Exercise 10.1
Multiple Choice Questions (MCQs)
Question. 1 To divide a line segment AB in the ratio 5:7, first a ray AX is drawn, so that ∠BAX is an acute angle and
then at equal distances points are marked on the ray AX such that the minimum number of these points is
(a) 8       
(b) 10      
(c) 11      
(d) 12
Answer. (d)
Explanation:
To divide a line segment AB in the ratio m : n, a ray AX making an acute ∠BAX, is drawn and then m + n points are marked at equal distances on the ray AX.
Here, m = 5, n = 7
Therefore, minimum number of points to be marked on AX = m + n = 5 + 7 = l2.
Question. 2 To divide a line segment AB in the ratio 4:7, a ray AX is drawn first such that∠BAX is an acute angle and then points are located at equal distances on the ray AX and the point B is joined to
(a) A₁₂           
(b) A₁₁           
(c) A₁₀    
(d) A₉
Answer. (b)
Explanation:
Since 4 + 7 = 11points are to be located on AX  at equal distantes, so B is joined to last point, A₁₁.
Question. 3 To divide a line segment AB in the ratio 5:6, draw a ray AX such that∠BAX is an acute angle, then draw a ray BY parallel to AX and the points and  are located to equal distances on ray AX and BY, respectively. Then, the points joined are
(a) A₅ and B₆                                                                 
(b) A₆ and B₅                                                                 
(c) A₄ and B₅                                                                  
(d) A₅ and B₄
Answer. (a)
Explanation:
To divide line segment AB in the ratio 5:6.

Steps of construction
1. Draw a ray AX making an acute ∠BAX.
2. Draw a ray BY parallel to AX by taking ∠ABY equal to ∠BAX.
3.Divide AX into five (m = 5) equal parts AA₁, A₁A₂, A₂A₃, A₃A₄ and A₄A₅
4. Divide BY into six (n = 6) equal parts and BB₁, B₁B₂, B₂B₃, B₃B₄, B₄B₅ and B₅B₆.
4. Join B₆A₅. Let it intersect AB at a point C.
Then,  AC : BC = 5 : 6
Question. 4 To construct a triangle similar to a given DABC with its sides 3/7  of the corresponding sides of DABC, first draw a ray BX such that ∠CBX is an acute angle and X lies on the opposite side of A with respect to BC. Then, locate points  on BX at equal distances and next step is to join
(a) B₁₀ to C           
(b) B₃ to C                                                                     
(c) B₇ to C                                                                      
(d) B₄ to C
Answer. (c)
Explanation:
As the sides of new triangle are 3/7  of the corresponding sides of DABC, so it will be smaller than DABC. So, after we locate points andon BX at equal distance, the next step is to join the last point B₇ to C so, that parallel line from third part of BX meet on BC without producing.
Question. 5 To construct a triangle similar to a given ΔABC with its sides 8/5of the corresponding sides of ΔABC
draw a ray BX such that ÐCBX is an acute angle and X is on the opposite side of A with respect to BC. The minimum number of points to be located at equal distances on ray BX is
(a) 5               
(b) 8            
(c) 13        
(d) 3
Answer. (b)
Explanation:
Toconstruct a triangle similar to a given triangle, with its sides m/n  of the corresponding sides of given triangle the minimum number of points to be located at equal distance is equal to the greater of m and n
in m/n. As, here m/n = 8/5.
So, the minimum number of points to be located at equal distances on ray BX is 8.
Question. 6 To draw a pair of tangents to a circle which are inclined to each other at an angle of 60°, it is required to draw tangents at end points of those two radii of the circle, the angle between them should be
(a) 135°                  
(b) 90°     
(c) 60°               
(d) 120°
Answer. (d)
Explanation:
Let PQ and PR be the two at points Q and R on circle with centre O.

Given ∠RPQ = 60^o
We know that the tangent at a point on a circle is perpendicular to the radius through that point.
∴ ∠P = ∠Q = 90°
In quadrilateral PQRS,
         ∠P +∠Q + ∠R + ∠O = 360°
⟹ 90° + 90° + 60°+  ∠O = 360°
⟹      ∠O = 360° − 240  = 120°
Hence, the required angle between two radii is 120°.
Exercise 10.2
Very Short Answer Type Questions:
Write True or False and give reason for your answer in each of the following:

Hence, it is possible to divide a line segment in the ratio 3:1 by the geometrical construction.
Question. 2 To construct a triangle similar to a given DABC with its sides 7/3 of the corresponding sides of DABC,
draw a ray BX making an acute angle with BC and X lies on the opposite side of A with respect of BC. The
points B₁, B₂,…, B₇are located at equal distances on BX, B₃is joined to C and then a line segment B₆C’ is drawn
parallel to B₃C, where C’ lies on BC produced. Finally line segment A’C’ is drawn parallel to AC.
Answer: False
Explanation:
As the sides of new new triangle will be equal to 7/3 of the corresponding sides of ΔABC. Therefore, it will be larger than ΔABC. Hence, B₇C’ (larger of 7/3) instead of B₆C’, is drawn parallel to B₃C.
Steps of construction are:
1. Draw the given ΔABC.
2. From B draw any ray BX downwards making an acute angle CBX.
3. Divide BX into 7 equal parts = B₂ = B₂B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B₇.
4. Join C and from B₇ draw a line B₇C’‖ B₃C intersecting the extended line segment BC at C’.
5. From point C’ draw C’A’||CA intersecting the extended line segment BA at A’.
6. Then, DA’BC’ is the required triangle with sides equal to 7/3 of the corresponding sides of DABC.

Thus, B₇C’ is parallel to B₃C.
Question. 3 A pair of tangents can be constructed from a point P to a circle of radius 3.5 cm situated at a distance of
3 cm from the centre.
Answer: False
Explanation:
Since, the point P is 3 cm away from the centre of a circle with radius 3.5 cm therefore, the point P lies inside the circle and we now that no tangent is possible from a point lying inside the circle. For a tangent to be drawn from a point, the distance of that point must be equal to or greater than the radius of circle. So, no tangent is possible here.
Question. 4 A pair of tangents can be constructed to a circle inclined at an angle of 170°.
Answer: True
Explanation:
It is possible to draw a pair of tangents to a circle, inclined at an angle between 0° to 180°. Hence, we can draw a pair of tangents to a circle inclined at an angle of 170°.
Exercise 10.3
Short Answer Type Questions:
Question. 1 Draw a line segment of length 7 cm. Find a point P on it which divides it in the ratio 3:5.
Solution.
Steps of construction:
1. Draw a line segment AB = 7 cm.
2. Draw a ray AX, making an acute ∠BAX.
3. Divede AX, in 3 + 5 = 8 equal parts AA₁, A₁A₂, A₂A₃, A₃A₄, A₄A₅, A₅A₆, A₆A₇ and A₇A₈.
4. Join A₈B.
5. From A₃, draw A₃C || A₈B intersecting AB at C.
6. Thus C is the required point on AB which divides it in the ratio 3:5.
i.e., AC : CB = 3 : 5

Question. 2 Draw a right ΔABC in which BC = 12 cm, AB = 5 cm and ∠B = 90°. Construct a triangle similar to it and of scale factor 2/3. Is the new triangle also a right triangle?
Solution.
Steps of construction:
First we draw the right ΔABC by following the steps given below:
1. Draw a line segment BC = 12 cm.
2. From B draw a line AB = 5 cm such that AB ⏊ BC.
3. Join AC.
Thusgiven right DABC is obtained.

Now we construct another triangle similar to DABC with its sides equal to 2/3 of the corresponding sides of DABC, by following the steps given below:
4. From B draw an acute ∠CBY such that A and Y are in opposite direction with respect to BC.
5. Divide BY into three equal parts, BB₁ = B₁B₂ = B₂B₃.
7.  From point B₂ draw B₂C’ || B₃C intersecting BC at C’.
8.From point C’ draw C’A’ || CA intersecting BA at A’.
DA’BC’ is the required triangle of scale factor 2/3.This is also a right angled triangle.
Question. 3  Draw a ΔABC in which BC = 6cm, CA =5 cm and AB =4 cm. Construct a triangle similar to it and of
scale factor.
Solution.
Steps of construction:
First we draw ΔABC by following steps given below:
1. Draw a line segment BC = 6 cm.
2. Taking B and C as centres, draw two arcs of radii 4 cm and 5 cm respectively, which intersect each other at A.
3. Join BA and CA. ΔABC is the required triangle.
Now we construct another triangle similar to DABC and of scale factor 5/3, by following the steps given below:
4.From B, draw an acute ∠CBY such that A and Y are in opposite direction with respect to BC.
5.Divide BY into five equal parts,

6. Join C and from B₅ draw B₅C’ || B₃C intersecting the extended line segment BC at C’.
7. Again from point C’ draw C’A’ || CA intersecting the extended line segment BA at A’.
Thus, ΔA’BC’ is the required triangle similar to ΔABC and of scale factor 5/3.
Question. 4 Construct a tangent to a circle of radius 4 cm from a point which is at a distance of 6 cm from its centre.
Solution.
Steps of construction
1. Draw a line-segment OP= 6 cm.
2. Draw a circle of radius 4 cm taking O as centre.
2. Draw a perpendicular bisector of OP intersecting OP at M.
4. Taking M as centre and MO as radius draw a circle to intersect circle C (O, 4) at two points, Q and R.
5. Join PQ and PR.

Thus, PQ and PR are the required tangents.
Exercise 10.4
Long Answer Type Questions:

Solution.
Steps of construction:
1. Draw a line segment AB = 5 cm.

2. Draw ∠BAZ = 60°.
3. With A as centre and radius equal to 7 cm draw an arc cutting the line AZ at C.
4. Draw a ray AX, making an acute ∠BAX.
5. Divide AX into four equal parts, AA₁ = A₁A₂ = A₂A₃ = A₃A₄.
6. Join A₄B.
7. Draw A₃P || A₄B meeting AB at P.
8. Thus, P is the point on AB such that AP = ¾ AB.
9. Next, draw a ray AY, making an acute ∠CAY.
10. Divide AY into four parts AB₁ = B₁B₂ = B₂B₃ = B₃B₄.
11. Join B₄C.
11. Draw B₁Q || B₄C meeting AC at Q.     
Then, Q is the point on AC such that AQ = ¼ AC.
12.   Join PQ and measure it.
13.   PQ = 3.25 cm.
Question. 2 Draw a parallelogram ABCD in which BC = 5 cm, AB = 3 cm and ∠ABC = 60°, divide it into triangles BCD and ABD by the diagonal BD. Construct the triangle BD’C’ similar to DBDC with scale factor
4/3. Draw the line segment D’A’ parallel to DA, where A’ lies on extended side BA. Is A’BC’D’ a parallelogram?
Solution.
Steps of construction:
First we draw the parallelogram ABCD by following the steps given below:
1. Draw a line segment AB = 3 cm.
2. Draw ∠ABY = 60°.
3. Taking B as centre and radius equal to 5 cm draw an arc cutting BY at C.
4. Draw CD || AB and AD || BC. Thus ABCD is the required parallelogram.
5. Join BD to get two triangles BCD and ABD.

Now to construct the ΔBD’C’ similar to DBDC with scale factor 4/3, follow the steps given below:
6. From B draw a ray BX making an acute ∠CBX.
7. Divide BX into four equal parts, BB₁ = B₁B₂ = B₂B₃ = B₃B₄.
8. Join B₃C and draw a line B₄C’ || B₃C intersecting the extended line segment BC at C’.
9.From point C’ draw C’D’ || CD intersecting the extended line segment BD at D’.
Thus, ΔBD’C’ is the required triangle similar to DBDC with scale factor 4/3.
10.Now draw a line segment D’A’ parallel to DA, where A’ lies on line segment BA at A’.
11. Thus, we obtain A’BC’D’, whichis a parallelogram.
Question. 3 Draw two concentric circles of radii 3 cm and 5 cm. Taking a point on outer circle construct the
pair of tangents to the other. Measure the length of a tangent and verify it by actual calculation.
Solution.
Steps of construction:
1. Draw two concentric circles with centre O and radii 3 cm and 5 cm.
2. Take a point P on outer circle and join OP.
3. Draw a perpendicular bisector of OP intersecting OP at M.Thus M is the mid-point of OP.
4. Now, taking M as centre and OM as radius draw a circle meeting the inner circle at points Q and R.
5. Join PQ  and PR. Thus, PQ and PR are the required tangents.
6. On measuring PQ  and PR, we get PQ = PR = 4 cm.

Actual calculation:
In DOPQ,
          ∠OQP = 90°          (Tangent at a point is perpendicular to the radius through that point)
Thus, using Pythagoras theorem in right angled DOPQ, we get:
∴         PQ² = OP² – OQ²                                
⟹       PQ² = (5)² – (3)² = 25 – 9 = 16
⟹       PQ = 4cm
Hence, verified.
Question. 4 Draw an isosceles triangle ABC in which AB = AC = 6 cm and BC = 5 cm. Construct a triangle PQR similar to ΔABC in which PQ = 8 cm. Also justify the construction.
Solution.              
To construct a ΔPQR similar to triangle ΔABC, we have to find the scale factor.

Steps of construction:
First we draw ΔABC by following the steps given below:
1.Draw a line segment BC = 5 cm.
2. Taking B and C as centers draw two arcs each of 6 cm, intersecting each other at point A.  
4. Join AB and AC.
Thus, ΔABC is the required isosceles triangle.

Now we draw ΔPQR similar to ΔABC with scale factor 4/3 by following the steps given below:
5. From B, draw a ray BX making an acute ∠CBX opposite to point A with respect to side BC.
6. Divide BX into four equal parts BB₁ = B₁B₂ = B₂B₃ = B₃B_4.
7. Join B₃C and from B₄ draw a line B₄R || B₃C intersecting the extended line segment BC at R.
8. From point R, draw RP || CA meeting the extended line segment BA at P.
9. Take point B as Q.
Then, ΔPQR is the required triangle.
Justification:

Hence, the ΔPQR is similar to ΔABC,  with its sides equal to 4/3  times of the corresponding sides of the ΔABC.
Question. 5 Draw a ΔABC in which AB = 5 cm, BC = 6 cm and ∠ABC = 60°. Construct a triangle similar to ΔABC with scale factor 5/7. Justify the construction.
Solution.
Steps of construction:
First we draw ΔABC by following the steps given below:
1. Draw a line segment AB = 5 cm.
2. Draw ∠ABY = 60°
3. Taking B as centre and radius = 6cm draw an arc cutting BY at C. Join AC.
ΔABC is the required triangle.

To draw another triangle similar to ΔABC with scale factor 5/7, we follow the steps given below:
4. From A, draw any ray AX making an acute angle BAX in the opposide side to vertex C with respect to side AB.
5. Divide AX into 7 equal parts AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = A₅A₆ = A₆A₇.  
6. Join A₇B and from A₅ draw A₅B’ || A₇B intersecting AB at B’.
7. Now from point B’ draw B’C’ || BC intersecting AC at C’.
Then, ΔAB’C’ isthe required triangle whose sides are equal to 5/7 of the corresponding sides of the ΔABC.
Justification:

Hence justified.
Question. 6 Draw a circle of radius 4 cm. Construct a pair of tangents to it, the angle between which is 60°. Also justify the construction. Measure the distance between the centre of the circle and the point of intersection of tangents.
Solution.              
Steps of construction
1. Draw a circle of with centre as O and radius OA = 4 cm.
2. Produce OA to B such that OA = AB = 4 cm.
3. Taking A as the centre and AO = 4 cm as radius draw a ciscle cutting the previous circle at points P and Q.
4. Join BP and BQ.

Thus BP and BQ are the desired tangents.
Justification:
Join OP.

Question. 7 Draw a ΔABC in which AB = 4 cm, BC = 6 cm and AC = 9 cm. Construct a triangle similar to ΔABC with scale factor 3/2. Justify the construction. Are the two triangles congruent? Note that, all the three angles and two sides of the two triangles are equal.
Solution.              
Steps of construction:
First we draw ΔABC by following the steps given below:
1. Draw a line segment BC = 6 cm.
2. Taking B and C as centres, draw two arcs of radii 4 cm and 9 cm respectively, which intersect each other at point A.
3. Join AB and AC.
Thus, ΔABC is the required triangle.

Now we draw another triangle similar to ΔABC with scale factor 3/2.
4. From B, draw a ray BX downwards making an acute angle CBX
5. Divide BX into 3 equal parts, BB₁ = B₁B₂ = B₂B₃.
6. Join B₂C and from B₃ draw B₃C’ || B₂C intersecting the extended line segment BC at C’.
7. From point C’, draw C’A’ || CA intersecting the extended line segment BA at A’.
Then, ΔA’BC’ is the required triangle whose sides are equal to 3/2 of the corresponding sides of the ΔABC.
Justification

Here, the two triangles, ABC and A’BC’are not congruent because, for the two triangles to be congruent, their corresponding sides and interior angles must be same. But here, all the three angles are same but three sides are not same.