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

∴ AP : PB = 1 : 2

Hence Choice (c) is correct

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 = \(\frac{B D}{B A}\) = \(\frac{2}{3}\)

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 = ?

(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 :

(a) x = 3

(b) x = 5

(c) x = 4

(d) x = 2.5

Answer:

(c) x = 4

Question 5.

In ∆ABC, DE || BC such that \(\frac{AD}{DB}\) = \(\frac{3}{5}\). If AC = 5.6 cm, then AE = ?

(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

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

⇒ (x – 2) (x + 2) = x (x – 1)

⇒ x^{2} – 4 = x^{2} – x

∴ x = 4

Question 2.

State whether the following quadrilaterals are similar or not 3 cm C

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°.

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,

⇒ EF is not parallel QR because converse of basic proportionality theorem is not satisfied.

(ii) In figure,

⇒ EF || QR because converse of basic proportionality of theorem is satisfied

(iii) In figure,

⇒ EF || QR because converse of basic proportionality theorem is satisfied

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.

Solution.

In ∆OAB and ∆OCD

∠AOB = ∠COD (vertically opposite angles)

∴ 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

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.

(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

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)

Question 8.

D is point on the side BC of a ∆ABC such that ∠ADC = ∠BAC. Show that CA^{2} = CB · CD.

Solution.

Draw a ∆ABC such that D is a point on BC and join AD.

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

⇒ CA^{2} = CB × CD

Question 9.

Let ∆ABC ~ ∆DEF and their areas be, 64 cm^{2} and 121 cm^{2}, respectively. If EF = 15.4 cm, find BC.

Solution.

∆ABC ~ ∆DEF (Given)

(Using property of area of similar triangles).

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.

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)

(Using property of area of similar triangles)

⇒ AB = PQ, BC = QR

and CA = PR

(SSS proportionality criterion)

⇒ ∆ABC ≅ ∆PQR.

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

Then, the diagonal, BD = a√2

We construct equilateral ∆PAB and ∆QBD.

∆PAB ~ ∆QBD

(Equilateral triangles are similar)

Question 13.

ABC is an isosceles triangle right angled at C. Prove that AB^{2} = 2AC^{2}.

Solution.

Draw ABC is an isosceles triangle right angled at C.

and AC = BC …..(i)

By Pythagoras theorem, we have

AB^{2} = AC^{2}+ BC^{2} = AC^{2} + AC^{2} = 2AC^{2}

[∵ BC = AC by Eq. (i)]

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.

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

AC^{2} + AB^{2} = BC^{2}

∴ x^{2} + (8)^{2} = (10)^{2}

⇒ x^{2} = 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).

Solution.

Question 2.

In ∆ABC, if the side AD is perpendicular to side BC and AD^{2} = BD × CD. Prove that ∆ABC is a right angle ∆.

Solution.

In ∆ABD and ∆ADC, ∠D = 90°

∴ AB^{2} = AD^{2} + BD^{2} …(i)

and AC^{2} = AD^{2} + CD^{2} … (ii)

Adding (i) & (ii)

AB^{2} + AC^{2} = 2AD^{2} +BD^{2}+CD^{2}

= 2(BD × CD) + BD^{2}+CD^{2}

= BD^{2} + 2BD × CD + CD^{2}

= 2(BD + CD)^{2}

(Given AD^{2} = BD × CD).

= BC^{2}

∴ ∠BAC = 90°

Hence ∆BAC is a right angle ∆.

Question 3.

In figure, AD ⊥ BC

Prove that AB^{2} + CD^{2} = BD^{2} + AC^{2}

Solution :

In ∆ADB, ∠D = 90°

∴ AB^{2} = AD^{2} + BD^{2} …(i)

again in ∆ADC, ∠D = 90°

AC^{2} = AD^{2} + CD^{2} …(ii)

Subtract (ii) from (i)

AB^{2 }– AC^{2} = BD^{2 }– CD^{2}

⇒ AB^{2} + CD^{2} = AC^{2} + BD^{2}

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.

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,

⇒ 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.

⇒ OE || CD(By converse of basic proportionality theorem)

Now, we have BA || OE and OE || CD = AB || CD

⇒ Quadrilateral ABCD is a trapezium.

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.

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

(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)

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)],

∠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

(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)

⇒ \(\frac{C A}{P A}=\frac{B C}{M P}\)

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

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(BL^{2} + CM)^{2} = 5 BC^{2}.

Solution.

Given,

In ∆BAC, ∠A = 90° and BL and CM are the medians.

T.P.T. – 4(BL^{2}+cm^{2}) = 5BC^{2}

Proof – In ∆BAC, ∠A = 90°

AB^{2} + AC^{2} = BC^{2} ….(i)

In ∆BAL, ∠A = 90°

BL^{2} = AB^{2} + AL^{2}

In ∆MAC, ∠A = 90°

CM^{2} = AM^{2} = AC^{2}

⇒ 4(BL^{2} + cm^{2}) = 5(AB^{2} + AC^{2})

= 5BC^{2} From Eq. (i)

(AB^{2} + AC^{2} = BC^{2})

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.

Question 13.

In figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O,

show that

Solution.

Draw AL ⊥ BC and DM ⊥ BC (See figure)

∠ALO = ∠DMO = 90°

and ∠AOL = ∠DOM

(Vertically opposite angle)

∴ ΔOLA ~ ΔΟΜD

(AAA similarity criterion)

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)

[∵ ar (∆CAB)= ar (∆ABC)]

Hence, the required ratio is 1 : 4.

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.

To prove. Area (∆PAB) + area (∆QBC) = area (∆RAC)

Proof. Since, triangles PAB, QBC and RAC are equilateral. Therefore, they are equiangular and hence similar.

Question 16.

In adjoining Fig. ABC and BCD are two triangles on the same base BC. If AD intersects BC at O, show that

Solution.

Given. ∆ABC and ABCD are on the same base and AD intersects BC at O.

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]

Question 17.

PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM^{2} = QM × MR.

Solution.

In ∆PQR and ∆MPQ,

∠1 + ∠2 = ∠2 + ∠4 (Each = 90°)

⇒ ∠1 = ∠4

Similarly, ∠2 = ∠3

and ∠PMR = ∠PMQ (each 90°)

∆QPM ~ ∆PRM (AAA criterion)

⇒ PM^{2} = QM × RM

or PM^{2} = QM × MR

Question 18.

ABC is an isosceles triangle with AC = BC. If AB^{2} = 2AC?, prove that ABC is a right triangle.

Solution.

Draw an isosceles ∆ABC with AC =BC.

In ∆ABC, we are given that

AC = BC …..(i)

and AB^{2} = 2AC^{2} …(ii)

Now, AC^{2} + BC^{2}-AC^{2} + AC^{2} [By Eq. (i)]

= 2AC^{2} = AB^{2} (By Eq. (ii)]

i.e., AC^{2} + BC^{2} = AB^{2}

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.

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

AB^{2} = AD^{2} + BD^{2}

⇒ (2a)^{2} = AD^{2} + a^{2}

⇒ AD^{2} = 3a^{2}

⇒ AD = √3a.

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)

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

OA^{2} + OB^{2} = AB^{2}

⇒ AC^{2} + BD^{2} = 4AB^{2}

or 4AB^{2} = AC^{2} + BD^{2}

⇒ AB^{2} + BC^{2} + CD^{2} + DA^{2} = AC^{2} + BD^{2}

(∵ 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

(i) OA^{2} + OB^{2}+OC^{2}-OD^{2}-OE^{2}-OF^{2} – AF^{2} + BD^{2} + CE^{2}

(ii) AF^{2}+ BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2}

Solution.

In ∆ABC, from point O join lines OB, OC and OA.

(i) In right angled ∆OFA,

OA^{2} = OF^{2} + AF^{2}

(By Pythagoras therorem)

⇒ OA^{2} – OF^{2} = AF^{2} …..(i)

Similarly, in ∆OBD,

OB^{2} – OD^{2} = BD^{2} …(ii)

and in ∆OCE,

OC^{2} – OE^{2} = CE^{2} …(iii)

On adding Eqs. (i), (ii) and (iii), we get

OA^{2} + OB^{2} + OC^{2} – OD^{2} – OE^{2} – OF^{2}

= AF^{2} + BD^{2} + CE^{2}

(ii) From part Eqs. (i), we get

OA^{2} + OB^{2} + OC^{2} – OD^{2} – OE^{2} – OF^{2}

= AF^{2} + BD^{2} + CE^{2} …(iv)

Similarly,

OA^{2} + OB^{2} + OC^{2} – OD^{2} – OE^{2} – OF^{2}

= BF^{2} + CD^{2} + AE^{2} …(v)

From Eqs. (iv) and (v), we have

AF^{2} + BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2}

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.

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., AC^{2} + AB^{2} = BC^{2}

∴ x^{2} + (18)^{2} = (24)^{2}

⇒ x^{2} = (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.

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

Let distance between tops of two poles DC m = x cm

Using Pythagoras theorem in ∆DEC, we get

i.e., DC^{2} = DE^{2} + CE^{2}

⇒ x^{2} = (12)^{2} + (5)^{2} = 169

⇒ x = 13

Hence, distance between their tops = 13 m.

Question 25.

D and E are points on the sides CA and CB, respectively of a ∆ABC right angled at C. Prove that AE^{2} + BD^{2} = AB^{2} + DE^{2}.

Solution.

Draw a right ∆ABC at C. Take D and E points on the sides CA and BC and join ED, BD and EA.

In right angled ∆ACE,

AE^{2} = CA^{2} + CE^{2} ……(i)

(By pythagoras theorem)

and in right angled ABCD,

BD^{2} = BC^{2} + CD^{2} …(ii)

On adding Eqs. (i) and (ii), we get

AE^{2} + BD^{2} = (Ca^{2} + CE^{2}) + (BC^{2} + CD^{2})

= (BC^{2} + Ca^{2}) + (CD^{2} + CE^{2})

(∵ In ∆ABC, Ba^{2} = BC^{2} + CA^{2} and in ∆ECD, DE^{2} = CD^{2} + CE^{2})

= BA^{2} + DE^{2}

(By Pythagoras theorem),

∴ AE^{2} + BD^{2} = AB^{2} + DE^{2}

Hence proved.

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)

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,

AB^{2} = AD^{2} + BD^{2}

⇒ a^{2} = x^{2} + (\(\frac {1}{2}\)a)^{2}

⇒ a^{2} = x^{2} + \(\frac {1}{4}\)a^{2}

⇒ 4a^{2} = ax^{2} + a^{2}

⇒ 3a^{2} = 4x^{2}

Hence proved.

### Triangles Class 10 Extra Questions Long Answer Type

Question 1.

In figure, if LM || CB and LN || CD,

prove that \(\frac{A M}{A B}=\frac{A N}{A D}\)

Solution.

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

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.

⇒ 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.

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.

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.

[∵ 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.

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.

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.

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.

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.

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.

Proof. Quadrilaterals ABEC and PQNR are parallelograms because their diagonals bisect each other at D and M, respectively.

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)

Question 8.

In figure, if ∆ABE ≅ ∆ACD, show that ∆ADE ~ ∆ABC.

Solution.

In figure,

∆ABE ≅ ∆ACD (Given)

⇒ AB = AC and AE = AD (CPCT)

and also,

∠DAE = ∠BAC (Each = ∠A)

∆ADE ~ ∆ABC (By SAS similarity criterion)

Hence proved.

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.

∆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),

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°

To Prove: AC^{2} = AB^{2} + BC^{2}

Construction : Draw BD ⊥ AC.

Pr00f : ∆ADB ~ ∆ABC(by A similarly criterion)

∴ \(\frac{A D}{A B}\) = \(\frac{A B}{A C}\)

⇒ AB^{2} = AD × AC

∆BDC ~ ∆ABC(by A similarty criterion)

∴ \(\frac{B C}{D C}\) + \(\frac{A C}{B C}\)

⇒ BC^{2} = AC × DC …(ii)

On adding (i) & (ii)

AB^{2} + BC^{2} = AD × AC + AC × DC

= AC(AD + DC)

= AC × AC

= AC^{2}

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.

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)

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 2AB^{2} = 2AC^{2} + BC^{2}.

Solution.

Given, DB = 3CD

= CD = \(\frac{1}{4}\)BC …(i)

and DB = \(\frac{3}{4}\) BC

In ∆ABD, AB^{2} = DB^{2} + AD^{2} …(ii)

In ∆ACD, AC^{2} = CD^{2} + AD^{2} …(iii)

(By Pythagoras as theorem)

On subtracting Eq. (iii) from Eq. (ii), we get

AB^{2} – AC^{2} = DB^{2} – CD^{2}

⇒ 2AB^{2} – 2AC^{2} = BC^{2}

⇒ 2AB^{2} = 2AC^{2} + BC^{2}

Hence proved.

Question 13.

In an equilateral ∆ABC, D is a point on side BC such that BD = \(\frac{1}{3}\)BC. Prove that 9AD^{2} = 7AB^{2}

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.

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,

AD^{2} = AE^{2} + DE^{2}

= AB^{2} – BE^{2} + DE^{2}

(∵ Right ∆ABE, AE^{2} = AB^{2} – BE^{2})

⇒ 9AD^{2} = 7AB^{2} Hence proved.