Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers

Here you will find Triangles Class 10 Extra Questions Maths Chapter 6 with Answers Solutions, Extra Questions for Class 10 Maths are solved by experts and will guide students in the right direction.

Extra Questions for Class 10 Maths Triangles with Answers Solutions

Extra Questions for Class 10 Maths Chapter 6 Triangles with Solutions Answers

Triangles Class 10 Extra Questions Objective Type

Question 1.
In the figure, a line segment PQ is drawn || to the base BC of ∆ABC If PQ : BC = 1: 3, then the ratio of AP and PB will be :
(a) 1 : 2
(b) 1 : 3
(c) 1 : 4
(d) 2 : 3.
Answer:
(c) 1 : 4
Solution.
Given PQ || BC and PQ : BC = 1 : 3
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 1
∴ AP : PB = 1 : 2
Hence Choice (c) is correct

Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers

Question 2.
In the figure AB = 3 cm, AC = 6 cm, BD = 2 cm and CD = 4 cm. Find the ratio ∠BAD and ∠CAD is
(a) 2 : 4
(b) 1 : 1
(c) 3 : 6
(d) 6 : 3
Answer:
(b) 1 : 1
Solution.
In ∆ADB, sin ∠BAD = \(\frac{B D}{B A}\) = \(\frac{2}{3}\)
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 2
In ∆ADC, sin ∠CAD =\(\frac{4}{6}\) = \(\frac{2}{3}\)
∴ ∠BAD = ∠CAD
∴ ratio is 1 : 1 choice (b) is correct

Question 3.
In a ∆ABC, if DE is drawn parallel to BC, cutting AB and ACat Dand E respectively such that AB = 7.2 cm, AC = 6.4 cm and AD = 4.5 cm. Then AE = ?
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 3
(a) 5.4 cm
(b) 4 cm
(c) 3.6 cm
(d) 3.2 cm
Answer:
(b) 4 cm

Question 4.
In ∆ABC, DE || BC so that AD = (7x – 4) cm, AE = (5x – 2) cm, DB = (3x + 4) cm and EC = 3x cm. Then, we have :
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 4
(a) x = 3
(b) x = 5
(c) x = 4
(d) x = 2.5
Answer:
(c) x = 4

Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers

Question 5.
In ∆ABC, DE || BC such that \(\frac{AD}{DB}\) = \(\frac{3}{5}\). If AC = 5.6 cm, then AE = ?
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 5
(a) 4.2 cm
(b) 3.1 cm
(c) 2.8 cm
(d) 2.1 cm
Answer:
(d) 2.1 cm

Question 6.
ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the area of ∆ABC and ∆BDE is :
(a) 2 : 1
(b) 1 : 2
(c) 4 : 1
(d) 1 : 4.
Answer:
(c) 4 : 1

Question 7.
Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio :
(a) 2 : 3
(b) 4 : 9
(c) 81 : 16
(d) 16 : 81.
Answer:
(d) 16 : 81.

Question 8.
Hari goes 18 m due east and then 24 m due north. The distance from the starting point is :
(a) 40 m
(b) 30 m
(c) 26 m
(d) 42 m.
Answer:
(b) 30 m

Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers

Question 9.
The length of the second diagonal of a rhombus, whose side is 5 cm and one of the diagonals is 6 cm is :
(a) 7 cm
(b) 8 cm
(c) 9 cm
(d) 12 cm.
Answer:
(b) 8 cm

Question 10.
In ∆ABC, AB = 6√3, AC = 12 cm and BC = 6 cm. The angle B is :
(a) 120°
(b) 60°
(c) 90°
(d) 45°
Answer:
(c) 90°

Triangles Class 10 Extra Questions Very Short Answer Type

Question 1.
In the figure, DE || BC. If AD = x, BD = x – 2, AE = x + 2 and EC = x – 1, find the value of x.
Solution.
In ∆ABC, DE || BC
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 6
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 7
⇒ (x – 2) (x + 2) = x (x – 1)
⇒ x2 – 4 = x2 – x
∴ x = 4

Question 2.
State whether the following quadrilaterals are similar or not 3 cm C
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 8
Solution.
The two quadrilaterals, in figure are not similar because their corresponding angles are not equal. It is clear from the figure that, ∠A is 90° but ∠P is not 90°.

Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers

Question 3.
E and Fare points on the sides PQ and PR respectively of a A PQR, for each of the following cases, state whether EF || QR
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm.
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
(iii) PQ = 1:28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm.
Solution.
(i) In figure,
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 9
⇒ EF is not parallel QR because converse of basic proportionality theorem is not satisfied.

(ii) In figure,
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 10
⇒ EF || QR because converse of basic proportionality of theorem is satisfied

(iii) In figure,
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 11
⇒ EF || QR because converse of basic proportionality theorem is satisfied

Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers

Question 4.
In the figure, if \(\frac{A O}{O C}=\frac{B O}{D O}=\frac{1}{2}\) and AB = 5 cm find the value of DC.
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 13
Solution.
In ∆OAB and ∆OCD
∠AOB = ∠COD (vertically opposite angles)
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 14
∴ DC = 5 × 2 = 10 cm.

Question 5.
Perimeters of two similar triangles are 40 cm and 60 cm respectively. Find ratio among their areas.
Solution.
Given that the triangles are similar
∴ ratio of the perimeters of triangle = 40 : 60
= 2 : 3
So the ratio of the area = (2)2 : (3)2
= 4 : 9

Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers

Question 6.
Using theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (recall that you done it in Class IX).
Solution.
In ∆ABC, D and E are midpoints of side AB and AC, respectively.
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 15
(By converse of basic proportionality theorem)

Question 7.
In figure, E is a point on side CB produced of an isosceles ∆ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ∆ABD ~ ∆ECF
Solution.
In figure, we are given that ∆ABC is isosceles
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 16
and AB = AC =
⇒ ∠B = ∠C …..(i)
For ∆ABD and ∆ECF,
∠ABD = ∠ECF [From Eq. (i)]
and ∠ADB = ∠EFC [Each = 90°]
⇒ ABD ~ AECF
(AAA similarity criterion)

Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers

Question 8.
D is point on the side BC of a ∆ABC such that ∠ADC = ∠BAC. Show that CA2 = CB · CD.
Solution.
Draw a ∆ABC such that D is a point on BC and join AD.
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 17
For ∆ABC and ∆DAC, we have
∠BAC = ∠ADC (Given)
and ∠ACB = ∠DCA (Common ∠C)
⇒ ∆ABC ~ ADAC
(AAA similarity criterion)
⇒ \(\frac{A C}{C B}\) \(\frac{C D}{C A}\)
⇒ \(\frac{C A}{C D}\) \(\frac{C B}{C A}\)
⇒ CA × CA = CB × CD
⇒ CA2 = CB × CD

Question 9.
Let ∆ABC ~ ∆DEF and their areas be, 64 cm2 and 121 cm2, respectively. If EF = 15.4 cm, find BC.
Solution.
∆ABC ~ ∆DEF (Given)
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 18
(Using property of area of similar triangles).
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 19

Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers

Question 10.
Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD. Find the ratio of the area of ∆AOB and ∆COD.
Solution.
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 20

Question 11.
If the areas of two similar triangles are equal, prove that they are congruent.
Solution.
Let ∆ABC ~ ∆PQR and ar
(∆ABC) = ar (∆PQR) (Given)
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 21
(Using property of area of similar triangles)
⇒ AB = PQ, BC = QR
and CA = PR
(SSS proportionality criterion)
⇒ ∆ABC ≅ ∆PQR.

Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers

Question 12.
Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.
Solution.
Draw ABCD is a square having sides of length = a
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 22
Then, the diagonal, BD = a√2
We construct equilateral ∆PAB and ∆QBD.
∆PAB ~ ∆QBD
(Equilateral triangles are similar)
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 23

Question 13.
ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2.
Solution.
Draw ABC is an isosceles triangle right angled at C.
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 24
and AC = BC …..(i)
By Pythagoras theorem, we have
AB2 = AC2+ BC2 = AC2 + AC2 = 2AC2
[∵ BC = AC by Eq. (i)]

Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers

Question 14.
A ladder 10 m long reaches a window 8 cm above the ground. Find the distance of the foot of the ladder from base of the wall.
Solution.
Let B be the position of the window and CB be the length of the ladder.
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 25
Then, AB = 8 cm (Height of window)
BC = 10 cm (Length of ladder)
Let AC = x m be the distance of the foot of the ladder from the base of the wall.
Using Pythagoras theorem in ∆ABC, we get
AC2 + AB2 = BC2
∴ x2 + (8)2 = (10)2
⇒ x2 = 100 – 64 = 36
⇒ x = 6, i.e., AC = 6 cm

Triangles Class 10 Extra Questions Short Answer Type

Question 1.
In figures, (i) and (ii), DE || BC. Find EC in figure (i) and AD in figure (ii).
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 26
Solution.
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 27

Question 2.
In ∆ABC, if the side AD is perpendicular to side BC and AD2 = BD × CD. Prove that ∆ABC is a right angle ∆.
Solution.
In ∆ABD and ∆ADC, ∠D = 90°
∴ AB2 = AD2 + BD2 …(i)
and AC2 = AD2 + CD2 … (ii)
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 28
Adding (i) & (ii)
AB2 + AC2 = 2AD2 +BD2+CD2
= 2(BD × CD) + BD2+CD2
= BD2 + 2BD × CD + CD2
= 2(BD + CD)2
(Given AD2 = BD × CD).
= BC2
∴ ∠BAC = 90°
Hence ∆BAC is a right angle ∆.

Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers

Question 3.
In figure, AD ⊥ BC
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 29
Prove that AB2 + CD2 = BD2 + AC2
Solution :
In ∆ADB, ∠D = 90°
∴ AB2 = AD2 + BD2 …(i)
again in ∆ADC, ∠D = 90°
AC2 = AD2 + CD2 …(ii)
Subtract (ii) from (i)
AB2 – AC2 = BD2 – CD2
⇒ AB2 + CD2 = AC2 + BD2

Question 4.
Using theorem, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. :
Solution.
In ∆ABC, D is the mid-point of AB.
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 30
i.e., \(\frac{A D}{D B}\) = 1 ……(i)
As straight line l || BC.
Line l is drawn through D and it meets AC at E.
By basic proportionality theorem,
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 31
⇒ E is the mid-point of AC.

Question 5.
The diagonals of a quadrilateral ABCD intersect each other at the point O. such \(\frac{A O}{B O}=\frac{C O}{D O}\) Show that ABCD is a trapezium.
Solution.
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 32
⇒ OE || CD(By converse of basic proportionality theorem)
Now, we have BA || OE and OE || CD = AB || CD
⇒ Quadrilateral ABCD is a trapezium.

Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers

Question 6.
S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS. Show that ∆RPQ ~ ∆RTS.
Solution.
Draw a ∆RPQ such that S and T are points on PR and QR and joining them.
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 33
In figure, we have ∆RPQ and ∆RTS in which
∠RPQ = ∠RTS (Given)
∠PRQ = ∠SRT (Each = ∠R)
Then, by AAA similarity criterion, we have
∆RPQ ~ ∆RTS
Note: If any two corresponding angles of the triangles are equal, then their third corresponding angles are also equal by AAA.

Question 7.
In figure, altitudes AD and CE of ∆ABC intersect each other at the point P. Show that
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 34
(i) ∆AEP ~ ∆CDP
(ii) ∆ABD ~ ∆CBE
(iii) ∆AEP ~ ∆ADB
(iv) ∆PDC ~ ∆BEC
Solution.
(i) In figure, ∠AEP = ∠CDP (Each = 90°)
and ∠APE = ∠CPD
(Vertically opposite angles)
⇒ ∠AEP ~ ∆CDP
(By AAA similarity criterion)

(ii) In figure,
∠ADB = ∠CEB (Each = 90°)
and ∠ABD = ∠CBE (Each = ∠B)
⇒ ∆АВD ~ ∆СВЕ
(By AAA similarity criterion)

(iii) In figure,
∠AEP = ∠ADB (Each = 90°)
and ∠PAE = ∠DAB (Common angle)
⇒ ∆AEP ~ ∆ADB
(By AAA similarity criterion)

(iv) In figure,
∠PDC = ∠BEC (Each = 90°)
and ∠PCD = ∠BCE (Common angle)
⇒ ΔPDC ~ ΔBEC
(By AAA similarity criterion)

Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers

Question 8.
E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ∆ABE ~ ∆CFB.
Solution.
Draw a parallelogram ABCD and produce a line AD to AE and joining BE. In parallelogram ABCD,
∠A = ∠C …(i)
Now, for ∆ABE and ∆CFB,
∠EAB = ∠BCF [From Eq. (i)],
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 35
∠ABE = ∠BFC
(Alternate angles as AB || FC)
∆ABE ~ ∆CFB (AAA similarity)

Question 9.
In figure, ABC and AMP are two right triangles, right angled at B and M, respectively. Prove that
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 36
(i) ∆ABC ~ ∆AMP
(ii) \(\frac{C A}{P A}=\frac{B C}{M P}\)
Solution.
(i) In figure, we have ∠ABC = ∠AMP (Each = 90°)
Because the ∆ABC and ∆AMP are right angled at B and M, respectively.
Also, ∠BAC = ∠PAM
(Common angle ∠A)
⇒ ABC – ∆AMP
(By AAA similarity criterion)

(ii) As ∆ABC ~ ∆AMP,
\(\frac{A C}{A P}=\frac{B C}{M P}\)
(Ratio of the corresponding sides of similar triangles)
⇒ \(\frac{C A}{P A}=\frac{B C}{M P}\)

Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers

Question 10.
In a triangle ABC, AD is the median of BC and E is mid-point of AD. If BE produced, it meets AC in figure. Prove that AF = \(\frac {1}{3}\) AC.
Solution.
Given: In ∆ABC, AD is the median and E is the mid-point of AD.BE is joined and produced intersect AC at F.
T.P.T. – AF = \(\frac {1}{3}\) AC
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 37
Construction-Draw a line parallel to BF through B, which intersect AC at G. EF || DG
Proof – In ∆ADC, EF || DG
∴ AF = FG (Mid-point theorem) …(i)
In ABFC, EF || DG
∴ FG = GC (Mid-point theorem) ….(ii)
From Eq. (i) & (ii)
AF = FG = GC Now
AC = AF + FG + GC
= AF + AF + AF
= 3AF
AF = \(\frac {1}{3}\)AC.

Question 11.
BL and CM are the medians of a right triangle ABC right angle at A. Prove that 4(BL2 + CM)2 = 5 BC2.
Solution.
Given,
In ∆BAC, ∠A = 90° and BL and CM are the medians.
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 38
T.P.T. – 4(BL2+cm2) = 5BC2
Proof – In ∆BAC, ∠A = 90°
AB2 + AC2 = BC2 ….(i)
In ∆BAL, ∠A = 90°
BL2 = AB2 + AL2
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 39
In ∆MAC,  ∠A = 90°
CM2 = AM2 = AC2
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 40
⇒ 4(BL2 + cm2) = 5(AB2 + AC2)
= 5BC2 From Eq. (i)
(AB2 + AC2 = BC2)

Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers

Question 12.
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.
Soultion.
In figure (i), AB is a pole behind it a Sun is risen which casts a shadow of length BC = 4 cm and makes an angle θ to the horizontal and in figure it, PM is a height of the tower and behind a Sun risen which casts a shadow of length, NM = 28 cm.
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 41
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 42

Question 13.
In figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O,
show that
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 43
Solution.
Draw AL ⊥ BC and DM ⊥ BC (See figure)
∠ALO = ∠DMO = 90°
and ∠AOL = ∠DOM
(Vertically opposite angle)
∴ ΔOLA ~ ΔΟΜD
(AAA similarity criterion)
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 44

Question 14.
D, E and Fare respectively the midpoint of sides AB, BC and CA of ∆ABC. Find the ratio of the areas of ∆DEF and ∆ABC.
Solution.
Draw a ∆ABC taking mid-points D, E and F on AB, BC and AC. Join them.
Here, DF = \(\frac {1}{2}\)BC, DE = \(\frac {1}{2}\)CA
and EF = \(\frac {1}{2}\)AB …….(i)
(∵ D, E and Fare mid-points of sides AB, BC and CA respectively)
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 45
[∵ ar (∆CAB)= ar (∆ABC)]
Hence, the required ratio is 1 : 4.

Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers

Question 15.
Equilateral triangles are drawn on the sides of a right angled triangle. Show that the area of the triangle on the hypotenuse is equal to the sum of the areas of triangles on the other two sides.
Solution.
Given. A right angled triangle ABC, with right angle at B. Equilateral triangles PAB, QBC and RAC are described on sides AB, BC and CA respectively.
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 46
To prove. Area (∆PAB) + area (∆QBC) = area (∆RAC)
Proof. Since, triangles PAB, QBC and RAC are equilateral. Therefore, they are equiangular and hence similar.
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 47
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 48

Question 16.
In adjoining Fig. ABC and BCD are two triangles on the same base BC. If AD intersects BC at O, show that
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 49
Solution.
Given. ∆ABC and ABCD are on the same base and AD intersects BC at O.
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 50
Construction. Draw AM ⊥ BC and DN
Proof. In ∆AMO and ∆DNO,
M = ∠N
[Each 90°, by construction]
∠AOM – ∠DON [same angles]
∆AMO ~ ∆DNO [AA corollary]
⇒ \(\frac{A O}{D O}=\frac{A M}{D N}\)
[Corresponding sides are proportional in similar triangles]
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 51

Question 17.
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM × MR.
Solution.
In ∆PQR and ∆MPQ,
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 52
∠1 + ∠2 = ∠2 + ∠4 (Each = 90°)
⇒ ∠1 = ∠4
Similarly, ∠2 = ∠3
and ∠PMR = ∠PMQ (each 90°)
∆QPM ~ ∆PRM (AAA criterion)
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 53
⇒ PM2 = QM × RM
or PM2 = QM × MR

Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers

Question 18.
ABC is an isosceles triangle with AC = BC. If AB2 = 2AC?, prove that ABC is a right triangle.
Solution.
Draw an isosceles ∆ABC with AC =BC.
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 54
In ∆ABC, we are given that
AC = BC …..(i)
and AB2 = 2AC2 …(ii)
Now, AC2 + BC2-AC2 + AC2 [By Eq. (i)]
= 2AC2 = AB2 (By Eq. (ii)]
i.e., AC2 + BC2 = AB2
Hence, by the converse of the Pythagoras theorem, we have ∆ABC is right angled at C.

Question 19.
ABC is an equilateral triangle of side 2a. Find each of its altitude.
Solution.
Draw equilateral ∆ABC, each side is 2a.
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 55
Also, draw AD ⊥ BC.
Where AD is an altitude.
In ∆ADB and ∆ADC
AD = AD (Common)
and ∠ADB = ∠ADC = 90°
∆ADB ≅ ∆ADC (RHS congruency)
BD = CD = \(\frac {1}{2}\)BC = a
(∵ In an equilateral triangle altitude AD is the perpendicular bisector of BC).
Now, from ∆ABD by Pythagoras theorem, we get
AB2 = AD2 + BD2
⇒ (2a)2 = AD2 + a2
⇒ AD2 = 3a2
⇒ AD = √3a.

Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers

Question 20.
Prove that the sum of the square of the sides of a rhombus is equal to the sum of the squares of its diagonals.
Solution.
Draw ABCD is a rhombus in which AB = BC = CD = DA = a (Say)
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 56
Its diagonal AC and BD are right angled bisector of each other at O.
In ∆OAB, ∠AOB = 90°,
OA = \(\frac {1}{2}\)AC and OB = \(\frac {1}{2}\)BD
In ∆AOB, use Pythagoras theorem, we have
OA2 + OB2 = AB2
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 57
⇒ AC2 + BD2 = 4AB2
or 4AB2 = AC2 + BD2
⇒ AB2 + BC2 + CD2 + DA2 = AC2 + BD2
(∵ AB = BC = CD = DA)
Hence proved.

Question 21.
In figure, O is a point in the interior of a ∆ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 58
(i) OA2 + OB2+OC2-OD2-OE2-OF2 – AF2 + BD2 + CE2
(ii) AF2+ BD2 + CE2 = AE2 + CD2 + BF2
Solution.
In ∆ABC, from point O join lines OB, OC and OA.
(i) In right angled ∆OFA,
OA2 = OF2 + AF2
(By Pythagoras therorem)
⇒ OA2 – OF2 = AF2 …..(i)
Similarly, in ∆OBD,
OB2 – OD2 = BD2 …(ii)
and in ∆OCE,
OC2 – OE2 = CE2 …(iii)
On adding Eqs. (i), (ii) and (iii), we get
OA2 + OB2 + OC2 – OD2 – OE2 – OF2
= AF2 + BD2 + CE2

(ii) From part Eqs. (i), we get
OA2 + OB2 + OC2 – OD2 – OE2 – OF2
= AF2 + BD2 + CE2 …(iv)
Similarly,
OA2 + OB2 + OC2 – OD2 – OE2 – OF2
= BF2 + CD2 + AE2 …(v)
From Eqs. (iv) and (v), we have
AF2 + BD2 + CE2 = AE2 + CD2 + BF2

Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers

Question 22.
A guy wire attached to a vertical pole of height 18m is 24m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
Solution.
Let AB be the vertical pole of height 18 m, A guy wire is of length BC = 24 m.
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 59
Let AC = x m be the distance of the stake from the base of the pole.
Using Pythagoras theorem in ∆ABC, we get
i.e., AC2 + AB2 = BC2
∴ x2 + (18)2 = (24)2
⇒ x2 = (24)2 – (18)2
= 576 – 324 = 252
⇒ x = √252 m
(∵ We take positive sign because cannot be negative)
⇒ x = 6 √7 m

Question 23.
An aeroplane leaves an airport and flies due North at a speed of 1000 kmh-1. At the same time, another aeroplane leaves the same airport and flies due West at a speed of 1200 kmh-1. How far apart will be two planes after 1\(\frac {1}{2}\)h?
Solution.
The first plane travels distance BC in the direction of North in 1\(\frac {1}{2}\)h at speed of 1000 km/h.
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 60

Question 24.
Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the f00t of the poles is 12 m, find the distance between their tops.
Solution.
Let BC and AD be the two poles of heights 11 m and 6 m.
Then, CE = BC – AD
= 11 – 6
= 5 cm
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 61
Let distance between tops of two poles DC m = x cm
Using Pythagoras theorem in ∆DEC, we get
i.e., DC2 = DE2 + CE2
⇒ x2 = (12)2 + (5)2 = 169
⇒ x = 13
Hence, distance between their tops = 13 m.

Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers

Question 25.
D and E are points on the sides CA and CB, respectively of a ∆ABC right angled at C. Prove that AE2 + BD2 = AB2 + DE2.
Solution.
Draw a right ∆ABC at C. Take D and E points on the sides CA and BC and join ED, BD and EA.
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 62
In right angled ∆ACE,
AE2 = CA2 + CE2 ……(i)
(By pythagoras theorem)
and in right angled ABCD,
BD2 = BC2 + CD2 …(ii)
On adding Eqs. (i) and (ii), we get
AE2 + BD2 = (Ca2 + CE2) + (BC2 + CD2)
= (BC2 + Ca2) + (CD2 + CE2)
(∵ In ∆ABC, Ba2 = BC2 + CA2 and in ∆ECD, DE2 = CD2 + CE2)
= BA2 + DE2
(By Pythagoras theorem),
∴ AE2 + BD2 = AB2 + DE2
Hence proved.

Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers

Question 26.
In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.
Solution.
Draw ∆ABC is an equilateral triangle of side a.(Say)
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 63
and AD ⊥ BC
Let AD = x
Now, BD = CD = \(\frac {1}{2}\)BC = \(\frac {1}{2}\)a
(In an equilateral triangle altitude AD is a perpendicular bisector of BC)
In right angled ∆ABD,
AB2 = AD2 + BD2
⇒ a2 = x2 + (\(\frac {1}{2}\)a)2
⇒ a2 = x2 + \(\frac {1}{4}\)a2
⇒ 4a2 = ax2 + a2
⇒ 3a2 = 4x2
Hence proved.

Triangles Class 10 Extra Questions Long Answer Type

Question 1.
In figure, if LM || CB and LN || CD,
prove that \(\frac{A M}{A B}=\frac{A N}{A D}\)
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 64
Solution.
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 65

Question 2.
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that \(\frac{A O}{B O}=\frac{C O}{D O}\).
Solution.
We draw,
EOF ||AB (Also || CD)
In ∆ACD, OE || CD
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 66

Question 3.
Prove that If the corresponding sides of two triangles are proportional then their corresponding angles are equal, and hence the two triangles are similar.
Solution.
Given. ∆ABC and ∆DEF in which
\(\frac{A B}{D E}\) = \(\frac{B C}{B F}\) = \(\frac{A C}{D F}\)
To Prove. ∆ABC ~ ∆DEF.
Construction. Let us take ∆ABC and ∆DEF such that
\(\frac{A B}{D E}\) = \(\frac{B C}{B F}\) = \(\frac{A C}{D F}\)(<1).
Cut DP = AB and DQ = AC. Join PQ.
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 67
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 68
⇒ BC = PQ
Thus, AB = DP, AC = DQ and DC = PQ
∴ ∆ABC ≅ ∆DPQ
[By SSS-congruence]
∴ ∠A = ∠D, ∠B = P = ∠E
and C = ∠Q = ∠F
⇒ ∠A = ∠D, ∠B = ∠E and
∠C = ∠F.
Thus, the given triangles are equiangular and hence similar.

Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers

Question 4.
Prove that 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.
Answer:
Given. ∆ABC and ∆DEF in which
∠A = ∠D and \(\frac{A B}{D E}\) = \(\frac{A C}{D F}\)
To prove. ∆ABC ~ ∆DEF.
Construction. Let us take ∆ABC and ∆DEF such that
\(\frac{A B}{D E}\) = \(\frac{A C}{D F}\) (<1) and ∠A = ∠D.
Cut DP = AB and DQ = AC. Join PQ.
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 69
Proof. In ∆ABC and ADPQ, we have :
AB = DP [By construction]
∠A = ∠D (Given)
AC = DQ [By construction]
∆ABC ≅ ∆DPQ [By SAS-congruence]
∴ ∠A = ∠D, ∠B = ∠P
and ∠C = ∠Q.
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 70
[∵ AB = DP and AC = DQ)
⇒ PQ || EF
[By the converse of Thales’ theorem)
⇒ ∠P = ∠E and ∠Q = ∠F
[Corresponding ∠s]
∴ ∠A = ∠D, ∠B = P = ∠E
and C = 2Q = ∠F.
Thus, ∠A = ∠D, ∠B = ∠E
and C = ∠F.
So, the given triangles are equiangular and hence similar.

Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers

Question 5.
Prove that If two triangles are equiangular, prove that the ratio of their corresponding sides is the same as the ratio of the corresponding medians.
Solution.
Given. ∆ABC and ADEF in which ∠A = ∠D, ∠B = ∠E and ∠C = ∠F, and AL and DM are the medians.
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 71
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 72

Question 6.
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other the point O. Using a similarity criterion for two triangles, show that
\(\frac{O A}{O C}=\frac{O B}{O D}\)
Solution.
Draw ABCD is a trapezium and AC and BD are diagonals intersect at O.
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 73
In figure, AB || DC (Given)
⇒ ∠1 = ∠3, ∠2 = 24
(Alternate interior angles)
Also, ∠DOC = ∠BOA
(Vertically opposite angles)
⇒ ∆OCD ~ ∆QAB (Similar triangle)
⇒ \(\frac{O C}{O A}=\frac{O D}{O B}\)
(Ratios of the corresponding sides of the similar triangles)
⇒ \(\frac{O A}{O C}=\frac{O B}{O D}\) (Taking reciprocals)
Hence proved.

Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers

Question 7.
Sides AB and AC and median AD of a ∆ABC are respectively proportional to sides PQ and PR and median PM of another ∆PQR. Show that ∆ABC ~ ∆PQR.
Solution.
Given, in ∆ABC and ∆PQR,
AD and PM are their medians, respectively.
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 74
To Prove. ∆ABC ~ ∆PQR
Construction. Produce AD to E such that AD = DE and produce PM to N such that PM = MN. Join BE, CE, QN and RN.
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 75
Proof. Quadrilaterals ABEC and PQNR are parallelograms because their diagonals bisect each other at D and M, respectively.
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 76
From Eqs. (ii) and (iii), we have
\(\frac{A B}{P Q}=\frac{B E}{Q N}=\frac{A E}{P N}\)
⇒ ΔΑΒΕ ~ ΔΡΩΝ
⇒ ∠1 = ∠2 ….(iv)
Similarly, we can prove that
∆ACE ~ ∆PRN
∠3 = ∠4 …..(v)
On adding Eqs. (iv) and (v), we have
∠1 + ∠3 = ∠2 + ∠4
⇒ ∠A = ∠P
⇒ ∆BC ~ ∆PQR
(SAS similarity criterion)

Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers

Question 8.
In figure, if ∆ABE ≅ ∆ACD, show that ∆ADE ~ ∆ABC.
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 77
Solution.
In figure,
∆ABE ≅ ∆ACD (Given)
⇒ AB = AC and AE = AD (CPCT)
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 78
and also,
∠DAE = ∠BAC (Each = ∠A)
∆ADE ~ ∆ABC (By SAS similarity criterion)
Hence proved.

Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers

Question 9.
If AD and PM are medians of ∆ABC and ∆PQR, respectively, where ∆ABC ~ ∆PQR,
prove that \(\frac{A B}{P Q}=\frac{A D}{P M}\)
Solution.
Draw two ∆ABC and APQR taking D and M points on BC and QR such that AD and PM are the medians of the ∆ABC and ∆PQR.
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 79
∆ABC – ∆PQR (Given)
⇒ \(\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R}\) ; ∠A = ∠P,
∠B = ∠Q, ∠C = ∠R …(i)
Now, BD = CD = \(\frac {1}{2}\)BC
andQM = RM = \(\frac {1}{2}\)QR …..(ii)
(∵ D is mid-point of BC and M is mid point of QR)
From Eq. (i),
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 80

Question 10.
Using the properties of similar triangles, prove that sum of square of two sides is equal to square of its hypotenuse in a right triangle.
Solution.
Given : In right angled ∆ABC, ∠B= 90°
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 81
To Prove: AC2 = AB2 + BC2
Construction : Draw BD ⊥ AC.
Pr00f : ∆ADB ~ ∆ABC(by A similarly criterion)
∴ \(\frac{A D}{A B}\) = \(\frac{A B}{A C}\)
⇒ AB2 = AD × AC
∆BDC ~ ∆ABC(by A similarty criterion)
∴ \(\frac{B C}{D C}\) + \(\frac{A C}{B C}\)
⇒ BC2 = AC × DC …(ii)
On adding (i) & (ii)
AB2 + BC2 = AD × AC + AC × DC
= AC(AD + DC)
= AC × AC
= AC2

Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers

Question 11.
Prove that the ratio of the area of two similar triangles is equal to the square of the ratio of their corresponding medians.
Solution.
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 82
In figure, AD is a median of ∆ABC and PM is a median of ∆PQR. Here, D is mid-point of BC and M is mid-point of QR. Now, we have,
∆ABC ~ APQR
⇒ ∠B = ∠Q
(Corresponding angles are equal) …(i)
Also, \(\frac{A B}{P Q}\) + \(\frac{B C}{Q R}\)
(Ratio of corresponding sides are equal)
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 83

Question 12.
The perpendicular from A on side BC of a ∆ABC intersects BC at D such that DB = 3 CD (see in figure). Prove that 2AB2 = 2AC2 + BC2.
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 84
Solution.
Given, DB = 3CD
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 85
= CD = \(\frac{1}{4}\)BC …(i)
and DB = \(\frac{3}{4}\) BC
In ∆ABD, AB2 = DB2 + AD2 …(ii)
In ∆ACD, AC2 = CD2 + AD2 …(iii)
(By Pythagoras as theorem)
On subtracting Eq. (iii) from Eq. (ii), we get
AB2 – AC2 = DB2 – CD2
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 86
⇒ 2AB2 – 2AC2 = BC2
⇒ 2AB2 = 2AC2 + BC2
Hence proved.

Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers

Question 13.
In an equilateral ∆ABC, D is a point on side BC such that BD = \(\frac{1}{3}\)BC. Prove that 9AD2 = 7AB2
Solution.
Draw ABC is an equilateral triangle, D is a point on side BC such that BD = \(\frac{1}{3}\)BC. Draw a line AE is perpendicular to BC.
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 87
AB = BC = CA = a (Say)
(By property of equilateral triangle)
BD = \(\frac{1}{3}\)BC = \(\frac{1}{3}\)a
⇒ CD = \(\frac{2}{3}\)BC = \(\frac{2}{3}\)a
∵ AE ⊥ BC
⇒ BE = EC = \(\frac{1}{2}\)a
(∵ In an equilateral triangle altitude AE is perpendicular bisector of BC.)
DE = BE – BD = \(\frac{1}{2}\)a – \(\frac{1}{3}\)a – \(\frac{1}{6}\)a
Using pythagoras theorem in ∆ADE,
AD2 = AE2 + DE2
= AB2 – BE2 + DE2
(∵ Right ∆ABE, AE2 = AB2 – BE2)
Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers 88
⇒ 9AD2 = 7AB2 Hence proved.

Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers

error: Content is protected !!