CBSE Class 7 Mathematics Practical Geometry Assignment Set 03

VII - Mathematics Assignment No-02 - Practical Geometry

Q1. Construct a right angled triangle ABC in which AB = 5.8 cm, BC = 4.5 cm and angle C = 90 degrees.
Answer:
Steps of construction:
1. Draw a line segment \(BC = 4.5\text{ cm}\).
2. At point \(C\), construct an angle of \(90^\circ\) using a protractor or compass and draw a ray \(CX\).
3. With center \(B\) and radius \(5.8\text{ cm}\), draw an arc to intersect the ray \(CX\) at point \(A\).
4. Join \(A\) to \(B\).
5. Triangle \(ABC\) is the required right-angled triangle.
In simple words: Draw a base line of 4.5 cm, make a 90-degree corner at point C, and cut an arc of 5.8 cm from point B to locate point A.

Exam Tip: Always use a sharp pencil for geometric constructions and label all vertices and measurements clearly.

Q2. Construct a triangle ABC in which angle A = 65 degrees, angle B = 75 degrees and BC = 4 cm.
Answer:
Steps of construction:
1. Find the third angle of the triangle using the angle sum property:
\(\text{Angle } C = 180^\circ - (65^\circ + 75^\circ) = 180^\circ - 140^\circ = 40^\circ\)
2. Draw a line segment \(BC = 4\text{ cm}\).
3. At point \(B\), construct an angle of \(75^\circ\) using a protractor.
4. At point \(C\), construct an angle of \(40^\circ\) using a protractor.
5. The rays of these angles intersect at point \(A\).
6. Triangle \(ABC\) is the required triangle.
In simple words: First calculate the missing angle C as 40 degrees, then draw the base 4 cm and construct the given angles at the ends.

Exam Tip: Always calculate the missing angle first if the two base angles provided are not adjacent to the given side.

Q3. Construct a triangle ABC in which BC = 5.1 cm, angle B = 50 degrees and angle C = 45 degrees.
Answer:
Steps of construction:
1. Draw a line segment \(BC = 5.1\text{ cm}\).
2. At point \(B\), construct an angle of \(50^\circ\) using a protractor and draw ray \(BX\).
3. At point \(C\), construct an angle of \(45^\circ\) using a protractor and draw ray \(CY\).
4. Mark the point of intersection of rays \(BX\) and \(CY\) as \(A\).
5. Triangle \(ABC\) is the required triangle.
In simple words: Draw the 5.1 cm base line BC, make a 50-degree angle at B and a 45-degree angle at C, and extend them until they meet at point A.

Exam Tip: Ensure your protractor is centered precisely on the vertex point when measuring angles.

Q4. Draw a triangle ABC such that AB = 5 cm, AC = 4.3 cm and angle A = 70 degrees.
Answer:
Steps of construction:
1. Draw a line segment \(AB = 5\text{ cm}\).
2. At point \(A\), construct an angle of \(70^\circ\) using a protractor and draw a ray \(AX\).
3. With center \(A\) and radius \(4.3\text{ cm}\), cut an arc on ray \(AX\) and label the intersection point as \(C\).
4. Join \(B\) to \(C\).
5. Triangle \(ABC\) is the required triangle.
In simple words: Draw side AB of 5 cm, make a 70-degree angle at vertex A, and cut off 4.3 cm along that ray to get point C, then connect B to C.

Exam Tip: When two sides and the included angle are given, always draw the included angle at the vertex common to both sides.

Q5. Construct an equilateral triangle ABC whose each side is 4.5 cm.
Answer:
Steps of construction:
1. Draw a line segment \(BC = 4.5\text{ cm}\).
2. With center \(B\) and radius \(4.5\text{ cm}\), draw an arc.
3. With center \(C\) and radius \(4.5\text{ cm}\), draw another arc to intersect the previous arc at point \(A\).
4. Join \(A\) to \(B\) and \(A\) to \(C\).
5. Triangle \(ABC\) is the required equilateral triangle.
In simple words: Draw a base of 4.5 cm and use a compass opened to 4.5 cm to draw intersecting arcs from both ends to find the top vertex.

Exam Tip: An equilateral triangle has all interior angles equal to 60 degrees, which can be used for verification.

Q6. Construct an isosceles triangle ABC whose base is 5 cm and vertical angle A = 40 degrees.
Answer:
Steps of construction:
1. Draw base \(BC = 5\text{ cm}\).
2. Calculate the base angles. Since triangle is isosceles with vertex angle \(A = 40^\circ\), the sum of base angles is \(180^\circ - 40^\circ = 140^\circ\).
3. Each base angle (\(\text{angle } B\) and \(\text{angle } C\)) is \(\frac{140^\circ}{2} = 70^\circ\).
4. At point \(B\), construct an angle of \(70^\circ\).
5. At point \(C\), construct an angle of \(70^\circ\).
6. The rays intersect at point \(A\). Triangle \(ABC\) is the required triangle.
In simple words: Find the two base angles as 70 degrees each by subtracting 40 from 180 and dividing by 2, then construct them on the 5 cm base.

Exam Tip: Vertical angle is the angle between the two equal sides of an isosceles triangle.

Q7. Construct a triangle PQR such that the side QR = 6 cm, PQ = 5 cm, PR = 4 cm.
Answer:
Steps of construction:
1. Draw a line segment \(QR = 6\text{ cm}\).
2. With center \(Q\) and radius \(5\text{ cm}\), draw an arc.
3. With center \(R\) and radius \(4\text{ cm}\), draw another arc to intersect the first arc at point \(P\).
4. Join \(P\) to \(Q\) and \(P\) to \(R\).
5. Triangle \(PQR\) is the required triangle.
In simple words: Draw base QR of 6 cm, then use compass radii of 5 cm from Q and 4 cm from R to find the intersecting point P.

Exam Tip: This is a SSS (Side-Side-Side) construction. Always verify that the sum of any two sides is greater than the third side.

Q8. Construct a triangle DEF in which side EF = 5 cm, angle DEF = 60 degrees, angle DFE = 30 degrees.
Answer:
Steps of construction:
1. Draw a line segment \(EF = 5\text{ cm}\).
2. At point \(E\), construct an angle of \(60^\circ\) using a protractor and draw ray \(EX\).
3. At point \(F\), construct an angle of \(30^\circ\) using a protractor and draw ray \(FY\).
4. Mark the intersection of rays \(EX\) and \(FY\) as \(D\).
5. Triangle \(DEF\) is the required triangle.
In simple words: Draw the 5 cm base EF, make a 60-degree angle at E and a 30-degree angle at F, and extend them to meet at point D.

Exam Tip: This is an ASA (Angle-Side-Angle) construction where the side is included between the two given angles.

Q9. Construct a triangle GHI such that HI = 5 cm, angle GHI = 45 degrees, GH = 4 cm.
Answer:
Steps of construction:
1. Draw a line segment \(HI = 5\text{ cm}\).
2. At point \(H\), construct an angle of \(45^\circ\) using a protractor and draw ray \(HX\).
3. With center \(H\) and radius \(4\text{ cm}\), cut an arc on ray \(HX\) and label the point as \(G\).
4. Join \(G\) to \(I\).
5. Triangle \(GHI\) is the required triangle.
In simple words: Draw base HI of 5 cm, construct a 45-degree angle at H, measure 4 cm along that ray to locate vertex G, and join G to I.

Exam Tip: This is a SAS (Side-Angle-Side) construction. The angle must be the one included between the two given sides.

Q10. Construct a triangle LMN in which MN = 5 cm, angle LMN = 60 degrees, angle MLN = 45 degrees.
Answer:
Steps of construction:
1. First, find the missing angle \(\text{angle } MNL\) using the angle sum property:
\(\text{angle } MNL = 180^\circ - (60^\circ + 45^\circ) = 180^\circ - 105^\circ = 75^\circ\)
2. Draw a line segment \(MN = 5\text{ cm}\).
3. At point \(M\), construct an angle of \(60^\circ\) using a protractor.
4. At point \(N\), construct an angle of \(75^\circ\) using a protractor.
5. The rays intersect at point \(L\). Triangle \(LMN\) is the required triangle.
In simple words: Calculate angle N as 75 degrees first, then draw the base MN of 5 cm and construct 60 degrees at M and 75 degrees at N.

Exam Tip: Always make sure the sum of all three angles inside the triangle equals 180 degrees before constructing.