NCERT Solutions for Class 10 Maths Chapter 6 - Triangles

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Chapter 6 - Triangles Exercise Ex. 6.4

Solution 1

Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

Let us assume two similar triangles as Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABC ~ Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesPQR. Let AD and PS be the   medians of these triangles.

Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesA = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesP, Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesB = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesQ, Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesC = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesR
Since, AD and PS are medians
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

Solution 2
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Solution 3
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

Since AB || CD
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesOAB = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesOCD            (Alternate interior angles)
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesOBA = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesODC            (Alternate interior angles)
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesAOB = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesCOD            (Vertically opposite angles)
Therefore Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesAOB ~ Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesCOD        (By AAA rule)
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Solution 4

Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Since Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABC and Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDBC are one same base,
Therefore ratio between their areas will be as ratio of their heights.
Let us draw two perpendiculars AP and DM on line BC.
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesAPO and Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDMO,
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesAPO = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDMO = 90
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesAOP = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDOM          (vertically opposite angles)
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesOAP = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesODM         (remaining angle)
Therefore Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesAPO ~  Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDMO    (By AAA rule)
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

Solution 5
Let us assume two similar triangles as Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABC ~ Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesPQR
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Solution 6
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

Since D and E are mid points of Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABC
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Solution 7
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

Let us assume two similar triangles as Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABC ~ Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesPQR. Let AD and PS be the   medians of these triangles.

Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesA = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesP, Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesB = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesQ, Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesC = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesR
Since, AD and PS are medians
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Solution 8
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

Let ABCD be a square of side a.
Therefore its diagonal Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles     
Two desired equilateral triangles are formed as Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABE and Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDBF
Side of an equilateral triangle Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABE described on one of its side = a
Side of an equilateral triangle Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDBF described on one of its diagonal
We know that equilateral triangles are having all its angles as 60º and all its sides of same length. So, all equilateral triangles are similar to each other. So, ratio between areas of these triangles will be equal to the square of the ratio between sides of these triangles.
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Solution 9
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

We know that equilateral triangles are having all its angles as 60º and all its sides of same length. So, all equilateral triangles are similar to each other. So, ratio between areas of these triangles will be equal to the square of the ratio between sides of these triangles.
Let side of Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABC = x
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles Hence, (c)
Solution 10
If, two triangles are similar to each other, ratio between areas of these triangles will be equal to the square of the ratio between sides of these triangles.
Given that sides are in the ratio 4:9.


Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Hence, (d).

Chapter 6 - Triangles Exercise Ex. 6.1

Solution 1
(i)    All circles are SIMILAR.
(ii)    All squares are SIMILAR.
(iii)    All EQUILATERAL triangles are similar.
(iv)    Two polygons of same number of sides are similar, if their corresponding angles are EQUAL and their corresponding sides are PROPORTIONAL.

Solution 2
(i)  Two equilateral triangles with sides 1 cm and 2 cm.
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Two squares with sides 1 cm and 2 cm
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

(ii)  Trapezium and Square
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Triangle and Parallelogram 
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Solution 3
Quadrilateral PQRS and ABCD are not similar as their corresponding sides are proportional i.e. 1:2 but their corresponding angles are not equal.

Chapter 6 - Triangles Exercise Ex. 6.2

Solution 1
(i)
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Let EC = x
Since DE || BC.
Therefore, by basic proportionality theorem,

Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles (ii)
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Let AD = x
Since DE || BC,
Therefore by basic proportionality theorem,
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Solution 2
(i)
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

Given that PE = 3.9, EQ = 3, PF = 3.6, FR = 2.4
Now,

Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
(ii)

Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
PE = 4, QE = 4.5, PF = 8, RF = 9

Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
(iii)
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

PQ = 1.28, PR = 2.56, PE = 0.18, PF = 0.36

Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Solution 3
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
In the given figure
Since LM || CB,
Therefore by basic proportionality theorem,
 Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Solution 4
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABC,
Since DE || AC
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Solution 5
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesPOQ
Since DE || OQ
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Solution 6
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Solution 7
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Consider the given figure
PQ is a line segment drawn through midpoint P of line AB such that PQ||BC
i.e. AP = PB
Now, by basic proportionality theorem
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
i.e. AQ = QC
Or, Q is midpoint of AC.

Solution 8
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Consider the given figure
PQ is a line segment joining midpoints P and Q of line AB and AC respectively.
i.e. AP = PB and AQ = QC
Now, we may observe that
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
And hence basic proportionality theorem is verified
So, PQ||BC
Solution 9
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Solution 10
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

Chapter 6 - Triangles Exercise Ex. 6.3

Solution 1

(i)    Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesA = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesP = 60°
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesB = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesQ = 80°
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesC = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesR = 40°
Therefore Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABC ~ Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesPQR     [by AAA rule]
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

(iii)   Triangles are not similar as the corresponding sides are not proportional.
    
(iv)    Triangles are not similar as the corresponding sides are not proportional.

(v)    Triangles are not similar as the corresponding sides are not proportional.

(vi)    In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDEF
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesD + Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesE + Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesF = 180°
(Sum of measures of angles of a triangle is 180)
70° + 80° + Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesF = 180°
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesF = 30°
Similarly in Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesPQR
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesP + Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesQ + Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesR = 180°
(Sum of measures of angles of a triangle is 180)
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesP + 80° +30° = 180°
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesP = 70°
Now In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDEF and Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesPQR
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesD = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesP = 70°
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesE = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesQ = 80°
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesF = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesR = 30°
Therefore Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDEF ~ Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesPQR     [by AAA rule]

Solution 2

Since DOB is a straight line
Therefore Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDOC + Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesCOB = 180°
Therefore Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDOC = 180° - 125°
                 = 55°
In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDOC,
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDCO + Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesCDO + Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDOC = 180°
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDCO + 70° + 55° = 180°
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDCO = 55°
Since Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesODC ~ Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesOBA,
Therefore  Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesOCD = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesOAB    [corresponding angles equal in similar triangles]

Therefore Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles OAB = 55°

Solution 3
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDOC and Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesBOA
AB || CD
Therefore Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesCDO = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABO    [Alternate interior angles]
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDCO = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesBAO         [Alternate interior angles]
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDOC = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesBOA        [Vertically opposite angles]

Therefore Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDOC ~ Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesBOA    [AAA rule)
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Solution 4
In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesPQR
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesPQR = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesPRQ
Therefore PQ = PR    (i)
Given,
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Solution 5
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesRPQ and Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesRST
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesRTS = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesQPS              [given]
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesR = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesR            [common angle]
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesRST = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesRQP                      [ Remaining angles]
Therefore Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesRPQ ~ Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesRTS    [by AAA rule]
Solution 6

Since Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABE Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesACD
Therefore     AB = AC               (1)
AD = AE                (2)
Now, in Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesADE and Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABC,
Dividing equation (2) by (1)
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

Solution 7

(i)
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesAEP and Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesCDP
Since Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesCDP = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesAEP = 90°
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesCPD = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesAPE     (vertically opposite angles)
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesPCD = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesPAE    (remaining angle)
Therefore by AAA rule,
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesAEP ~ Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesCDP
(ii) Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABD and Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesCBE
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesADB = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesCEB = 90°
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABD = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesCBE     (common angle)
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDAB = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesECB    (remaining angle)
Therefore by AAA rule
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABD ~ Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesCBE
(iii)Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesAEP and Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesADB
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesAEP = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesADB = 90°
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesPAE = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDAB    (common angle)
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesAPE = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABD    (remaining angle)
Therefore by AAA rule
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesAEP ~ Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesADB
(iv)Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesPDC and Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesBEC
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesPDC = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesBEC = 90°
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesPCD = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesBCE    (common angle)
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesCPD = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesCBE
Therefore by AAA rule
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesPDC ~ Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesBEC

Solution 8

Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles


Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABE and Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesCFB

Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesA = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesC             (opposite angles of a parallelogram)
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesAEB = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesCBF         (Alternate interior angles AE || BC)
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABE = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesCFB         (remaining angle)
Therefore Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABE ~ Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesCFB    (by AAA rule)

Solution 9

In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABC and Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesAMP
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABC = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesAMP = 90
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesA = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesA                 (common angle)
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesACB = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesAPM            (remaining angle)
Therefore Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABC ~ Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesAMP        (by AAA rule)
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

Solution 10

Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

Since Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABC ~ Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesFEG
Therefore Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesA = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesF        
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesB = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesE
As, Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesACB = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesFGE
Therefore Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesACD = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesFGH    (angle bisector)  
And Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDCB = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesHGE        (angle bisector)
Therefore Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesACD ~ Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesFGH    (by AAA rule)

And Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDCB ~ Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesHGE        (by AAA rule)

Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

and Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesACD = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesFGH 

Therefore Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDCA ~ Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesHGF    (by SAS rule)

Solution 11

In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABD and Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesECF,
Given that AB = AC        (isosceles triangles)
So, Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABD = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesECF    
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesADB = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesEFC = 90
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesBAD = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesCEF
Therefore Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABD ~ Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesECF     (by AAA rule)

Solution 12
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles 

Median divides opposite side.
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Therefore Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABD ~ Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesPQM    (by SSS rule)
Therefore Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABD = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesPQM    (corresponding angles of similar triangles)
Therefore Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABC ~ Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesPQR     (by SAS rule)

Solution 13
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesADC and Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesBAC
Given that Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesADC = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesBAC
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesACD = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesBCA            (common angle)
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesCAD = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesCBA            (remaining angle)
Hence,  Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesADC ~ Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesBAC        [by AAA rule]
So, corresponding sides of similar triangles will be proportional to each other
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Solution 14
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Solution 15
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

Let AB be a tower
CD be a pole
Shadow of AB is BE
Shadow of CD is DF

The  light rays from sun will fall on tower and pole at same angle and at the same time.


So, Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDCF = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesBAE
 And Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDFC = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesBEA
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesCDF = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABE        (tower and pole are vertical to ground)    
Therefore Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABE ~ Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesCDF

Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles


So, height of tower will be 42 meters.

Solution 16
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

Since Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABC ~ Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesPQR
So, their respective sides will be in proportion
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Also, Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesA = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesP, Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesB = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesQ, Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesC = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesR                     (2)
Since, AD and PM are medians so they will divide their opposite sides in equal halves.
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
From equation (1) and (3)
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
So, we had observed that two respective sides are in same proportion in both triangles and also angle included between them is respectively equal  
Hence, Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABD ~ Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesPQM             (by SAS rule)
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

Chapter 6 - Triangles Exercise Ex. 6.5

Solution 1

i.Given that sides are 7 cm, 24 cm, and 25 cm. Squaring the lengths of these sides we get 49, 576, and 625.

Clearly, 49 + 576 = 625 or 72 + 242 = 252 .

Therefore, given triangle is satisfying Pythagoras theorem. So, it is a right triangle. The longest side in a right angled triangle is the hypotenuse.

Therefore length of hypotenuse of this triangle = 25 cm.

ii.Given that sides are 3 cm, 8 cm, and 6 cm. Squaring the lengths of these sides we may get 9, 64, and 36. Clearly, sum of squares of lengths of two sides is not equal to square of length of third side. Therefore given triangle is not satisfying Pythagoras theorem. So, it is not a right triangle

iii.Given that sides are 50 cm, 80 cm, and 100 cm. Squaring the lengths of these sides we may get 2500, 6400, and 10000. Clearly, sum of squares of lengths of two sides is not equal to square of length of third side. Therefore given triangle is not satisfying Pythagoras theorem. So, it is not a right triangle.

iv.Given that sides are 13 cm, 12 cm, and 5 cm. Squaring the lengths of these sides we may get 169, 144, and 25. Clearly, 144 +25 = 169 Or, 122 + 52 = 132.

Therefore given triangle is satisfying Pythagoras theorem. So, it is a right triangle.

The longest side in a right angled triangle is the hypotenuse.

Therefore length of hypotenuse of this triangle = 13 cm.

Solution 2
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Solution 3
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles 
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
iii.     In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDCA & Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDAB

Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDCA = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDAB = 90º
         Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesCDA = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesADB            (common angle)
         Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDAC = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDBA            (remaining angle)
        Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Solution 4
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

Given that Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABC is an isosceles triangle.
Therefore AC = CB
Applying Pythagoras theorem in ABC (i.e. right angled at point C)
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Solution 5
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Solution 6
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

Let AD be the altitude in given equilateral triangle Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABC.    
We know that altitude bisects the opposite side.
So, BD = DC = a
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Since in an equilateral triangle, all the altitudes are equal in length.
So, length of each altitude will be Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Solution 7
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles 

In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesAOB, Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesBOC, Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesCOD, Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesAOD
Applying Pythagoras theorem
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Solution 8
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Solution 9
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Solution 10
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles 

Let OB be the pole and AB be the wire.
Therefore by Pythagoras theorem,

Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Solution 11
 
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Let these distances are represented by OA and OB respectively.
Now applying Pythagoras theorem

Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Solution 12
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Let CD and AB be the poles of height 11 and 6 m.
Therefore CP = 11 -  6 = 5 m
From the figure we may observe that AP = 12m
In Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles  APC, by applying Pythagoras theorem
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Therefore distance between their tops = 13 m.
Solution 13
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

          In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesACE,

Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Solution 14
 Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles  
 
Solution 15

Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles 

Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Solution 16
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

Let side of equilateral triangle be a. And AE be the altitude of Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABC
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Now, in Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABE by applying Pythagoras theorem
AB2 = AE2 + BE2
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Solution 17
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

Given that AB = Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles cm, AC = 12 cm and BC = 6 cm
We may observe that
AB2 = 108
AC2 = 144
And BC2 = 36
AB2 +BC2 = AC2

Thus the given triangle Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABC is satisfying Pythagoras theorem

Therefore triangle is a right angled triangle right angled at B
Therefore Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesB = 90°.

Hence, (c).

Chapter 6 - Triangles Exercise Ex. 6.6

Solution 1
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

Let us draw a line segment RT parallel to SP which intersects extended line segment QP at point T.
Given that PS is angle bisector of Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles QPR.
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles QPS = Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles SPR                                    (1)
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles SPR = Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles PRT        (As PS || TR)          (2)
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles QPS = Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles QTR        (As PS || TR)         (3)

Using these equations we may find
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesPRT = Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles QTR                                     from  (2) and (3)
So, PT = PR             (Since Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles PTR is isosceles triangle)
    Now in  Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesQPS and Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesQTR
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles QSP = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesQRT        (As PS || TR)
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles QPS = Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles QTR        (As PS || TR)
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles Q is common
 Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesQPS  ~ Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesQTR
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Solution 2

(i).  Let us join DB.
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

DN || CB
DM || AB
So, DN = MB
DM = NB

The condition to be proved is the case when  DNBM is a square or D is the  midpoint of side AC.
Then Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesCDB = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesADB = 90°
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles2 + Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles3 = 90°        (1)
In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesCDM
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles 1 + Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles 2 + Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles DMC = 180°
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles 1 + Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles 2 = 90°        (2)
In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDMB
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles 3 + Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles DMB + Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles 4 = 180°
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles 3 + Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles 4 = 90°        (3)
From equation (1) and (2)
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles 1 = Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles 3
From equation (1) and (3)
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles 2 = Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles 4
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesBDM ~ Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles DCM
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

(ii).  Similarly in Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDBN
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles 4 + Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles 3 = 90°            (4)
In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDAN
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles 5 + Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles 6 = 90°            (5)
In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDAB
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles4 + Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles 5 = 90°            (6)
From equation (4) and (6)
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles3 = Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles 5
From equation (5) and (6)
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles4 = Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles 6
 Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDNA   ~ Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesBND
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

Solution 3
In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesADB applying Pythagoras theorem
AB2 = AD2 + DB2        (1)
In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesACD applying Pythagoras theorem
AC2 = AD2 + DC2
AC2 = AD2 + (DB + BC)2
AC2 = AD2 + DB2 + BC2 + 2DB  x  BC
Now using equation (1)
AC2 = AB2 + BC2 + 2BC . BD

Solution 4
In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesADB applying Pythagoras theorem
AD2 + DB2 = AB2
AD2 = AB2 - DB2                (1)
In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesADC applying Pythagoras theorem
AD2 + DC2 = AC2                (2)
Now using equation (1)
AB2 -  BD2 + DC2 = AC2
AB2 -  BD2 + (BC -  BD)2 = AC2
AC2 = AB2 -  BD2 + BC2 + BD2 - 2BC. BD
      = AB2 + BC2 - 2BC. BD

Solution 5
(i).  In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesAMD
AM2 + MD2 = AD2            (1)
In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesAMC
AM2 + MC2 = AC2            (2)
AM2 + (MD + DC)2 = AC2
(AM2 + MD2) + DC2 + 2MD.DC = AC2
Using equation (1) we may get
AD2 + DC2 + 2MD.DC = AC2
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

(ii).  In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABM applying Pythagoras theorem
AB2 = AM2 + MB2
= (AD2 - DM2) + MB2
= (AD2 -  DM2) + (BD -  MD)2
= AD2 -  DM2 + BD2 + MD2 - 2BD.MD
= AD2 + BD2 - 2BD.MD
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

(iii).  InNcert Solutions Cbse Class 10 Mathematics Chapter - TrianglesAMB
AM2 + MB2 = AB2        (1)
In Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles AMC    
AM2 + MC2 = AC2        (2)
Adding equation (1) and (2)
2AM2 + MB2 + MC2 = AB2 + AC2
2AM2 + (BD - DM)2 + (MD + DC)2 = AB2 + AC2
2AM2+BD2 + DM2 - 2BD.DM + MD2 + DC2 + 2MD.DC = AB2 + AC2
2AM2 + 2MD2 + BD2 + DC2 + 2MD (-BD + DC) = AB2 + AC2
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Solution 6
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

Let Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles ABCD be a parallelogram
Let us draw perpendicular DE on extended side AB and AF on side DC.
In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDEA
DE2 + EA2 = DA2                      (i)
In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDEB
DE2 + EB2 = DB2
DE2 + (EA + AB)2 = DB2
(DE2 + EA2) + AB2 + 2EA. AB = DB2
DA2 + AB2 + 2EA.AB = DB2        (ii)
In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesADF
AD2 = AF2 + FD2
In Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles AFC
AC2 = AF2 + FC2
= AF2 + (DC - FD)2
= AF2 + DC2 + FD2 - 2DC - FD
= (AF2 + FD2) + DC2 - 2DC . FD
AC2 = AD2 + DC2 - 2DC  FD        (iii)
Since ABCD is a parallelogram
AB = CD                 (iii)
And BC = AD           (iv)
In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDEA and Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesADF
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles DEA =Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles AFD
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles EAD = Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles FDA        (EA || DF)
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles EDA = Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles FAD        (AF || ED)
AD is common in both triangles.
Since respective angles are same and respective sides are same
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDEA Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesAFD
So EA = DF            (v)
Adding equation (ii) and (iii)
DA2 + AB2 + 2EA.AB + AD2 + DC2 - 2DC.FD = DB2 + AC2
DA2 + AB2 + AD2 + DC2 + 2EA.AB - 2DC.FD = DB2 + AC2
BC2 + AB2 + AD2 + DC2 + 2EA.AB-2AB.EA = DB2 + AC2
AB2 + BC2 + CD2 + DA2 = AC2 + BD2


Solution 7
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

Let us join CB
(i)    In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesAPC and Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDPB
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesAPC = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDPB    {Vertically opposite angles}
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesCAP = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesBDP    {Angles in same segment for chord CB}
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesAPC ~ Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDPB    {BY AA similarly criterion}

    
(ii)    We know that corresponding sides of similar triangles are proportional
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Solution 8

(i)    In Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesPAC and Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesPDB
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles P = Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles P        (common)
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles PAC = Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles PDB    (exterior angle of a cyclic quadrilateral is equal to opposite interior angle)

Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles PCA = Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesPBD
Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesPAC ~ Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesPDB

(ii)    We know that corresponding sides of similar triangles are proportional.
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

Solution 9
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
AD = AD        (common)
So, Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDBA ~ Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesDCA    (By SSS)
Now, corresponding angles of similar triangles will be equal.
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles BAD = Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles CAD
 AD is angle bisector of  Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles BAC

Solution 10

Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles

Let AB be the height of tip of fishing rod from water surface. Let BC be the horizontal distance of fly from the tip of fishing rod.
Then, AC is the length of string.
AC can be found by applying Pythagoras theorem in Ncert Solutions Cbse Class 10 Mathematics Chapter - TrianglesABC
AC2 = AB2 + BC2
AC2 = (1.8)2 + (2.4)2  
AC2 = 3.24 + 5.76  
AC2 = 9.00

AC = Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles = 3

Thus, length of string out is 3 m. 
Now, she pulls string at rate of 5 cm per second.
So, string pulled in 12 seconds = 12 x 5 = 60 cm = 0.6 m
Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles
Let after 12 second Fly be at point D.
Length of string out after 12 second is AD
AD = AC - string pulled by Nazima in 12 seconds
= 3.00 - 0.6
= 2.4
In Ncert Solutions Cbse Class 10 Mathematics Chapter - Triangles ADB
AB2 + BD2 = AD2
(1.8)2 + BD2 = (2.4)2
BD2 = 5.76 - 3.24 = 2.52
BD = 1.587
Horizontal distance of fly = BD + 1.2
= 1.587 + 1.2
= 2.787
= 2.79 m

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