CBSE Class 7 Mathematics Practical Geometry Assignment Set 02

VII - Mathematics Assignment No - 03 - Practical Geometry

Q1. Construct a triangle ABC in which BC = 4 cm, AB = 5 cm, and AC = 4.5 cm.
Answer: Steps of construction:
1. Draw a line segment BC of length 4 cm.
2. With B as center and radius 5 cm, draw an arc.
3. With C as center and radius 4.5 cm, draw another arc to intersect the previous arc at point A.
4. Join AB and AC. Triangle ABC is the required triangle.
In simple words: Draw the base first, then use a compass with the given side lengths to find the third vertex.

Exam Tip: Always use a sharp pencil and a well-calibrated ruler for accurate construction of sides.

Q2. Construct a triangle PQR in which PQ = PR = 5 cm and QR = 6 cm.
Answer: Steps of construction:
1. Draw a line segment QR of length 6 cm.
2. With Q as center and radius 5 cm, draw an arc.
3. With R as center and radius 5 cm, draw another arc to intersect the previous arc at point P.
4. Join PQ and PR. Triangle PQR is the required isosceles triangle.
In simple words: Draw the base QR and make equal arcs from both ends with radius 5 cm to locate the top vertex P.

Exam Tip: Since two sides are equal (PQ = PR), this is an isosceles triangle.

Q3. Construct a triangle ABC in which BC = 7.2 cm, AC = 6 cm, and angle C = 120 degrees.
Answer: Steps of construction:
1. Draw a line segment BC of length 7.2 cm.
2. At point C, construct an angle of \(120^\circ\) using a protractor or compass.
3. Along the terminal ray of the angle, cut off an arc of length AC = 6 cm.
4. Join AB. Triangle ABC is the required triangle.
In simple words: Draw the base, construct an obtuse angle at one end, measure the given side along that ray, and connect the ends.

Exam Tip: Ensure the angle is measured properly from the baseline; a 120 degree angle opens wide towards the interior.

Q4. Construct a triangle ABC such that AB = 5 cm, AC = 4.8 cm, and angle A = 90 degrees.
Answer: Steps of construction:
1. Draw a line segment AB of length 5 cm.
2. At point A, construct a right angle (\(90^\circ\)) using a compass or protractor.
3. Along the ray from A, cut off an arc of length AC = 4.8 cm.
4. Join BC. Triangle ABC is the required right-angled triangle.
In simple words: Form a perpendicular at the vertex A between the two given sides and connect the third side.

Exam Tip: A 90-degree angle can be constructed by bisecting the angle between a \(60^\circ\) and a \(120^\circ\) arc.

Q5. Construct a triangle ABC in which AB = AC = 5.3 cm and angle A = 90 degrees.
Answer: Steps of construction:
1. Draw a line segment AB of length 5.3 cm.
2. At point A, construct a right angle (\(90^\circ\)).
3. Along the perpendicular ray from A, cut off an arc AC = 5.3 cm.
4. Join BC. Triangle ABC is the required isosceles right-angled triangle.
In simple words: Draw one side, make a 90-degree corner, measure an equal length on the perpendicular arm, and complete the triangle.

Exam Tip: This is a special isosceles right-angled triangle where the two perpendicular sides are equal.

Q6. Construct a triangle ABC such that BC = 6.4 cm, angle B = 45 degrees, and angle C = 60 degrees.
Answer: Steps of construction:
1. Draw a line segment BC of length 6.4 cm.
2. At point B, construct an angle of \(45^\circ\).
3. At point C, construct an angle of \(60^\circ\).
4. The rays of these two angles intersect at point A. Triangle ABC is the required triangle.
In simple words: Draw the base and make the two given angles at each end point; where their arms meet is the third vertex.

Exam Tip: Make sure your angle rays are drawn long enough so that they clearly intersect each other.

Q7. Construct an equilateral triangle PQR whose each side is 5 cm.
Answer: Steps of construction:
1. Draw a line segment QR of length 5 cm.
2. With Q as center and radius 5 cm, draw an arc.
3. With R as center and radius 5 cm, draw another arc intersecting the first arc at point P.
4. Join PQ and PR. Triangle PQR is the required equilateral triangle.
In simple words: All three sides are equal, so use the same compass opening of 5 cm from both base ends to find the top vertex.

Exam Tip: In an equilateral triangle, all interior angles are automatically \(60^\circ\).

Q8. Construct an isosceles triangle PQR such that QR = 5 cm and each base angle is 65 degrees.
Answer: Steps of construction:
1. Draw a line segment QR of length 5 cm.
2. At point Q, construct an angle of \(65^\circ\) using a protractor.
3. At point R, construct an angle of \(65^\circ\) using a protractor.
4. The intersection of these two angle rays is point P. Triangle PQR is the required isosceles triangle.
In simple words: Draw the base and construct equal angles of 65 degrees at both base vertices.

Exam Tip: Equal base angles guarantee equal opposite sides, forming an isosceles triangle.

Q9. Construct a triangle ABC whose sides are 3, 4, 5 cm. State the type of triangle.
Answer: Steps of construction:
1. Draw a line segment BC of length 4 cm (or any base side, e.g., 5 cm). Let BC = 5 cm.
2. With B as center and radius 3 cm, draw an arc.
3. With C as center and radius 4 cm, draw an arc intersecting at point A.
4. Join AB and AC.
Type of triangle: Since \(3^2 + 4^2 = 9 + 16 = 25 = 5^2\), it satisfies Pythagoras theorem. Therefore, it is a right-angled triangle.
In simple words: Construct the triangle using side lengths 3, 4, and 5 cm. It is a right-angled triangle because the sum of squares of the two smaller sides equals the square of the longest side.

Exam Tip: A triangle with sides 3, 4, and 5 cm is the classic Pythagorean triplet right triangle.

Q10. Is it possible to construct a triangle ABC of sides 4 cm, 5 cm, 10 cm? If not, why not?
Answer: No, it is not possible to construct such a triangle.
Reason: According to the triangle inequality property, the sum of the lengths of any two sides of a triangle must always be greater than the length of the third side.
Here, taking the two smaller sides: \(4 + 5 = 9\), which is less than the third side \((10)\). Since \(9 < 10\), the arcs drawn from the endpoints of the 10 cm base with radii 4 cm and 5 cm will never intersect.
In simple words: No, because the sum of any two sides must be greater than the third side, but here \(4 + 5 = 9\), which is smaller than 10.

Exam Tip: Always check the triangle inequality condition (\(a + b > c\)) before attempting side-side-side constructions.