Chapter 6 Triangles NCERT Exemplar Solutions Exercise 6.4 Class 10 Maths
Chapter Name  NCERT Maths Exemplar Solutions for Chapter 6 Triangles Exercise 6.4 
Book Name  NCERT Exemplar for Class 10 Maths 
Other Exercises 

Related Study  NCERT Solutions for Class 10 Maths 
Exercise 6.4 Solutions
Long Answer Questions
1. In Fig., if ∠A = ∠C, AB = 6 cm, BP = 15 cm, AP = 12 cm and CP = 4 cm, then find the lengths of PD and CD.
Solution
We have,
∠A = ∠C,
AB = 6 cm,
BP = 15 cm,
AP = 12 cm and
CP = 4 cm
In ∆APB and ∆CPD,
∠A = ∠C [given]
∠APB = ∠CPD [vertically opposite angles]
∆APB ~ ∆CPD [by AA similarity criterion]
AP/CP = PB/PD = AB/CD
⇒ 12/4 = 15/PD = 6/CD
So,
12/4 = 15/PD
⇒ PD = 5 cm
Also,
12/4 = 6/CD
⇒ CD = 2 cm
Therefore, length of PD is 5 cm and length of CD is 2 cm.
2. It is given that ∆ ABC ~ ∆ EDF such that AB = 5 cm, AC = 7 cm, DF= 15 cm and DE = 12 cm. Find the lengths of the remaining sides of the triangles.
Solution
We have,
∆ABC ~ ∆EDF, so the corresponding sides of ∆ABC and ∆EDF are in the same ratio
AB/ED = AC/EF = BC/DF ...(i)
Also, we have,
AB = 5 cm,
AC = 7 cm,
DF= 15 cm and
DE = 12 cm
Putting value in AB/ED = AC/EF = BC/DF ,
5/12 = 7/EF = BC/15
So,
5/12 = 7/EF
⇒ EF = 16.8 cm
Also,
5/12 = BC/15
⇒ BC = 6.25 cm
So, lengths of the remaining sides of the triangles are EF = 16.8 cm and
BC = 6.25 cm.
3. Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides, then the two sides are divided in the same ratio.
Solution
Let us take ∆ABC in which a line DE parallel to BC intersects AB at D and AC at E.
To prove: DE divides the two sides in the same ratio.
AD/DB = AE/EC
Construction :
Join BE and CD
EF ⊥ AB
DG ⊥ AC
Proof :
ar(ADE)/ar(BDE) = (1/2 × AD × EF)/(1/2 × DB × EF)
= AD/DB ...(i)
Also,
ar(ADE)/ar(DEC) = (1/2 × AE × GD)/(1/2 × EC × GD)
= AE/EC ...(ii)
As,
∆BDE and ∆DEC lie between the same parallel lines DE and BC and on the same base DE
So,
ar(∆BDE) = ar(∆DEC) ...(iii)
From Eqs. (i), (ii) and (iii),
AD/DB = AE/EC
Hence proved
4. In Fig., if PQRS is a parallelogram and AB PS, then prove that OC  SR.
Solution
We have,
PQRS is a parallelogram, so PQ  SR and PS  QR.
Also,
AB  PS.
To prove :
OC  SR
Proof : In Î”OPS and Î”OAB, PS  AB
∠POS = ∠AOB [common angle]
∠OSP = ∠OBA [corresponding angles]
∆OPS ~ ∆OAB [by AA similarity criterion]
Then PS/AB = OS/OB ...(i)
In ∆CQE and ∆CAB, QR  PS  AB
∠QCR = ∠ACB [common angle]
∠CRQ = ∠CBA [corresponding angles]
So,
∆CQR ~ ∆CAB
QR/AB = CR/OB
PS/AB = CR/OB ...(ii)
(PS = QR, opposite sides of parallelogram)
From (i) and (ii)
OS/OB = CR/CB
or,
OB/OS = CB/CR
Subtracting 1 from both sides, we get,
(OB/OS)  1 = (CB/CR)  1
⇒ (OB  OS)/OS = (CB  CR)/CR
⇒ BS/OS = BR/CR
By using converse of basic proportionality theorem, SR  OC.
Hence, proved
5. A 5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4 m high. If the foot of the ladder is moved 1.6 m towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall.
Solution
Taking AC be the ladder of length 5 m and BC = 4 m be the height of the wall, which ladder is placed.
If the foot of the ladder is moved 1.6 m towards the wall so, AD = 1.6 m, then the ladder is slide upward i.e., CE = x m.
In right angled ∆ABC,
AC^{2} = AB^{2} + BC^{2} [Using Pythagoras theorem]
⇒ (5)^{2} = (AB)^{2} + (4)^{2}
⇒ AB^{2} = 25  16 = 9
⇒ AB = 3m
Now,
DB = AB  AD
= 3  1.6
= 1.4 m
In right angled ∆EBD,
ED^{2} = EB^{2} + BD^{2}
[using Pythagoras theorem]
(5)^{2} = (EB)^{2} + (1.4)^{2}
[ ∵ BD = 1.4 m]
25 = (EB)^{2} + 1.96
⇒ (EB)^{2} = 25 – 1.96 = 23.04
⇒ EB = 4.8
Now,
EC = EB – BC
= 4.8 – 4
= 0.8
Therefore, the top of the ladder would slide upwards on the wall at distance is 0.8 m.
6. For going to a city B from city A, there is a route via city C such that AC⊥CB, AC = 2 x km and CB = 2 (x + 7) km. It is proposed to construct a 26 km highway which directly connects the two cities A and B. Find how much distance will be saved in reaching city B from city A after the construction of the highway.
Solution
We have,
AC ⊥ CB,
AC = 2 × km,
CB = 2(x + 7) km and AB = 26 km
On drawing the figure, we get the right angle ∆ACB right angle at C.
Now,
In ∆ACB, by Pythagoras theorem,
AB^{2} = AC^{2} + BC^{2}
⇒ (26)^{2} = (2x)^{2} + {2(x + 7)}^{2}
⇒ 676 = 4x^{2} + 4(x^{2} + 49 + 11x)
⇒ 676 = 4x^{2} + 4x^{2} + 196 + 56x
⇒ 676 = 8x^{2} + 56x + 196
⇒ 8x^{2} + 56x – 480 = 0
On dividing by 8, we get,
x^{2} + 7x – 60 = 0
⇒ x^{2} + 12x5x60 = 0
⇒ x(x + 12) – 5(x + 12) = 0
⇒ (x + 12)(x – 5) = 0
⇒ x = 12 or, x = 5
As, distance cannot be negative.
x = 5 [∵ x ≠ 12]
Now,
AC = 2x
= 10 km and
BC = 2(x + 7)
= 2(5 + 7)
= 24 km
The distance covered to reach city B from city A via city C = AC + BC = 10 + 24 = 34 km
Distance covered to reach city B from city A after the construction of the highway is
BA = 26 km
So, the required saved distance is 34 – 26= 8 km.
7. A flag pole 18 m high casts a shadow 9.6 m long. Find the distance of the top of the pole from the far end of the shadow.
Solution
Let BC = 18 m be the flag pole and its shadow be AB = 9.6 m. The distance of the top of the pole, C from the far end which is A of the shadow is AC
In right angled ∆ABC
AC^{2} = AB^{2} + BC^{2} [using Pythagoras theorem]
⇒ AC^{2} = (9.6)^{2} + (18)^{2}
⇒ AC^{2} = 92.16 + 324
⇒ AC^{2} = 416.16
⇒ AC = 20.4 m
So, the required distance is 20.4 m.
8. A street light bulb is fixed on a pole 6 m above the level of the street. If a woman of height 1.5 m casts a shadow of 3m, find how far she is away from the base of the pole.
Solution
Taking A be the position of the street bulb fixed on a pole AB = 6 m and CD = 1.5 m be the height of a woman and her shadow be ED = 3 m. And distance between pole and woman be x m.
In this question, woman and pole both are standing vertically
So,
CD  AB
In ∆CDE and ∆ABE,
∠E = ∠E [common angle]
∠ABE = ∠CDE [each equal to 90°]
∆CDE ~ ∆ABE [by AA similarity criterion]
Then,
ED/EB = CD/AB
⇒ 3/(3 +x) = 1.5/6
⇒ 3×6 = 1.5(3 + x)
⇒ 18 = 1.5 × 3 + 1.5x
⇒ 1.5x = 18 – 4.5
⇒ x = 9 m
So, she is at the distance of 9 m from the base of the pole.
9. In Fig., ABC is a triangle right angled at B and BD ⊥ AC. If AD = 4 cm, and CD = 5 cm, find BD and AB.
Solution
Given,
∆ABC in which ∠B = 90° and
BD ⊥ AC
Also, AD = 4 cm and
CD = 5 cm
In ∆DBA and ∆DCB,
∠ADB = ∠CDB [each equal to 90°]
and
∠BAD = ∠DBC [each equal to 90° – ∠C]
∆DBA ~ ∆DCB [by AA similarity criterion]
So,
DB/DA = DC/DB
⇒ DB^{2} = DA × DC = 4 × 5
⇒ DB = 2√5 cm
In Î”BDC,
BC^{2} = BD^{2} + CD^{2} (Using Pythagoras theorem)
= (2√5)^{2} + 5^{2}
= 3√5
Also,
Î”DBA ~ Î”DBC
DB/DC = BA/BC
⇒ (2√5)/5 = BA/3√5
⇒ AB = 6 cm
10. 10. In Fig., PQR is a right triangle right angled at Q and QS ⊥ PR . If PQ = 6 cm and PS = 4 cm, find QS, RS and QR.
Solution
We have,
In ∆PQR,
∠Q = 90°,
QS ⊥ PR and
PQ = 6 cm,
PS = 4 cm
In ∆SQP and ∆SRQ,
∠PSQ = ∠RSQ [each equal to 90°]
∠SPQ = ∠SQR [each equal to 90° – ∠R]
∆SQP ~ ∆SRQ [By AA similarity criterion]
Then, SQ/PS = SR/SQ
SQ^{2} = PS × SR
In right angled Î”PSQ,
PQ^{2} = PS^{2} + QS^{2} [using Pythagoras theorem]
⇒ (6)^{2} = (4)^{2} + QS^{2}
⇒ 36 = 16 + QS^{2}
⇒ QS^{2} = 36 – 16 = 20
⇒ QS = 2√5 cm
From eq. (i),
Putting value of PS and QS we get,
RS = 5cm
Now, In QSR,
QR^{2} = QS^{2} +SR^{2}
So, putting value of QS and SR we get,
QR = 3√5cm
11. In ∆PQR, PD ⊥ QR such that D lies on QR. If PQ = a, PR = b, QD = c and DR = d, prove that (a + b)(a  b) = (c + d)(c  d).
Solution
Given:
In ∆PQR, PD ⊥ QR,
PQ = a,
PR = b,
QD = c and
DR = d
To prove: (a + b)(a b) = (c + d)(c – d)
Proof:
In right angled ∆PDQ,
PQ^{2} = PD^{2} + QD^{2} [using Pythagoras theorem]
a^{2} = PD^{2} + c^{2}
PD^{2} = a^{2}  c^{2} ...(i)
In right angled ∆PDR,
PR^{2} = PD^{2} + DR^{2} [Using Pythagoras theorem]
b^{2} = PD^{2} + d^{2}
PD^{2} = b^{2}  d^{2} ...(ii)
From Eq. (i) and (ii)
a^{2}  c^{2} = b^{2}  d^{2}
⇒ a^{2}  b^{2} = c^{2}  d^{2}
⇒ (a  b)(a + b) = (c  d)(c + d)
Hence proved.
12. In a quadrilateral ABCD, ∠A + ∠D = 90° . Prove that AC^{2} + BD^{2} = AD^{2} + BC^{2}
[Hint : Produce AB and DC to meet at E.]
Solution
Given :
Quadrilateral ABCD
∠A + ∠D = 90°
To prove : AC^{2} + BD^{2} = AD^{2} + BC^{2}
Construct : Produce AB and CD to meet at E
Also join AC and BD
Proof:
In ∆AED,
∠A + ∠D = 90° [given]
∠E = 180° – (∠A + ∠D)
= 90° [sum of angles of a triangle = 180°]
So, by Pythagoras theorem,
AD^{2} = AE^{2} + DE^{2}
In ∆BEC, by Pythagoras theorem,
BC^{2} = BE^{2} + EC^{2}
Adding both equations, we get
AD^{2} + BC^{2} = AE^{2} + DE^{2} + BE^{2} + CE^{2} ...(i)
In ∆AEC, by Pythagoras theorem,
AC^{2} = AE^{2} + CE^{2}
In ∆BED, by Pythagoras theorem,
BD^{2} = BE^{2} + DE^{2}
Adding both equations, we get
AC^{2} + BD^{2} = AE^{2} + CE^{2} + BE^{2} + DE^{2} ...(ii)
From Eqs. (i) and (ii)
AC^{2} + BD^{2} = AD^{2} + BC^{2}
Hence proved.
13. In Fig., l  m and line segments AB, CD and EF are concurrent at point P.
Prove that : AE/BF = AC/BD = CE/FD
Solution
We have, l  m and line segment AB, CD and EF are concurrent at point P
To Prove,
AE/BF = AC/BD = CE/FD
In Î”APC and Î”BPD,
APC = BPD (vertically opposite angles)
PAC =PBD (Alternate angles)
So,
Î”APC:Î”BPD (By AA Similarity)
AP/PB = AC/BD = PC/PD
Now,
In Î”APE and Î”BPF,
APE =BPF (vertically opposite angles)
PAE =PBF (Alternate angles)
so,
Î”APE:Î”BPF (By AA Similarity)
Now,
In Î”PEC and Î”PFD,
APC = BPD (vertically opposite angles)
PAC = PBD (Alternate angles)
so,
Î”PEC:Î”PDF (By AA Similarity)
PC/PD = PE/PF = EC/FD
So, from above equations,
AP/PB = AC/BD = PE/PF = EC/FD = AE/BF
AE/BF = AC/BD = EC/FD
Hence, proved.
14. In Fig., PA, QB, RC and SD are all perpendiculars to a line l, AB = 6 cm, BC = 9 cm, CD = 12 cm and SP = 36 cm. Find PQ, QR and RS.
Solution
We have,
AB = 6 cm,
BC = 9 cm,
CD = 12 cm and
SP = 36 cm
Also, PA, QB, RC and SD are all perpendiculars to line l,
PA  QB  RC  SD
Using Basic proportionality theorem,
PQ : QR : RS = AB : BC : CD
= 6 : 9 : 12
Taking,
PQ = 6x,
QR = 9x and
RS = 12x
As,
Length of PS = 36 cm
PQ + QR + RS = 36
⇒ 6x + 9x + 12x = 36
⇒ 27x = 36
⇒ x = 4/3
Now,
PQ = 6x
= 6 × 4/3
= 8 cm
QR = 9x
= 9 × 4/3
= 12 cm
RS = 12x
= 12 × 4/3
= 16 cm
15. O is the point of intersection of the diagonals AC and BD of a trapezium ABCD with AB  DC. Through O, a line segment PQ is drawn parallel to AB meeting AD in P and BC in Q. Prove that PO = QO.
Solution
To prove : PO = QO
OP/AD = PD/AD
In ∆ABC and ∆OQC,
OQ  AB
OQ/AB = QC/BC
Also, In Î”ADC,
OP  DC
AP/PD = OA/OC
In Î”ABC,OQ  AB
BQ/QC = OA/OC
Therefore,
AP/PD = BQ/QC
Adding 1 on both sides,
(AP/PD) + 1 = (BQ/QC) + 1
(AP+ PD)/PD = (BQ + QC)/QC
AD/PD = BC/QC
Or,
PD/AD = QC/BC
Also,
OP/AB = QC/BC and OP/AB = OQ/AB
Therefore,
OP = OQ
[Hint : Take point G on AB such that CG  DF.]
CE = AE ...(i)
Now, in ∆BCG and ∆BDF, CG  DF
BC/CD = BG/GF
⇒ BC/CD = (BF  GF)/GF
⇒ BC/CD = (BF/GF)  1
⇒ (BC/CD) + 1 = BF/CE [from (ii)]
⇒ (BC + CD)/CD = BF/CE
⇒ BD/CD = BF/CE
AC^{2} = AB^{2} + BC^{2}
⇒ AC^{2} = y^{2} + x^{2}
⇒ AC = √(y^{2} + x^{2} )
A_{3} = A_{1} + A_{2}
BC^{2} = AC^{2} + AB^{2}
BC^{2} = y^{2} + x^{2}
BC = √(y^{2} + x^{2} )
Also Area of equilateral triangle = (√3/4)a^{2}
Now we calculate the area A_{1} , A_{2} and A_{3} respectively