7th Standard CBSE Mathematics Unit 10 Practical Geometry Model Question Paper | QB365

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7th Standard CBSE Mathematics Unit 10 Practical Geometry Model Question Paper

Published on: 14/08/2019

Practical Geometry

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Std: 7th Standard CBSE - Mathematics

chapter: - Model Question paper

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Questions:

Multiple Choice Question

5 x 1 = 5

We can construct a triangle only when the sum of any two sides is

(a)

less than the third side

(b)

greater than the third side

(c)

equal to the third side

(d)

equal to the perimeter of the triangle

Which of the following cannot be the side of a triangle?

(a)

8 cm,5 cm, 4 cm

(b)

10 cm, 10 cm, 4 cm

(c)

10 cm, 6 cm, 4 cm

(d)

9 cm, 8 cm, 7 cm.

In the given figure, AB = AC = BC, then the value of each angle is equal to

(a)

40°

(b)

70°

(c)

90°

(d)

60°

Which of the following sets of triangles could be the lengths of the sides of a right angled triangle?

(a)

3 cm, 4 cm, 6 cm

(b)

9 cm, 16 cm, 26 cm

(c)

1.5 cm, 3.6 cm, 3.9 cm

(d)

7 cm, 24 cm, 26 cm

A triangle can be constructed by taking its sides as.

(a)

1.8 cm, 2.6 cm, 4.4 cm

(b)

2 cm, 3 cm, 4 cm

(c)

2.4 cm, 2.4 cm, 6.4 cm

(d)

3.2 cm, 2.3 cm, 5.5 cm

Generic 3

4 x 3 = 12

Construct a triangle ABC when AB = 5.5 cm, BC = 4.5 cm and \(\angle\)B = 60°.

In the following constructed figure, find the value of \(\angle X.\)

The Measures of certain sides and angles ot triangles. Identity those which cannot be constructed and say why you cannot construct them? Construct rest ot the triangles.

Triangle	Given Measurements
ΔABC	mㄥA=70⁰	mㄥB=50⁰	AC=3 cm

In the given construction, can you draw any other line through A that would be also parallel to the line l?

Generic 2

6 x 2 = 12

10.

Draw a triangle whose sides are of lengths 4 cm, 5 cm and 7 cm.

11.

Construct an angle bisector of 90°.

12.

If a perpendicular bisector is drawn on a ray. Then, find the angle made by perpendicular bisector.

13.

The Measures of certain sides and angles ot triangles. Identity those which cannot be constructed and say why you cannot construct them? Construct rest ot the triangles.

Triangle	Given Measurements
ΔABC	BC=2 cm	AB=4 cm	AC=2 cm

14.

The Measures of certain sides and angles ot triangles. Identity those which cannot be constructed and say why you cannot construct them? Construct rest ot the triangles.

Triangle	Given Measurements
ΔABC	mㄥA=85⁰	mㄥB=115⁰	AB=5 cm

15.

Study the following problem. ΔABC, if AC = 7 cm, m ㄥA = 60⁰ and m ㄥB = 50°, can you draw the triangle?

Generic 5

2 x 5 = 10

16.

Draw a line AB of length 5 cm and draw a line PQ parallel to AB at a distance of 3 cm from AB.

17.

Construct the right angled ΔPQR, where mㄥQ = 90°, QR = 8 cm and PR = 10 cm.

Fill up / 1 Marks

5 x 1 = 5

18.

The side opposite to \(\angle\)Q is____________

19.

The angle made by perpendicular bisector of line is equal to__________.

20.

We can construct a right angled triangle, if the value of__________of the angle is given.

21.

A triangle can be constructed only if the sum of its any two sides is_________than the third side.

22.

The bisector of a line segment, divide the line segment in two___________parts.

True or False

3 x 1 = 3

23.

A right angled triangle can be constructed, if the given angles are 90°, 60° and 70°.

(a)

True

(b)

False

24.

In a right angled triangle, the square of hypotenuse is greater than the sum of square of base and perpendicular length.

(a)

True

(b)

False

25.

The distance between the two parallel lines is the same everywhere.

(a)

False

(b)

True

1 Marks

5 x 1 = 5

26.

Can 45°,60°, and 40° be angles of an acute angled triangle?

27.

What is the relation between the difference of any two sides and the third side?

28.

Write the angle measures of an isosceles right angled triangle.

29.

Can 45°, 60° and 40° be the angles of an acute angled mangle?

30.

Can we draw a mangle having two right angles?

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Answers Key:

Multiple Choice Question

5 x 1 = 5

(b)
greater than the third side

(d)
60°

(b)
2 cm, 3 cm, 4 cm

Generic 3

4 x 3 = 12

Steps of construction:
I. Draw a line segment BC = 4.5 cm.
II. Construct \(\angle\)CBX = 60° at B.
III. From BX, cut off line segment BA = 5.5 cm,
IV. Join AC
Thus, ABC is the required triangle.

In the given figure, angle made by an arc is equal to 60°.
∴ The sum of all three angles in a triangle is equal to 180°.
So, 60° + 60° + x = 180° \(\Rightarrow\) 120° + x = 180°
\(\Rightarrow\) x = 180° - 120° \(\Rightarrow\) x = 60°

In ΔABC, we have mㄥA = 70°, mㄥB = 50° and AC=3 cm.
Here, two angles and one side AC are given.
So, to construct the triangle, we need mㄥC
By angle sum property of a triangle,
mㄥA + mㄥB + mㄥC = 180°
70° + 50° + mㄥC = 180° ⇒ 120° + mㄥC = 180°
⇒ mㄥC = 180° -120° = 60°

No, we cannot draw any other line through A that would be parallel to given line I.
Through A, only a unique line can be drawn parallel to I because there exists a unique pair of alternate interior angles, \(\angle ABC,\angle BAH,\) such that \(\angle BAH=\angle ABC\).

Generic 2

6 x 2 = 12

10.

11.

12.

Angle made by perpendicular bisector on a ray is equal to 90°.

13.

In ΔABC, we have
BC = 2 cm, AB = 4 cm and AC = 2 cm
We observe that BC + AC = 2 + 2 cm = 4 cm = AB,
i.e. the sum of two sides of MBC is equal to the third side.
But we know that, in a triangle, the sum of the lengths of any two sides is always greater than the third side.
So, ΔABC with given measures cannot be constructed.

14.

In ΔABC, we have,
mㄥA = 85°, mㄥB = 115° and AB = 5 cm
Here, mㄥA + mㄥB = 85° + 115° = 200° > 180°
which is not possible because the sum of all three angles of a triangle should be 180°.
So, ΔABC cannot be constructed.

15.

Yes,we can draw ΔABC.
Here, the side AC, ㄥA and ㄥB of ΔABC are given. But to draw the triangle, we required ㄥC.
In ΔABC, by angle sum property, we have
ㄥA + ㄥB + ㄥC = 180° ⇒ 60° + 50° + ㄥC = 180°
⇒ 110° + ㄥC = 180° ⇒ ㄥC = 180° -110° = 70°
Now, we have side AC, m ⇒ A and mㄥC.
So, we can draw MBC, by ASA criterion.

Generic 5

2 x 5 = 10

16.

Steps of construction
(i) Draw a line AB and mark a point C on it.
(ii) Construct an angle equal to 90° at C and draw CX at C.
(iii) Mark a point M on CX such that CM = 3 cm.
(iv) At M, construct an angle equal to 90° to draw a line PQ perpendicular to CX.

Hence, PQ is the required line which is 3 cm from AB and parallel to AB.

17.

Given, two sides and an angle of ΔPQR are QR = 8 cm, PR = 10 cm and mㄥQ = 90°.
To construct a triangle with these two sides and one right angle, we use the following steps:
Steps of construction
Step I Firstly, we draw a rough sketch with measures marked on it.

Step II Draw a line segment QR = 8 cm.

Step III At point Q, draw QX 丄 QR.

Step IV With R as centre and radius 10 cm, draw an arc which intersects ray QX at P.

Step V Join PR.

Thus, ΔPQR is the required triangle.