## RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.4

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.4

**Other Exercises**

- RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.1
- RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.2
- RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.3
- RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.4
- RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles VSAQS
- RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS

**Question 1.**

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

**Solution:**

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

Let ∠1 = 3x

Then ∠2 = 2x

But ∠1 + ∠2 = 180° (Linear pair)

∴ 3x + 2x = 180° ⇒ 5x = 180°

⇒ x = \(\frac { { 180 }^{ \circ } }{ 5 }\) = 36°

∴ ∠1 = 3x = 3 x 36° = 108°

∠2 = 2x = 2 x 36° = 72°

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

∴ ∠3 = 108° and ∠4 = 72°

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

∴ ∠5 = 108°, ∠6 = 72°

Similarly, ∠4 = ∠8 and

∠3 = ∠7

∴ ∠8 = 72° and ∠7 = 108°

Hence, ∠1 = 108°, ∠2= 72°

∠3 = 108°, ∠4 = 72°

∠5 = 108°, ∠6 = 72°

∠7 = 108°, ∠8 = 12°

**Question 2.**

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

**Solution:**

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

∠2 = ∠4 (Alternate angles)

∴ ∠2 = 120°

But ∠3 + ∠4 = 180° (Linear pair)

⇒ ∠3 + 120° = 180°

⇒ ∠3 = 180° – 120°

∴ ∠3 = 60°

But ∠l = ∠3 (Corresponding angles)

∴ ∠l = 60°

Hence ∠l = 60°, ∠2 = 120°, ∠3 = 60°

**Question 3.**

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

**Solution:**

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

∠BAC = 70°, ∠CEF = 130°

∵ EF || CD

∴ ∠ECD + ∠CEF = 180° (Co-interior angles)

⇒ ∠ECD + 130° = 180°

∴ ∠ECD = 180° – 130° = 50°

∵ BA || CD

∴ ∠BAC = ∠ACD (Alternate angles)

∴ ∠ACD = 70° (∵ ∠BAC = 70°)

∵ ∠ACE = ∠ACD – ∠ECD = 70° – 50° = 20°

**Question 4.**

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

**Solution:**

In the figure,

∵ ∠ACD = ∠CDE = 100°

But they are alternate angles

∴ AC || DE

**Question 5.**

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

**Solution:**

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

∵ n || p

∴ ∠1 = ∠3 (Corresponding anlges)

But ∠1 = 85°

∴ ∠3 = 85°

∵ m || 1

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

⇒ 85° + ∠2 = 180°

⇒ ∠2 = 180° – 85° = 95°

**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,

∠A and ∠B are unequal

and ∠A : ∠B = 2 : 3

Let ∠A = 2x, then

∠B = 3x

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

∴ 2x + 3x = 180°

⇒ 5x = 180°

⇒ x = \(\frac { { 180 }^{ \circ } }{ 5 }\) = 36°

∴ ∠A = 2x = 2 x 36° = 72°

∠B = 3x = 3 x 36° = 108°

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

∴ ∠C = 72° and ∠D = 108°

Hence ∠A = 72°, ∠B = 108°, ∠C = 72°, ∠D = 108°

**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 ⊥ line l and CD ⊥ line l

∴ ∠B = 90° and ∠D = 90°

∴ ∠B = ∠D

But there are corresponding angles

∴ AB || CD

**Question 9.**

**In the figure, ∠1 = 60° and ∠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

∠1 = 60° and

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

= \(\frac { 2 }{ 3 }\) x 90° = 60°

∴ ∠1 = ∠2

But there are corresponding angles

∴ l || m

**Question 10.**

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

**Solution:**

In the figure,

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

∠1 = 60°

∴ ∠1 = ∠3 (Corresponding angles)

∴ ∠3 = 60°

But ∠3 + ∠4 = 180° (Linear pair)

60° + ∠4 = 180° ⇒ ∠4 = 180° – 60°

∴ ∠4 = 120°

But ∠2 = ∠4 (Alternate angles)

∴ ∠2 = 120°

**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 ⊥ l and CD ⊥ l

**Question 12.**

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

**Solution:**

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

∵ AD || BC and AB || DC

∴ ABCD is a parallelogram

∴ ∠A + ∠B = 180° (Co-interior angles)

60° + ∠B = 180°

⇒ ∠B = 180°-60°= 120°

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

∴ ∠C = 60° and ∠D = 120°

Hence ∠B = 120°, ∠C = 60° and ∠D = 120°

**Question 13.**

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

**Solution:**

Two lines AB and CD intersect at O

and ∠AOC + ∠COB + ∠BOD = 270°

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

∴ 270° + ∠DOA = 360°

⇒ ∠DOA = 360° – 270° = 90°

But ∠DOA = ∠BOC (Vertically opposite angles)

∴ ∠BOC = 90°

But ∠DOA + ∠BOD = 180° (Linear pair)

⇒ 90° + ∠BOD = 180°

∴ ∠BOD= 180°-90° = 90° ,

But ∠BOD = ∠AOC (Vertically opposite angles)

∴ ∠AOC = 90°

Hence ∠AOC = 90°,

∠COB = 90°,

∠BOD = 90° and ∠DOA = 90°

**Question 14.**

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

**Solution:**

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

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

∠2 = 120°, and ∠5 = 60°

To prove : m || n

Proof : ∠2 + ∠3 = 180° (Linear pair)

⇒ 120°+ ∠3 = 180°

⇒ ∠3 = 180°- 120° = 60°

But ∠5 = 60°

∴ ∠3 = ∠5

But there are alternate angles

∴ m || n

**Question 15.**

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

**Solution:**

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

∠4 = 110° and ∠7 = 65°

**To prove :** Whether m || n or not

**Proof :** ∠4 = 110° and ∠7 = 65°

∠7 = ∠5 (Vertically opposite angles)

∴ ∠5 = 65°

Now ∠4 + ∠5 = 110° + 65° = 175°

∵ Sum of co-interior angles ∠4 and ∠5 is not 180°.

∴ m is not parallel to n

**Question 16.**

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

**Solution:**

**Given :** In the figure, ∠A = 115°, ∠B = 65°, ∠C = 115° and ∠D = 65°

∵ ∠A + ∠B = 115°+ 65°= 180°

But these are co-interior angles,

∴ AD || BC

Similarly, ∠A + ∠D = 115° + 65° = 180°

∴ AB || DC

**Question 17.**

**If l, m, n are three lines such that l ||m and n ⊥ l, prove that n ⊥ m.**

**Solution:**

**Given :** l, m, n are three lines such that l || m and n ⊥ l

**To prove :** n ⊥ m

**Proof :** ∵ l || m and n is the transversal.

∴ ∠l = ∠2 (Corresponding angles)

But ∠1 = 90° (∵ n⊥l)

∴ ∠2 = 90°

∴ n ⊥ 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 180° 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 180°, 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 180°, then the lines are **parallel.**

**Question 20.**

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

**Solution:**

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

∵ AB || CD and HK is transversal

∴ ∠1 = 25° (Alternate angles)

∠3 = 60° (Corresponding angles)

and ∠3 = ∠4 (Corresponding angles)

= 60°

But ∠4 + ∠5 = 180° (Linear pair)

⇒ 60° + ∠5 = 180°

⇒ ∠5 = 180° – 60° = 120°

∴ ∠HKL = ∠1 + ∠5 = 25° + 120° = 145°

**Question 21.**

**In the figure, show that AB || EF.**

**Solution:**

**Given :** In the figure, AB || EF

∠BAC = 57°, ∠ACE = 22°

∠ECD = 35° and ∠CEF =145°

**To prove :** AB || EF,

**Proof :** ∠ECD + ∠CEF = 35° + 145°

= 180°

But these are co-interior angles

∴ EF || CD

But AB || CD

∴ AB || EF

**Question 22.**

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

**Solution:**

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

∠QPR = 102°

Produce BA to meet PR at D

∵ PQ || AB or DB

∴ ∠QPR = ∠ADR (Corresponding angles)

∴∠ADR = 102° or ∠BDR = 102°

∵ PR || BC

∴ ∠BDR + ∠DBC = 180°

(Sum of co-interior angles) ⇒ 102° + ∠DBC = 180°

⇒ ∠DBC = 180° – 102° = 78°

⇒ ∠ABC = 78°

**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 ∠ABC and ∠DEF AB ⊥ DE and BC ⊥ EF

**To prove:** ∠ABC + ∠DEF = 180° or ∠ABC = ∠DEF

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

**Proof:** In figure (i) BGEH is a quadrilateral

∠BHE = 90° and ∠BGE = 90°

But sum of angles of a quadrilateral is 360°

∴ ∠HBG + ∠HEG = 360° – (90° + 90°)

= 360° – 180°= 180°

∴ ∠ABC and ∠DEF are supplementary

In figure (if) in quadrilateral BGEH,

∠BHE = 90° and ∠HEG = 90°

∴ ∠HBG + ∠HEG = 360° – (90° + 90°)

= 360°- 180° = 180° …(i)

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

From (i) and (ii)

∴ ∠HEF = ∠HBG

⇒ ∠DEF = ∠ABC

Hence ∠ABC and ∠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 ∠ABP + ∠CDP = ∠DPB**.

**Solution:**

**Given :** In the figure, AB || CD

P is a point between AB and CD PD

and PB are joined

**To prove :** ∠APB + ∠CDP = ∠DPB

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

**Proof:** ∵ AB || PQ

∴ ∠ABP = BPQ …(i) (Alternate angles)

Similarly,

CD || PQ

∴ ∠CDP = ∠DPQ …(ii)

(Alternate angles)

Adding (i) and (ii)

∠ABP + ∠CDP = ∠BPQ + ∠DPQ

Hence ∠ABP + ∠CDP = ∠DPB

**Question 25.**

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

**∠ABP + ∠BPD + ∠CDP = 360°**

**Solution:**

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

**To prove :** ∠ABP + ∠BPD + ∠CDP = 360°

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

**Proof :** ∵ AB || PQ

∴ ∠ABP+ ∠BPQ= 180° ……(i) (Sum of co-interior angles)

Similarly, CD || PQ

∴ ∠QPD + ∠CDP = 180° …(ii)

Adding (i) and (ii)

∠ABP + ∠BPQ + ∠QPD + ∠CDP

= 180°+ 180° = 360°

⇒ ∠ABP + ∠BPD + ∠CDP = 360°

**Question 26.**

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

**Solution:**

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

**To prove :** ∠ABC = ∠DEF

**Construction :** Produce BC to meet DE at G

**Proof:** AB || DE

∴ ∠ABC = ∠DGH…(i) (Corresponding angles)

BC or BH || EF

∴ ∠DGH = ∠DEF (ii) (Corresponding angles)

From (i) and (ii)

∠ABC = ∠DEF

**Question 27.**

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

**Solution:**

**Given:** In ∠ABC = ∠DEF

BA || ED and BC || EF

**To prove:** ∠ABC = ∠DEF = 180°

**Construction :** Produce BC to H intersecting ED at G

**Proof:** ∵ AB || ED

∴ ∠ABC = ∠EGH …(i) (Corresponding angles)

∵ BC or BH || EF

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

⇒ ∠ABC + ∠DEF = 180° [From (i)]

Hence proved.

Hope given RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.4 are helpful to complete your math homework.

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