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

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

Ans : Let âˆ 1 = 3x and âˆ 2 = 2x

âˆ 1 and âˆ 2 are linear pair of angle

Now, âˆ 1 and âˆ 2

â‡’ 3x + 2x = 180

â‡’ 5x = 180

â‡’ x = 180 / 5

â‡’ x = 36

âˆ 1 = 3x = 108^{âˆ˜},âˆ 2 = 2x = 72^{âˆ˜}

We know, Vertically opposite angles are equal

âˆ 1=âˆ 3 = 108^{âˆ˜}

âˆ 2 =âˆ 4 = 72^{âˆ˜}

âˆ 6 =âˆ 7=108^{âˆ˜}

âˆ 5 =âˆ 8 = 72^{âˆ˜}

We also know, corresponding angles are equal

âˆ 1=âˆ 5 = 108^{âˆ˜}

âˆ 2 = âˆ 6 = 72^{âˆ˜}

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

Ans : From the given figure :

âˆ 3+âˆ mYZ = 180^{âˆ˜} [Linear pair]

â‡’ âˆ 3 =180â€“120

â‡’ âˆ 3 = 60^{âˆ˜}

Now line I parallel to m

âˆ 1 =âˆ 3 [Corresponding angles]

âˆ 1 = 60^{âˆ˜}

Also m parallel to n

â‡’ âˆ 2 = 120^{âˆ˜} [Alternative interior angle]

Hence, âˆ 1 =âˆ 3 = 60^{âˆ˜}

âˆ 2 = 120^{âˆ˜}

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

Ans : Produce LK to meet GF at N.

Now, alternative angles are equal

âˆ CHG = âˆ HGN = 60âˆ˜

âˆ HGN = âˆ KNF = 60âˆ˜ [Corresponding angles]

Hence, âˆ KNG = 180â€“60 = 120

â‡’ âˆ GNK =âˆ AKL = 120^{âˆ˜} [Corresponding angles]

âˆ AKH =âˆ KHD = 25^{âˆ˜} [alternative angles]

Therefore, âˆ HKL=âˆ AKH+âˆ AKL = 25+120 = 145^{âˆ˜}

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

Ans : Produce EF to intersect AC at K.

Now, âˆ DCE+âˆ CEF = 35+145 = 180âˆ˜

Therefore, EF || CD (Since Sum of Co-interior angles is 180) â€”â€“(1)

Now, âˆ BAC =âˆ ACD = 57^{âˆ˜}

â‡’ BA || EF [Alternative angles are equal] â€”â€“(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 âˆ ACE.**

Ans : Since EF || CD

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

â‡’ ECD = 180 â€“ 130 = 50

Also BA || CD

â‡’ BAC = ACD = 70 [alternative angles]

But, ACE + ECD =70

â‡’ ACE = 70 â€” 50 = 20

**Q 6 : In the below fig, PQ || AB and PR || BC. If âˆ QPR = 102 ^{âˆ˜}, determine âˆ ABC Give reasons.**

Ans : AB is produce to meet PR at K

Since PQ || AB

âˆ QPR = âˆ BKR = 102^{âˆ˜} [corresponding angles]

Since PR || BC

âˆ RKB+âˆ CBK = 180^{âˆ˜} [ Since Corresponding angles are supplementary]

âˆ CKB=180â€“102 = 78

âˆ´ âˆ CKB = 78Â°

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

Ans : Vertically opposite angles are equal

â‡’ âˆ EOC = âˆ DOK = 100âˆ˜

âˆ DOK = âˆ ACO = 100âˆ˜

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 âˆ 1 = 85Â°. find âˆ 2.**

Ans : Corresponding angles are equal

â‡’ âˆ 1=âˆ 3 = 85^{âˆ˜}

By using the property of co-interior angles are supplementary

âˆ 2+âˆ 3 = 180^{âˆ˜}

âˆ 2+55 = 180

âˆ 2 = 180â€“85

âˆ 2 = 95^{âˆ˜}

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

**Ans : **Given m perpendicular t and l perpendicular to t

âˆ 1=âˆ 2 = 90^{âˆ˜}

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.**

**Ans :** Consider be angles AOB and ACB

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

To prove : âˆ AOB+âˆ ACB=180âˆ˜ (or) âˆ AOB+âˆ ACB = 180^{âˆ˜}

Proof : In a quadrilateral = âˆ A+âˆ O+âˆ B+âˆ C = 360^{âˆ˜}

[ Sum of angles of quadrilateral is 360 ]

â‡’ 180+ O +B + C = 360

â‡’ O+ C = 360 â€“180

Hence AOB + ACB = 180 â€”â€“(1)

Also, B + ACB = 180

â‡’ ACB =180 â€“ 90 = ACES = 90Â° â€”â€“(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 âˆ ABP+âˆ CDP = âˆ DPB.**

**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

âˆ ABP = âˆ BPF

Alternative angles are equal

âˆ CDP = âˆ DPF

âˆ ABP+âˆ CDP = âˆ BPF+âˆ DPF

âˆ ABP+âˆ CDP = âˆ DPB

Hence proved

AB parallel to CD, P is any point

To prove: âˆ ABP+âˆ BPD+âˆ CDP = 360^{âˆ˜}

Construction : Draw EF || AB passing through P

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

âˆ ABP+âˆ EPB = 180^{âˆ˜} [Sum of co-interior angles is180)

âˆ EPD+âˆ COP = 180^{âˆ˜} â€”â€“(1) [Sum of co-interior angles is180)

âˆ EPD+âˆ CDP = 180^{âˆ˜} â€”â€“(2)

By adding (1) end (2)

âˆ ABP+âˆ EPB+âˆ EPD+âˆ CDP = (180+180)^{âˆ˜}

âˆ ABP+âˆ EPB+âˆ COP = 360^{âˆ˜}

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

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

Now,

A8 || PM â‡’ ABP + BPM = 180

And

CD||PM = MPD + CDP = 180

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

â‡’ ABP + BPD + CDP = 360

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

Ans : Let A = 2x and B = 3x

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

2x + 3x â€“ 180 [AD II BC and AB is the transversal)

â‡’ 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?**

**Ans :**

Let AB and CD be perpendicular to MN

ABD = 90 [AB perpendicular to MN] â€”â€“ (i)

CON = 90 [CO perpendicular to MN] â€”â€“ (ii)

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

AB parallel to CD,

Since corresponding angles are equal