# Free Practical Geometry Subjective Test 02 Practice Test - 7th grade

How many right angles are there in a cubical room?  [1 MARK]

#### SOLUTION

Solution :

Each corner has three mutually perpendicular lines and a cuboid has 8 corners. Hence, the total number of right angles in a cubical room is 8×3=24.

What is the criterion of SAS and ASA congruency?  [2 MARKS]

#### SOLUTION

Solution :

Each part: 1 Mark

For SAS, the criterion is as follows:

a) Lengths of any two sides.

b) The measure of the angle between these sides.

For ASA, the criterion are as follows

a) Measures of any two angles.

b) Lengths of the side between these angles.

(a) In the figure shown below, what are the conditions on NOB and MPC that assure that line segment AB is parallel to the line segment CD?  [2 MARKS] (b) Given line 'm' is parallel to line 'l'. Find x?

#### SOLUTION

Solution :

Each part: 1 Mark

(a) For the lines to be parallel, NOB  has to be equal to MPC (Alternate Interior Angles)

(b)  If two parallel lines are cut by a transversal then the corresponding angles are equal.

So, x=35.

(a) For constructing a triangle, we must know two sides and included angle. Which triangle can be uniquely constructed by knowing two sides and any other angle?
(b) Which of the following congruency conditions is preferred to draw a triangle exactly congruent to a right-angled triangle? [2 MARKS]

#### SOLUTION

Solution :

Each part: 1 Mark

(a) We only need to know R (Right angle), H (Hypotenuse) and S (Side) to construct a right-angled triangle. The right angle is not included between the hypotenuse and the other side.

(b) If the hypotenuse and any one leg of one right-angled triangle are equal to the corresponding hypotenuse and leg of another right-angled triangle, the two triangles are congruent.

It means that if we have two right-angled triangles with the same length of the hypotenuse and the same length for one of the other two legs, then both right triangles are identical.

Check if a triangle with sides 5 cm, 19 cm and 20 cm can be constructed.  [2 MARKS]

#### SOLUTION

Solution : Steps: 1 Mark

Adding up the sides, taking two at a time, we get

5 cm + 19 cm = 24 cm > 20 cm

5 cm + 20 cm = 25 cm > 19 cm

19 cm + 20 cm = 39 cm > 5 cm

So we see that the sum of the pair of sides when added is greater than the third side.

Therefore, we can construct the triangle with these measurements.

Construct a triangle PQR, given that PQ = 5 cm and QPR=60and QRP=45.  [3 MARKS]

#### SOLUTION

Solution :

Construction: 1 Mark
Steps: 2 Marks

Draw a line segment PQ of length 5 cm.

At P, draw a ray PX by making an angle of 60 with PQ (according to the given condition, R lies on ray PX).

At Q, draw a ray QY by making an angle of 45 with PQ (according to the given condition, R also lies on ray QY).

The intersection point of ray PX and QY is the point where R lies. Draw a line l. Then draw a perpendicular on l at any point. On this perpendicular, choose a point X, 5.5 cm away from l. Through X, draw a line m parallel to l.  [3 MARKS]

#### SOLUTION

Solution :

Construction: 3 Marks

Draw a line l.

Choose any point on l and draw a perpendicular at that point.

On the perpendicular cut a point X which is 5.5cm away from l.

Then through X construct a line parallel to l. Construct a right-angled triangle ABC such that side ¯¯¯¯¯¯¯¯BC=48 cm and hypotenuse ¯¯¯¯¯¯¯¯AC=50 cm.  [3 MARKS]

#### SOLUTION

Solution : Constructions: 2 Marks
Steps: 1 Mark

Step 1: Draw a line segment ¯¯¯¯¯¯¯¯BC=48 cm Step 2: Draw CBD=90 at B using protractor or compass.

Step 3: With C as centre and radius equal to 50 cm, draw an arc to cut the ray ¯¯¯¯¯¯¯¯¯BD at A and join ¯¯¯¯¯¯¯¯AC

ABC is the required triangle.

Construct ΔPQR, given ¯¯¯¯¯¯¯¯PQ=6.5 cm,¯¯¯¯¯¯¯¯PR=7 cm, and ¯¯¯¯¯¯¯¯¯QR=8 cm  [3 MARKS]

#### SOLUTION

Solution : Construction: 2 Marks
Steps: 1 Mark

1) Draw ¯¯¯¯¯¯¯¯PQ=6.5 cm

2) With P as centre, draw an arc with radius equal to 7 cm.

3) With Q as centre, draw another arc with radius 8 cm to cut the previous arc at R.

4) Join R to P and Q. PQR is the required triangle. Construct a triangle whose sides are 13 cm, 12 cm and 5 cm.  [4 MARKS]

#### SOLUTION

Solution :

Construction: 1 Mark
Steps: 3 Marks

Let assume that AB = 5, BC = 12 and AC = 13

1) Draw a line segment BC of length 12 cm.

2) From B, point A is at a distance of 5 cm. So, with B as the centre, draw an arc of radius 5 cm. (Now A will be somewhere on this arc.)

3) From C, point A is at a distance of 13 cm. So, with C as the centre, draw an arc of radius 13 cm. (A will be somewhere on this arc)

4) A has to be on both the arcs drawn. So, it is the point of intersection of arcs. Mark the point of intersection of arcs as A. Join AB and AC. ΔABC is now ready. Construct a triangle PQR, given that PQ = 5 cm, QR = 8 cm and PQR=45.  [4 MARKS]

#### SOLUTION

Solution :

Construction: 1 Mark
Steps: 3 Marks

1) Draw a line segment QR of length 8 cm

2) At Q, Draw QX making 45 with QR.(The point P must be somewhere on this line)

3) The distance QP has been given. So cut an arc of 5 cm taking Q as a centre.

4) Join the points P and R, you will get triangle PQR. (a) Construct a right-angled triangle with hypotenuse 7.6 cm and one of the sides 5.2 cm.

(b)
Construct ΔABC, where  AB = 6 cm, BC = 7 cm and CA = 9 cm. Is it a right angled triangle? [4 MARKS]

#### SOLUTION

Solution :

Each part: 2 Marks

(a) Steps of construction:

Step 1: Assume base of the triangle AB = 5.2 cm.

Step 2: Draw a perpendicular on AB at point A, Extend the ray AP.

Step 3: From point B, take an arc of radius 7.6 cm, taking B as a center.

Step 4: The point where the arc and ray intersect is point C. (b) Steps of construction:

Step 1: Draw line segment BC = 7 cm.

Step 2: Draw an arc with B as the centre and the radius equal to 6 cm.

Step 3: Draw an arc with C as the centre and the radius equal to 9 cm.

Step 4: Name the point of intersection of these two arcs as A.

Step 5: Join points A and B, and points A and C.

Triangle ABC is the required triangle. The triangle is not a right-angled triangle. It can be checked by a protractor and also by using Pythagoras Theorem. The sides do not satisfy the theorem.

(a) Draw a line l. Draw a perpendicular on l at any point. On this perpendicular, choose a point X, 4 cm away from l. Through X, draw a line m parallel to l.

(b) Line l and m are parallel to each other. A transversal intersects them at points A and B. i) An arc of suitable measure is taken from B cutting ‘l’ at D and transversal at M (as shown in the figure). ii) Now, with the same radius and A as the centre an arc FG is drawn (as shown in the figure). What is the relation between the length of arc DM and arc FG?
[4 MARKS]

#### SOLUTION

Solution : Each part: 2 Marks

(a) To construct: A line parallel to given line when the perpendicular line is also given. Draw a line l and take a point P on it.

At point P, draw a perpendicular line n.

Take PX = 4 cm on line n.

At point X, again draw the line m.

(b) Since both the angles are alternate interior angles, they will be equal to each other. Also, since both arcs are drawn with the same radius, they are part of two equal circles. This means that they both have same shape and size. Thus, the two arcs DM and FG are congruent which makes their lengths equal.

Draw a line, say AB. Take a point C outside it. Through C, draw a line parallel to AB using ruler and compass only.  [4 MARKS]

#### SOLUTION

Solution : Construction: 2 Marks
Steps: 2 Marks To construct: A line parallel to given line by using ruler and compass.

Steps of construction:

1) Draw a line segment AB and take a point C outside AB.

2) Take any point D on AB and join C to D.

3) With D as the centre and taking convenient radius, draw an arc cutting AB at E and CD at F.

4) With C as the centre and same radius as in step 3, draw an arc GH cutting CD at I.

5) With the same arc EF, draw the equal arc cutting GH at J.

6) Join JC to draw a line l.

Check if the following triangles can be constructed.

a) 80 cm, 5 cm, 5 cm

b) 13 cm, 6 cm, 7 cm  [4 MARKS]

#### SOLUTION

Solution : Each part: 2 Marks

(a) Adding up the sides by taking two at a time, we get:

5 cm + 5 cm = 10 cm < 80 cm

80 cm + 5 cm = 85 cm > 5 cm

So we see that the sum of one pair of sides when added is not greater than the third side.

Therefore, we cannot draw a triangle with these measurements.

(b) Adding up the sides by taking two at a time, we get

13 cm + 6 cm = 19 cm > 7 cm

13 cm + 7 cm = 20 cm > 6 cm

6 cm + 7 cm = 13 cm = 13 cm

So in each case, the sum of two sides is equal to the third side.

Hence, this triangle cannot be constructed.