Mathematics Part II Solutions Solutions for Class 10 Math Chapter 4 Geometric Constructions are provided here with simple step-by-step explanations. These solutions for Geometric Constructions are extremely popular among Class 10 students for Math Geometric Constructions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part II Solutions Book of Class 10 Math Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part II Solutions Solutions. All Mathematics Part II Solutions Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

#### Page No 96:

#### Question 1:

∆ABC ~ ∆LMN. In ∆ABC, AB = 5.5 cm, BC = 6 cm, CA = 4.5 cm. Construct ∆ABC and ∆LMN such that $\frac{\mathrm{BC}}{\mathrm{MN}}=\frac{5}{4}.$

#### Answer:

Construct ∆ABC such that AB = 5.5 cm, BC = 6 cm and CA = 4.5 cm.

∆ABC and ∆LMN are similar.

Therefore, their corresponding sides are proportional.

$\therefore \frac{\mathrm{AB}}{\mathrm{LM}}=\frac{\mathrm{BC}}{\mathrm{MN}}=\frac{\mathrm{CA}}{\mathrm{NL}}=\frac{5}{4}$

$\Rightarrow \frac{5.5}{\mathrm{LM}}=\frac{6}{\mathrm{MN}}=\frac{4.5}{\mathrm{NL}}=\frac{5}{4}$

$\Rightarrow \mathrm{LM}=\frac{4}{5}\times 5.5=4.4\mathrm{cm},\mathrm{MN}=\frac{4}{5}\times 6=4.8\mathrm{cm},\mathrm{NL}=\frac{4}{5}\times 4.5=3.6\mathrm{cm}$

Now, construct ∆LMN such that LM = 4.4 cm, MN = 4.8 cm and NL = 3.6 cm.

#### Page No 96:

#### Question 2:

∆PQR ~ ∆LTR. In ∆PQR, PQ = 4.2 cm, QR = 5.4 cm, PR = 4.8 cm. Construct ∆PQR and ∆LTR, such that $\frac{\mathrm{PQ}}{\mathrm{LT}}=\frac{3}{4}.$

#### Answer:

**Steps of Construction**:

**Step 1**: Draw ∆PQR such that PQ = 4.2 cm, QR = 5.4 cm and PR = 4.8 cm.

**Step 2**: Divide segment RQ into 3 equal parts. Fix point T on ray RQ such that length of RT is 4 times of each equal part of segment BC.

**Step 3**: Draw a line parallel to side PQ, through T. The point of intersection of the line and ray RP is L.

Here, ∆PQR ~ ∆LTR.

#### Page No 96:

#### Question 3:

∆RST ~ ∆XYZ. In ∆RST, RS = 4.5 cm, ∠RST = 40°, ST = 5.7 cm Construct ∆RST and ∆XYZ, such that $\frac{\mathrm{RS}}{\mathrm{XY}}=\frac{3}{5}.$

#### Answer:

Construct ∆RST such that RS = 4.5 cm, ∠RST = 40° and ST = 5.7 cm.

∆RST and ∆XYZ are similar.

So, their corresponding sides are proportional and corresponding angles are equal.

$\therefore \frac{\mathrm{RS}}{\mathrm{XY}}=\frac{\mathrm{ST}}{\mathrm{YZ}}=\frac{\mathrm{RT}}{\mathrm{XZ}}=\frac{3}{5}$

$\Rightarrow \frac{4.5}{\mathrm{XY}}=\frac{5.7}{\mathrm{YZ}}=\frac{3}{5}$

$\Rightarrow \mathrm{XY}=\frac{5}{3}\times 4.5=7.5\mathrm{cm},\mathrm{YZ}=\frac{5}{3}\times 5.7=9.5\mathrm{cm}$

Also, ∠XYZ = ∠RST = 40°

Now, construct ∆XYZ such that XY = 7.5 cm, ∠XYZ = 40° and YZ = 9.5 cm.

#### Page No 96:

#### Question 4:

∆AMT ~ ∆AHE. In ∆AMT, AM = 6.3 cm, ∠TAM = 50°, AT = 5.6 cm. $\frac{\mathrm{AM}}{\mathrm{AH}}=\frac{7}{5}.$ Construct ∆AHE.

#### Answer:

**Steps of Construction**:

**Step 1**: Draw ∆AMT such that AM = 6.3 cm, ∠TAM = 50°, AT = 5.6 cm.

**Step 2**: Divide segment AM into 7 equal parts.

**Step 3**: Name the end point of the 5th part of segment AM as H.

**Step 4**: Draw a line parallel to side TM, through H. Name the point where the parallel line intersects AT as E.

Here, ∆AMT ~ ∆AHE.

#### Page No 98:

#### Question 1:

Construct a tangent to a circle with centre P and radius 3.2 cm at any point M on it.

#### Answer:

**Steps of Construction:**

**Step 1**: Draw a circle with centre P and radius 3.2 cm.

**Step 2**: Take any point M on the circle.

**Step 3**: Draw ray PM.

**Step 4**: Draw line *l* perpendicular to ray PX through point M.

Here, line *l* is the required tangent to the circle at the point M.

#### Page No 98:

#### Question 2:

Draw a circle of radius 2.7 cm. Draw a tangent to the circle at any point on it.

#### Answer:

**Steps of Construction:**

**Step 1**: Draw a circle with centre C and radius 2.7 cm.

**Step 2**: Take any point P on the circle.

**Step 3**: Draw ray CX.

**Step 4**: Draw line *l* perpendicular to ray CX through point P.

Here, line *l* is the required tangent to the circle at the point P.

#### Page No 98:

#### Question 3:

Draw a circle of radius 3.6 cm. Draw a tangent to the circle at any point on it without using the centre.

#### Answer:

**Steps of Construction:**

**Step 1**: Draw a circle with radius 3.6 cm. Mark any point C on it.

**Step 2**: Draw chord CB and an inscribed ∠CAB.

**Step 3**: With the centre A and any convenient radius draw an arc intersecting the sides of ∠BAC in points M and N.

**Step 4**: Using the same radius and centre C, draw an arc intersecting the chord CB at point R.

**Step 5**: Taking the radius equal to d(MN) and centre R, draw an arc intersecting the arc drawn in step 4. Suppose D be the point of intersection of these arcs. Draw line CD.

Here, line CD is the required tangent to the circle.

#### Page No 98:

#### Question 4:

Draw a circle of radius 3.3 cm Draw a chord PQ of length 6.6 cm. Draw tangents to the circle at points P and Q. Write your obeservation about the tangents.

#### Answer:

**Steps of Construction:**

**Step 1**: Draw a circle with radius 3.3 cm. Mark any point P on it.

**Step 2**: Draw chord PQ = 6.6 cm (PQ is the diameter of the circle).

**Step 3**: Draw rays CX and CY.

**Step 4**: Draw line *l* perpendicular to ray CX through point P.

**Step 4**: Draw line *m* perpendicular to ray CY through point Q.

Here, line *l* and line *m* are the required tangents to the circle at points P and Q, respectively.

It can be observed that the tangents *l* and *m* are parallel to each other.

#### Page No 99:

#### Question 5:

Draw a circle with radius 3.4 cm. Draw a chord MN of length 5.7 cm in it. construct tangents at point M and N to the circle.

#### Answer:

**Steps of Construction:**

**Step 1**: Draw a circle of radius 3.4 cm with centre C. Mark any point M on it.

**Step 2**: Draw chord MN = 5.7 cm and an inscribed ∠NOM.

**Step 3**: With the centre O and any convenient radius draw an arc intersecting the sides of ∠MPN in points P and Q.

**Step 4**: Using the same radius and centre M, draw an arc intersecting the chord MN at point R.

**Step 5**: Taking the radius equal to d(PQ) and centre R, draw an arc intersecting the arc drawn in step 4. Suppose S be the point of intersection of these arcs. Draw line MS.

**Step 6**: With the centre M and any convenient radius draw an arc intersecting the sides of ∠OMN in points T and U.

**Step 7**: Using the same radius and centre N, draw an arc intersecting the chord MN at point V.

**Step 8**: Taking the radius equal to d(TU) and centre V, draw an arc intersecting the arc drawn in step 7. Suppose W be the point of intersection of these arcs. Draw line NW.

Here, line MS and NW are the required tangents to the circle at points M and N, respectively.

#### Page No 99:

#### Question 6:

Draw a circle with centre P and radius 3.4 cm. Take point Q at a distance 5.5 cm from the centre. Construct tangents to the circle from point Q.

#### Answer:

**Steps of Construction:**

**Step 1**: Construct a circle of radius 3.4 cm with centre P.

**Step 2**: Take a point Q in the exterior of the circle such that PQ = 5.5 cm

**Step 3**: Draw segment PQ. Draw perpendicular bisector of segment PQ to get its midpoint M.

**Step 4**: Draw a circle with radius PM and centre M.

**Step 5**: Name the point of intersection of the two circles as A and B.

**Step 6**: Draw line QA and line QB.

Here, line QA and line QB are tangents to the circle from point Q.

#### Page No 99:

#### Question 7:

Draw a circle with radius 4.1 cm. Construct tangents to the circle from a point at a distance 7.3 cm from the centre.

#### Answer:

**Steps of Construction:**

**Step 1**: Construct a circle of radius 4.1 cm with centre P.

**Step 2**: Take a point Q in the exterior of the circle such that PQ = 7.3 cm

**Step 3**: Draw segment PQ. Draw perpendicular bisector of segment PQ to get its midpoint M.

**Step 4**: Draw a circle with radius PM and centre M.

**Step 5**: Name the point of intersection of the two circles as A and B.

**Step 6**: Draw line QA and line QB.

Here, line QA and line QB are tangents to the circle from point Q.

#### Page No 99:

#### Question 1:

Select the correct alternative for each of the following questions.

(1) The number of tangents that can be drawn to a circle at a point on the circle is ............... .

(2) The maximum number of tangents that can be drawn to a circle from a point out side it is .............. .

(3) If ∆ABC ~ ∆PQR and $\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{7}{5}$, then ...............

(B) ∆PQR is bigger.

(C) Both triangles will be equal.

(D) Can not be decided.

#### Answer:

(1) The number of tangents that can be drawn to a circle at a point on the circle is __1__.

Hence, the correct answer is option (C).

(2) The maximum number of tangents that can be drawn to a circle from a point outside it is __2__.

Hence, the correct answer is option (A).

(3) If ∆ABC ~ ∆PQR and $\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{7}{5}$, then __∆ABC is bigger__.

∆ABC and ∆PQR are similar. Therefore, their corresponding sides are proportional.

$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{CA}}{\mathrm{RP}}=\frac{7}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AB}=\frac{7}{5}\mathrm{PQ},\mathrm{BC}=\frac{7}{5}\mathrm{QR},\mathrm{CA}=\frac{7}{5}\mathrm{RP}$

Here, sides of ∆ABC are $\frac{7}{5}$ times that of the corresponding sides of ∆PQR. So, ∆ABC is bigger than ∆PQR.

Hence, the correct answer is option (A).

#### Page No 99:

#### Question 2:

Draw a circle with centre O and radius 3.5 cm. Take point P at a distance 5.7 cm from the centre. Draw tangents to the circle from point P.

#### Answer:

**Steps of Construction:**

**Step 1**: Construct a circle of radius 3.5 cm with centre O.

**Step 2**: Take a point P in the exterior of the circle such that OP = 5.7 cm

**Step 3**: Draw segment OP. Draw perpendicular bisector of segment OP to get its midpoint M.

**Step 4**: Draw a circle with radius OM and centre M.

**Step 5**: Name the point of intersection of the two circles as A and B.

**Step 6**: Draw line PA and line PB.

Here, line PA and line PB are tangents to the circle from point P.

#### Page No 99:

#### Question 3:

Draw any circle. Take any point A on it and construct tangent at A without using the centre of the circle.

#### Answer:

**Steps of Construction:**

**Step 1**: Draw a circle with any radius and mark any point A on it.

**Step 2**: Draw chord AB and an inscribed ∠BCA.

**Step 3**: With the centre C and any convenient radius draw an arc intersecting the sides of ∠ACB in points M and N.

**Step 4**: Using the same radius and centre A, draw an arc intersecting the chord AB at point O.

**Step 5**: Taking the radius equal to d(MN) and centre O, draw an arc intersecting the arc drawn in step 4. Suppose D be the point of intersection of these arcs. Draw line AD.

Here, line AD is the required tangent to the circle.

#### Page No 99:

#### Question 4:

Draw a circle of diameter 6.4 cm. Take a point R at a distance equal to its diameter from the centre. Draw tangents from point R.

#### Answer:

**Steps of Construction:**

**Step 1**: Construct a circle of radius = $\frac{6.4}{2}$ = 3.2 cm with centre O.

**Step 2**: Take a point R in the exterior of the circle such that OR = 6.4 cm.

**Step 3**: Draw segment OR. Draw perpendicular bisector of segment OR to get its midpoint M.

**Step 4**: Draw a circle with radius OM and centre M.

**Step 5**: Name the point of intersection of the two circles as A and B.

**Step 6**: Draw line RA and line RB.

Here, line RA and line RB are tangents to the circle from point R.

#### Page No 99:

#### Question 5:

Draw a circle with centre P. Draw an arc AB of 100° measure. Draw tangents to the circle at point A and point B.

#### Answer:

**Steps of Construction:**

**Step 1**: Draw a circle with centre P and any radius.

**Step 2**: Take any point A on the circle.

**Step 3**: Construct ∠APB = 100°, such that point B is on the circle.

**Step 4**: Draw rays PA and PB.

**Step 5**: Draw line *l* perpendicular to ray PX through point A.

**Step 6**: Draw line *m* perpendicular to ray PY through point B.

Here, line *l* and line *m* are the required tangents to the circle at the point A and point B, respectively.

#### Page No 99:

#### Question 6:

Draw a circle of radius 3.4 cm and centre E. Take a point F on the circle. Take another point A such that E-F-A and FA = 4.1 cm. Draw tangents to the circle from point A.

#### Answer:

**Steps of Construction:**

**Step 1**: Construct a circle of radius 3.4 cm with centre E.

**Step 2**: Mark a point F on the circle.

**Step 3**: Take a point A in the exterior of the circle such that FA = 4.1 cm

**Step 3**: Draw segment FA. Draw perpendicular bisector of segment EA to get its midpoint M.

**Step 4**: Draw a circle with radius EM and centre M.

**Step 5**: Name the point of intersection of the two circles as X and Y.

**Step 6**: Draw line AX and line AY.

Here, line AX and line AY are tangents to the circle from point A.

#### Page No 99:

#### Question 7:

∆ABC ~ ∆LBN. In ∆ABC, AB = 5.1cm, ∠B = 40°, BC = 4.8 cm, $\frac{\mathrm{AC}}{\mathrm{LN}}=\frac{4}{7}$. Construct ∆ABC and ∆LBN.

#### Answer:

**Steps of Construction**:

**Step 1**: Draw ∆ABC such that AB = 5.1 cm, ∠B = 40°, BC = 4.8 cm.

**Step 2**: Divide segment BC into 4 equal parts.

**Step 3**: Mark point N on ray BC such that length of BN is seven times of each part of segment BC.

**Step 4**: Draw a line parallel to side AC, through N. Name the point where the parallel line intersects ray BA as L.

Here, ∆ABC ~ ∆LBN.

#### Page No 99:

#### Question 8:

Construct ∆PYQ such that, PY = 6.3 cm, YQ = 7.2 cm, PQ = 5.8 cm. If $\frac{\mathrm{YZ}}{\mathrm{YQ}}=\frac{6}{5},$ then construct ∆XYZ similar to ∆PYQ.

#### Answer:

**Steps of Construction**:

**Step 1**: Draw ∆PYQ such that PY = 6.3 cm, YQ = 7.2 cm and PQ = 5.8 cm.

**Step 2**: Divide segment YQ into 5 equal parts.

**Step 3**: Mark point Z on ray YQ such that length of YZ is six times of each part of segment YQ.

**Step 4**: Draw a line parallel to side QP, through Z. Name the point where the parallel line intersects ray YP as X.

Here, ∆PYQ ~ ∆XYZ.

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