Triangles Class 10 Important Questions
The start of class 10 marks the beginning of the foundation for class 11 and class 12. It is very important study the basics in order to understand each and every chapter properly. In this page, we have provided all the important question for cbse class 9 that could be asked in the examination. Students also need to study the ncert solutions for class 10 in order to gain more knowledge and understanding the lessons. Questions and Answers are way to learn the new things in a proper way. NCERT textbooks downloads for class 9 in pdf are also available for the students if they need more help. By downloading these books, they can study from it. Our experts also prepared revision notes for class 9 so that students should see the details of each and every chapters. Class 9 important questions are the best to revise all the chapters in the best way.
Question 18.
In the given figure, CD || LA and DE || AC. Find the length of CL if BE = 4 cm and EC = 2 cm. (2012)
Solution:
In ∆ABL, CD || LA
Question 19.
If a line segment intersects sides AB and AC of a ∆ABC at D and E respectively and is parallel to BC, prove that ADAB=AEAC. (2013)
Solution:
Given. In ∆ABC, DE || BC
To prove. ADAB=AEAC
Proof.
In ∆ADE and ∆ABC
∠1 = ∠1 … Common
∠2 = ∠3 … [Corresponding angles
∆ADE ~ ∆ABC …[AA similarity
∴ ADAB=AEAC
…[In ~∆s corresponding sides are proportional
Question 20.
In a ∆ABC, DE || BC with D on AB and E on AC. If ADDB=34 , find BCDE. (2013)
Solution:
Given: In a ∆ABC, DE || BC with D on AB and E
on AC and ADDB=34
To find: BCDE
Proof. Let AD = 3k,
DB = 4k
∴ AB = 3k + 4k = 7k
In ∆ADE and ∆ABC,
∠1 = ∠1 …[Common
∠2 = ∠3 … [Corresponding angles
∴ ∆ADE ~ ∆ABC …[AA similarity
Question 21.
In the figure, if DE || OB and EF || BC, then prove that DF || OC. (2014)
Solution:
Given. In ∆ABC, DE || OB and EF || BC
To prove. DF || OC
Proof. In ∆AOB, DE || OB … [Given
Question 22.
If the perimeters of two similar triangles ABC and DEF are 50 cm and 70 cm respectively and one side of ∆ABC = 20 cm, then find the corresponding side of ∆DEF. (2014)
Solution:
Given. ∆ABC ~ ∆DEF,
Perimeter(∆ABC) = 50 cm
Perimeter(∆DEF) = 70 cm
One side of ∆ABC = 20 cm
To Find. Corresponding side of ∆DEF (i.e.,) DE. ∆ABC ~ ∆DEF …[Given
∴ The corresponding side of ADEF = 28 cm
Question 23.
A vertical pole of length 8 m casts a shadow 6 cm long on the ground and at the same time a tower casts a shadow 30 m long. Find the height of tower. (2014)
Solution:
Let BC be the pole and EF be the tower Shadow AB = 6 m and DE = 30 m.
In ∆ABC and ∆DEF,
∠2 = ∠4 … [Each 90°
∠1 = ∠3 … [Sun’s angle of elevation at the same time
∆ABC ~ ∆DEF …[AA similarity
ABDE=BCEF … [In -As corresponding sides are proportional
⇒ 630=8EF ∴ EF = 40 m
Question 24.
In given figure, EB ⊥ AC, BG ⊥ AE and CF ⊥ AE (2015)
Prove that:
(a) ∆ABG ~ ∆DCB
(b) BCBD=BEBA
Solution:
Given: EB ⊥ AC, BG ⊥ AE and CF ⊥ AE.
To prove: (a) ∆ABG – ∆DCB,
(b) BCBD=BEBA
Proof: (a) In ∆ABG and ∆DCB,
∠2 = ∠5 … [each 90°
∠6 = ∠4 … [corresponding angles
∴ ∆ABG ~ ∆DCB … [By AA similarity
(Hence Proved)
∴ ∠1 = ∠3 …(CPCT … [In ~∆s, corresponding angles are equal
(b) In ∆ABE and ∆DBC,
∠1 = ∠3 …(proved above
∠ABE = ∠5 … [each is 90°, EB ⊥ AC (Given)
∆ABE ~ ∆DBC … [By AA similarity
BCBD=BEBA
… [In ~∆s, corresponding sides are proportional
∴ BCBD=BEBA (Hence Proved)
Question 25.
∆ABC ~ ∆PQR. AD is the median to BC and PM is the median to QR. Prove that ABPQ=ADPM. (2017D)
Solution:
∆ABC ~ ∆PQR … [Given
∠1 = ∠2 … [In ~∆s corresponding angles are equal
Question 26.
State whether the given pairs of triangles are similar or not. In case of similarity mention the criterion. (2015)
Solution:
(b) In ∆PQR, ∠P + ∠Q + ∠ZR = 180° …[Angle-Sum Property of a ∆
45° + 78° + ∠R = 180°
∠R = 180° – 45° – 78° = 57°
In ∆LMN, ∠L + ∠M + ∠N = 180° …[Angle-Sum Property of a ∆
57° + 45° + ∠N = 180°
∠N = 180° – 57 – 45° = 78°
∠P = ∠M … (each = 45°
∠Q = ∠N … (each = 78°
∠R = ∠L …(each = 57°
∴ ∆PQR – ∆MNL …[By AAA similarity theorem
Question 27.
In the figure of ∆ABC, D divides CA in the ratio 4 : 3. If DE || BC, then find ar (BCDE) : ar (∆ABC). (2015)
Solution:
Given:
D divides CA in 4 : 3
CD = 4K
DA = 3K
DE || BC …[Given
In ∆AED and ∆ABC,
∠1 = ∠1 …[common
∠2 = ∠3 … corresponding angles
∴ ∆AED – ∆ABC …(AA similarity
⇒ ar(△AED)ar(△ABC)=(ADAC)2
… [The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides
⇒ frac(3K)2(7K)2=9K249K2=ar(△AED)ar(△ABC)=949
Let ar(∆AED) = 9p
and ar(∆ABC) = 49p
ar(BCDE) = ar (∆ABC) – ar (∆ADE)
= 49p – 9p = 40p
∴ ar(BCDE)ar(△ABC)=40p49p
∴ ar (BCDE) : ar(AABC) = 40 : 49
Question 28.
In the given figure, DE || BC and AD : DB = 7 : 5, find \frac { ar\left( \triangle DEF \right) }{ ar\left( \triangle CFB \right) } [/latex] (2017OD)
Solution:
Given: In ∆ABC, DE || BC and AD : DB = 7 : 5.
To find: ar(△DEF)ar(△CFB) = ?
Proof: Let AD = 7k
and BD = 5k then
AB = 7k + 5k = 12k
In ∆ADE and ∆ABC,
∠1 = ∠1 …(Common
∠2 = ∠ABC … [Corresponding angles
Question 29.
In the given figure, the line segment XY is parallel to the side AC of ∆ABC and it divides the triangle into two parts of equal areas. Find the ratio AXAB. (2017OD)
Solution:
We have XY || AC … [Given
So, ∠BXY = ∠A and ∠BYX = ∠C …[Corresponding angles
∴ ∆ABC ~ ∆XBY …[AA similarity criterion
Question 30.
In the given figure, AD ⊥ BC and BD = 13CD. Prove that 2AC2 = 2AB2 + BC2. (2012)
Solution:
BC = BD + DC = BD + 3BD = 4BD
∴ BC4 = BD
In rt. ∆ADB, AD2 = AB2 – BD2 ….(ii)
In rt. ∆ADC, AD2 = AC2 – CD2 …(iii)
From (ii) and (iii), we get
AC2 – CD2 = AB2 – BD2
AC2 = AB2 – BD2 + CD2
∴ 2AC2 = 2AB2 + BC2 (Hence proved)
Question 31.
In the given figure, ∆ABC is right-angled at C and DE ⊥ AB. Prove that ∆ABC ~ ∆ADE and hence find the lengths of AE and DE. (2012, 2017D)
Solution:
Given: ∆ABC is rt. ∠ed at C and DE ⊥ AB.
AD = 3 cm, DC = 2 cm, BC = 12 cm
To prove:
(i) ∆ABC ~ ∆ADE; (ii) AE = ? and DE = ?
Proof. (i) In ∆ABC and ∆ADE,
∠ACB = ∠AED … [Each 90°
∠BAC = ∠DAE …(Common .
∴ ∆ABC ~ ∆ADE …[AA Similarity Criterion
(ii) ∴ ABAD=BCDE=ACAE … [side are proportional
AB3=12DE=3+2AE
…..[In rt. ∆ACB, … AB2 = AC2 + BC2 (By Pythagoras’ theorem)
= (5)2 + (12)2 = 169
∴ AB = 13 cm
Question 32.
In ∆ABC, if AP ⊥ BC and AC2 = BC2 – AB2, then prove that PA2 = PB × CP. (2015)
Solution:
AC2 = BC2 – AB2 …Given
AC2 + AB2 = BC2
∴ ∠BAC = 90° … [By converse of Pythagoras’ theorem
∆APB ~ ∆CPA
[If a perpendicular is drawn from the vertex of the right angle of a triangle to the hypotenuse then As on both sides of the perpendicular are similar to the whole triangle and to each other.
∴ APCP=PBPA … [In ~∆s, corresponding sides are proportional
∴ PA2 = PB. CP (Hence Proved)
Question 33.
ABCD is a rhombus. Prove that AB2 + BC2 + CD2 + DA2 = AC2 + BD2. (2013)
Solution:
Given. In rhombus ABCD, diagonals AC and BD intersect at O.
To prove: AB2 + BC2 + CD2 + DA2 = AC2 + BD2
Proof: AC ⊥ BD [∵ Diagonals of a rhombus bisect each other at right angles
∴ OA = OC and
OB = OD
In rt. ∆AOB,
AB2 = OA2 + OB2 … [Pythagoras’ theorem
AB2 = (AC2)2+(BD2)2
AB2 = (AC2)2+(BD2)2
4AB2 = AC2 + BD2
AB2 + AB2 + AB2 + AB2 = AC2 + BD2
∴ AB2 + BC2 + CD2 + DA2 = AC2 + BD2
…[∵ In a rhombus, all sides are equal
Question 34.
The diagonals of trapezium ABCD intersect each other at point o. If AB = 2CD, find the ratio of area of the ∆AOB to area of ∆COD. (2013)
Solution:
In ∆AOB and ∆COD, … [Alternate int. ∠s
∠1 = ∆3
∠2 = ∠4
Question 35.
The diagonals of a quadrilateral ABCD intersect each other at the point O such that AOBO=CODO. Show that ABCD is a trapezium. (2014)
Solution:
1st method.
Given: Quadrilateral ABCD in which
AC and BD intersect each other at 0.
Such that AOBO=CODO
To prove: ABCD is a trapezium
Const.: From O, draw OE || CD.
But these are alternate interior angles
∴ AB || DC Quad. ABCD is a trapezium.
Triangles Class 10 Important Questions Long Answer (4 Marks).
Question 36.
In a rectangle ABCD, E is middle point of AD. If AD = 40 m and AB = 48 m, then find EB. (2014D)
Solution:
E is the mid-point of AD …[Given
AE = 402 = 20 m
∠A = 90° …[Angle of a rectangle
In rt. ∆BAE,
EB2 = AB2 + AE2 …[Pythagoras’ theorem
= (48)2 + (20)2
= 2304 + 400 = 2704
∴ EB = 2704−−−−√ = 52 m
Question 37.
Let ABC be a triangle and D and E be two points on side AB such that AD = BE. If DP || BC and EQ || AC, then prove that PQ || AB. (2013)
Solution:
In ∆ABC,
DP || BC
and EQ || AC … [Given
Now, in ∆ABC, P and Q divide sides CA and CB respectively in the same ratio.
∴ PQ || AB
Question 38.
In the figure, ∠BED = ∠BDE & E divides BC in the ratio 2 : 1.
Prove that AF × BE = 2 AD × CF. (2015)
Solution:
Construction:
Draw CG || DF
Proof: E divides
BC in 2 : 1.
BEEC=21 …(i)
Question 39.
In the given figure, AD = 3 cm, AE = 5 cm, BD = 4 cm, CE = 4 cm, CF = 2 cm, BF = 2.5 cm, then find the pair of parallel lines and hence their lengths. (2015)
Solution:
Question 40.
If sides AB, BC and median AD of AABC are proportional to the corresponding sides PQ, QR and median PM of PQR, show that ∆ABC ~ ∆PQR. (2017OD)
Solution:
Question 41.
Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. (2012)
Solution:
Given: ∆ABC ~ ∆DEF
Question 42.
State and prove converse of Pythagoras theorem. Using the above theorem, solve the following: In ∆ABC, AB = 63–√ cm, BC = 6 cm and AC = 12 cm, find ∠B. (2015)
Solution:
Part I:
Statement: Prove that, in a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.
To prove: ∠ABC = 90°
Const.: Draw a right angle ∆DEF in which DE = BC and EF = AB.
Proof: In rt. ∆ABC,
AB2 + BC2 = AC2 …(i) Given
In rt. ∆DEF
DE2 + EF2 = DF2 … [By Pythagoras’ theorem
BC2 + AB2 = DF2…(ii)…[∵ DE = BC; EF = AB
From (i) and (ii), we get
AC2 = DF2 = AC = DF
Now, DE = BC …[By construction
EF = AB …[By construction
DF = AC … [Proved above :
∴ ∆DEF = ∆ABC … (SSS congruence :
∴ ∠DEF = ∠ABC …[c.p.c.t.
∵ ∠DEF = 90° ∴ ∠ABC = 90°
Given: In rt. ∆ABC,
AB2 + BC2 = AC2
AB2 + BC2 = (63–√)2 + (6)2
= 108 + 36 = 144 = (12)2
AB2 + BC2 = AC2 ∴ ∠B = 90° … [Above theorem
Question 43.
In the given figure, BL and CM are medians of a triangle ABC, right angled at A. Prove that: 4(BL2 + CM2) = 5BC2 (2012)
Solution:
Given: BL and CM are medians of ∆ABC, right angled at A.
To prove: 4(BL2 + CM2) = 5 BC2
Proof: In ∆ABC, BC2 = BA2 + CA2 …(i)
In ∆BAL,
BL2 = BA2 + AL2 …[Pythagoras’ theorem
BL2 = BA2 + (CA2)2
BL2 = BA2+ CA24
⇒ 4BL2 = 4BA2 + CA2 …(ii)
Now, In ∆MCA,
MC2 = CA2 + MA2 …[Pythagoras’ theorem
MC2 = CA22 + (BA2)2
MC2 = CA2 + BA24
4MC2 = 4CA2 + BA2
Adding (ii) and (iii), we get
4BL2 + 4MC2 = 4BA2 + CA2 + 4CA2+ BA2 …[From (ii) & (iii)
4(BL2 + MC2) = 5BA2 + 5CA2
4(BL2 + MC2) = 5(BA2 + CA2)
∴ 4(BL2 + MC2) = 5BC2 … [Using (1)
Hence proved.
Question 44.
In the given figure, AD is median of ∆ABC and AE ⊥ BC. (2013)
Prove that b2 + c2 = 2p2 + 12 a2.
Solution:
Proof. Let ED = x
BD = DC = BC2=a2 = …[∵ AD is the median
In rt. ∆AEC, AC2 = AE2 + EC2 …..[By Pythagoras’ theorem
b2 = h2 + (ED + DC)2
b2 = (p2 – x2) + (x = a2)2
…[∵ In rt. ∆AED, x2 + h2 = p2 ⇒ h2 = p2 – x2 …(i)
b2 = p2 – x2 + x2 + (a2)22+ 2(x)(a2)
b2 = p2 + ax + a24 …(ii)
In rt. ∆AEB, AB2 = AE2 + BE2 … [By Pythagoras’ theorem
Question 45.
In a ∆ABC, the perpendicular from A on the side BC of a AABC intersects BC at D such that DB = 3 CD. Prove that 2 AB2 = 2 AC2 + BC2. (2013; 2017OD)
Solution:
In rt. ∆ADB,
AD2 = AB2 – BD2 …(i) [Pythagoras’ theorem
In rt. ∆ADC,
AD2 = AC2 – DC2 …(ii) [Pythagoras’ theorem
From (i) and (ii), we get
AB2 – BD2 = AC2 – DC2
AB2 = AC2 + BD2 – DC2
Now, BC = BD + DC
= 3CD + CD = 4 CD …[∵ BD = 3CD (Given)
⇒ BC2 = 16 CD2 …(iv) [Squaring
Now, AB2 = AC2 + BD2 – DC2 …[From (iii)
= AC2 + 9 DC2 – DC2 ….[∵ BD = 3 CD ⇒ BD2 = 9 CD2
= AC2 + 8 DC2
= AC2 + 16DC22
= AC2 + BC22 … [From (iv)
∴ 2AB2 = 2AC2 + BC2 … [Proved
Question 46.
In ∆ABC, altitudes AD and CE intersect each other at the point P. Prove that: (2014)
(i) ∆APE ~ ∆CPD
(ii) AP × PD = CP × PE
(iii) ∆ADB ~ ∆CEB
(iv) AB × CE = BC × AD
Solution:
Given. In ∆ABC, AD ⊥ BC & CE ⊥ AB.
To prove. (i) ∆APE ~ ∆CPD
(ii) AP × PD = CP × PE
(iii) ∆ADB ~ ∆CEB
(iv) AB × CE = BC × AD
Proof: (i) In ∆APE and ∆CPD,
∠1 = ∠4 …[Each 90°
∠2 = ∠3 …[Vertically opposite angles
∴ ∆APE ~ ∆CPD …[AA similarity
(ii) APCP=PEPD … [In ~ ∆s corresponding sides are proportional
∴ AP × PD = CP × PE
(iii) In ∆ADB and ∆CEB,
∠5 = ∠7 …[Each 90°
∠6 = ∠6 …(Common
∴ ∆ADB ~ ∆CEB …[AA similarity
(iv) ∴ ABCB=ADCE … [In ~ ∆s corresponding sides are proportional
∴ AB × CE = BC × AD
Question 47.
In the figure, PQR and QST are two right triangles, right angled at R and T resepctively. Prove that QR × QS = QP × QT. (2014)
Solution:
Given: Two rt. ∆’s PQR and QST.
To prove: QR × QS = QP × QT
Proof: In ∆PRQ and ∆STQ,
∠1 = ∠1 … [Common
∠2 = ∠3 … [Each 90°
∆PRQ ~ ∆STO …(AA similarity
∴ QRQT=QPQS ..[In -∆s corresponding sides are proportional
∴ QR × QS = QP × QT (Hence proved)
Question 48.
In the given figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that ar(ABC)ar(DBC)=AODO. (2012)
Solution:
Given: ABC and DBC are two As on the same base BC. AD intersects BC at O.
To prove:
Question 49.
Hypotenuse of a right triangle is 25 cm and out of the remaining two sides, one is longer than the other by 5 cm. Find the lengths of the other
two sides. (2013)
Solution:
Let Base, AB = x cm
Then altitude, BC = (x + 5) cm
In rt. ∆,
By Pythagoras’ theorem
AB2 + BC2 = AC2
⇒ (x)2 + (x + 5)2 = 252
⇒ x22 + x2 + 10x + 25 – 625 = 0
⇒ 2x2 + 10x – 600 = 0
⇒ x2 + 5x – 300 = 0 … [Dividing both sides by 2
⇒ x2 + 20x – 15x – 300 = 0
⇒ x(x + 20) – 15(x + 20) = 0
(x – 15)(x + 20) = 0
x – 15 = 0 or x + 20 = 0
x = 15 or x = -20
Base cannot be -ve
∴ x = 15 cm
∴ Length of the other side = 15 + 5 = 20 cm
Two sides are = 15 cm and 20 cm
Question 50.
In Figure, AB ⊥ BC, FG ⊥ BC and DE ⊥ AC. Prove that ∆ADE ~ ∆GCF. (2016 OD)
Solution:
In rt. ∆ABC,
∠A + ∠C = 90° …(i)
In rt. ∆AED,
∠A + ∠2 = 90°
From (i) and (ii), ∠C = ∠2
Similarly, ∠A = ∠1
Now in ∆ADE & ∆GCF
∠A = 1 … [Proved
∠C = 2 … [Proved
∠AED = ∠GFC … [rt. ∠s
∴ ∆ADE – ∆GCF …(Hence Proved)
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Important MCQs- Triangles
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