KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Additional Questions

KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Additional Questions

Karnataka State Syllabus Class 10 Maths Chapter 2 Triangles Additional Questions

I Multiple choice questions:

Question 1.
Two triangles are said to be similar if their corresponding sides are
a) Proportional
b) Equal
c) Paralled
d) Perpendicular
Answer:
a) Proportional

Question 2.
Congruent triangles are always,
a) Equal
b) similar
c) Not similar
d) Not equal
Answer:
a) Equal

Question 3.
Which of the following is can form similar triangles i) 3, 6, 9 ii) 9,18, 27 iii) 2,4, 6 iv) 5, 6, 7
a) i and ii
b) ii and iii
c) i and iii
d) i and iv
Answer:
a) i and ii

Question 4.
Basic proportional theorem is proposed by
a) Pythagoras
b) Thales
c) aryabhata
d) newton
Answer:
b) Thales

Question 5.
If straight line is drawn parallel to one side of a triangle then it divides the other two sides proportionally is called.
a) thales theorem
b) converse of thales theorem
c) Corollary of thales
d) Pythagoras theorem
Answer:
a) thales theorem

Question 6.
If straight-line divides two sides of a triangle proportionally then the straight line is parallel to the third side is called.
a) converse of thales theorem
b) pythogoral theorem
c) thales theorem
d) converse of Pythagoras theorem
Answer:
a) converse of thales theorem

Question 7.
In the given fig. which of following is

Answer:

 

Question 8.
In the given fig. which of the following Is true


Answer:

Question 9.
In the following figure DE || BC. AD = 4 cm, DB = 6 cm and AE =5 cm, then the value of EC is

a) 6.5 cm
c) 7.5 cm
b) 7 cm
d) 8.0 cm
Answer:
c) 7.5 cm

Question 10.
In the following figure DE || BC then the value of ‘x’ is

Answer:

Question 11.
In a trapezium the line joining the midpoints of non-parallel sides is.
a) Parallel to the parallel sides.
b) Half of the sum of the parallel side
c) both a and b
d) perpendicular to each other
Answer:
c) both a and b

Question 12.
If two triangles are equiangular then their corresponding sides are
a) proportional
b) equal
e) non-parallel
d) perpendicular
Answer:
a) proportional

Question 13.
If the three sides of a triangle are proportional to the corresponding three sides of another triangle then their corresponding angles are
a) equal
b) proportional
e) not equal
d) not proportional
Answer:
a) equal

Question 14.
In the given figure ∆ ADB ~ ∆ BDC then BD² is

a) AC . CD
b) AC . AD
c) AD . CD
d) A13 . BC
Answer:
a) AC . CD

Question 16.
In the given figure ∆ ADB ~ ∆ BDC then B² is

a) AC . AD
c) AD . CD
b) AC . CD
d) AB . BC
Answer:
a) AC . AD

Question 17.
If the area of two similar triangles are equal then they are
a) similar
b) congruent
e) proportional
d) equal
Answer:
d) equal

Question 18.
Area of similar triangles are proportional to
a) squares on their corresponding sides
b) squares on their corresponding altitude
e) squares on their corresponding medians
d) all the above
Answer:
d) all the above

Question 19.
The corresponding altitudes of two similar triangles are 9 cm and 15cm respectively then the ratio between their areas.
a) 9 : 15
b) 100 : 25
c) 81 : 225
d) 3 : 15
Answer:
c) 81 : 225

Question 20.
In a trapezium ABCD AB || CD and Its diagonals intersect at 0. if AB = 6 cm and DC = 3 cm then the ratio of the area of AOB and COD
a) 4 : 1
b) 1 : 2
c) 2 : 1
d) 1 : 4
Answer:
a) 4 : 1

II. Short Answer Questions:

Question 1.
State thales theorem (Basic proportionality theorem).
Answer:
A straight drawn parallel to one side of triangle then other two sides divides proportionally.

Question 2.
State converse of thales theorem.
Answer:
If two sides divides proportionally then straight line drawn parallel to given side.

Question 3.
What is the ratio of areas of two similar triangles whose sides are in the ratio 3 : 4
Answer:

∴ Ratio of areas are 9 : 16

Question 4.
If two triangles are similar such that the ratio of their areas 36 : 121, then what is the ratio of their corresponding medians.
Answer:

∴ ratio of their corresponding medians 6 : 11

 

Question 5.
Is the triangle with sides 12 cm, 16 cm and 18 cm a right triangle? give reason.
Answer:
122 + 162 = 182
144 + 256 = 324
400 ≠ 324
LHS ≠ RHS
∴ Given sides not form right angle ∆^le.

Question 6.
What is the name given to the longest side of a right-angled triangle
Answer:
hypotenuse

Question 7.
State pythagoras theorem
Answer:
In a right-angled triangle square on the hypotenuse is equal to sum of the squares of other sides.

Question 8.
State converse of pythagoras theorem.
Answer:
“If the square on the longest side of a triangle is equal to the sum of the squares on the other two sides, then those two sides contain a right angle”.

Question 9.
Verify m² – n², 2mn, m² + n² form the sides of a right-angled triangle.
Answer:
(m² + n²)² = (m² – n²)² + (2mn)²
m⁴ + 2m²n² + n⁴ = m⁴ – 2m²n² + n⁴ + 4m²n²
m⁴ + n⁴ + 2m²n² m – n⁴ + 2m²n²
LHS = RHS
given sides form right angle ∆^le.

Question 10.
In figure ∠ABC = 90°, BD ⊥ AC if BD = 8 cm, AD = 4 cm, find CD

Answer:
BD² = AD × CD
(8)² = 4 × CD
64 = 4 CD

III. Long answer questions:

Question 2.
ABC is an equilateral triangle of side 4a. Find each of its altitude.
Answer:

Let ABC be an equilateral triangle of side 4a units, We draw AD ⊥ BC. Then D is the mid-point of BC.

Now ABD is a right ∆^le right angle at D.
∴ AB² = AD² + (2a)². [By Pythagoras theorem]
= (4a)² = AD² + 4a²
= AD² = 16a² – 4a² = 12a²

Question 4.
ABC is a triangle in which AB = AC and D is a point on AC such that BC² = AC × CD Prove that BD = BC.
Answer:

Question 5.
A door of width 6 meter has an arc above it having a height of 2 meters find the radius of the arc?

Let the radius of the semi – circle
OA = OC = OB
= (x + 2)m
=> OA = OB. In ∆ OAB
∆ OAB is an isosceles ∆^le
∴ OD ⊥ AB.
=> OD bisects AB
∴ AD = BD = 3 cm.
In ∆^le OAD, ∠PDA = 90°
AO² = AD² + OD²
(x+2)² = 3² + x²
x² + 4x + 4 = 9 + x²
4x = 9 – 4 = 5
x = 5/4
Hence, radius = x + 2

Question 6.
In an equilateral ∆^le ABC, AD is drawn ⊥^r to BC meeting BC in D, prove that AD² = 3BD².

Answer:
In right ∆ ABD, by pythogoras theorem.
∴ AB² = AD² + BD²
In right ∆ ABC, by pythagoras theorem.
∴ AC² = AD² + DC²
Subtracting (ii) from (i) we get,
∴ AB² – AC² = BD² – DC²

Answer:
Given: ABC is right angled at B and D is the mid-point of B C
BD = DC = 1/2 BC
In ∆ ABD, AD² = AB² + BD²
[pythogoras theorem] – (1)
In ∆ ABC, AC² = AB² + BC²
[pythogoras theorem] – (2)
From eqn (1) and (2)

=> 4AD² = 4AB² + BC²
=> BC² = 4AD² – 4AB² – (3)
Using this eqn in (2)
AC² = AB² + 4AD² – 4AB²
AC² = 4AD² – 3AB².

Question 9.
Through the midpoint M of the sides of parallelogram ABCD. The line BM is drawn intersecting AC at L and AD produced to E. Prove that EL = 2BL.
Answer:

Question 11.
A ladder of length 2.6 m is lended against a wall. When it is at a distance of 2.4 m from the foot of the wall, the top of the ladder touches the bottom edge of the window in the wall, if the foot of the ladder is moved 1.4 m towards the wall it touches the top edge of the window, find the height of the window.
Answer:

Length of ladder AC = DE = 2.6 m
Height of window AD =?
BC = 2.4 m
EC = 1.4 m
BE = 1 m
In ∆^le ABC
AC² = AB² + BC²
AB² = AC² – BC²
AB² = (2.6)² – (2.4)²
= (2.6 + 2.4) x (2.6 – 2.4)
= 5 x 0.2 = lm
AB = 1m
In ∆^le DBE
DE² = BD² + BE²
(2.6)² – 1² = BD²
BD2 = (2.6 + 1) (2.6 – 1)
= 3.6 x 1.6
BD = 
BD = 2.4 m
∴ Height of window
AD = BD – AB
= 2.4 – 1 = 1.4
Height of window = 1.4m

 

Question 12.
In a ∆^le ABC, D and E be two points on side AB such that AD = BE : IF DE || BC and EQ || AC, then prove that PQ || AB.
Answer:

From fig, given BE = AD
EA = AD + DE
= BE + ED
= BD
Then (2) becomes

∴ PQ || AB [converse of BPT]

Question 13.
A tree 32 m tall broke due to gale and Its top fell at a distance of 16m from its foot at what height above the ground did the tree break?
Answer:

Height of the tree AB = 32 m
AD is broken part of tree = x
BC is distance between foot to the top oftreeBC = 16m
BD is the remaining height of tree = 32 – x
In Ale BDC ∠B = 90°
(CD)² = BD² + BC²
x² = (32 -x)² + (16)²
x² = 1024 + x² – 64x + 256
x² – x² = 1280 – 64x
64x = 1280
x = 1280/64
x = 20m
Remaining part of a tree
BD = (32 – x) = 32 – 20 = 12m

Question 14.
In ∆^le ABC, i) ∠ACB = 60°, ii) AD ⊥^ar BC. Derive an expression for AB in terms of AC and BC.
Answer:

In ∆^le ABC
i) ∠ACB = 60° and
ii) AD ⊥^r BC
Construction:
Draw DE such that ∠EDC = 60°
∴ Ale DEC is an equilateral
(Data and construction)
DE = DC = EC
Ale DEA is an isosceles Ale [∴ AED = 120 and ∠ ADE = 30° = ∠ DAE]
DE = EA
=> DC = EA = EC [∴ Axiom – 1]
∴ AC = 2DC
Now in Ale ADB, AB² = AD² + BD² [pythagoras theorem]
AB² = (AC² – DC²) + (BC -CD)²
AB² = 4DC² – DC² + BC² – 2BC DC + DC² [∴ AC = 2DC]
AB² = 4DC² + BC² – 2BC . DC
AB² = 4DC² – 2BC . DC + BC²
AB² = 2DC (2DC – BC) + BC²
= AC (AC – BC) + BC² [2DC = AC]
AB2 = AC² – AC . BC + BC²
AB² = AC² + BC² – AC.BC

Question 15.
ABCD is a rhombus. Prove that AC² + BD² = 4AB²
Answer:

Given: In rhombus ABCD, AC and BD are diagonals, AB = BC = CD = DA
∴ AO = 1/2 AC and BO = 1/2 BD
To prove: AC² + BD² = 4AB²
In right angle ∆^le AOB, ∠O = 90°
AB² = OA² + OB²
AB² = (1/2 AC)² + (1/2 BD)²

4AB² = AC² + BD²

Question 16.
AD is the altitude from A to BC in the ∆ ABC and DB:CD = 3:1 prove that BC² = 2(AB² – AC²)
Answer:

In ∆^le ABD ∠D = 90°
AB² = AD² + BD²
AD² = AB² – BD² – (1)
In ∆^le ADC ∠D = 90°
AC² = AD² + CD²
AD² = AC² – CD² – (2)
From (1) and (2)
AB² – BD² = AC² – CD²
AB² – AC² = BD² – CD²


BC² = 2(AB² – AC²)

 

Question 17.
In the given figure in ∆ ABC, AD ⊥ BC and AD² = BD × CD. Prove that the ∆ ABC is a right angle triangle.
Answer:

∆ ABC, AD ⊥ BC and AD² = BD × CD
To prove: ∆ ABC is a right angle triangle
Proof: In ∆^le ABD ∠D = 90°
AB2 = BD² + AD² – (1)
In ∆^le ADC ∠D = 90°
AC² = AD² + DC² – (2)
Adding eqn (1) and (2)
AB² + AC² = BD² + AD² + AD² + DC²
= 2 AD² + BD² + DC²
= 2(BD x CD) + BD² + DC²
= (BD + DC)²
∴ AB² + AC² = BC²
∆ ABC is a right triangle, right angled at A.

Question 18.
In trapezium ABCD, AB || CD and DC = 2AB. EF || AB, where E and F lie on BC and AD respectively such BE/EC=4/3 . Diagonal BD intersect EF at G. Prove that 7EF = 11 AB.
Answer:
In trapezium ABCD, AB || CD and

Question 20.
In ∆ MGN, MP ⊥ GN. If MG = a units, MN = b units, GP = c units and PN = d unit. Prove that (a + b) (a – b) = (c + d) (c – d).
Answer:

In ∆^le MGP ∠P = 90°
MG² = GP² + MP²
(a)² = (c)² + MP²
MP² = a² – c² – (1)
In ∆^le MNP ∠P = 90°
MN² = PM² + PN²
(b)² = d² + MP²
MP² = b² – d² – (2)
From (1) and (2)
a2 – c² = b² – d²
a² – b² = c2 – d²
(a + b) (a – b) = (c + d) (c – d).

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