**Steps of Construction**

**Step 1**. Draw a circle with centre O and radius 4 cm.

**Step 2**. Draw any diameter AOB of the circle.

**Step 3**. At A, draw $\angle $OAX = 90º. Produce XA to Y.

**Step 4**. At B, draw $\angle $OBX' = 90º. Produce X'B to Y'.

Here, XAY and X'BY' are the tangents to the circle at the end points of the diameter AB.

#### Page No 530:

#### Answer:

**Steps of Construction**

**Step 1**. Draw a circle with the help of a bangle.

**Step 2**. Mark a point P outside the circle.

**Step 3**. Through P, draw a secant PAB to intersect the circle at A and B.

**Step 4**. Produce AP to C such that PA = PC.

**Step 5**. Draw a semicircle with CB as diameter.

**Step 6**. Draw PD ⊥ BC, intersecting the semicircle at D.

**Step 7**. With P as centre and PD as radius, draw arcs to intersect the circle at T and T'.

**Step 8**. Join PT and PT'.

Here, PT and PT' are the required pair of tangents.

#### Page No 530:

#### Answer:

**Steps of Construction**

**Step 1**. Draw a line segment AB = 8 cm.

**Step 2**. With A as centre and radius 4 cm, draw a circle.

**Step 3**. With B as centre and radius 3 cm, draw another circle.

**Step 4**. Draw the perpendicular bisector XY of AB, cutting AB at C.

**Step 5**. With C as centre and radius AC (or BC), draw a circle intersecting the circle with centre A at P and P'; and the circle with centre B at Q and Q'.

**Step 6**. Join BP and BP'. Also, join AQ and AQ'.

Here, AQ and AQ' are the tangents from A to the circle with centre B. Also, BP and BP' are the tangents from B to the circle with centre A.

#### Page No 530:

#### Answer:

Steps of Construction:

Step 1. Draw a circle with centre O and radius = 4.2 cm.

Step 2. Draw any diameter AOB of this circle.

Step 3. Construct $\angle BOC=45\xb0$, such that the radius OC meets the circle at C.

Step 4. Draw AM $\perp AB\mathrm{and}CN\perp OC.$

AM and CN intersect at P.

Thus, PA and PC are the required tangents to the given circle inclined at an angle of 45^{o}.

#### Page No 530:

#### Answer:

**Steps of Construction**

**Step 1**. Draw a circle with centre O and radius 3 cm.

**Step 2**. Draw any diameter AOB of the circle.

**Step 3**. Construct $\angle $BOC = 60º such that radius OC cuts the circle at C.

**Step 4**. Draw AM ⊥ AB and CN ⊥ OC. Suppose AM and CN intersect each other at P.

Here, AP and CP are the pair of tangents to the circle inclined to each other at an angle of 60º.

#### Page No 530:

#### Answer:

Steps Of construction:

Step 1. Draw a circle with centre O and radius 3 cm.

Step 2. Draw radius OA and produce it to B.

Step 3. Make $\angle AOP=60\xb0$.

Step 4. Draw PQ$\perp OP$, meeting OB at Q.

Step 5. Then, PQ is the desired tangent, such that $\angle OQP=30\xb0$.

#### Page No 530:

#### Answer:

**Steps of Construction**

**Step 1**. Mark a point O on the paper.

**Step 2**. With O as centre and radii 4 cm and 6 cm, draw two concentric circles.

**Step 3**. Mark a point P on the outer circle.

**Step 4**. Join OP.

**Step 5**. Draw the perpendicular bisector XY of OP, cutting OP at Q.

**Step 6**. Draw a circle with Q as centre and radius OQ (or PQ), to intersect the inner circle in points T and T'.

**Step 7**. Join PT and PT'.

Here, PT and PT' are the required tangents.

PT = PT' = 4.5 cm (Approx)

**Verification by actual calculation**

Join OT to form a right ∆OTP. (Radius is perpendicular to the tangent at the point of contact)

In right ∆OTP,

${\mathrm{OP}}^{2}={\mathrm{OT}}^{2}+{\mathrm{PT}}^{2}\left(\mathrm{Pythagoras}\mathrm{Theorem}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{PT}=\sqrt{{\mathrm{OP}}^{2}-{\mathrm{OT}}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{PT}=\sqrt{{6}^{2}-{4}^{2}}=\sqrt{36-16}=\sqrt{20}\approx 4.5\mathrm{cm}\left(\mathrm{OP}=6\mathrm{cm}\mathrm{and}\mathrm{OT}=4\mathrm{cm}\right)$

#### Page No 530:

#### Answer:

**Steps of Construction**

**Step 1**. Mark a point O on paper.

**Step 2**. Taking O as a centre, draw two concentric circles of radius 3 cm and 5 cm. Mark any random point P on outer circle.

**Step 3**. Join OP and draw its perpendicular bisector which meets OP at M.

**Step 4**. Draw a circle with M as centre and radius PM (or OM), to intersect the inner circle at the points A and B.

**Step 5**. Join PA and PB.

Here, PA and PB are the required tangents.

#### Page No 530:

#### Answer:

**Steps of Construction**

**Step 1**. Draw a circle with O as centre and some radius.

**Step 2**. Mark a point P outside the circle. Join OP.

**Step 3**. Draw the perpendicular bisector of OP cutting it at M.

**Step 4**. Draw another circle with M as centre and radius MP (or OM), to intersect the given circle at the points A and B.

**Step 5**. Join PA and PB.

Here, PA and PB are the required tangents.

#### Page No 530:

#### Answer:

Steps of Construction :

Step 1 . Draw a line segment AB = 5.4 cm.

Step 2. Draw a ray AX, making an acute angle, $\angle BAX$.

Step 3. Along AX, mark 6 points A_{1}, A_{2}_{,} A_{3}_{, }A_{4}_{,} A_{5}, A_{6}_{ }such that,

AA_{1} = A_{1}A_{2} = A_{2}A_{3} = A_{3}A_{4} = A_{4}A_{5}_{ }= A_{5}A_{6} .

Step 4. Join A_{6}B.

Step 5. Draw A_{1}C _{2}D, A_{3}D, A_{4}F and A_{5}G .

Thus, AB is divided into six equal parts.

#### Page No 530:

#### Answer:

Steps of Construction :

Step 1 . Draw a line segment AB = 6.5 cm.

Step 2. Draw a ray AX, making an acute angle $\angle BAX$.

Step 3. Along AX, mark (4+7) =11 points A_{1}, A_{2}_{,} A_{3}_{, }A_{4}_{,} A_{5}, A_{6}, A_{7}, A_{8}, A_{9}, A_{10}, A_{11}_{, }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}

Step 4. Join A_{11}B.

Step 5. From A_{4}_{, }draw A_{4}C $\parallel $A_{11}B, meeting AB at C.

Thus, C is the point on AB, which divides it in the ratio 4:7.

Thus, AC : CB = 4:7

From the figure, AC = 2.36 cm

CB = 4.14 cm

#### Page No 531:

#### Answer:

Steps of Construction :

Step 1. Draw a line segment BC = 6.5 cm.

Step 2. With B as centre, draw an angle of 60^{o}.

Step 3. With B as centre and radius equal to 4.5 cm, draw an arc, cutting the angle at A.

Step 4. Join AB and AC.

Thus, △ ABC is obtained .

Step 5. Below BC, draw an acute $\angle CBX$.

Step 6. Along BX, mark off four points B_{1}, B_{2}, B_{3}, B_{4}, such that BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4}_{ .}

Step 7. Join B_{4}C.

Step 8. From B_{3}, draw B_{3}D ∥ B_{4}C, meeting BC at D.

Step 9. From D, draw DE ∥ CA, meeting AB at E.

Thus, △ EBD is the required triangle, each of whose sides is$\frac{3}{4}$ the corresponding sides of ∆*ABC*.

#### Page No 531:

#### Answer:

Steps of Construction :

Step 1. Draw a line* l* .

Step 2. Draw an angle of 90^{o} at M on *l* .

Step 3. Cut an arc of radius 3 cm on the perpendicular. Mark the point as A.

Step 4. With A as centre, make an angle of 30^{o}^{ }and let it cut *l *at C. We get $\angle ACB=60\xb0$.

Step 5. Cut an arc of 5 cm from C on* l *and mark the point as B.

Step 6. Join AB.

Thus, △ABC is obtained .

Step 7. Extend AB to D, such that BD =

Step 8. Draw DE

Then, △ADE is the required triangle, each of whose sides is $\frac{3}{2}$of the corresponding sides of △ABC.

#### Page No 531:

#### Answer:

Steps of Construction :

Step 1. Draw a line segment BC = 9 cm.

Step 2. With B as centre, draw an arc each above and below BC.

Step 3. With C as centre, draw an arc each above and below BC.

Step 4. Join their points of intersection to obtain the perpendicular bisector of BC. Let it intersect BC at D.

Step 5. From D, cut an arc of radius 5 cm and mark the point as A.

Step 6. Join AB and AC.

Thus, △ABC is obtained .

Step 5. Below BC, make an acute $\angle CBX$.

Step 6. Along BX, mark off four points B_{1}, B_{2}, B_{3}, B_{4}, such that BB_{1}=B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4}_{.}

Step 7. Join B_{4}C.

Step 8. From B_{3}, draw B_{3}E ∥ B_{4}C, meeting BC at E.

Step 9. From E, draw EF ∥ CA, meeting AB at F.

Thus, △FBE is the required triangle, each of whose sides is $\frac{3}{4}$ the corresponding sides of the first triangle.

#### Page No 531:

#### Answer:

Steps of Construction :

Step 1. Draw a line segment BC = 4 cm.

Step 2. With B as centre, draw an angle of 90^{o}.

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, △ ABC is obtained .

Step 5. Extend BC to D, such that BD = $\frac{7}{5}$ BC = $\frac{7}{5}$ (4) cm = 5.6 cm.

Step 6. Draw DE ∥ CA, cutting AB produced to E.

Thus, △EBD is the required triangle, each of whose sides is $\frac{7}{5}$ the corresponding sides of ∆*ABC*.

#### Page No 531:

#### Answer:

Steps of Construction :

Step 1. Draw a circle of radius 4.8 cm.

Step 2. Mark a point P on it.

Step 3. Draw any chord PQ.

Step 4. Take a point R on the major arc QP.

Step 5. Join PR and RQ.

Step 6. Draw $\angle QPT=\angle PRQ$

Step 7. Produce TP to T', as shown in the figure.

T'PT is the required tangent.

#### Page No 531:

#### Answer:

Steps of Construction:

Step 1. Draw a circle with centre O and radius = 3.5 cm.

Step 2. Draw any diameter AOB of this circle.

Step 3. Construct

Step 4. Draw MA

Let AM and CN intersect at P.

Then, PA and PC are the required tangents to the given circle that are inclined at an angle of 60^{o}.

#### Page No 531:

#### Answer:

Steps Of construction:

Step 1. Draw a circle with centre O and radius 4 cm.

Step 2. Draw radius OA and produce it to B.

Step 3. Make $\angle AOP=30\xb0$

Step 4. Draw PQ $\perp OP$ , meeting OB at Q.

Step 5. Then, PQ is the desired tangent, such that

∠OQP = 30°

#### Page No 531:

#### Answer:

Steps of Construction :

Step 1. Draw a circle with O as centre and radius 6 cm.

Step 2. Draw another circle with O as centre and radius 4 cm.

Step 2 . Mark a point P on the circle with radius 6 cm.

Step 3. Join OP and bisect it at M.

Step 4. Draw a circle with M as centre and radius equal to MP to intersect the given circle with radius 4 cm at points T and T^{'}.

Step 5. Join PT and P T^{'}.

Thus, PT or P T^{'}are the required tangents and measure 4.4 cm each.

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