Class 9 Exam  >  Class 9 Notes  >  RD Sharma Solutions for Class 9 Mathematics  >  RD Sharma Solutions Ex-8.4, (Part -1), Lines And Angles, Class 9, Maths

RD Sharma Solutions Ex-8.4, (Part -1), Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics PDF Download

Q 1:  In below fig. AB CD and ∠1and∠2 are in the ratio 3 : 2. Determine all angles from 1 to 8.

RD Sharma Solutions Ex-8.4, (Part -1), Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Ans :     Let ∠1 = 3x and ∠2 = 2x

∠1 and ∠2 are linear pair of angle

Now, ∠1 and ∠2

⇒  3x + 2x = 180

⇒ 5x  = 180

⇒ x = 180 / 5

⇒ x = 36

∠1 = 3x = 108,∠2 = 2x = 72

We know, Vertically opposite angles are equal

∠1=∠3 = 108

∠2 =∠4 = 72

∠6 =∠7=108

∠5 =∠8 = 72

We also know, corresponding angles are equal

∠1=∠5 = 108

∠2 = ∠6 = 72


Q 2: In the below fig, I, m and n are parallel lines intersected by transversal p at X. Y and Z respectively. Find ∠1,∠2 and ∠3

RD Sharma Solutions Ex-8.4, (Part -1), Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Ans : From the given figure :

∠3+∠mYZ = 180  [Linear pair]

⇒ ∠3 =180–120

⇒ ∠3 = 60

Now line I parallel to m

∠1 =∠3 [Corresponding angles]

∠1 = 60

Also m parallel to n

⇒ ∠2 = 120  [Alternative interior angle]

Hence, ∠1 =∠3 = 60

∠2 = 120

3. In the below fig, AB || CD || EF and GH || KL Find ∠HKL

RD Sharma Solutions Ex-8.4, (Part -1), Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Ans  :    Produce LK to meet GF at N.

Now, alternative angles are equal

∠CHG = ∠HGN = 60∘

∠HGN = ∠KNF = 60∘        [Corresponding angles]

Hence, ∠KNG = 180–60 = 120

⇒ ∠GNK =∠AKL = 120 [Corresponding angles]

∠AKH =∠KHD = 25        [alternative angles]

Therefore, ∠HKL=∠AKH+∠AKL = 25+120 = 145


Q 4 : In the below fig, show that AB || EF

RD Sharma Solutions Ex-8.4, (Part -1), Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Ans : Produce EF to intersect AC at K.

Now, ∠DCE+∠CEF = 35+145 = 180∘

Therefore, EF || CD       (Since Sum of Co-interior angles is 180) —–(1)

Now, ∠BAC =∠ACD = 57

⇒ BA || EF               [Alternative angles are equal]        —–(2)

From (1) and (2)

AB || EF [Since, Lines parallel to the same line are parallel to each other]

Hence proved.


Q 5 : If below fig.if AB || CD and CD || EF, find ∠ACE.

RD Sharma Solutions Ex-8.4, (Part -1), Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Ans :  Since EF || CD

Therefore,  EFC + ECD = 180       [co-interior angles are supplementary]

⇒ ECD = 180 – 130 = 50

Also BA || CD

⇒ BAC = ACD = 70                [alternative angles]

But, ACE + ECD =70

⇒ ACE = 70 — 50 = 20

Q 6 : In the below fig, PQ || AB and PR || BC. If ∠QPR = 102, determine ∠ABC Give reasons.

RD Sharma Solutions Ex-8.4, (Part -1), Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Ans :  AB is produce to meet PR at K

Since PQ || AB

∠QPR = ∠BKR = 102              [corresponding angles]

Since PR || BC

∠RKB+∠CBK = 180               [ Since Corresponding angles are supplementary]

∠CKB=180–102 = 78

∴ ∠CKB = 78°


Q 7 : In the below fig, state Which lines are parallel and why?

RD Sharma Solutions Ex-8.4, (Part -1), Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Ans :  Vertically opposite angles are equal

⇒ ∠EOC = ∠DOK = 100∘

∠DOK = ∠ACO = 100∘

Here two lines EK and CA cut by a third line and the corresponding angles to it are equal Therefore, EK || AC.


8. In the below fig. if l||m, n || p and ∠1 = 85°. find ∠2.

RD Sharma Solutions Ex-8.4, (Part -1), Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Ans :  Corresponding angles are equal

⇒ ∠1=∠3 = 85

By using the property of co-interior angles are supplementary

∠2+∠3 = 180

∠2+55 = 180

∠2 = 180–85

∠2 = 95


Q 9 : If two straight lines are perpendicular to the same line, prove that they are parallel to each other.

RD Sharma Solutions Ex-8.4, (Part -1), Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Ans : Given m perpendicular t and l perpendicular to t

∠1=∠2 = 90

Since,  I and m are two lines and it is transversal and the corresponding angles are equal

L || M

Hence proved


Q 10 : 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.

RD Sharma Solutions Ex-8.4, (Part -1), Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Ans : Consider be angles AOB and ACB

Given  0A perpendicular to A0, also 0B perpendicular to BO

To prove : ∠AOB+∠ACB=180∘ (or) ∠AOB+∠ACB = 180

Proof :  In a quadrilateral = ∠A+∠O+∠B+∠C = 360

[ Sum of angles of quadrilateral is 360 ]

⇒ 180+ O +B + C = 360

⇒ O+ C = 360 –180

Hence AOB + ACB = 180                     —–(1)

Also, B + ACB = 180

⇒ ACB =180 – 90 = ACES = 90°        —–(2)

From (i) and (ii), ACB = A0B = 90

Hence, the angles are equal as well as supplementary.


Q 11 : In the below fig, lines AB and CD are parallel and P is any point as shown in the figure. Show that ∠ABP+∠CDP = ∠DPB.

RD Sharma Solutions Ex-8.4, (Part -1), Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Ans :

Given that AB ||CD

Let EF be the parallel line to AB and CD which passes through P

It can be seen from the figure

Alternative angles are equal

∠ABP = ∠BPF

Alternative angles are equal

∠CDP = ∠DPF

∠ABP+∠CDP = ∠BPF+∠DPF

∠ABP+∠CDP = ∠DPB

Hence proved

AB parallel to CD, P is any point

To prove: ∠ABP+∠BPD+∠CDP = 360

Construction : Draw EF || AB passing through P

RD Sharma Solutions Ex-8.4, (Part -1), Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Proof : Since AB  ||EF and AB || CD, Therefore EF || CD                     [Lines parallel to the same line are parallel to each other)

∠ABP+∠EPB = 180      [Sum of co-interior angles is180)

∠EPD+∠COP = 180       —–(1) [Sum of co-interior angles is180)

∠EPD+∠CDP = 180       —–(2)

By adding (1) end (2)

∠ABP+∠EPB+∠EPD+∠CDP = (180+180)

∠ABP+∠EPB+∠COP = 360


Q 12 : In the below fig, AB || CD and P is any point shown in the figure. Prove that : ∠ABP+∠BPD+∠CDP = 360

RD Sharma Solutions Ex-8.4, (Part -1), Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

 

Ans : Through P, draw a line PM parallel to AB or CD.

Now,

A8 || PM ⇒ ABP + BPM = 180

And

CD||PM = MPD + CDP = 180

Adding (i) and (ii), we get A8P + (BPM + MPD) CDP = 360

⇒ ABP + BPD + CDP = 360


Q 13 : Two unequal angles of a parallelogram are in the ratio 2 : 3. Find all its angles in degrees.

RD Sharma Solutions Ex-8.4, (Part -1), Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

 Ans : Let A = 2x and B = 3x

Now, A +B = 180 [Co-interior angles are supplementary]

2x  + 3x – 180 [AD II BC and AB is the transversal)

⇒ 5x = 180

x = 180/5

x = 36

Therefore, A = 2 x 36 = 72

b = 3 x 36  = 108

Now, A = C = 72  [Opposite side angles of a parallelogram are equal)

B = D = 108


Q 14 :  If each of the two lines is perpendicular to the same line, what kind of lines are they to each other?

RD Sharma Solutions Ex-8.4, (Part -1), Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Ans :

Let AB and CD be perpendicular to MN

ABD =  90 [AB perpendicular to MN]             —– (i)

CON = 90 [CO perpendicular to MN]             —– (ii)

Now, ABD = CDN = 90   (From (i) and (ii))

AB parallel to CD,

Since corresponding angles are equal

The document RD Sharma Solutions Ex-8.4, (Part -1), Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
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FAQs on RD Sharma Solutions Ex-8.4, (Part -1), Lines And Angles, Class 9, Maths - RD Sharma Solutions for Class 9 Mathematics

1. What are the different types of angles in geometry?
Ans. In geometry, there are several types of angles. Some of the commonly known types include acute angles, right angles, obtuse angles, straight angles, and reflex angles.
2. How do you identify corresponding angles?
Ans. Corresponding angles are formed when a pair of parallel lines is intersected by a transversal line. They are located on the same side of the transversal and in the same position relative to the parallel lines. Corresponding angles are congruent, meaning they have the same measure.
3. What is the sum of interior angles in a triangle?
Ans. The sum of the interior angles in a triangle is always 180 degrees. This property is known as the Triangle Sum Theorem. It holds true for all types of triangles, whether they are equilateral, isosceles, or scalene.
4. How do you find the measure of an angle in a straight line?
Ans. When two lines intersect, the angles opposite each other are known as vertically opposite angles. In the case of a straight line, these angles are always congruent, meaning they have the same measure. Therefore, to find the measure of an angle in a straight line, you can simply divide the total measure of 180 degrees by the number of angles formed.
5. What is the relationship between alternate interior angles?
Ans. Alternate interior angles are formed when a pair of parallel lines is intersected by a transversal line. They are located on the opposite sides of the transversal and inside the parallel lines. The main property of alternate interior angles is that they are congruent, meaning they have the same measure. This relationship holds true regardless of the angle's position or orientation.
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