## RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles

### RD Sharma Class 9 Solutions Chapter 10 Congruent TrianglesÂ Ex 10.1

Question 1.

Write the complement of each of the following angles:

(i) 20Â°

(ii) 35Â°

(iii) 90Â°

(iv) 77Â°

(v) 30Â°

Solution:

We know that two angles are complement to each other if their sum is 90Â°. Therefore,

(i) Complement of 20Â° is (90Â° – 20Â°) = 70Â°

(ii) Complement of 35Â° is (90Â° – 35Â°) = 55Â°

(iii) Complement of 90Â° is (90Â° – 90Â°) = 0Â°

(iv) Complement of 77Â° is (90Â° – 77Â°) = 13Â°

(v) Complement of 30Â° is (90Â° – 30Â°) = 60Â°

Question 2.

Write the supplement of each of the following angles:

(i) 54Â°

(ii) 132Â°

(iii) 138Â°

Solution:

We know that two angles are supplement to each other if their sum if 180Â°. Therefore,

(i) Supplement of 54Â° is (180Â° – 54Â°) = 126Â°

(ii) Supplement of 132Â° is (180Â° – 132Â°) = 48Â°

(iii) Supplement of 138Â° is (180Â° – 138Â°) = 42Â°

Question 3.

If an angle is 28Â° less than its complement, find its measure.

Solution:

Let required angle = x, then

Its complement = x + 28Â°

âˆ´Â x + x + 28Â° = 90Â° â‡’Â 2x = 90Â° – 28Â° = 62Â°

âˆ´ x =Â \(\frac { { 62 }^{ \circ } }{ 2 }\)Â = 31Â°

âˆ´ Required angle = 31Â°

Question 4.

If an angle is 30Â° more than one half of its complement, find the measure of the angle.

Solution:

Let the measure of the required angle = x

âˆ´Â Its complement =Â 90Â° – x

âˆ´Â x = \(\frac { 1 }{ 2 }\) (90Â° – x) + 30Â°

2x = 90Â° – x + 60Â°

â‡’ 2x + x = 90Â° + 60Â°

â‡’Â 3x = 150Â°

â‡’ x = Â \(\frac { { 150 }^{ \circ } }{ 3 }\)Â = 50Â°

âˆ´ Required angle = 50Â°

Question 5.

Two supplementary angles are in the ratio 4 : 5. Find the angles.

Solution:

Ratio in two supplementary angles = 4 : 5

Let first angle = 4x

Then second angle = 5x

âˆ´Â 4x + 5x = 180

â‡’Â 9x = 180Â°

âˆ´ xÂ = \(\frac { { 180 }^{ \circ } }{ 9 }\) = 20Â°

âˆ´Â First angle = 4x = 4 x 20Â° = 80Â°

and second angle = 5x

= 5 x 20Â° = 100Â°

Question 6.

Two supplementary angles differ by 48Â°. Find the angles.

Solution:

Let first angle = xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ”

Then second angle = x + 48Â°

âˆ´Â x + x + 48Â° = 180Â°â‡’Â 2x + 48Â° = 180Â°

â‡’Â 2x = 180Â° – 48Â° = 132Â°

x= \(\frac { { 132 }^{ \circ } }{ 2 }\) =66Â°

âˆ´Â First angle = 66Â°

and second angle = x + 48Â° = 66Â° + 48Â° = 114Â°

âˆ´ Angles are 66Â°, 114Â°

Question 7.

An angle is equal to 8 times its complement. Determine its measure.

Solution:

Let the required angle = x

Then its complement angle = 90Â° – x

âˆ´ x = 8(90Â° – x)

â‡’ x = 720Â° – 8x â‡’Â x + 8x = 720Â°

â‡’ 9x = 720Â°Â â‡’ x =Â \(\frac { { 720 }^{ \circ } }{ 9 }\) = 80Â°

âˆ´Â Required angle = 80Â°

Question 8.

If the angles (2x – 10)Â° and (x – 5)Â° are complementary angles, find x.

Solution:

First complementary angle = (2x – 10Â°) and second = (x – 5)Â°

âˆ´ 2x – 10Â° + x – 5Â° = 90Â°

â‡’ 3x – 15Â° = 90Â° â‡’Â 3x = 90Â° + 15Â° = 105Â°

âˆ´ x =Â \(\frac { { 105 }^{ \circ } }{ 3 }\) = 35Â°

âˆ´Â First angle = 2x – 10Â° = 2 x 35Â° – 10Â°

= 70Â° – 10Â° = 60Â°

and second angle = x – 5 = 35Â° – 5 = 30Â°

Question 9.

If an angle differ from its complement by 10Â°, find the angle.

Solution:

Let required angle = xÂ°

Then its complement angle = 90Â° – xÂ°

âˆ´ x – (90Â° – x) = 10

â‡’Â x – 90Â° + x = 10Â°â‡’Â 2x = 10Â° + 90Â° = 100Â° 100Â°

â‡’ x =Â \(\frac { { 100 }^{ \circ } }{ 2 }\) = 50Â°

âˆ´ Required angle = 50Â°

Question 10.

If the supplement of an angle is two-third of itself Determine the angle and its supplement.

Solution:

Let required angle = x

Then its supplement angle = 180Â° – x

âˆ´Â (180Â°-x)= \(\frac { 2 }{ 3 }\)x

540Â° – 3x = 2x â‡’ 2x + 3x = 540Â°

â‡’ 5x = 540Â°â‡’Â x =Â \(\frac { { 540 }^{ \circ } }{ 5 }\) = 108Â°

-. Supplement angle = 180Â° – 108Â° = 72Â°

Question 11.

An angle is 14Â° more than its complementary angle. What is its measure?

Solution:

Let required angle = x

Then its complementary angle = 90Â° – x

âˆ´Â x + 14Â° = 90Â° – x

x + x = 90Â° – 14Â° â‡’Â 2x = 76Â°

â‡’ x =Â \(\frac { { 76 }^{ \circ } }{ 2 }\) = 38Â°

âˆ´Â Required angle = 38Â°

Question 12.

The measure of an angle is twice the measure of its supplementary angle. Find its measure.

Solution:

Let the required angle = x

âˆ´Â Its supplementary angle = 180Â° – x

âˆ´Â x = 2(180Â°-x) = 360Â°-2x

â‡’Â x +Â 2x = 360Â°

â‡’ 3x = 360Â°

â‡’Â x = \(\frac { { 360 }^{ \circ } }{ 3 }\) = 120Â°

âˆ´Â Required angle = 120Â°

Question 13.

If the complement of an angle is equal to the supplement of the thrice of it. Find the measure of the angle.

Solution:

Let required angle = x

Then its complement angle = 90Â° – x

and supplement angle = 180Â° – x

âˆ´Â 3(90Â° – x) = 180Â° – x

â‡’ 270Â° – 3x = 180Â° – x

â‡’270Â° – 180Â° = -x + 3x => 2x = 90Â°

â‡’ x = 45Â°

âˆ´Â Required angle = 45Â°

Question 14.

If the supplement of an angle is three times its complement, find the angle.

Solution:

Let required angle = x

Then its complement = 90Â°-Â x

and supplement = 180Â° – x

âˆ´Â 180Â°-x = 3(90Â°-x)

â‡’Â 180Â° – x = 270Â° – 3x

â‡’Â -x + 3x = 270Â° – 180Â°

â‡’ 2x = 90Â° â‡’ x = \(\frac { { 90 }^{ \circ } }{ 2 }\) =45Â°

âˆ´ Required angle = 45Â°

### Class 9 RD Sharma Solutions Chapter 10 Congruent Triangles Ex 10.2

Question 1.

In the figure, OA and OB are opposite rays:

(i) If x = 25Â°, what is the value of y?

(ii) If y = 35Â°, what is the value of x?

Solution:

(i) If x = 25Â°

âˆ´ 3x = 3 x 25Â° = 75Â°

But âˆ AOC + âˆ BOC = 180Â° (Linear pair)

â‡’ âˆ AOC + 75Â° = 180Â°

â‡’ âˆ AOC = 180Â° – 75Â°

â‡’ âˆ AOC = 105Â°

âˆ´ 2y + 5 = 105Â° â‡’ 2y= 105Â° – 5Â° = 100Â°

â‡’ y = \(\frac { { 100 }^{ \circ } }{ 2 }\)Â = 50Â°

âˆ´ If x = 25Â° then y = 50Â°

(ii) If y = 35Â°, then âˆ AOC = 2y + 5

âˆ´ 2y + 5 = 2 x 35Â° + 5 = 70Â° + 5 = 75Â°

But âˆ AOC + âˆ BOC = 180Â° (Linear pair)

â‡’ 75Â° + âˆ BOC = 180Â°

â‡’ âˆ BOC = 180Â°-75Â°= 105Â°

âˆ´ 3x = 105Â° â‡’ x = \(\frac { { 105 }^{ \circ } }{ 3 }\) = 35Â°

âˆ´ x = 35Â°

Question 2.

In the figure, write all pairs of adjacent angles and all the linear pairs.

Solution:

In the given figure,

(i) âˆ AOD, âˆ COD; âˆ BOC, âˆ COD; âˆ AOD, âˆ BOD and âˆ AOC, âˆ BOC are the pairs of adjacent angles.

(ii) âˆ AOD, âˆ BOD and âˆ AOC, âˆ BOC are the pairs of linear pairs.

Question 3.

In the figure, find x, further find âˆ BOC, âˆ COD and âˆ AOD.

Solution:

In the figure,

AOB is a straight line

âˆ´ âˆ AOD + âˆ DOB = 180Â° (Linear pair)

â‡’ âˆ AOD + âˆ DOC + âˆ COB = 180Â°

â‡’ x+ 10Â° + x + x + 20 = 180Â°

â‡’ 3x + 30Â° = 180Â°

â‡’ 3x= 180Â° -30Â°= 150Â°

â‡’ x = \(\frac { { 150 }^{ \circ } }{ 3 }\) = 50Â°

âˆ´ x = 50Â°

Now âˆ BOC =x + 20Â° = 50Â° + 20Â° = 70Â°

âˆ COD = x = 50Â°

and âˆ AOD = x + 10Â° = 50Â° + 10Â° = 60Â°

Question 4.

In the figure, rays OA, OB, OC, OD and OE have the common end point O. Show that âˆ AOB + âˆ BOC + âˆ COD + âˆ DOE + âˆ EOA = 360Â°.

Solution:

Produce AO to F such that AOF is a straight line

Now âˆ AOB + âˆ BOF = 180Â° (Linear pair)

â‡’ âˆ AOB + âˆ BOC + âˆ COF = 180Â° …(i)

Similarly, âˆ AOE + âˆ EOF = 180Â°

â‡’ âˆ AOE + âˆ EOD + âˆ DOF = 180Â° …(ii)

Adding (i) and (ii)

âˆ AOB + âˆ BOC + âˆ COF + âˆ DOF + âˆ EOD + âˆ AOE = 180Â° + 180Â°

â‡’ âˆ AOB + âˆ BOC + âˆ COD + âˆ DOE + âˆ EOA = 360Â°

Hence proved.

Question 5.

In the figure, âˆ AOC and âˆ BOC form a linear pair. If a – 2b = 30Â°, find a and b.

Solution:

âˆ AOC + âˆ BOC = 180Â° (Linear pair)

â‡’ a + 6 = 180Â° …(i)

and a – 2b = 30Â° …(ii)

Subtracting (ii) from (i), 3b = 150Â°

â‡’ b = \(\frac { { 150 }^{ \circ } }{ 3 }\) = 50Â°

and a + 50Â° = 180Â°

â‡’ a= 180Â°-50Â°= 130Â°

âˆ´ a = 130Â°, b = 50Â°

Question 6.

How many pairs of adjacent angles are formed when two lines intersect in a point?

Solution:

If two lines AB and CD intersect each other at a point O, then four pairs of linear pairs are formed i.e.

âˆ AOC, âˆ BOC; âˆ BOC, âˆ BOD; âˆ BOD, âˆ AOD and âˆ AOD, âˆ AOC

Question 7.

How many pairs of adjacent angles, in all, can you name in the figure.

Solution:

In the given figure 10 pairs of adjacent angles are formed as given below:

âˆ AOB, âˆ BOC; âˆ AOB, âˆ BOD; âˆ AOC, âˆ COD; âˆ AOD, âˆ BOE; âˆ AOB, âˆ BOE; âˆ AOC, âˆ COF; âˆ BOC, âˆ COD; âˆ BOC, âˆ COE; âˆ COD, âˆ DOE and âˆ BOD, âˆ DOE

Question 8.

In the figure, determine the value of x.

Solution:

In the figure

âˆ AOC + âˆ COB + âˆ BOD +âˆ AOD = 360 (angles at a point )

â‡’ 3x + 3x + x + 150Â° = 360Â°

â‡’ 7x – 360Â° – 150Â° = 210Â°

â‡’ x = \(\frac { { 210 }^{ \circ } }{ 7 }\) = 30Â°

âˆ´ x = 30Â°

Question 9.

In the figure, AOC is a line, find x.

Solution:

In the figure,

âˆ AOB + âˆ BOC = 180Â° (Linear pair)

â‡’ 70Â° + 2x = 180Â°

â‡’ 2x = 180Â° – 70Â°

â‡’ 2x = 110Â°â‡’x = \(\frac { { 110 }^{ \circ } }{ 2 }\) = 55Â°

âˆ´ x = 55Â°

Question 10.

In the figure, POS is a line, find x.

Solution:

In the figure, POS is a line

âˆ´ âˆ POQ + âˆ QOS = 180Â° (Linear pair)

â‡’ âˆ POQ + âˆ QOR + âˆ ROS = 180Â°

â‡’ 60Â° + 4x + 40Â° = 180Â°

â‡’ 4x + 100Â° – 180Â°

â‡’ 4x = 180Â° – 100Â° = 80Â°

â‡’ x = \(\frac { { 80 }^{ \circ } }{ 4 }\) =20Â°

âˆ´ x = 20Â°

Question 11.

In the figure, ACB is a line such that âˆ DCA = 5x and âˆ DCB = 4x. Find the value of x.

Solution:

ACB is a line, âˆ DCA = 5x and âˆ DCB = 4x

âˆ ACD + âˆ DCB = 180Â° (Linear pair)

â‡’ 5x + 4x = 180Â° â‡’ 9x = 180Â°

â‡’ x = \(\frac { { 180 }^{ \circ } }{ 9 }\) = 20Â°

âˆ´ x = 20Â°

âˆ´ âˆ ACD = 5x = 5 x 20Â° = 100Â° and âˆ DCB = 4x = 4 x 20Â° = 80Â°

Question 12.

Given âˆ POR = 3x and âˆ QOR = 2x + 10Â°, find the value ofx for which POQ will be a line.

Solution:

âˆ POR = 3x and âˆ QOR = 2x + 10Â°

If POQ is a line, then

âˆ POR + âˆ QOR = 180Â° (Linear pair)

â‡’ 3x + 2x + 10Â° = 180Â°

â‡’ 5x = 180Â° – 10Â° = 170Â°

â‡’ x = \(\frac { { 170 }^{ \circ } }{ 5 }\) = 34Â°

âˆ´ x = 34Â°

Question 13.

What value ofy would make AOB, a line in the figure, if âˆ AOC = 4y and âˆ BOC = (6y + 30).

Solution:

In the figure,

AOB is a line if

âˆ AOC + âˆ BOC = 180Â°

â‡’ 6y + 30Â° + 4y= 180Â°

â‡’ 10y= 180Â°-30Â°= 150Â°

150Â°

â‡’ y = \(\frac { { 150 }^{ \circ } }{ 10 }\) = 15Â°

âˆ´ y = 15Â°

Question 14.

In the figure, OP, OQ, OR and OS are four rays. Prove that: âˆ POQ + âˆ QOR + âˆ SOR + âˆ POS = 360Â° [NCERT]

Solution:

In the figure, OP, OQ, OR and OS are the rays from O

To prove : âˆ POQ + âˆ QOR + âˆ SOR + âˆ POS = 360Â°

Construction : Produce PO to E

Proof: âˆ POQ + âˆ QOE = 180Â° (Linear pair)

Similarly, âˆ EOS + âˆ POS = 180Â°

Adding we get,

âˆ POQ + âˆ QOR + âˆ ROE + âˆ EOS + âˆ POS = 180Â° + 180Â° ,

â‡’ âˆ POQ + âˆ QOR + âˆ ROS + âˆ POS = 360Â° Hence âˆ POQ + âˆ QOR + âˆ SOR + âˆ POS = 360Â°

Question 15.

In the figure, ray OS stand on a line POQ. Ray OR and ray OT are angle bisectors of âˆ POS and âˆ SOQ respectively. If âˆ POS = x, find âˆ ROT. [NCERT]

Solution:

Ray OR stands on a line POQ forming âˆ POS and âˆ QOS

OR and OT the angle bisects of âˆ POS and âˆ QOS respectively. âˆ POS = x

But âˆ POS + âˆ QOS = 180Â° (Linear pair)

â‡’ x + âˆ QOS = 180Â°

â‡’ âˆ QOS = 180Â° – x

âˆµ OR and OT are the bisectors of angle

Question 16.

In the figure, lines PQ and RS intersect each other at point O. If âˆ POR : âˆ ROQ = 5 : 7. Find all the angles. [NCERT]

Solution:

Lines PQ and PR, intersect each other at O

âˆµ Vertically opposite angles are equal

âˆ´ âˆ POR = âˆ QOS and âˆ ROQ = âˆ POS

âˆ POR : âˆ ROQ = 5:7

Let âˆ POR = 5x and âˆ ROQ = 7x

But âˆ POR + âˆ ROQ = 180Â° (Linear pair)

âˆ´ 5x + 7x = 180Â° â‡’ 12x = 180Â°

â‡’ x = \(\frac { { 180 }^{ \circ } }{ 12 }\) = 15Â°

âˆ´ âˆ POR = 5x = 5 x 15Â° = 75Â°

and âˆ ROQ = 7x = 7 x 15Â° = 105Â°

But âˆ QOS = POR = 75Â° (Vertically opposite angles)

and âˆ POS = âˆ ROQ = 105Â°

Question 17.

In the figure, a is greater than b by one third of a right-angle. Find the values of a and b.

Solution:

In the figure,

âˆ AOC + âˆ BOC = 180Â° (Linear pair)

â‡’ a + b =180Â° …(i)

But a = b + \(\frac { 1 }{ 3 }\) x 90Â° = b + 30Â°

â‡’ a – b = 30Â° …(ii)

Adding (i) and (ii)

210Â°

2a = 210Â° â‡’ a = \(\frac { { 210 }^{ \circ } }{ 2 }\) = 105Â°

and 105Â° + b = 180Â°

â‡’ b = 180Â° – 105Â°

âˆ´ b = 75Â°

Hence a = 105Â°, b = 75Â°

Question 18.

In the figure, âˆ AOF and âˆ FOG form a linear pair.

âˆ EOB = âˆ FOC = 90Â° and âˆ DOC = âˆ FOG = âˆ AOB = 30Â°

(i) Find the measures of âˆ FOE, âˆ COB and âˆ DOE.

(ii) Name all the right angles.

(iii) Name three pairs of adjacent complementary angles.

(iv) Name three pairs of adjacent supplementary angles.

(v) Name three pairs of adjacent angles.

Solution:

In the figure,

âˆ AOF and âˆ FOG form a linear pair

âˆ EOB = âˆ FOC = 90Â°

âˆ DOC = âˆ FOG = âˆ AOB = 30Â°

(i) âˆ BOE = 90Â°, âˆ AOB = 30Â°

But âˆ BOE + âˆ AOB + âˆ EOG = 180Â°

â‡’ 30Â° + 90Â° + âˆ EOG = 180Â°

âˆ´âˆ EOG = 180Â° – 30Â° – 90Â° = 60Â°

But âˆ FOG = 30Â°

âˆ´ âˆ FOE = 60Â° – 30Â° = 30Â°

âˆ COD = 30Â°, âˆ COF = 90Â°

âˆ´ âˆ DOF = 90Â° – 30Â° = 60Â°

âˆ DOE = âˆ DOF – âˆ EOF

= 60Â° – 30Â° = 30Â°

âˆ BOC = BOE – âˆ COE

= 90Â° – 30Â° – 30Â° = 90Â° – 60Â° = 30Â°

(ii) Right angles are,

âˆ AOD = 30Â° + 30Â° + 30Â° = 90Â°

âˆ BOE = 30Â° + 30Â° + 30Â° = 90Â°

âˆ COF = 30Â° + 30Â° + 30Â° = 90Â°

and âˆ DOG = 30Â° + 30Â° + 30Â° = 90Â°

(iii) Pairs of adjacent complementaiy angles are âˆ AOB, âˆ BOD; âˆ AOC, âˆ COD; âˆ BOC, âˆ COE

(iv) Pairs of adjacent supplementary angles are âˆ AOB, âˆ BOG; âˆ AOC, âˆ COG and âˆ AOD, âˆ DOG

(v) Pairs of adjacent angles are âˆ BOC, âˆ COD; âˆ COD, âˆ DOE and âˆ DOE, âˆ EOF.

Question 19.

In the figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that âˆ ROS = \(\frac { 1 }{ 2 }\)(âˆ QOS – âˆ POS). [NCERT]

Solution:

Ray OR âŠ¥ ROQ. OS is another ray lying between OP and OR.

To prove : âˆ ROS = \(\frac { 1 }{ 2 }\)(âˆ QOS – âˆ POS)

Proof : âˆµ RO âŠ¥ POQ

âˆ´ âˆ POR = 90Â°

â‡’ âˆ POS + âˆ ROS = 90Â° …(i)

â‡’ âˆ ROS = 90Â° – âˆ POS

But âˆ POS + âˆ QOS = 180Â° (Linear pair)

= 2(âˆ POS + âˆ ROS) [From (i)]

âˆ POS + âˆ QOS = 2âˆ ROS + 2âˆ POS

â‡’ 2âˆ ROS = âˆ POS + âˆ QOS – 2âˆ POS

= âˆ QOS – âˆ POS

âˆ´ ROS = \(\frac { 1 }{ 2 }\) (âˆ QOS – âˆ POS)

### Class 9 Maths Chapter 10 Congruent Triangles RD Sharma Solutions Ex 10.3

Question 1.

In the figure, lines l_{1} and l_{2} intersect at O, forming angles as shown in the figure. If x = 45, find the values of y, z and n.

Solution:

Two lines l_{1} and l_{2} intersect each other at O âˆ x = 45Â°

âˆµ âˆ z = âˆ x (Vertically opposite angles)

= 45Â°

But x + y = 180Â° (Linear pair)

â‡’45Â° + y= 180Â°

â‡’ y= 180Â°-45Â°= 135Â°

But u = y (Vertically opposite angles)

âˆ´ u = 135Â°

Hence y = 135Â°, z = 45Â° and u = 135Â°

Question 2.

In the figure, three coplanar lines intersect at a point O, forming angles as shown in the figure. Find the values of x, y, z and u.

Solution:

Three lines AB, CD and EF intersect at O

âˆ BOD = 90Â°, âˆ DOF = 50Â°

âˆµ AB is a line

âˆ´ âˆ BOD + âˆ DOF + FOA = 180Â°

â‡’ 90Â° + 50Â° + u = 180Â°

â‡’ 140Â° + w = 180Â°

âˆ´ u= 180Â°- 140Â° = 40Â°

But x = u (Vertically opposite angles)

âˆ´ x = 40Â°

Similarly, y = 50Â° and z = 90Â°

Hence x = 40Â°, y = 50Â°, z = 90Â° and u = 40Â°

Question 3.

In the figure, find the values of x, y and z.

Solution:

Two lines l_{1} and l_{2} intersect each other at O

âˆ´ Vertically opposite angles are equal,

âˆ´ y = 25Â° and x = z

Now 25Â° + x = 180Â° (Linear pair)

â‡’ x= 180Â°-25Â°= 155Â°

âˆ´ z = x = 155Â°

Hence x = 155Â°, y = 25Â°, z = 155Â°

Question 4.

In the figure, find the value of x.

Solution:

âˆµ EF and CD intersect each other at O

âˆ´ Vertically opposite angles are equal,

âˆ´ âˆ 1 = 2x

AB is a line

3x + âˆ 1 + 5x = 180Â° (Angles on the same side of a line)

â‡’ 3x + 2x + 5x = 180Â°

â‡’ 10x = 180Â° â‡’ x = \(\frac { { 180 }^{ \circ } }{ 10 }\)Â = 18Â°

Hence x = 18Â°

Question 5.

If one of the four angles formed by two intersecting lines is a right angle, then show that each of the four angles is a right angle.

Solution:

Given : Two lines AB and CD intersect each other at O. âˆ AOC = 90Â°

To prove: âˆ AOD = âˆ BOC = âˆ BOD = 90Â°

Proof : âˆµ AB and CD intersect each other at O

âˆ´ âˆ AOC = âˆ BOD and âˆ BOC = âˆ AOD (Vertically opposite angles)

âˆ´ But âˆ AOC = 90Â°

âˆ´ âˆ BOD = 90Â°

âˆ´ âˆ AOC + âˆ BOC = 180Â° (Linear pair)

â‡’ 90Â° + âˆ BOC = 180Â°

âˆ´ âˆ BOC = 180Â° -90Â° = 90Â°

âˆ´ âˆ AOD = âˆ BOC = 90Â°

âˆ´ âˆ AOD = âˆ BOC = âˆ BOD = 90Â°

Question 6.

In the figure, rays AB and CD intersect at O.

(i) Determine y when x = 60Â°

(ii) Determine x when y = 40Â°

Solution:

In the figure,

AB is a line

âˆ´ 2x + y = 180Â° (Linear pair)

(i) If x = 60Â°, then

2 x 60Â° + y = 180Â°

â‡’ 120Â° +y= 180Â°

âˆ´ y= 180Â°- 120Â° = 60Â°

(ii) If y = 40Â°, then

2x + 40Â° = 180Â°

â‡’ 2x = 180Â° – 40Â° = 140Â°

â‡’ x= \(\frac { { 140 }^{ \circ } }{ 2 }\)Â =70Â°

âˆ´ x = 70Â°

Question 7.

In the figure, lines AB, CD and EF intersect at O. Find the measures of âˆ AOC, âˆ COF, âˆ DOE and âˆ BOF.

Solution:

Three lines AB, CD and EF intersect each other at O

âˆ AOE = 40Â° and âˆ BOD = 35Â°

(i) âˆ AOC = âˆ BOD (Vertically opposite angles)

= 35Â°

AB is a line

âˆ´ âˆ AOE + âˆ DOE + âˆ BOD = 180Â°

â‡’ 40Â° + âˆ DOE + 35Â° = 180Â°

â‡’ 75Â° + âˆ DOE = 180Â°

â‡’ âˆ DOE = 180Â°-75Â° = 105Â°

But âˆ COF = âˆ DOE (Vertically opposite angles)

âˆ´ âˆ COF = 105Â°

Similarly, âˆ BOF = âˆ AOE (Vertically opposite angles)

â‡’ âˆ BOF = 40Â°

Hence âˆ AOC = 35Â°, âˆ COF = 105Â°, âˆ DOE = 105Â° and âˆ BOF = 40Â°

Question 8.

AB, CD and EF are three concurrent lines passing through the point O such that OF bisects âˆ BOD. If âˆ BOF = 35Â°, find âˆ BOC and âˆ AOD.

Solution:

AB, CD and EF intersect at O. Such that OF is the bisector of

âˆ BOD âˆ BOF = 35Â°

âˆµ OF bisects âˆ BOD,

âˆ´ âˆ DOF = âˆ BOF = 35Â° (Vertically opposite angles)

âˆ´ âˆ BOD = 35Â° + 35Â° = 70Â°

But âˆ BOC + âˆ BOD = 180Â° (Linear pair)

â‡’ âˆ BOC + 70Â° = 180Â°

â‡’ âˆ BOC = 180Â°-70Â°= 110Â°

But âˆ AOD = âˆ BOC (Vertically opposite angles)

= 110Â°

Hence âˆ BOC = 110Â° and âˆ AOD =110Â°

Question 9.

In the figure, lines AB and CD intersect at O. If âˆ AOC + âˆ BOE = 70Â° and âˆ BOD = 40Â°, find âˆ BOE and reflex âˆ COE.

Solution:

In the figure, AB and CD intersect each other at O

âˆ AOC + âˆ BOE = 70Â°

âˆ BOD = 40Â°

AB is a line

âˆ´ âˆ AOC + âˆ BOE + âˆ COE = 180Â° (Angles on one side of a line)

â‡’ 70Â° + âˆ COE = 180Â°

â‡’ âˆ COE = 180Â°-70Â°= 110Â°

and âˆ AOC = âˆ BOD (Vertically opposite angles)

â‡’ âˆ AOC = 40Â°

âˆ´ âˆ BOE = 70Â° – 40Â° = 30Â°

and reflex âˆ COE = 360Â° – âˆ COE

= 360Â°- 110Â° = 250Â°

Question 10.

Which of the following statements are true (T) and which are false (F)?

(i) Angles forming a linear pair are supplementary.

(ii) If two adjacent angles are equal, then each angle measures 90Â°.

(iii) Angles forming a linear pair can both be acute angles.

(iv) If angles forming a linear pair are equal, then each of these angles is of measure 90Â°.

Solution:

(i) True.

(ii) False. It can be possible if they are a linear pair.

(iii) False. In a linear pair, if one is acute, then the other will be obtuse.

(iv) True.

Question 11.

Fill in the blanks so as to make the following statements true:

(i) If one angle of a linear pair is acute, then its other angle will be …….. .

(ii) A ray stands on a line, then the sum of the two adjacent angles so formed is ……… .

(iii) If the sum of two adjacent angles is 180Â°, then the …… arms of the two angles are opposite rays.

Solution:

(i) If one angle of a linear pair is acute, then its other angle will be obtuse.

(ii) A ray stands on a line, then the sum of the two adjacent angles so formed is 180Â°.

(iii) If the sum of two adjacent angles is 180Â°, then the uncommon arms of the two angles are opposite rays.

Question 12.

Prove that the bisectors of a pair of vertically opposite angles are in the same straight line.

Solution:

Given : Lines AB and CD intersect each other at O.

OE and OF are the bisectors of âˆ AOC and âˆ BOD respectively

To prove : OE and OF are in the same line

Proof : âˆµ âˆ AOC = âˆ BOD (Vertically opposite angles)

âˆµ OE and OF are the bisectors of âˆ AOC and âˆ BOD

âˆ´ âˆ 1 = âˆ 2 and âˆ 3 = âˆ 4

â‡’ âˆ 1 = âˆ 2 = \(\frac { 1 }{ 2 }\) âˆ AOC and

âˆ 3 = âˆ 4 = \(\frac { 1 }{ 2 }\) âˆ BOD

âˆ´ âˆ 1 = âˆ 2 = âˆ 3 = âˆ 4

âˆµ AOB is a line

âˆ´ âˆ BOD + âˆ AOD = 180Â° (Linear pair)

â‡’ âˆ 3 + âˆ 4 + âˆ AOD = 180Â°

â‡’ âˆ 3 + âˆ 1 + âˆ AOD = 180Â° (âˆµ âˆ 1 = âˆ 4)

âˆ´ EOF is a straight line

Question 13.

If two straight lines intersect each other, prove that the ray opposite to the bisector of one of the angles thus formed bisects the vertically opposite angle.

Solution:

Given : AB and CD intersect each other at O. OE is the bisector of âˆ AOD and EO is produced to F.

To prove : OF is the bisector of âˆ BOC

Proof : âˆµ AB and CD intersect each other at O

âˆ´ âˆ AOD = âˆ BOC (Vertically opposite angles)

âˆµOE is the bisector of âˆ AOD

âˆ´ âˆ 1 = âˆ 2

âˆµ AB and EF intersect each other at O

âˆ´âˆ 1 = âˆ 4 (Vertically opposite angles) Similarly, CD and EF intersect each other at O

âˆ´ âˆ 2 = âˆ 3

But âˆ 1 = âˆ 2

âˆ´ âˆ 3 = âˆ 4

OF is the bisector of âˆ BOC

### RD Sharma Class 9 Maths Book Questions Chapter 10 Congruent Triangles Ex 10.4

Question 1.

In the figure, AB || CD and âˆ 1 and âˆ 2 are in the ratio 3 : 2. Determine all angles from 1 to 8.

Solution:

AB || CD and l is transversal âˆ 1 : âˆ 2 = 3 : 2

Let âˆ 1 = 3x

Then âˆ 2 = 2x

But âˆ 1 + âˆ 2 = 180Â° (Linear pair)

âˆ´ 3x + 2x = 180Â° â‡’ 5x = 180Â°

â‡’ x = \(\frac { { 180 }^{ \circ } }{ 5 }\)Â = 36Â°

âˆ´ âˆ 1 = 3x = 3 x 36Â° = 108Â°

âˆ 2 = 2x = 2 x 36Â° = 72Â°

Now âˆ 1 = âˆ 3 and âˆ 2 = âˆ 4 (Vertically opposite angles)

âˆ´ âˆ 3 = 108Â° and âˆ 4 = 72Â°

âˆ 1 = âˆ 5 and âˆ 2 = âˆ 6 (Corresponding angles)

âˆ´ âˆ 5 = 108Â°, âˆ 6 = 72Â°

Similarly, âˆ 4 = âˆ 8 and

âˆ 3 = âˆ 7

âˆ´ âˆ 8 = 72Â° and âˆ 7 = 108Â°

Hence, âˆ 1 = 108Â°, âˆ 2= 72Â°

âˆ 3 = 108Â°, âˆ 4 = 72Â°

âˆ 5 = 108Â°, âˆ 6 = 72Â°

âˆ 7 = 108Â°, âˆ 8 = 12Â°

Question 2.

In the figure, l, m and n are parallel lines intersected by transversal p at X, Y and Z respectively. Find âˆ l, âˆ 2 and âˆ 3.

Solution:

l || m || n and p is then transversal which intersects then at X, Y and Z respectively âˆ 4 = 120Â°

âˆ 2 = âˆ 4 (Alternate angles)

âˆ´ âˆ 2 = 120Â°

But âˆ 3 + âˆ 4 = 180Â° (Linear pair)

â‡’ âˆ 3 + 120Â° = 180Â°

â‡’ âˆ 3 = 180Â° – 120Â°

âˆ´ âˆ 3 = 60Â°

But âˆ l = âˆ 3 (Corresponding angles)

âˆ´ âˆ l = 60Â°

Hence âˆ l = 60Â°, âˆ 2 = 120Â°, âˆ 3 = 60Â°

Question 3.

In the figure, if AB || CD and CD || EF, find âˆ ACE.

Solution:

Given : In the figure, AB || CD and CD || EF

âˆ BAC = 70Â°, âˆ CEF = 130Â°

âˆµ EF || CD

âˆ´ âˆ ECD + âˆ CEF = 180Â° (Co-interior angles)

â‡’ âˆ ECD + 130Â° = 180Â°

âˆ´ âˆ ECD = 180Â° – 130Â° = 50Â°

âˆµ BA || CD

âˆ´ âˆ BAC = âˆ ACD (Alternate angles)

âˆ´ âˆ ACD = 70Â° (âˆµ âˆ BAC = 70Â°)

âˆµ âˆ ACE = âˆ ACD – âˆ ECD = 70Â° – 50Â° = 20Â°

Question 4.

In the figure, state which lines are parallel and why.

Solution:

In the figure,

âˆµ âˆ ACD = âˆ CDE = 100Â°

But they are alternate angles

âˆ´ AC || DE

Question 5.

In the figure, if l || m,n|| p and âˆ 1 = 85Â°, find âˆ 2.

Solution:

In the figure, l || m, n|| p and âˆ 1 = 85Â°

âˆµ n || p

âˆ´ âˆ 1 = âˆ 3 (Corresponding anlges)

But âˆ 1 = 85Â°

âˆ´ âˆ 3 = 85Â°

âˆµ m || 1

âˆ 3 + âˆ 2 = 180Â° (Sum of co-interior angles)

â‡’ 85Â° + âˆ 2 = 180Â°

â‡’ âˆ 2 = 180Â° – 85Â° = 95Â°

Question 6.

If two straight lines are perpendicular to the same line, prove that they are parallel to each other.

Solution:

Question 7.

Two unequal angles of a parallelogram are in the ratio 2:3. Find all its angles in degrees.

Solution:

In ||gm ABCD,

âˆ A and âˆ B are unequal

and âˆ A : âˆ B = 2 : 3

Let âˆ A = 2x, then

âˆ B = 3x

But âˆ A + âˆ B = 180Â° (Co-interior angles)

âˆ´ 2x + 3x = 180Â°

â‡’ 5x = 180Â°

â‡’ x = \(\frac { { 180 }^{ \circ } }{ 5 }\)Â = 36Â°

âˆ´ âˆ A = 2x = 2 x 36Â° = 72Â°

âˆ B = 3x = 3 x 36Â° = 108Â°

But âˆ A = âˆ C and âˆ B = âˆ D (Opposite angles of a ||gm)

âˆ´ âˆ C = 72Â° and âˆ D = 108Â°

Hence âˆ A = 72Â°, âˆ B = 108Â°, âˆ C = 72Â°, âˆ D = 108Â°

Question 8.

In each of the two lines is perpendicular to the same line, what kind of lines are they to each other?

Solution:

AB âŠ¥ line l and CD âŠ¥ line l

âˆ´ âˆ B = 90Â° and âˆ D = 90Â°

âˆ´ âˆ B = âˆ D

But there are corresponding angles

âˆ´ AB || CD

Question 9.

In the figure, âˆ 1 = 60Â° and âˆ 2 = (\(\frac { 2 }{ 3 }\))3 a right angle. Prove that l || m.

Solution:

In the figure, a transversal n intersects two lines l and m

âˆ 1 = 60Â° and

âˆ 2 = \(\frac { 2 }{ 3 }\) rd of a right angle 2

= \(\frac { 2 }{ 3 }\) x 90Â° = 60Â°

âˆ´ âˆ 1 = âˆ 2

But there are corresponding angles

âˆ´ l || m

Question 10.

In the figure, if l || m || n and âˆ 1 = 60Â°, find âˆ 2.

Solution:

In the figure,

l || m || n and a transversal p, intersects them at P, Q and R respectively

âˆ 1 = 60Â°

âˆ´ âˆ 1 = âˆ 3 (Corresponding angles)

âˆ´ âˆ 3 = 60Â°

But âˆ 3 + âˆ 4 = 180Â° (Linear pair)

60Â° + âˆ 4 = 180Â° â‡’ âˆ 4 = 180Â° – 60Â°

âˆ´ âˆ 4 = 120Â°

But âˆ 2 = âˆ 4 (Alternate angles)

âˆ´ âˆ 2 = 120Â°

Question 11.

Prove that the straight lines perpendicular to the same straight line are parallel to one another.

Solution:

Given : l is a line, AB âŠ¥ l and CD âŠ¥ l

Question 12.

The opposite sides of a quadrilateral are parallel. If one angle of the quadrilateral is 60Â°, find the other angles.

Solution:

In quadrilateral ABCD, AB || DC and AD || BC and âˆ A = 60Â°

âˆµ AD || BC and AB || DC

âˆ´ ABCD is a parallelogram

âˆ´ âˆ A + âˆ B = 180Â° (Co-interior angles)

60Â° + âˆ B = 180Â°

â‡’ âˆ B = 180Â°-60Â°= 120Â°

But âˆ A = âˆ C and âˆ B = âˆ D (Opposite angles of a ||gm)

âˆ´ âˆ C = 60Â° and âˆ D = 120Â°

Hence âˆ B = 120Â°, âˆ C = 60Â° and âˆ D = 120Â°

Question 13.

Two lines AB and CD intersect at O. If âˆ AOC + âˆ COB + âˆ BOD = 270Â°, find the measure of âˆ AOC, âˆ COB, âˆ BOD and âˆ DOA.

Solution:

Two lines AB and CD intersect at O

and âˆ AOC + âˆ COB + âˆ BOD = 270Â°

But âˆ AOC + âˆ COB + âˆ BOD + âˆ DOA = 360Â° (Angles at a point)

âˆ´ 270Â° + âˆ DOA = 360Â°

â‡’ âˆ DOA = 360Â° – 270Â° = 90Â°

But âˆ DOA = âˆ BOC (Vertically opposite angles)

âˆ´ âˆ BOC = 90Â°

But âˆ DOA + âˆ BOD = 180Â° (Linear pair)

â‡’ 90Â° + âˆ BOD = 180Â°

âˆ´ âˆ BOD= 180Â°-90Â° = 90Â° ,

But âˆ BOD = âˆ AOC (Vertically opposite angles)

âˆ´ âˆ AOC = 90Â°

Hence âˆ AOC = 90Â°,

âˆ COB = 90Â°,

âˆ BOD = 90Â° and âˆ DOA = 90Â°

Question 14.

In the figure, p is a transversal to lines m and n, âˆ 2 = 120Â° and âˆ 5 = 60Â°. Prove that m || n.

Solution:

Given : p is a transversal to the lines m and n

Forming âˆ l, âˆ 2, âˆ 3, âˆ 4, âˆ 5, âˆ 6, âˆ 7 and âˆ 8

âˆ 2 = 120Â°, and âˆ 5 = 60Â°

To prove : m || n

Proof : âˆ 2 + âˆ 3 = 180Â° (Linear pair)

â‡’ 120Â°+ âˆ 3 = 180Â°

â‡’ âˆ 3 = 180Â°- 120Â° = 60Â°

But âˆ 5 = 60Â°

âˆ´ âˆ 3 = âˆ 5

But there are alternate angles

âˆ´ m || n

Question 15.

In the figure, transversal l, intersects two lines m and n, âˆ 4 = 110Â° and âˆ 7 = 65Â°. Is m || n?

Solution:

A transversal l, intersects two lines m and n, forming âˆ 1, âˆ 2, âˆ 3, âˆ 4, âˆ 5, âˆ 6, âˆ 7 and âˆ 8

âˆ 4 = 110Â° and âˆ 7 = 65Â°

To prove : Whether m || n or not

Proof : âˆ 4 = 110Â° and âˆ 7 = 65Â°

âˆ 7 = âˆ 5 (Vertically opposite angles)

âˆ´ âˆ 5 = 65Â°

Now âˆ 4 + âˆ 5 = 110Â° + 65Â° = 175Â°

âˆµ Sum of co-interior angles âˆ 4 and âˆ 5 is not 180Â°.

âˆ´ m is not parallel to n

Question 16.

Which pair of lines in the figure are parallel? Give reasons.

Solution:

Given : In the figure, âˆ A = 115Â°, âˆ B = 65Â°, âˆ C = 115Â° and âˆ D = 65Â°

âˆµ âˆ A + âˆ B = 115Â°+ 65Â°= 180Â°

But these are co-interior angles,

âˆ´ AD || BC

Similarly, âˆ A + âˆ D = 115Â° + 65Â° = 180Â°

âˆ´ AB || DC

Question 17.

If l, m, n are three lines such that l ||m and n âŠ¥ l, prove that n âŠ¥ m.

Solution:

Given : l, m, n are three lines such that l || m and n âŠ¥ l

To prove : n âŠ¥ m

Proof : âˆµ l || m and n is the transversal.

âˆ´ âˆ l = âˆ 2 (Corresponding angles)

But âˆ 1 = 90Â° (âˆµ nâŠ¥l)

âˆ´ âˆ 2 = 90Â°

âˆ´ n âŠ¥ m

Question 18.

Which of the following statements are true (T) and which are false (F)? Give reasons.

(i) If two lines are intersected by a transversal, then corresponding angles are equal.

(ii) If two parallel lines are intersected by a transversal, then alternate interior angles are equal.

(iii) Two lines perpendicular to the same line are perpendicular to each other.

(iv) Two lines parallel to the same line are parallel to each other.

(v) If two parallel lines are intersected by a transversal, then the interior angles on the same side of the transversal are equal.

Solution:

(i) False. Because if lines are parallel, then it is possible.

(ii) True.

(iii) False. Not perpendicular but parallel to each other.

(iv) True.

(v) False. Sum of interior angles on the same side is 180Â° not are equal.

Question 19.

Fill in the blanks in each of the following to make the statement true:

(i) If two parallel lines are intersected by a transversal then each pair of corresponding angles are ……..

(ii) If two parallel lines are intersected by a transversal, then interior angles on the same side of the transversal are …….

(iii) Two lines perpendicular to the same line are ……… to each other.

(iv) Two lines parallel to the same line are ……… to each other.

(v) If a transversal intersects a pair of lines in such away that a pair of alternate angles are equal, then the lines are …….

(vi) If a transversal intersects a pair of lines in such away that the sum of interior angles on the same side of transversal is 180Â°, then the lines are …….

Solution:

(i) If two parallel lines are intersected by a transversal, then each pair of corresponding angles are equal.

(ii) If two parallel lines are intersected by a transversal, then interior angles on the same side of the transversal are supplementary.

(iii) Two lines perpendicular to the same line are parallel to each other.

(iv) Two lines parallel to the same line are parallel to each other.

(v) If a transversal intersects a pair of lines in such away that a pair of alternate angles are equal, then the lines are parallel.

(vi) If a transversal intersects a pair of lines in such away that the sum of interior angles on the same side of transversal is 180Â°, then the lines are parallel.

Question 20.

In the figure, AB || CD || EF and GH || KL. Find âˆ HKL.

Solution:

In the figure, AB || CD || EF and KL || HG Produce LK and GH

âˆµ AB || CD and HK is transversal

âˆ´ âˆ 1 = 25Â° (Alternate angles)

âˆ 3 = 60Â° (Corresponding angles)

and âˆ 3 = âˆ 4 (Corresponding angles)

= 60Â°

But âˆ 4 + âˆ 5 = 180Â° (Linear pair)

â‡’ 60Â° + âˆ 5 = 180Â°

â‡’ âˆ 5 = 180Â° – 60Â° = 120Â°

âˆ´ âˆ HKL = âˆ 1 + âˆ 5 = 25Â° + 120Â° = 145Â°

Question 21.

In the figure, show that AB || EF.

Solution:

Given : In the figure, AB || EF

âˆ BAC = 57Â°, âˆ ACE = 22Â°

âˆ ECD = 35Â° and âˆ CEF =145Â°

To prove : AB || EF,

Proof : âˆ ECD + âˆ CEF = 35Â° + 145Â°

= 180Â°

But these are co-interior angles

âˆ´ EF || CD

But AB || CD

âˆ´ AB || EF

Question 22.

In the figure, PQ || AB and PR || BC. If âˆ QPR = 102Â°. Determine âˆ ABC. Give reasons.

Solution:

In the figure, PQ || AB and PR || BC

âˆ QPR = 102Â°

Produce BA to meet PR at D

âˆµ PQ || AB or DB

âˆ´ âˆ QPR = âˆ ADR (Corresponding angles)

âˆ´âˆ ADR = 102Â° or âˆ BDR = 102Â°

âˆµ PR || BC

âˆ´ âˆ BDR + âˆ DBC = 180Â°

(Sum of co-interior angles) â‡’ 102Â° + âˆ DBC = 180Â°

â‡’ âˆ DBC = 180Â° – 102Â° = 78Â°

â‡’ âˆ ABC = 78Â°

Question 23.

Prove that if the two arms of an angle are perpendicular to the two arms of another angle, then the angles are either equal or supplementary.

Solution:

Given : In two angles âˆ ABC and âˆ DEF AB âŠ¥ DE and BC âŠ¥ EF

To prove: âˆ ABC + âˆ DEF = 180Â° or âˆ ABC = âˆ DEF

Construction : Produce the sides DE and EF of âˆ DEF, to meet the sides of âˆ ABC at H and G.

Proof: In figure (i) BGEH is a quadrilateral

âˆ BHE = 90Â° and âˆ BGE = 90Â°

But sum of angles of a quadrilateral is 360Â°

âˆ´ âˆ HBG + âˆ HEG = 360Â° – (90Â° + 90Â°)

= 360Â° – 180Â°= 180Â°

âˆ´ âˆ ABC and âˆ DEF are supplementary

In figure (if) in quadrilateral BGEH,

âˆ BHE = 90Â° and âˆ HEG = 90Â°

âˆ´ âˆ HBG + âˆ HEG = 360Â° – (90Â° + 90Â°)

= 360Â°- 180Â° = 180Â° …(i)

But âˆ HEF + âˆ HEG = 180Â° …(ii) (Linear pair)

From (i) and (ii)

âˆ´ âˆ HEF = âˆ HBG

â‡’ âˆ DEF = âˆ ABC

Hence âˆ ABC and âˆ DEF are equal or supplementary

Question 24.

In the figure, lines AB and CD are parallel and P is any point as shown in the figure. Show that âˆ ABP + âˆ CDP = âˆ DPB.

Solution:

Given : In the figure, AB || CD

P is a point between AB and CD PD

and PB are joined

To prove : âˆ APB + âˆ CDP = âˆ DPB

Construction : Through P, draw PQ || AB or CD

Proof: âˆµ AB || PQ

âˆ´ âˆ ABP = BPQ …(i) (Alternate angles)

Similarly,

CD || PQ

âˆ´ âˆ CDP = âˆ DPQ …(ii)

(Alternate angles)

Adding (i) and (ii)

âˆ ABP + âˆ CDP = âˆ BPQ + âˆ DPQ

Hence âˆ ABP + âˆ CDP = âˆ DPB

Question 25.

In the figure, AB || CD and P is any point shown in the figure. Prove that:

âˆ ABP + âˆ BPD + âˆ CDP = 360Â°

Solution:

Given : AB || CD and P is any point as shown in the figure

To prove : âˆ ABP + âˆ BPD + âˆ CDP = 360Â°

Construction : Through P, draw PQ || AB and CD

Proof : âˆµ AB || PQ

âˆ´ âˆ ABP+ âˆ BPQ= 180Â° ……(i) (Sum of co-interior angles)

Similarly, CD || PQ

âˆ´ âˆ QPD + âˆ CDP = 180Â° …(ii)

Adding (i) and (ii)

âˆ ABP + âˆ BPQ + âˆ QPD + âˆ CDP

= 180Â°+ 180Â° = 360Â°

â‡’ âˆ ABP + âˆ BPD + âˆ CDP = 360Â°

Question 26.

In the figure, arms BA and BC of âˆ ABC are respectively parallel to arms ED and EF of âˆ DEF. Prove that âˆ ABC = âˆ DEF.

Solution:

Given : In âˆ ABC and âˆ DEF. Their arms are parallel such that BA || ED and BC || EF

To prove : âˆ ABC = âˆ DEF

Construction : Produce BC to meet DE at G

Proof: AB || DE

âˆ´ âˆ ABC = âˆ DGH…(i) (Corresponding angles)

BC or BH || EF

âˆ´ âˆ DGH = âˆ DEF (ii) (Corresponding angles)

From (i) and (ii)

âˆ ABC = âˆ DEF

Question 27.

In the figure, arms BA and BC of âˆ ABC are respectively parallel to arms ED and EF of âˆ DEF. Prove that âˆ ABC + âˆ DEF = 180Â°.

Solution:

Given: In âˆ ABC = âˆ DEF

BA || ED and BC || EF

To prove: âˆ ABC = âˆ DEF = 180Â°

Construction : Produce BC to H intersecting ED at G

Proof: âˆµ AB || ED

âˆ´ âˆ ABC = âˆ EGH …(i) (Corresponding angles)

âˆµ BC or BH || EF

âˆ EGH || âˆ DEF = 180Â° (Sum of co-interior angles)

â‡’ âˆ ABC + âˆ DEF = 180Â° [From (i)]

Hence proved.

### RD Sharma Class 9 Solution Chapter 10 Congruent Triangles VSAQS

Question 1.

Define complementary angles.

Solution:

Two angles whose sum is 90Â°, are called complementary angles.

Question 2.

Define supplementary angles.

Solution:

Two angles whose sum is 180Â°, are called supplementary angles.

Question 3.

Define adjacent angles.

Solution:

Two angles which have common vertex and one arm common are called adjacent angles.

Question 4.

The complement of an acute angles is…….

Solution:

The complement of an acute angles is an acute angle.

Question 5.

The supplement of an acute angles is………

Solution:

The supplement of an acute angles is a obtuse angle.

Question 6.

The supplement of a right angle is…….

Solution:

The supplement of a right angle is a right angle.

Question 7.

Write the complement of an angle of measure xÂ°.

Solution:

The complement of xÂ° is (90Â° – x)Â°

Question 8.

Write the supplement of an angle of measure 2yÂ°.

Solution:

The supplement of 2yÂ° is (180Â° – 2y)Â°

Question 9.

If a wheel has six spokes equally spaced, then find the measure of the angle between two adjacent spokes.

Solution:

Total measure of angle around a point = 360Â°

Number of spokes = 6

âˆ´ Angle between the two adjacent spokes =Â \(\frac { { 360 }^{ \circ } }{ 6 }\) = 60Â°

Question 10.

An angle is equal to its supplement. Determine its measure.

Solution:

Let required angle = xÂ°

Then its supplement angle = 180Â° – x

x = 180Â° – x

â‡’ x + x = 180Â°

â‡’Â 2x = 180Â° â‡’Â x = \(\frac { { 180 }^{ \circ } }{ 2 }\) = 90Â°

âˆ´ Required angle = 90Â°

Question 11.

An angle is equal to five times its complement. Determine its measure.

Solution:

Let required measure of angle = xÂ°

âˆ´Â Its complement angle = 90Â° – x

âˆ´Â x = 5(90Â° – x)

â‡’Â x = 450Â° – 5x

â‡’Â x + 5x = 450Â°

â‡’Â 6x = 450Â°

â‡’ x = \(\frac { { 450 }^{ \circ } }{ 6 }\) = 75Â°

âˆ´ Required angle = 75Â°

Question 12.

How many pairs of adjacent angles are formed when two lines intersect in a point?

Solution:

If two lines AB and CD intersect at a point O, then pairs of two adjacent angles are, âˆ AOC and âˆ COB, âˆ COB and âˆ BOD, âˆ BOD and DOA, âˆ DOA and âˆ ZAOC

i.e, 4 pairs

### RD Sharma Solutions Class 9 Chapter 10 Congruent Triangles MCQS

Mark the correct alternative in each of the following:

Question 1.

One angle is equal to three times its supplement. The measure of the angle is

(a) 130Â°

(b) 135Â°

(c) 90Â°

(d) 120Â°

Solution:

Let required angle = x

Then its supplement = (180Â° – x)

x = 3(180Â° – x) = 540Â° – 3x

â‡’ x + 3x = 540Â°

â‡’ 4x = 540Â°

â‡’ x = \(\frac { { 540 }^{ \circ } }{ 4 }\)Â = 135Â°

âˆ´ Required angle = 135Â° (b)

Question 2.

Two straight lines AB and CD intersect one another at the point O. If âˆ AOC + âˆ COB + âˆ BOD = 274Â°, then âˆ AOD =

(a) 86Â°

(b) 90Â°

(c) 94Â°

(d) 137Â°

Solution:

Sum of angles at a point O = 360Â°

Sum of three angles âˆ AOC + âˆ COB + âˆ BOD = 274Â°

âˆ´ Fourth angle âˆ AOD = 360Â° – 274Â°

= 86Â° (a)

Question 3.

Two straight lines AB and CD cut each other at O. If âˆ BOD = 63Â°, then âˆ BOC =

(a) 63Â°

(b) 117Â°

(c) 17Â°

(d) 153Â°

Solution:

CD is a line

âˆ´ âˆ BOD + âˆ BOC = 180Â° (Linear pair)

â‡’ 63Â° + âˆ BOC = 180Â°

â‡’ âˆ BOC = 180Â° – 63Â°

âˆ´ âˆ BOC =117Â° (b)

Question 4.

Consider the following statements:

When two straight lines intersect:

(i) adjacent angles are complementary

(ii) adjacent angles are supplementary

(iii) opposite angles are equal

(iv) opposite angles are supplementary Of these statements

(a) (i) and (iii) are correct

(b) (ii) and (iii) are correct

(c) (i) and (iv) are correct

(d) (ii) and (iv) are correct

Solution:

Only (ii) and (iii) arc true. (b)

Question 5.

Given âˆ POR = 3x and âˆ QOR = 2x + 10Â°. If POQ is a striaght line, then the value of x is

(a) 30Â°

(b) 34Â°

(c) 36Â°

(d) none of these

Solution:

âˆµ POQ is a straight line

âˆ´ âˆ POR + âˆ QOR = 180Â° (Linear pair)

â‡’ 3x + 2x + 10Â° = 180Â°

â‡’ 5x = 180 – 10Â° = 170Â°

âˆ´ x = \(\frac { { 170 }^{ \circ } }{ 5 }\)Â = 34Â° (b)

Question 6.

In the figure, AOB is a straight line. If âˆ AOC + âˆ BOD = 85Â°, then âˆ COD =

(a) 85Â°

(b) 90Â°

(c) 95Â°

(d) 100Â°

Solution:

AOB is a straight line,

OC and OD are rays on it

and âˆ AOC + âˆ BOD = 85Â°

But âˆ AOC + âˆ BOD + âˆ COD = 180Â°

â‡’ 85Â° + âˆ COD = 180Â°

âˆ COD = 180Â° – 85Â° = 95Â° (c)

Question 7.

In the figure, the value of y is

(a) 20Â°

(b) 30Â°

(c) 45Â°

(d) 60Â°

Solution:

In the figure,

y = x (Vertically opposite angles)

âˆ 1 = 3x

âˆ 2 = 3x

âˆ´ 2(x + 3x + 2x) = 360Â° (Angles at a point)

2x + 6x + 4x = 360Â°

12x = 360Â° â‡’ x = \(\frac { { 360 }^{ \circ } }{ 12 }\)Â = 30Â°

âˆ´ y = x = 30Â° (b)

Question 8.

In the figure, the value of x is

(a) 12

(b) 15

(c) 20

(d) 30

Solution:

âˆ 1 = 3x+ 10 (Vertically opposite angles)

But x + âˆ 1 + âˆ 2 = 180Â°

â‡’ x + 3x + 10Â° + 90Â° = 180Â°

â‡’ 4x = 180Â° – 10Â° – 90Â° = 80Â°

x = \(\frac { { 80 }^{ \circ } }{ 4 }\) = 20Â Â (c)

Question 9.

In the figure, which of the following statements must be true?

(i) a + b = d + c

(ii) a + c + e = 180Â°

(iii) b + f= c + e

(a) (i) only

(b) (ii) only

(c) (iii) only

(d) (ii) and (iii) only

Solution:

In the figure,

(i) a + b = d + c

aÂ° = dÂ°

bÂ° = eÂ°

cÂ°= fÂ°

(ii) a + b + e = 180Â°

a + e + c = 180Â°

â‡’ a + c + e = 180Â°

(iii) b + f= e + c

âˆ´ (ii) and (iii) are true statements (d)

Question 10.

If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio 2:3, then the measure of the larger angle is

(a) 54Â°

(b) 120Â°

(c) 108Â°

(d) 136Â°

Solution:

In figure, l || m and p is transversal

= \(\frac { 3 }{ 5 }\) x 180Â° = 108Â° (c)

Question 11.

In the figure, if AB || CD, then the value of x is

(a) 20Â°

(b) 30Â°

(c) 45Â°

(d) 60Â°

Solution:

In the figure, AB || CD,

and / is transversal

âˆ 1 = x (Vertically opposite angles)

and 120Â° + x + âˆ 1 = 180Â° (Co-interior angles)

Question 12.

Two lines AB and CD intersect at O. If âˆ AOC + âˆ COB + âˆ BOD = 270Â°, then âˆ AOC =

(a) 70Â°

(b) 80Â°

(c) 90Â°

(d) 180Â°

Solution:

Two lines AB and CD intersect at O

âˆ AOC + âˆ COB + âˆ BOD = 270Â° …(i)

But âˆ AOC + âˆ COB + âˆ BOD + âˆ DOA = 360Â° …(ii)

Subtracting (i) from (ii),

âˆ DOA = 360Â° – 270Â° = 90Â°

But âˆ DOA + âˆ AOC = 180Â°

âˆ´ âˆ AOC = 180Â° – 90Â° = 90Â° (c)

Question 13.

In the figure, PQ || RS, âˆ AEF = 95Â°, âˆ BHS = 110Â° and âˆ ABC = xÂ°. Then the value of x is

(a) 15Â°

(b) 25Â°

(c) 70Â°

(d) 35Â°

Solution:

In the figure,

PQ || RS, âˆ AEF = 95Â°

âˆ BHS = 110Â°, âˆ ABC = x

âˆµ PQ || RS,

âˆ´ âˆ AEF = âˆ 1 = 95Â° (Corresponding anlges)

But âˆ 1 + âˆ 2 = 180Â° (Linear pair)

â‡’ âˆ 2 = 180Â° – âˆ 1 = 180Â° – 95Â° = 85Â°

In âˆ†AGH,

Ext. âˆ BHS = âˆ 2 +x

â‡’ 110Â° = 85Â° + x

â‡’ x= 110Â°-85Â° = 25Â° (b)

Question 14.

In the figure, if l1 || l2, what is the value of x?

(a) 90Â°

(b) 85Â°

(c) 75Â°

(d) 70Â°

Solution:

In the figure,

âˆ 1 = 58Â° (Vertically opposite angles)

Similarly, âˆ 2 = 37Â°

âˆµ l_{1} || l_{2}, EF is transversal

âˆ GEF + EFD = 180Â° (Co-interior angles)

â‡’ âˆ 2 + âˆ l +x = 180Â°

â‡’ 37Â° + 58Â° + x = 180Â°

â‡’ 95Â° + x= 180Â°

x = 180Â°-95Â° = 85Â° (b)

Question 15.

In the figure, if l_{1} || l_{2}, what is x + y in terms of w and z?

(a) 180-w + z

(b) 180Â° + w- z

(c) 180 -w- z

(d) 180 + w + z

Solution:

In the figure, l_{1} || l_{2}

p and q are transversals

âˆ´ w + x = 180Â° â‡’ x = 180Â° – w (Co-interior angle)

z = y (Alternate angles)

âˆ´ x + y = 180Â° – w + z (a)

Question 16.

In the figure, if l_{1} || l_{2}, what is the value of y?

(a) 100

(b) 120

(c) 135

(d) 150

Solution:

In the figure, l_{1} || l_{2} and l_{3}Â is the transversal

Question 17.

In the figure, if l1 || l2 and l3 || l4 what is y in terms of x?

(a) 90 + x

(b) 90 + 2x

(c) 90 – \(\frac { x }{ 2 }\)

(d) 90 – 2x

Solution:

In the figure,

l_{1} || l_{2} and l_{3} || l_{4} and m is the angle bisector

âˆ´ âˆ 2 = âˆ 3 = y

âˆµ l_{1} || l_{2}

âˆ 1 = x (Corresponding angles)

âˆµ l_{3} || l_{4}

âˆ´ âˆ 1 + (âˆ 2 + âˆ 3) = 180Â° (Co-interior angles)

â‡’ x + 2y= 180Â°

â‡’ 2y= 180Â°-x

â‡’ y = \(\frac { { 540 }^{ \circ }-x }{ 4 }\)

= 90Â° – \(\frac { x }{ 2 }\) (c)

Question 18.

In the figure, if 11| m, what is the value of x?

(a) 60

(b) 50

(c) 45

d) 30

Solution:

In the figure, l || m and n is the transversal

â‡’ y = 25Â°

But 2y + 25Â° = x+ 15Â°

(Vertically opposite angles) â‡’ x = 2y + 25Â° – 15Â° = 2y+ 10Â°

= 2 x 25Â°+10Â° = 50Â°+10Â° = 60Â° (a)

Question 19.

In the figure, if AB || HF and DE || FG, then the measure of âˆ FDE is

(a) 108Â°

(b) 80Â°

(c) 100Â°

(d) 90Â°

Solution:

In the figure,

AB || HF, DE || FG

âˆ´ HF || AB

âˆ 1 =28Â° (Corresponding angles)

But âˆ 1 + âˆ FDE + 72Â° – 180Â° (Angles of a straight line)

â‡’ 28Â° + âˆ FDE + 72Â° = 180Â°

â‡’ âˆ FDE + 100Â° = 180Â°

â‡’ âˆ FDE = 180Â° – 100 = 80Â° (b)

Question 20.

In the figure, if lines l and m are parallel, then x =

(a) 20Â°

(b) 45Â°

(c) 65Â°

(d) 85Â°

Solution:

In the figure, l || m

âˆ´ âˆ 1 =65Â° (Corresponding angles)

In âˆ†BCD,

Ext. âˆ 1 = x + 20Â°

â‡’ 65Â° = x + 20Â°

â‡’ x = 65Â° – 20Â°

â‡’ x = 45Â° (b)

Question 21.

In the figure, if AB || CD, then x =

(a) 100Â°

(b) 105Â°

(c) 110Â°

(d) 115Â°

Solution:

In the figure, AB || CD

Through P, draw PQ || AB or CD

âˆ A + âˆ 1 = 180Â° (Co-interior angles)

â‡’ 132Â° + âˆ 1 = 180Â°

â‡’ âˆ 1 = 180Â°- 132Â° = 48Â°

âˆ´ âˆ 2 = 148Â° – âˆ 1 = 148Â° – 48Â° = 100Â°

âˆµ DQ || CP

âˆ´ âˆ 2 = x (Corresponding angles)

âˆ´ x = 100Â° (a)

Question 22.

In tlie figure, if lines l and in are parallel lines, then x =

(a) 70Â°

(b) 100Â°

(c) 40Â°

(d) 30Â°

Solution:

In the figure, l || m

âˆ l =70Â° (Corresponding angles)

In âˆ†DEF,

Ext. âˆ l = x + 30Â°

â‡’ 70Â° = x + 30Â°

â‡’ x = 70Â° – 30Â° = 40Â° (c)

Question 23.

In the figure, if l || m, then x =

(a) 105Â°

(b) 65Â°

(c) 40Â°

(d) 25Â°

Solution:

In the figure,

l || m and n is the transversal

âˆ 1 = 65Â° (Alternate angles)

In âˆ†GHF,

Ext. x = âˆ 1 + 40Â° = 65Â° + 40Â°

â‡’ x = 105Â°

âˆ´ x = 105Â° (a)

Question 24.

In the figure, if lines l and m are parallel, then the value of x is

(a) 35Â°

(b) 55Â°

(c) 65Â°

(d) 75Â°

Solution:

In the figure, l || m

and PQ is the transversal

âˆ 1 = 90Â°

In âˆ†EFG,

Ext. âˆ G = âˆ E + âˆ F

â‡’ 125Â° = x + âˆ 1 = x + 90Â°

â‡’ x = 125Â° – 90Â° = 35Â° (a)

Question 25.

Two complementary angles are such that two times the measure of one is equal to three times the measure of the other. The measure of the smaller angle is

(a) 45Â°

(b) 30Â°

(c) 36Â°

(d) none of these

Solution:

Let first angle = x

Then its complementary angle = 90Â° – x

âˆ´ 2x = 3(90Â° – x)

â‡’ 2x = 270Â° – 3x

â‡’ 2x + 3x = 270Â°

â‡’ 5x = 270Â°

â‡’ x = \(\frac { { 270 }^{ \circ } }{ 5 }\)Â = 54Â°

âˆ´ second angle = 90Â° – 54Â° = 36Â°

âˆ´ smaller angle = 36Â° (c)

Question 26.

Solution:

Question 27.

In the figure, AB || CD || EF and GH || KL.

The measure of âˆ HKL is

(a) 85Â°

(b) 135Â°

(c) 145Â°

(d) 215Â°

Solution:

In the figure, AB || CD || EF and GH || KL and GH is product to meet AB in L.

âˆµ AB || CD

âˆ´ âˆ 1 = 25Â° (Alternate angle)

and GH || KL

âˆ´ âˆ 4 = 60Â° (Corresponding angles)

âˆ 5 = âˆ 4 = 60Â° (Vertically opposite angle)

âˆ 5 + âˆ 2 = 180Â° (Co-interior anlges)

âˆ´ â‡’ 60Â° + âˆ 2 = 180Â°

âˆ 2 = 180Â° – 60Â° = 120Â°

Now âˆ HKL = âˆ 1 + âˆ 2 = 25Â° + 120Â°

= 145Â° (c)

Question 28.

AB and CD are two parallel lines. PQ cuts AB and CD at E and F respectively. EL is the bisector of âˆ FEB. If âˆ LEB = 35Â°, then âˆ CFQ will be

(a) 55Â°

(b) 70Â°

(c) 110Â°

(d) 130Â°

Solution:

AB || CD and PQ is the transversal EL is the bisector of âˆ FEB and âˆ LEB = 35Â°

âˆ´ âˆ FEB = 2 x 35Â° = 70Â°

âˆµ AB || CD

âˆ´ âˆ FEB + âˆ EFD = 180Â°

(Co-interior angles)

70Â° + âˆ EFD = 180Â°

âˆ´ âˆ EFD = 180Â°-70Â°= 110Â°

But âˆ CFQ = âˆ EFD

(Vertically opposite angles)

âˆ´ âˆ CFQ =110Â° (c)

Question 29.

In the figure, if line segment AB is parallel to the line segment CD, what is the value of y?

(a) 12

(b) 15

(c) 18

(d) 20

Solution:

In the figure, AB || CD

BD is transversal

âˆ´ âˆ ABD + âˆ BDC = 180Â° (Co-interior angles)

â‡’y + 2y+y + 5y = 180Â°

â‡’ 9y = 180Â° â‡’ y = \(\frac { { 180 }^{ \circ } }{ 9 }\)Â = 20Â° (d)

Question 30.

In the figure, if CP || DQ, then the measure of x is

(a) 130Â°

(b) 105Â°

(c) 175Â°

(d) 125Â°

Solution:

In the figure, CP || DQ

BA is transversal

Produce PC to meet BA at D

âˆµ QB || PD

âˆ´ âˆ D = 105Â° (Corresponding angles)

In âˆ†ADC,

Ext. âˆ ACP = âˆ CDA + âˆ DAC

â‡’ x = âˆ 1 + 25Â°

= 105Â° + 25Â° = 130Â° (a)