ADDITIONAL QUESTIONS AND ANSWERS:

ADDITIONAL QUESTIONS AND ANSWERS:

1) 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?
Sol:  Since the perimeters and two sides are proportional
∴ The third side is proportional to the corresponding third side.
So the two triangles will be similar by SSS criterion.

2) A and B are respectively the points on the sides PQ and PR of a ∆PQR such that PQ = 12.5 cm, PA = 5 cm, BR = 6 cm, and PB = 4 cm.
Is AB || QR? Give reason.
Sol:

3) If ∆ABC ~ ∆QRP,  ar(ΔABC) /ar(ΔPQR) = 9/4, AB = 18 cm and BC = 15 cm, then find the length of RP.
Sol:

4) If it is given that ∆ABC ~ ∆PQR with BC/QR = 1/3, then
find ar(ΔPQR)/ ar(ΔABC)
Sol:

5) Is the triangle with sides 12 cm, 16 cm and 18 cm a right triangle? Give reason.
Sol:  In a right angled triangle (hypotenuse)² = (side)² + (side)²
Here, 12²+ 16² = 144 + 256 = 400 ≠ 18²
∴ The given triangle is not a right triangle.

6) In triangles PQR and TSM, ∠P = 55°, ∠Q = 25°, ∠M = 100°, and ∠S = 25°.
Is ∆QPR ~ ∆TSM? Why?
Sol:  Şince, ∠R = 180° – (∠P + ∠Q)
= 180° – (55° + 25°) = 100° = ∠M
∠Q = ∠S = 25° (Given)

Here ∠P=∠T = 55° ,∠ Q = ∠S = 25° and ∠R = ∠M = 100°
∆QPR ~ ∆STM
  ∆QPR is similar to ∆TSM.

7) If ΔABC and ΔDEF are similar triangles such that ∠A = 47° and ∠E = 63°, then the measures of ∠C = 70°. Is it true? Give reason.
Sol:  Since ∆ABC ~ ∆DEF
∴ ∠A = ∠D = 47°
∠B = ∠E = 63°
∴ ∠C = 180° – (∠A + ∠B) = 180° – (47° + 63°) = 70°
∴ The given statement is true.

8) ABC is an isosceles triangle right-angled at C. Prove that AB² = 2AC².
Sol:  ∆ABC is right-angled at C.
∴ AB² = AC² + BC² [By Pythagoras theorem]
⇒ AB² = AC² + AC²   [∵ AC = BC as it is an isosceles Δ]
⇒ AB² = 2AC²

9) Sides of triangle are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.
(i) 7cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
Sol: (i) Let a = 7 cm, b = 24 cm and c = 25 cm.
Here, largest side, c = 25 cm
We have, a² + b² = (7)² + (24)² = 49 + 576 = 625 = c²     [∵ c = 25]
So, the triangle is a right triangle.

(ii) Let a = 3 cm, b = 8 cm and c = 6 cm
Here, largest side, b = 8 cm
We have, a² + c² = (3)² + (6)² = 9 + 36 = 45 ≠ b²
So, the triangle is not a right triangle.

10) If triangle ABC is similar to triangle DEF such that 2AB = DE and BC = 8 cm. Then find the length of EF.
Sol:  ∆ABC ~ ∆DEF (Given)

11) If the ratio of the perimeter of two similar triangles is 4 : 25,
then find the ratio of the areas of the similar triangles.
Sol:  Ratio of perimeter of 2 ∆’s = 4 : 25
 Ratio of corresponding sides of the two ∆’s = 4 : 25
Now, the ratio of area of 2 ∆’s = Ratio of square of its corresponding sides.
= (4)²/(25)² = 16/625
∴ The ratios of the areas is 16:625

12) In an isosceles ∆ABC, if AC = BC and AB² = 2AC², then find ∠C.
Sol:

AB² = 2AC² (Given)
AB² = AC² + AC²
AB² = AC² + BC²   (∵ AC = BC as it is an isosceles Δ)
Hence AB is the hypotenuse and ∆ABC is a right angle A.
So, ∠C = 90°

13) The length of the diagonals of a rhombus are 16 cm and 12 cm. Find the length of side of the rhombus.

Sol: The diagonals of rhombus bisect each other at 90°.
∴ In the right angle ∆BOC
BO = 8 cm
CO = 6 cm
∴ By Pythagoras Theorem
BC² = BO² + CO² = 64 + 36
∴ BC² = 100
BC = 10 cm

14) A man goes 24 m towards West and then 10 m towards North. How far is he from the starting point?
Sol:

By Pythagoras Theorem
AC² = AB² + BC² = (24)² + (10)²
AC² = 676
AC = 26 m
∴ The man is 26 m away from the starting point.

15) ∆ABC ~ ∆DEF such that AB = 9.1 cm and DE = 6.5 cm. If the perimeter of ∆DEF is 25 cm. What is the perimeter of ∆ABC?
Sol: Since ∆ABC ~ ∆DEF.

16) ∆ABC ~ ∆PQR; if area of ∆ABC = 81 cm², area of ∆PQR = 169 cm² and AC = 7.2 cm. Find the length of PR.
Sol: Since ∆ABC ~ ∆PQR

17) E and F are points on the sides PQ and PR respectively of a ∆PQR. Show that EF ||QR  if PQ = 1.28 cm, PR= 2.56 cm, PE = 0.18 cm and PF = 0.36 cm.
Sol:

We have, PQ = 1.28 cm, PR = 2.56 cm
PE = 0.18 cm, PF = 0.36 cm
Now, EQ = PQ−PE = 1.28 – 0.18 = 1.10 cm and
FR = PR – PF = 2.56 – 0.36 = 2.20 cm

Therefore, EF || QR [By the converse of Basic Proportionality Theorem]

18) A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Sol: Let AB be a vertical pole of length 6m and BC be its shadow and DE be tower and EF be its shadow. Join AC and DF.
Now, in ∆ABC and ∆DEF, we have

Hence, height of tower, DE = 42m

19) In Fig. 7.13, if LM || CB and LN || CD, prove that ΔMAB = ΔNAD
Sol:

Firstly, in ∆ABC, we have
LM || CB (given)
Therefore, by Basic Proportionality Theorem, we have

20) In Fig. 7.14, DE || OQ and DF || OR. Show that EF || QR.
Sol: In  ΔPOQ, we have
DE || OQ (Given)

[Applying the converse of Basic Proportionality Theorem in ∆PQR]

21) Using converse of Basic Proportionality Theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.
Sol:

 

∆ABC in which D and E are the mid-points of sides AB and AC respectively(given).
We have to prove that  DE || BC
Proof: Since D and E are the mid-points of AB and AC respectively
∴ AD = DB and AE = EC

Therefore, DE || BC (By the converse of Basic Proportionality Theorem)

22) State which pairs of triangles in the following figures are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form.

Sol:
i) In ∆ABC and ∆QRP, we have

∆NML is not similar to ∆PQR.

23) In Fig. 7.17, AO/OC = BO/OD = 12 and AB = 5 cm.
Find the value of DC.
Sol:

⇒ DC = 10 cm.

24) S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS. Show that ∆RPQ ~ ∆RTS.
Sol:

In ∆RPQ and ∆RTS, we have
∠RPQ = ∠RTS (given)
∠PRQ = ∠TRS = ∠R (Common angle )
∴ ∆RPQ ~ ∆RTS   (By AA criterion of similarity)

25) D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC.
Show that CA² = CB.CD.
Sol:

26) ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Sol:

Let ABC be an equilateral triangle of side 2a units.
We draw AD ⊥ BC. Then D is the mid-point of BC.
⇒ BC/2 = 2a/2 = a
Now, ΔABD is a right triangle right-angled at D.
⇒ AB² = AD² + BD²   [By Pythagoras Theorem]
⇒ (2a)² = AD² + a²
⇒ AD² = 4a² – a² = 3a²
⇒ AD = √3a
Hence, each altitude = √3a unit.

27) In Fig. 7.26, if ∆ABC ~ ∆DEF and their sides are of lengths (in cm) as marked along with them, then find the lengths of the sides of each triangle.
Sol:  ∆ABC ~ ∆DEF (given)

⇒ 4x – 2 = 18
⇒ x = 5
∴ AB = 2 × 5 – 1 = 9,    BC = 2 × 5 + 2 = 12
CA = 3 × 5 = 15,     DE = 18,    EF = 3 × 5 + 9 = 24
and FD = 6 × 5 = 30
Hence, AB = 9 cm, BC = 12 cm, CA = 15 cm
DE = 18 cm, EF = 24 cm, FD = 30 cm

28) In ΔABC, it is given that AB/AC = BD/DC . If ∠B = 70° and ∠C = 50° then find ∠BAD.
Sol:

In ∆ABC
∵ ∠A + ∠B + 2C = 180° (Angle sum property)
∠A + 70° + 50° = 180°
⇒ ∠A = 180° – 120°
⇒ ∠A = 60°
∵ AB/AC = BD/DC    (given)
∴ ∠1 = ∠2
[Because if a line through one vertex of a triangle divides the opposite sides in the ratio of the other two sides, then the line bisects the angle at the vertex.]
But ∠1 + ∠2 = 60° …(ii)
From (i) and (ii) we get,
2∠1 = 60°
⇒ ∠1 = 60°/2 = 30°
Hence, ∠BAD = 30°

29) ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO.
Sol: Given: ABCD is a trapezium, in which AB || DC and
its diagonals intersect each other at point O.

30) The perpendicular from A on side BC of a ∆ABC intersects BC at D such that DB = 3CD (see Fig. 7.42). Prove that 2AB² = 2AC² + BC²
Sol:

We have, DB = 3CD
Now,
BC = BD + CD
⇒ BC = 3CD + CD = 4CD    (given DB = 3CD)
∴ CD = 14 BC
and DB = 3CD = 14BC
Now, in right angled triangle ABD using Pythagoras Theorem we have
AB² = AD² + DB² …(i)
Again, in right angled triangle ∆ADC, we have
AC² = AD² + CD² …(ii)
Subtracting (ii) from (i), we have
AB² – AC² = DB² – CD²

∴ 2AB² – 2AC² = BC²
⇒ 2AB² = 2AC² + BC²