7th Maths : Term 1 Unit 5 : Geometry : Exercise 5.6 : Miscellaneous Practice problems, Challenge Problems, Text Book Back Exercises Questions with Answers, Solution

**Exercise
5.6**

** **

__Miscellaneous
Practice problems__

** **

**1.
Find the value of ***x*** if ****∠***AOB*** is a right angle.**

**Answer: **

∠AOC + ∠COB = 90°

2*x* + 3*x* = 90°

5*x* = 90°

*x* = 90° / 5 = 18°

* x* = 18°

** **

**2.
In the given figure, find the value of ***x***.**

**Answer: **

∠BOC + ∠AOC = 180°

2*x* + 23 + 3*x* – 48 = 180°

5*x* – 25 = 180°

5*x* = 180° + 25^{o}

= 205^{o}

* x* = 205^{o}
/ 5 = 41°

* x* = 41°

** **

**3.
Find the value of ***x, y*** and ***z*

**Answer: **

∠BOD + ∠BOC = 180°

*x* + 3*x* + 48 = 180°

4*x* + 40 = 180°

4*x* = 180° – 40^{o}

= 140^{o}

*x* = 140^{o} / ~~4~~ = 35°

* x* = 35°

∠BOD + ∠AOD = 180°

*x* + *y* + 30° = 180°

35° + *y* + 30° = 180°

* y* = 180° – 65^{o}

* y* = 115^{o}

∠AOD + ∠AOC = 180°

*y* + 30° + *z* + 10° =
180°

115° + *z* + 40° = 180°

* z* + 155° = 180°

*z* = 180° – 155° = 25°

*z *= 25°

** **

**4.
Two angles are in the ratio 11: 25. If they are linear pair, find the angles.**

**Answer: **

Two angles are in the ratio 11:25

Let the two angles be 11*x*,
25*x*

Sum of two angles 11*x*
+ 25*x* = 180°

36*x* = 180°

* x* = 180^{o}
/ 36 = 5°

* x* = 5°

11*x* = 11 × 5° = 55°

25*x* = 25 × 5° = 125°

The two angles are 55^{o}, 125^{o}

** **

**5.
Using the figure, answer the following questions and justify your answer.**

**(i)
Is ****∠****1 adjacent to ****∠****2?**

**(ii)
Is ****∠****AOB adjacent to ****∠****BOE?**

**(iii)
Does ****∠****BOC and ****∠****BOD form a linear
pair?**

**(iv)
Are the angles ****∠****COD and ****∠****BOD supplementary?**

**(v)
Is ****∠****3 vertically opposite to ****∠****1?**

**(i) Is ****∠****l adjacent to ****∠****2?**

Yes. They have a common vertex, a common arm and their interiors
do not overlap.

**(ii) Is ****∠****AOB adjacent to ****∠****BOE?**

No. They have overlapping interiors.

**(iii) Do ****∠****BOC and ****∠****BOD form a linear pair?**

No. Since ∠BOC is a straight angle, their sum
will exceed 180°.

**(iv) Are ****∠****COD and ****∠****BOD supplementary?**

Yes. They are linear pair.

**(v) Is ****∠****3 vertically opposite to ****∠****1 ?**

No. They are not formed by intersecting lines.

** **

**6.
In the figure POQ, ROS and TOU are straight lines. Find the x°.**

**Answer: **

∠UOV + ∠VOP + ∠POR + ∠ROT = 180°

*x* + 45° + 47° + 36° = 180°

*x* + 128° = 180°

*x* = 180° – 128° = 52°

*x* = 52^{o}

** **

**7.
In the figure AB is parallel to DC. Find the value of **

**Answer: **

From the figure, ∠DCE = ∠1 (alternate angles)

30° = ∠l

∠l = 30°

∠2 = ∠DEB (alternate angles)

∠2 = 80°

** **

**8.
In the figure AB is parallel to CD. Find x,y and z.**

**Answer: **

From the figure,

∠GED = ∠CEF (vertically opposite angles)

*y* = 42°

∠CEF = ∠EFB (alternate angles)

42° = ∠*z*

∠*z* = 42°

∠AFC + ∠CFE + ∠EFB = 180°

*x*° + 63°+ *z*° = 180°

*x*° + 63° + 42° = 180°

*x*° + 105° =180°

*x*° = 180°– 105° = 75°

*x*° = 75°

*x*° = 75°, *y* = 42°, *z* = 42°

** **

**9.
Draw two parallel lines and a transversal. Mark two alternate interior angles G and H. If they are supplementary, what is the measure of each angle?**

**Answer: **

Angles G and H are supplementary

∠G and ∠H are alternate interior angles.

∴ ∠G = ∠H = 90°

** **

**10.
A plumber must install pipe 2 parallel to pipe 1. He knows that ****∠****1 is 53. What is the measure of ****∠****2?**

**Answer: **

∠1 + ∠2 = 180°

53° + ∠2 = 180°

∠2 = 180° – 53° = 127°

∠2 = 127°

** **

__Challenge
Problems__

** **

**11.
Find the value of ***y***.**

**Answer: **

∠POT + ∠TOS + ∠SOR + ∠ROQ = 180^{o}

60° + 3y – 20 + *y *+ *y *+ 10 = 180°

60° + 10 – 20 + 5*y* =
180°

70 – 20 + 5*y* = 180°

50 + 5*y* = 180°

5y = 180° – 50° = 130°

* y* = 130° / 5 = 26^{o}

* y* = 26°

** **

**12.
Find the value of z.**

**Answer: **

∠POQ + ∠QOM + ∠MON + ∠PON = 360^{o}

3*z* + 4*z* –25 + *z* + 10 +2*z* – 5 = 360^{o}

3*z* + 4*z* + *z* + 2*z*
+ 10 – 25 – 5 = 360^{o}

10*z* – 20 = 360^{o}

10*z* = 360^{o} + 20^{o} = 380^{o}

* z = *380^{o} /
10 = 38^{o}

* z *= 38^{o}

** **

**13.
Find the value of ***x*** and ***y ***if***
RS ***is parallel to*** PQ***.**

**Answer: **

From the figure,

∠POR = ∠ORS (alternate angles)

∠POR = 25°

∠POR = ∠QOU (vertically opposite angles)

∠QOU = 25°

* y* = 25°

∠POT = ∠SOQ (vertically opposite angles)

40° = ∠SOQ

∠RSO = ∠SOQ (alternate angles)

*x*° = 40°

*y*° = 25°

** **

**14.
Two parallel lines are cut by a transversal. For each pair of interior angles on
the same side of the transversal, if one angle exceeds the twice of the other angle
by 48°. Find the angles.**

**Answer: **

Given one angle exceeds twice of other angle by 48°

= 2*x* + 48

Other angle = *x*°

Sum of interior angles = *x*°
+ 2*x*° + 48° = 180°

3*x* + 48° = 180°

3*x* = 180° – 48° = 132°

*x* = 132^{o }/ 3 = 44^{o}

*x* = 44°

2*x* + 48 = 2 × 44 + 48

= 88 + 48° = 136°

The angles are 44°, 136°.

** **

**15.
In the figure, the lines GH and IJ are parallel. If ****∠****1=108° and ****∠****2 = 123°, find
the value of ***x, y*** and ***z***.**

**Answer: **

∠1 = 108°

Interior angle ∠G = 180^{o} – 108^{o} = 72^{o}

*x*° = Interior angle ∠G (corresponding angles)

*x*° = 72°

∠2 = 123°

Interior angle
∠H = 180° – 123° = 57°

*y*° = Interior angle ∠H = (corresponding angles)

*y* = 57°

*x*° + *y*° + 2° = 180°
(sum of the triangles)

72° + 57° + *z*° = 180°

129° + *z*° = 180°

*z*° = 180° – 129° = 51°

**x****° = 72°, y° = 57°, z = 51°**

** **

**16.
In the parking lot shown, the lines that mark the width of each space are parallel.
If ****∠****1 = (2***x***–3***y***)°, ****∠****2 = (***x*** +39)°, find ***x*** and ***y***.**

**Answer: **

∠2 + 65° = 180° (interior angles)

(*x* + 39) + 65° = 180°

*x* + 104° = 180°

* x* = 180° – 104° = 76°

* x* = 76°

∠1= 65° (vertically opposite angles)

(2*x* – 3*y*)^{o }= 65°

2(76^{o}) – 3*y*^{o
}= 65°

152^{o} – 3*y*^{o}
= 65^{o}

–3*y*° = 65° – 152° = –87^{o}

*y *= –87 / –3 = 29°

*y *= 29°

**x = 76° , y = 29°**

** **

**17.
Draw two parallel lines and a transversal. Mark two corresponding angles ***A*** and ***B.*** If ****∠****A = 4***x*** and ****∠***B
***= 3***x ***+ 7, find the value
of*** x***. Explain..**

**Answer: **

A and B are corresponding angles.

∠A = ∠B

4*x* = 3*x* + 7

4*x *– 3*x* = 7

*x* = 7°

** **

**18.**** ****In the figure ***AB*** is parallel to
***CD***. Find ***x***˚, ***y***˚ and ***z***˚.**

**Answer: **

∠A = ∠D (alternate angles)

*x*° = 48°

∠C = ∠B (alternate angles)

*y*° = 60°

from ΔCDE,

∠C + ∠D + ∠E = 180° (sum of the three
angles of a triangle)

60° + 48° + ∠E = 180°

108° + ∠E = 180°

∠E = 180° – 108°

∠E = 72°

∠E + Z° = 180°∠ (straight angles)

72° + Z° = 180°

Z° = 180° – 72° = 108°

Z° =108°

*x*** = 48° y = 60°
z = 108° **

** **

**19.
Two parallel lines are cut by a transversal. If one angle of a pair of corresponding
angles can be represented by 42˚ less than three times the other. Find the corresponding
angles.**

**Answer: **

One angle = *x*

Other angle = 42 less than 3 times the other

= 3*x* – 42

The corresponding angles are equal

3*x* – 42 = *x*

3*x* – x = 42

2*x* = 42

*x* = 21

**The corresponding
angles = 21**

** **

**20.
In the given figure, ****∠****8 = 107˚, what
is the sum of the ****∠****2 and ****∠****4?**

**Answer: **

∠8 = ∠4 (Corresponding angles)

∠8 = 107°

∴ ∠4 = 107°

∠4 = ∠2 (vertically opposite angles)

∠2 = 107°

The sum of ∠2 and ∠4 = 107° + 107° = 214°.

** **

**ANSWERS**

**Exercise 5.6**

1. 18°

2. 41°

3. 35°, 115°, 25°

4. 55°, 125°

5. (i) Yes. They have a common
vertex, a common arm and their interiors do not overlap.

(ii) No. They have overlapping
interiors.

(iii) No. Since ∠BOC is a straight angle, their
sum will exceed 180°.

(iv) Yes. They are linear pair.

(v) No. They are not formed by
intersecting lines

6. 52°

7. 30°, 80°

8. 75° , 42° , 42°

9. 90°

10. 127°

**Challenge Problems **

11. 26°

12. 38°

13. 40°,25°

14. 44°

15. 72°,57°,51°

16. 76°,29°

17. 7°

18. 48°,60°,108°

19. 21°

20. 214°

** **

Tags : Questions with Answers, Solution | Geometry | Term 1 Chapter 5 | 7th Maths , 7th Maths : Term 1 Unit 5 : Geometry

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7th Maths : Term 1 Unit 5 : Geometry : Exercise 5.6 | Questions with Answers, Solution | Geometry | Term 1 Chapter 5 | 7th Maths

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