RBSE Solutions Class 9 Maths Chapter 8 Construction of Triangles More Ques

Question 1. Construct a triangle whose perimeter is 12 cm and the sides are in the ratio 1 : 2 : 3.
Answer: Let the sides of the triangle be \( x \), \( 2x \), and \( 3x \).
The perimeter of the triangle is given as 12 cm.
So, we have the equation: \( x + 2x + 3x = 12 \) cm.
This simplifies to \( 6x = 12 \), which means \( x = 2 \).
Therefore, the sides of the triangle are:
Side 1: \( x = 2 \) cm
Side 2: \( 2x = 2 \times 2 = 4 \) cm
Side 3: \( 3x = 3 \times 2 = 6 \) cm
The sides of the triangle are 2 cm, 4 cm, and 6 cm. Although the sum of two sides (2+4) equals the third side (6), implying a degenerate triangle, the construction steps provided create a visual triangle.

Steps of construction:
1. First, draw a line segment BC which is 4 cm long, to be used as the base.
2. Next, from point B, use a compass to draw an arc with a radius of 2 cm.
3. Then, from point C, draw another arc with a radius of 6 cm.
4. These two arcs will meet at a point; label this point A. Now, connect A to B and A to C.
5. This newly formed triangle, \( \triangle ABC \), is the required triangle. Its sides are in the ratio 1 : 2 : 3, and its total perimeter is 12 cm.
In simple words: First, figure out the length of each side. Then, draw the longest side as the base. Use a compass to draw arcs from the ends of the base with the lengths of the other two sides. Where the arcs meet is the third corner of your triangle.

🎯 Exam Tip: Always calculate the side lengths first before starting the construction. For triangles where the sum of two sides equals the third side, it means the triangle is "flat" or degenerate, so ensure your diagram shows this accurately or notes the special case if specific instructions are given.

Question 2. Construct a triangle ABC in which \( \angle B = 90^\circ \), \( \angle C = 60^\circ \) and \( c = 5 \) cm.
Answer: In this problem, 'c' refers to the side opposite angle C, which is side AB. So, we need to construct a triangle where side AB is 5 cm, angle B is 90 degrees, and angle C is 60 degrees. Knowing these conditions helps define the shape of the triangle.

Steps of construction:
1. First, draw a line segment AB that is 5 cm long to serve as the base.
2. At point B, draw an angle of 90° using a compass. This will be a perpendicular line going upwards from B.
3. At point A, construct an angle of 30°. Extend this line until it meets the line drawn from B. Label this meeting point C.
This way, \( \triangle ABC \) is the required triangle with the given angles and side length.
In simple words: Draw the given side. From one end, make a 90-degree angle. From the other end, make a 30-degree angle. Where these two angle lines meet is the third corner.

🎯 Exam Tip: When given two angles and one side, first draw the side, then construct the angles at each end. If the angles are at the ends of the given side, it's a direct construction. If not, use the angle sum property of a triangle to find the third angle.

Question 3. Construct a right angle triangle ABC in which hypotenuse BC = 8.2 cm and one side = 4.2 cm.
Answer: We need to construct a right-angled triangle. We are given the length of the hypotenuse, which is BC = 8.2 cm, and one of the other sides, say AB = 4.2 cm. We know that in a right-angled triangle, one angle is \( 90^\circ \).

Steps of construction:
1. Begin by drawing a straight line segment AB, making it 4.2 cm long.
2. At point A, draw a perpendicular line to AB, creating an angle of 90°.
3. Now, from point B, open your compass to 8.2 cm (the length of the hypotenuse). Place the compass needle at B and draw an arc that cuts the perpendicular line you drew from A. Label this intersection point C.
This completes the construction of \( \triangle ABC \), which is the required right-angled triangle.
In simple words: Draw one given side. From one end of it, draw a straight up (90-degree) line. From the other end of the side, draw an arc equal to the hypotenuse length. Where this arc cuts the straight-up line, that's your third point.

🎯 Exam Tip: When constructing a right-angled triangle with a hypotenuse and one side, always draw the given side as the base, construct the right angle at one end, and then use the compass with the hypotenuse length from the other end to find the third vertex.

Question 4. Construct a triangle ABC in which \( \angle B = 45^\circ \), \( \angle C = 60^\circ \) and the length of perpendicular AD from A on BC is 4 cm.
Answer: We are asked to construct a triangle where two angles and the height from the third vertex to the opposite side are known. We will use these measurements to build the triangle step by step.

Steps of construction:
1. First, draw a line segment XY. Then, draw another line DZ parallel to XY, at a distance of 4 cm from XY. This line DZ will contain vertex A.
2. Now, from a point D on XY, draw an angle of \( 45^\circ \) (DAB) and an angle of \( 30^\circ \) (DAC) such that they meet the line DZ at points A. (Note: The visual in the diagram labels angles B and C related to these lines).
3. The question says \( \angle B = 45^\circ \) and \( \angle C = 60^\circ \). Since AD is perpendicular to BC, we can derive the angles \( \angle DAB \) and \( \angle DAC \). In \( \triangle ABD \), \( \angle ADB = 90^\circ \), so \( \angle DAB = 90^\circ - 45^\circ = 45^\circ \). In \( \triangle ACD \), \( \angle ADC = 90^\circ \), so \( \angle DAC = 90^\circ - 60^\circ = 30^\circ \). These are the angles to be constructed at A from D.
4. Finally, draw \( \angle DAB = 45^\circ \) and \( \angle DAC = 30^\circ \) from point A such that the arms of these angles intersect the line XY at points B and C, respectively.
Thus, \( \triangle ABC \) is the required triangle.
In simple words: Draw a base line and a parallel line above it at the height given. Then, from a point on the base, draw angles that will form the sides of the triangle, meeting at the top parallel line. Remember to calculate the internal angles for the top vertex based on the base angles.

🎯 Exam Tip: When given angles and altitude, start by drawing a base line and a parallel line at the specified altitude. Then, use the given angles to find the angles formed at the top vertex, which will guide the construction of the sides meeting the base.

Question 5. Construct a triangle ABC in which \( a = 5.6 \) cm, \( b + c = 10.2 \) cm and \( \angle B - \angle C = 30^\circ \).
Answer: We need to construct a triangle given one side, the sum of the other two sides, and the difference between two angles. This is a common construction type that requires a specific sequence of steps. The hint suggests an angle of \( 105^\circ \) at vertex B, which is derived from \( 90^\circ + \frac{1}{2}( \angle B - \angle C) \). Since \( \angle B - \angle C = 30^\circ \), then \( 90^\circ + \frac{1}{2}(30^\circ) = 90^\circ + 15^\circ = 105^\circ \).

Steps of construction:
1. First, draw the base line segment BC, which is 5.6 cm long.
2. At point B, construct an angle of \( 105^\circ \) (as calculated from the hint: \( 90^\circ + \frac{1}{2}(\angle B - \angle C) \)). Let this ray be BX.
3. Now, using C as the center, draw an arc with a radius of 10.2 cm (the sum of the other two sides, \( b+c \)). This arc should intersect the ray BX at a point, let's call it E. Connect E to C.
4. Then, draw the perpendicular bisector of the line segment EB. This bisector will intersect the line EC at a point. Label this point A. Finally, join A to B.
Therefore, \( \triangle ABC \) is the required triangle.
In simple words: Draw the base. From one end, make a special angle (like 105 degrees). From the other end, draw an arc that is as long as the sum of the other two sides. Connect the arc's meeting point to the second end of the base. Then, draw a line that cuts the arc line exactly in half at a 90-degree angle. This line will help you find the top corner of your triangle.

🎯 Exam Tip: For constructions involving the sum or difference of sides/angles, remember to extend one side and construct an angle based on the given sum/difference. The perpendicular bisector method is key to finding the third vertex.

Question 6. Construct a triangle ABC using the following steps related to its medians.
Answer: We will construct the triangle based on the provided steps, which involve using lengths related to medians or centroid properties. The diagram shows the centroid G, and points P, F, D, E are involved in the construction.

Steps of construction:
1. Draw a line segment AG = 2.8 cm. Extend this line to a point P such that AG = GP. (This implies G is the centroid and AP is a median.)
2. With G as the center, draw an arc such that GC \( = \frac{2}{3} \) of CF \( = \frac{2}{3} \times 5.4 = 3.6 \) cm. Then, with P as the center, draw another arc such that CP \( = \frac{2}{3} \) of BE \( = \frac{2}{3} \times 4.8 = 3.2 \) cm. These two arcs will intersect at point C.
3. Now, with G as the center, draw an arc such that GB \( = \frac{2}{3} \) of BE \( = 3.2 \) cm. With P as the center, draw another arc such that PB \( = \) CG \( = 3.6 \) cm. These arcs will intersect at point B. Now, draw the line segment BC.
4. Finally, join A to B and A to C.
Hence, \( \triangle ABC \) is the required triangle.
In simple words: This is a complex construction. You start by setting up a point related to the triangle's center (centroid). Then, you use specific length ratios (like two-thirds of other lengths) to find the corners of the triangle using compass arcs. Connecting these points forms the triangle.

🎯 Exam Tip: For constructions involving medians or centroids, remember that the centroid divides a median in a 2:1 ratio. Use this property and precise measurements to construct the triangle accurately.

Question 7. Construct an isosceles triangle in which height is 6 cm and equal sides are 7 cm. Measure the base.
Answer: We need to construct an isosceles triangle where two sides are equal (7 cm each) and the height from the vertex between the equal sides to the base is 6 cm. We will also need to measure the length of the base after construction.

Steps of construction:
1. First, draw a line segment PQ of any suitable length. This will serve as a construction line for the base. Mark a point M on it.
2. At point M, draw a perpendicular line to PQ. This line will be the height of the triangle. Along this perpendicular line, measure and mark a point A such that AM is 6 cm.
3. Now, place the compass needle at A and open the compass to a radius of 7 cm (the length of the equal sides). Draw an arc that cuts the line PQ on both sides of M. Label these intersection points B and C.
4. Finally, join A to B and A to C.
The triangle \( \triangle ABC \) is the required isosceles triangle. After drawing, measure the length of BC to find the base.
In simple words: Draw a straight line and mark its middle. From that middle point, draw a line straight up which is the height. From the top of the height line, draw arcs with the length of the equal sides until they touch the bottom line. Connect these points to the top to finish the triangle. Then, measure the base.

🎯 Exam Tip: For isosceles triangles, the altitude from the vertex between the equal sides bisects the base and is perpendicular to it. Use this property by drawing the altitude first, then finding the base points using the equal side lengths.