# Free Practical Geometry Subjective Test 01 Practice Test - 7th grade

### Question 1

A triangle is constructed by joining the minute's hand position when it was showing 12 and 3. Find the angle between the positions and what type of triangle is it? [1 MARK]

#### SOLUTION

Solution :Angle: 0.5 Mark

Type: 0.5 Mark

The angle between the two hands is 90 degrees.

It is an isosceles right-angled triangle.

### Question 2

Given a line l and a point M on it, draw a perpendicular MP to l where MP = 5.2 cm and a line q parallel to l through P. [2 MARKS]

#### SOLUTION

Solution :Steps: 1 Mark

Construction: 1 Mark

Steps of construction:

(i) Draw a line l.

(ii) Take a point M on it.

(iii) Draw an angle of 90∘ at M with l which is perpendicular to l at M.

(iv) With M as centre and radius 5.2 cm, draw an arc which intersects the above perpendicular at point P. MP is the required perpendicular.

(v) At P, draw an angle of 90∘ with PM and make a line q.

Line q is the required line parallel to line l.

### Question 3

The sides of the triangle are 3 m, 5 m and 9 m. Construct a triangle. [2 MARKS]

#### SOLUTION

Solution :Correct answer: 2 Marks

With the given lengths, a triangle cannot be constructed because the sum of two sides of a triangle is less than the third side.

3 + 5 < 9

5 + 9 > 3

3 + 9 > 5

### Question 4

Construct a triangle whose sides are AB = 5 m, BC = 9 m and ∠CAB=60∘. [2 MARKS]

#### SOLUTION

Solution :Steps: 1 Mark

Construction: 1 Mark

Draw a line segment AB = 5 m

Draw AC making an angle of 60 degrees with AB.

Taking B as centre, draw a circle of radius 9 m.

The point where the circle meets the line will be the other point of the triangle.

### Question 5

Construct a triangle ABC, where ∠BAC=∠ACB. Let D be any point on the line AB, such that the ∠CBD=120∘, AB = 3 m. [2 MARKS]

#### SOLUTION

Solution :Steps: 1 Mark

Construction: 1 Mark

Draw a line segment AB = 3 m and extend it.

Sum of two interior opposite angles is equal to the exterior angle.

∠BAC+∠ACB=120∘

Hence, ∠ABC=60∘

Hence, the given triangle is an equilateral triangle.

Draw AB = 3 m

Draw a line AC of length 3 m making an angle of 60 degrees with AB. Join BC.

### Question 6

A man runs in the east direction and suddenly changes his direction towards the north. After running for some time, he comes back to his original position. Given that the man's path of running is a triangle and one of its angles is 90 degrees, find the sum of the other two angles. [3 MARKS]

#### SOLUTION

Solution :Properties: 1 Mark

Sum: 2 Marks

The given conditions make a right-angled triangle because one of its angles is90∘

Using the angle sum property of a triangle,

∠O+∠A+∠B=180∘

∠O+90∘+∠B=180∘

∠O+∠B=90∘

Sum of the other two angles will be 90∘

### Question 7

Construct a triangle, given that the lengths of its two sides are 3 m and 8 m and the third side makes an angle of 60 degrees with the 3 m side. [3 MARKS]

#### SOLUTION

Solution :Steps: 1 Mark

Construction: 2 Marks

Draw a line with given length of 3 m, let it be AB.

Draw a line making 60 degrees with the line AB, let it be AC.

With B as centre, draw a circle of radius 8 m.

The drawn arc meets the line AC at c.

Join all the points and it will form a triangle with side 3 m and 8 m with an angle of 60∘.

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### Question 8

Construct a triangle whose side length AB is 5 m and ∠BAC and ∠CBA are 45∘ and 50∘ respectively. [3 MARKS]

#### SOLUTION

Solution :Steps: 1 Mark

Construction: 2 Marks

Draw a line segment AB = 5 m

Draw AC making an angle of 45∘ with AB.

Draw BC making an angle of 50∘ with BA.

The point where the two lines meet is C.

### Question 9

Construct a right-angled triangle whose hypotenuse is 5 m and one of its lengths is 3 m. [3 MARKS]

#### SOLUTION

Solution :Steps: 2 Marks

Construction: 1 Mark

Given that it is a right-angled triangle.

Sum of the squares of the two sides of the triangle is equal to the square of the hypotenuse

Let the length of the other side be X m

32+X2=52

X = 4

Draw a line segment AB = 3 m

Draw a line segment AC of length 4 m perpendicular to AB

Join BC.

### Question 10

Which of the triangles whose three sides are given below can be constructed?

(a) 9 m, 5 m, 3 m

(b) 10 cm, 10 cm, 4 cm [4 MARKS]

#### SOLUTION

Solution :Each part: 2 Marks

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

9 m + 5 m = 14 m > 3 m

9 m + 3 m = 12 m > 5 m

but, 5 m + 3 m = 8 m < 9 m

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, taking two at a time, we get

10 cm + 10 cm = 20 cm > 4 cm

10 cm + 4 cm = 14 cm > 10 cm

4 cm + 10 cm = 14 cm > 10 cm

So in each case, the sum of two sides is greater than the third side.

Hence, this triangle can be constructed.

### Question 11

(a) Construct ΔPQR, given ¯¯¯¯¯¯¯¯PQ=3.5 cm,¯¯¯¯¯¯¯¯PR=4.5 cm, and ¯¯¯¯¯¯¯¯¯QR=5.5 cm

(b) Construct ΔPQR, when PQ = 4 cm, QR = 6 cm and ∠PQR = 60∘. [4 MARKS]

#### SOLUTION

Solution :Each part: 2 Marks

(a) Steps of construction:

Draw ¯¯¯¯¯¯¯¯PQ=3.5 cm

With P as the centre, draw an arc with the radius equal to 4.5 cm.

With Q as the centre, draw another arc with radius 5.5 cm to cut the previous arc at R.

Join R to P and Q.

PQR is the required triangle.

Thus, we see that if three sides of a triangle are given and the sum of any two sides is always more than the third side, then a triangle can be constructed. This is known as SSS criterion for construction of triangles.

(b) Steps of construction:

Step 1: Draw a line segment QR = 6 cm.

Step 2: Construct an angle of 60∘ at point Q.

Step 3: Draw an arc on the ray QX with Q as the centre and the radius equal to 4 cm.

Step 4: Name the point where the arc cuts the ray QX, as P.

Step 5: Join points P and R.

PQR is the required triangle.

### Question 12

Construct ΔPQR, given ¯¯¯¯¯¯¯¯PQ=3.5 cm,¯¯¯¯¯¯¯¯PR=3 cm, and ∠RPQ=120∘ [4 MARKS]

#### SOLUTION

Solution :Construction: 2 Marks

Steps: 2 Marks

Draw ¯¯¯¯¯¯¯¯PQ=3.5 cm

Construct ∠XPQ=120∘ with your compass.

With P as centre, draw an arc with radius equal to 3 cm that intersects ¯¯¯¯¯¯¯¯¯XP at point R.

Join ¯¯¯¯¯¯¯¯¯RQ

PQR is the required triangle.

Thus, we see that if two sides of a triangle are given and the measure of the angle between these two sides is also given, then we can construct a triangle. This is known as SAS criterion for construction of triangles.

### Question 13

Construct ΔPQR, given ∠RPQ=30∘,¯¯¯¯¯¯¯¯PQ=4.5 cm, and ∠PQR=45∘ [4 MARKS]

#### SOLUTION

Solution :Construction: 2 Marks

Steps: 2 Marks

Step 1: Draw ¯¯¯¯¯¯¯¯PQ=4.5 cm

Step 2: Construct ∠XPQ=30∘ and ∠PQY=45∘

Step 3: Mark the point of intersection of ¯¯¯¯¯¯¯¯¯XP and ¯¯¯¯¯¯¯¯¯YQ as R.

PQR is the required triangle.

Thus, we see that if two angles of a triangle and the side included between the two are given, then we can draw a triangle. This is known as ASA criterion of construction of triangles.

### Question 14

(a) Construct a right-angled triangle ABC such that side ¯¯¯¯¯¯¯¯BC=8 cm and hypotenuse ¯¯¯¯¯¯¯¯AC=10 cm.

(b) Draw a right-angled triangle at C in which AB = 7 cm and BC = 5 cm. [4 MARKS]

#### SOLUTION

Solution :Each part: 2 Marks

(a) Draw a line segment ¯¯¯¯¯¯¯¯BC=8 cm

Draw ∠CBD=90∘ at B using compass.

With C as centre and radius equal to 10 cm, draw an arc to cut the ray ¯¯¯¯¯¯¯¯¯BD at A and join ¯¯¯¯¯¯¯¯AC.

ABC is the required triangle.

(b) Steps of Construction:(i) Draw BC = 5 cm.

(ii) Construct ∠BCX = 90∘.

(iii) With B as centre and radius 7 cm, cut BA at 7 cm.

(iv) Join AB.

Thus, ABC is the required triangle.

### Question 15

Let l be a line and P be a point not on l. Through P, draw a line m parallel to l. Now join P to any point Q on l. Choose any other point R on m. Through R, draw a line parallel to PQ. Let this meet l at S. What shape do the two sets of parallel lines enclose? [4 MARKS]

#### SOLUTION

Solution :Steps: 2 Marks

Construction: 2 Marks

To construct: A pair of parallel lines intersecting other parts of parallel lines.

Steps of construction:

Draw a line l and take a point P outside l.

Take point Q on line l and join PQ.

Make equal angle at point P such that ∠Q=∠P.

Extend line at P to get line m.

Similarly, taking a point R on line m, at point R, draw angles such that ∠P=∠R.

The extended line at R which intersects at S on line l, Draw line RS. Thus, we get a parallelogram PQRS.