Exercise 6.1
Multiple Choice Questions (Q. No. 1-6)
Question. 1 In figure, if ∠BAC = 90^o and AD ^ BC. Then,

Answer. (c)
Explanation:
Consider the following diagram:

Question. 2 If the Lengths of the diagonals of rhombus are 16 cm and 12 cm. Then, the Length of the sides of the rhombus is
(a) 9cm                               
(b) 10cm                            
(c) 8cm                               
(d) 20cm
Answer. (b)
Explanation:
Given, AC = 16cm and BD = 12cm

As the diagonals of a rhombus are perpendicular bisectors of each other.
⟹ AO = 8cm, BO = 6cm and ÐAOB = 90°   ….(i)
Using Pythagoras theorem in right angled ΔAOB, w have
            AB² = AO² + OB²       
⟹       AB² = 8² + 6²               [Using (i)]
⟹       AB= 10cm.
Thus length of each side of rhombus is 10 cm.
Question. 3 If DABC ~ DEDF and DABC is not similar to DDEF, then which of the following is not true?
(a) BC.EF = AC.FD              
(b) AB.EF = AC.DE
(c) BC.DE = AB.EF  
(d)BC.DE = AB.FD
Answer.  (c)
Explanation:

Hence, option (b) is also true.

This expression shows that sides of one triangle DCAB are proportional to the sides of the other triangle DPQR and their corresponding angles are also equal.
Therefore, DPQR ~ DCAB.
Question. 5 In figure, two line segments AC and BD intersect each other at the point P such that PA = 6cm, PB = 3cm, PC = 2.5cm, PD = 5cm, ÐAPB = 50° and ÐCDP = 30°. Then, ÐPBA is equal to

Answer. (d)
Explanation:

Thus, ÐPBA = 100° verifies the option (d).
Question. 6 If in two triangles DEF and PQR, ÐD =ÐQ and ÐR =ÐE, then which of the following is not true?

Answer.(b)
Explanation:

Exercise 6.1
Multiple Choice Questions (Q. No. 7-12)
Question. 7 In ΔABC and ΔDEF, ∠B = ∠E, ∠F = ∠C and AB = 3DE then, the two triangles are
(a) congruent but not similar
(b) similar but notcongruent
(c) neither congruent nor similar
(d) congruent as well as similar
Answer. (b)
Explanation:

Here, two triangles DABC and DDEF do not satisfy any rule of congruency, (SAS, ASA, SSS), so both are not congruent.

Answer. (a)
Explanation:

Question. 9 If ΔABC ~ ΔDFE, ∠A = 30°, ∠C = 50°, AB = 5cm, AC = 8cm and DF = 7.5cm. Then, which of the following is true?
(a) DE =12cm, ∠F =50°    
(b) DE =12cm, ∠F =100°
(c) EF =12cm, ∠D =100°  
(d) EF =12cm, ∠D =30°
Answer. (b)
Explanation:


Answer. (c)
Explanation:

And, angle formed by DE and FD is ∠D.
So,          ∠B = ∠D
⟹           DABC ~ DEDF     (By SAS criterion)

Answer. (a)
Explanation:

As the ratio of area of two similar triangles is equal to the ratio of square of their corresponding sides.
Therefore, we get:

Question. 12 If S is a point on side PQ of a ΔPQR such that PS = QS = RS, then
(a) PR . QR = RS²
(b) QS² + RS² =QR²
(c) PP² + QP² = PQ²
(d) PS² + RS² = PR²
Answer. (c)          
Explanation:


Exercise 6.2
Very Short Answer Type Questions
Write whether True or False and justify your answer:
Question. 1 Is the triangle with sides 25cm, 5cm and 24cm a right triangle? Give reason for your answer.
Answer. False
Solution:
For given triangle to be a right triangle, it must satisfy Pythagoras theorem.
^{i.e.,           252 =  52 + 242
⟹            625 = 25 + 576
Which is not true.
So, given triangle is not a right triangle.}
Question. 2 It is given that ΔDEF ~ ΔRPQ. Is it true to say that ∠ D = ∠R and ∠F = ∠P? Why?
Answer. False
We know that, if two triangles are similar, then their corresponding angles are equal,
Thus, in ΔDEF and ΔRPQ, we have:
            ∠D = ∠R, ∠E = ∠P and ∠F = ∠Q
Question. 3 A and B are respectively the points on the sides PQ and PR of a ΔPQR such that PQ = 12.5cm, PA = 5cm, BR = 6cm and PB = 4cm. Is AB || QR? Give reason for your answer.
Answer. False
Solution:
By converse of BPT, AB will be parallel to QR if AB, divides PQ and PR in the same ratio i.e.,

Question. 4 In figure, BD and CE intersect each other at the point P. Is ΔPBC ~ ΔPDE?Why?

Question. 5 In ΔPQR and ΔMST, ∠P = 55°, ∠Q =25°, ∠M = 100° and ∠S = 25°. Is ΔQPR ~ ΔTSM? Why?
Answer. False
Solution:
We know that, the sum of three angles of a triangle is 180°
Thus, in ΔPQP,          
            ∠P + ∠Q + ∠R = 180°
⟹        55° + 25° + ÐR = 180°
⟹        ∠R = 180° – (55° + 25°) = 180° – 80° =100°
Also. in ΔTSM,
∠T + ∠S + ∠M = 180°
⟹        ∠T + ∠25° + 100° = 180°
⟹       ∠T = 180^o – (25° + 100°)
              = 180°-125° = 55°

Question. 6 Is the following statement true? Why? “Two quadrilaterals are similar, if their corresponding angles are equal”.
Answer. False
Solution:
Two quadrilaterals will be similar, if their corresponding angles are equal and the ratio of their corresponding sides area also equal.
Question. 7 Two sides and the perimeter of one triangle are respectively three times the corresponding sides and the perimeter of the other triangle. Are the two triangles similar? Why?
Answer. True
Solution:
Here, the corresponding two sides and the perimeters of two triangles are proportional, then third side of both triangles will also be in proportion.
Question. 8 If in two right triangles, one of the acute angles of one triangle is equal to an acute angle of the other triangle. Can you say that two triangles will be similar? Why?
Answer. True
Solution:

Thus, ΔABC ~ ΔPQR             [By AAA similarity criterion]

So, given statement is not correct.
 Question. 10  D is a point on side QR of ΔPQR such that PD ⏊ QR. Will it be correct to say that ΔPQD ~ ΔRPD? Why?
Answer. False
Solution:

Here, no other sides or angles are equal, so ΔPQD is not similar to ΔRPD.
Question. 11   In figure, if ∠D = ∠C, then it is true that ΔADE ~ΔACB?Why?

Question. 12 Is it true to say that, if in two triangles, an angle of one triangle is equal to an angle of another triangle and two sides of one triangle are proportional to the two sides of the other triangle, then the triangles are similar? Give reason for your answer.
Answer. False
Solution:
Considering the SAS similarity criterion, if one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar. But here, one angle and two sides of two triangles are equal but these sides not including equal angle, so given statement is not correct.
Exercise 6.3
Short Answer Type Questions (Q. NO. 1-8):
Question. 1 In a ΔPQR, PR²– PQ² = QR²and M is a point on side PR such that QM ⏊ PR. Prove that QM² = PM × MR.
Solution.
Given: ln ΔPQR,
            PR² – PQ² = QR²
And     QM ⏊ PR
To prove: QM² =PM × MR
Proof: Since, PR² – PQ² = QR²
⟹       PR² = PQ²+ QR²
⟹       ΔPQA is right angled triangle at Q with PR being its hypotenuse.
Also,    QM ⏊ PR
⟹      ∠QMR = ∠QMP = 90^o

Question. 2 Find the value of x for which ΔEAB in given figure.

Hence, the required value of x is 2
Question. 3 In figure, if ∠l = ∠2 and ΔNSQ≅ΔMTR, then prove that ΔPTS ~ ΔPRQ.


Hence proved.
Question. 4 Diagonals of a trapezium PQRS intersect each other at the point O, PQ || RS and PQ = 3RS. Find the ratio of the areas of ΔPOQ and ΔROS.
Solution.

Hence, the required ratio is 9:1
Question. 5 In figure, if AB || DC and AC, PQ intersect each other at the point O. Prove that OA.CQ = OC.AP.   

Hence proved.
Question. 6 Find the altitude of an equilateral triangle of side 8cm.
Solution.
Let ABC be an equilateral triangle.

Question. 7 If ΔABC ~ ΔDEF, AB = 4cm, DE = 6, EF = 9cm and FD = 12cm, then find the perimeter of ΔABC.
Solution.
Given,  ΔABC~ΔDEF
           AB = 4cm, DE = 6cm
And     EF = 9cm, FD = 12cm.

Thus, perimeter of D ABC = AB + BC + AC = 4 + 6 + 8 = 18cm
Question. 8 In figure, if DE || BC, then find the ratio of ar (ΔADE) and ar (DECB).

Solution.
Given, in ΔABC
   DE || BC
   DE = 6cm and BC =12cm
Now, in ΔABC and ΔADE,
            ∠ABC = ∠ADE                      [Corresponding angles]
            ∠ACB = ∠AED                      [Corresponding angles]
∴ ΔABC ~ ΔAED                               [By AAsimilarity criterion]
Using the property, “The ratio of areas of similar triangles is equal to the ratio of the squares of their corresponding sides,” we have:

Exercise 6.3
Short Answer Type Questions (Q. NO. 9-15):
Question. 9 ABCD is a trapezium in which AB ||DC and P,Q are points on AD and BC
respectively, such that PQ || DC, if PD = 18cm, BQ = 35cm and QC = 15cm, find AD.
Solution.


Question. 10 Corresponding sides of two similar triangles are in the ratio of 2:3. If the area of the smaller triangle is 48cm², then find the area of the larger triangle.
Solution.
Let the two triangles be ABC and PQR, where ΔABC is smaller of the two.
Given, ΔABC~ ΔDEF
            AB : DE = 2:3
Area of smaller ΔABC = 48cm²
Since ΔABC~ ΔDEF
Using the property, “The ratio of areas of similar triangles is equal to the ratio of the squares of their corresponding sides,” we have:

Question. 11 In a  ΔPQR, N is a point on PR, such that QN ^ PR. If PN.NR = QN², then prove that ∠PQR = 90°.
Solution.


Question. 12 Areas of two similar triangles are 36cm² and 100cm². If the length of a side of the larger triangle is 20cm. Find the length of the corresponding side of the smaller triangle.
Solution.
Let the two triangles be, ΔABC and ΔDEF where ΔABC is the smaller of the two and
          Area of ΔABC = 36cm²
And,    Area ΔDEF = 100cm²
Also, DE = 20cm
Let length of the corresponding side of ΔABC, AB = x cm
As ΔABC ~ ΔDEF
Therefore, using property, “The ratio of areas of similar triangles is equal to the ratio of the squares of their corresponding sides,” we have:

Hence, the length of corresponding side of the smaller triangle is 12cm.
Question. 13 In given figure, if ÐACB =ÐCDA, AC = 8cm and AD = 3cm, then find BD.


Question. 14 A 15 metres high tower casts a shadow 24 metres long at a certain time and at the same time, a telephone pole casts a shadow 16 metres long. Find the height of the telephone pole.
Solution.
Let BC = 15m be the tower and its shadow AB is 24m. Again, let EF = h be telephone pole and DE = 16m be its shadow.

Question. 15 Foot of a 10m long ladder leaning against a vertical wall is 6m away from the base of the wall Find the height of the point on the wall where the top of the Ladder reaches.
Solution.
Let BC be the vertical wall and AC = 10m is a ladder.
Distance of ladder from the base of the wall BC is 6m

Hence, the height of the point on the wall where the top of the ladder reaches is 8cm.
Exercise 6.4
Long Answer Type Questions (Q. NO. 1-6):
Question. 1 In given figure, if ÐA = ÐC, AB = 6cm, BP = 15cm, AP = 12cm and CP = 4cm, then find the lengths of PD and CD.


Question. 2 It is given that DABC ~ DEDF such that AB = 5 cm, AC = 7cm, DF = 15cm and DE = 12cm. Find the lengths of the remaining sides of the triangles.
Solution.

Hence, the lengths of remaining sides of given triangles are : EF = 16.8cm and BC = 6.25cm
Question. 3 Prove that, if a line is drawn parallel to one side of a triangle to intersect the other two sides, then the two sides are divided in the same ratio.
Solution. Let a  DABC in which a line DE is drawn parallel to BC intersecting AB at D and AC at E.


Hence proved.
Question. 4 In the given figure, if PQRS is a parallelogram and AB || PS then prove that OC || SR.
Solution.



Question. 5 A 5m long Ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4m high. If the foot of the ladder is moved 1.6m towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall.
Solution.
Let AC be the ladder and BC be the wall.
Let the two positions of ladder be represented by AC and DE.


Hence, the top of the ladder would slide upwards on the wall by a distance equal to 0.8m.
Question. 6 For going to a city B from city A there is a route via city C such that AC ⏊CB, AC = 2x km and CB = 2(x+ 7) km. It is proposed to construct a 26km highway which directly connects the two cities A and B. Find how much distance will be saved in reaching city B from city A after the construction of the highway.
Solution.        

Using Pythagoras theorem in right angled DACB, we have:
            AB² = AC² + BC²
⟹        (26)² = (2x)² + {2(x + 7)²
⟹        676 = 4x² + 4(x² + 49 + 14x)
⟹        676 = 4x² + 4x² + 196 + 56
⟹        676 = 8x² + 56x+ 196
⟹      8x² + 56x – 480 = 0
⟹           x² + 7x – 60 = 0
⟹       x²+ 12x – 5x – 60= 0              [Solving by factorization method]
⟹       x(x + 12) – 5(x + 12) = 0
⟹        (x + 12)(x – 5) = 0
⟹         x = -12 or 5
Ignoring x = -12, as distance cannot be negative, we have:
            x = 5
∴ AC = 2x = 10km
And BC = 2(x + 7) = 2(5 + 7) = 24km
So, distance covered from A to B via C= AC + BC = 10 + 24 = 34km
And, distance covered directly from A to B = AB = 26km
Hence, the distance saved = 34km – 26km =8km.