Chapter 6 Triangles NCERT Exemplar Solutions Exercise 6.3 Class 10 Maths
Chapter Name  NCERT Maths Exemplar Solutions for Chapter 6 Triangles Exercise 6.3 
Book Name  NCERT Exemplar for Class 10 Maths 
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Related Study  NCERT Solutions for Class 10 Maths 
Exercise 6.3 Solutions
Short Answer Questions
1. In a Î”PQR, PR^{2}  PQ^{2} = QR^{2} and M is a point on side PR such that QM ⊥ PR.
Prove that :
QM^{2} = PM × MR.
Solution
In ∆PQR,
PR^{2} = QR^{2} and
QM ⊥ PR
Using Pythagoras theorem, we have,
PR^{2} = PQ^{2} + QR^{2}
∆PQR is right angled triangle at Q.
From ∆QMR and ∆PMQ, we get,
∠M = ∠M
∠MQR = ∠QPM [each 90°∠R]
So, using the AAA similarity criteria,
We have,
∆QMR ∼ ∆PMQ
Also,
Area of triangles = 1/2 × base × height
So, by property of area of similar triangles,
QM^{2} = PM × RM
Hence proved.
2. Find the value of x for which DE  AB in given figure.
Solution
As given in the question,
DE  AB
Using basic proportionality theorem,
CD/AD = CE/BE
If a line is drawn parallel to one side of a triangle such that it intersects the other sides at distinct points, then, the other two sides are divided in the same ratio. Therefore, we can conclude that, the line drawn is equal to the third side of the triangle.
(x +3)/(3x + 19) = x/(3x + 4)
⇒ (x + 3)(3x + 4) = x(3x + 19)
⇒ 3x^{2} + 4x + 9x + 12 = 3x^{2} + 19x
⇒ 19x  13x = 12
⇒ 6x = 12
⇒ x = 2
3. In figure, if ∠1 =∠2 and ∆ NSQ ≅ ∆MTR, then prove that Î”PTS ~ Î”PRQ.
Solution
As given in the question,
Î”NSQ ≅ Î”MTR
∠1 = ∠2
As,
∆NSQ = ∆MTR
So,
SQ = TR ...(i)
Also,
∠1 = ∠2
So,
PT = PS ...(ii)
[As, sides opposite to equal angles are also equal]
Using Equation (i) and (ii).
PS/SQ = PT/TR
So,
ST  QR
By converse of basic proportionality theorem, If a line is drawn parallel to one side of a triangle to intersect the other sides in distinct points, the other two sides are divided in the same ratio.
∠1 = PQR and∠2 = ∠PRQ
Now, In ∆PTS and ∆PRQ.
∠P = ∠P [Common angles]
∠1 = ∠PQR (proved)
∠2 = ∠PRQ (proved)
∆PTS – ∆PRQ [By AAA similarity criteria]
Hence proved.
4. Diagonals of a trapezium PQRS intersect each other at the point 0, PQ  RS and PQ = 3 RS. Find the ratio of the areas of Î” POQ and Î” ROS.
Solution
As given in the question,
PQRS is a trapezium in which PQ  RS and PQ = 3RS
PQ/RS = 3/2 ...(i)
In ∆POQ and ∆ROS,
∠SOR = ∠QOP [vertically opposite angles]
∠SRP = ∠RPQ [alternate angles]
∆POQ ∼ ∆ROS [by AAA similarity criterion]
Using property of area of similar triangle,
So, the required ratio = 9 : 1.
5. In figure, if AB  DC and AC, PQ intersect each other at the point O. Prove that OA.CQ = OC.AP.
Solution
As given in the question,
AC and PQ intersect each other at the point O and ABDC.
Using ∆AOP and ∆COQ,
∠AOP = ∠COQ [as they are vertically opposite angles]
∠APO = ∠CQO [since, ABDC and PQ is transversal, Angles are alternate angles]
So,
∆AOP ∼ ∆COQ [using AAA similarity criterion]
As, corresponding sides are proportional
We have,
OA/OC = AP/CQ
OA × CQ = OC × AP
Hence, Proved.
⇒ (8)^{2} = AD^{2} + (4)^{2}
⇒ 64 = AD^{2} + 16
⇒ AD = √48
⇒ AD = 4√3 cm.
AB/ED = BC/EF = AC/DF
4/6 = BC/9 = AC/12
Now, 4/6 = BC/9
BC = 6 cm
Similarly,
AC/12 = 4/6
⇒ AD AC = 8cm
Perimeter of ∆ABC = AB + BC + AC
= 4 + 6 + 8 = 18 cm
So, the perimeter of the triangle is 18 cm.
DE  BC,
DE = 6 cm and
BC = 12 cm
In Î”ABC and Î”ADE,
∠ABC = ∠ADE [corresponding angle]
and
∠A = ∠A [common side]
= 6^{2} /12^{2} = 1/4
Taking,
ar(∆ADE) = k, then
⇒ ar(∆ABC) = 4k
Now,
ar(∆ECB) = ar(ABC)  ar(ADE)
= 4k  k = 3k
So,
Required ratio = ar(ADE) : ar(DECB)
= k : 3k
= 1 : 3
So, AB  PQ  DC.
In ∆ABD,
PO  AB
DP/AP = DO/OB ...(i)
In ∆BDC,
OQ  DC
BQ/QC = OB/OD
or, QC/BQ = DO/OB ...(ii)
So, from (i) and (ii),
DP/AP = QC/BQ
⇒ 18/AP = 15/35
AP = 42 cm
Also,
AD = AP + PD
= 42 + 18 = 60
So, AD = 60 cm
In Î”PQR, N is a point on PR, such that QN ⊥ PR and PN .
Also,
⇒ x = 12 cm
AC/AD = AB/AC
⇒ 8/3 = AB/8
⇒ AB = 64/3 cm
Also,
AB = BD + AD
⇒ 64/3 = BD + 3
⇒ BD = 55/3 cm
So by AA rule,
AB/DE = BC/EF
⇒ 24/16 = 15/h
⇒ h = 10
Hence, the height of the point on he wall where the top of the ladder reaches is 8 m.
In right angled ∆ABC
AC^{2 } = AB^{2 } + BC^{2 }
⇒ (10)^{2 } = AB^{2 } + (6)^{2 }
⇒ 100 = AB^{2 } + 36
⇒ AB^{2 } = 100  36 = 64
⇒ AB = 8 m
Therefore, the height of the point on the wall where the top of the ladder reaches is 8 m.