Samacheer Kalvi Class 8 Maths Solutions Chapter 5 Geometry Exercise 5.4

I. Construct the following quadrilaterals with the given measurements and also find their area.

 

Question 1. Construct quadrilateral ABCD with AB = 5 cm, BC = 4.5 cm, CD = 3.8 cm, DA = 4.4 cm and AC = 6.2 cm. Also, find its area.
Answer: First, we draw a rough diagram to plan the construction. Then, we follow these steps to construct the quadrilateral ABCD:

• Draw a line segment \(AB = 5\) cm.
• With A as the center, draw an arc with a radius of \(6.2\) cm.
• With B as the center, draw another arc with a radius of \(4.5\) cm. These arcs will intersect at point C.
• Join AC and BC to form triangle ABC.
• Now, with A as the center, draw an arc with a radius of \(4.4\) cm.
• With C as the center, draw another arc with a radius of \(3.8\) cm. These arcs will intersect at point D.
• Join AD and CD to complete the quadrilateral.
• ABCD is the required quadrilateral.

Next, we calculate the area. The diagonal AC divides the quadrilateral into two triangles. To find the area, we need the perpendicular heights from B to AC (\(h_1\)) and from D to AC (\(h_2\)). For construction purposes, these heights are typically measured from the constructed diagram. We can use the formula for the area of a quadrilateral when a diagonal and perpendiculars to it are known. The area of a quadrilateral is half the product of the diagonal and the sum of the perpendiculars drawn to it from the opposite vertices. The calculation for the area is as follows (assuming measured heights from the diagram):
Area of the quadrilateral ABCD \( = \frac { 1 }{ 2 } \times d \times (h_1 + h_2) \) sq. units
\( = \frac { 1 }{ 2 } \times 6.2 \times (2.6 + 3.6)\) cm\(^2\)
\( = \frac { 1 }{ 2 } \times 6.2 \times 6.2\) cm\(^2\)
\( = 3.1 \times 6.2 = 19.22 \) cm\(^2\)
The area of the constructed quadrilateral is \(19.22\) cm\(^2\). This method is useful for finding the area of complex shapes by breaking them into simpler triangles.In simple words: First, draw a rough sketch and then build the shape step by step using a ruler and compass. Once the quadrilateral is drawn, measure the perpendicular lines from the other two corners to the diagonal, and then use the formula to find the total area.

🎯 Exam Tip: Always start with a rough diagram to visualize the construction. Clearly label all vertices and measurements in your final diagram. When calculating area using diagonals and perpendiculars, ensure accurate measurement of heights from the diagram.

 

Question 2. Construct quadrilateral PLAY with PL = 7cm, LA = 6cm, AY = 6cm, PA = 8cm and LY = 7cm. Also, find its area.
Answer: First, we draw a rough diagram to plan the construction. Then, we follow these steps to construct the quadrilateral PLAY:

• Draw a line segment \(PL = 7\) cm.
• With P as the center, draw an arc with a radius of \(8\) cm.
• With L as the center, draw another arc with a radius of \(6\) cm. These arcs will intersect at point A.
• Join PA and LA to form triangle PLA.
• Now, with L as the center, draw an arc with a radius of \(7\) cm.
• With A as the center, draw another arc with a radius of \(6\) cm. These arcs will intersect at point Y.
• Join LY, PY and AY to complete the quadrilateral.
• PLAY is the required quadrilateral.

Next, we calculate the area. The diagonal LA divides the quadrilateral into two triangles. We use the formula for the area of a quadrilateral when a diagonal and perpendiculars to it are known. The calculation for the area is as follows (assuming measured heights from the diagram):
Area of the quadrilateral PLAY \( = \frac { 1 }{ 2 } \times d \times (h_1 + h_2) \) sq. units
\( = \frac { 1 }{ 2 } \times 8 \times (5.1 + 1.4)\) cm\(^2\)
\( = \frac { 1 }{ 2 } \times 8 \times 6.5\) cm\(^2\)
\( = 4 \times 6.5 = 26 \) cm\(^2\)
The area of the constructed quadrilateral is \(26\) cm\(^2\). This step-by-step construction ensures all given measurements are accurately represented.In simple words: Draw the first side, then use compasses from its ends to find the next point. Repeat this for all points to complete the shape. To find the area, use the main diagonal and the heights from the other corners to that diagonal in a special formula.

🎯 Exam Tip: When constructing quadrilaterals with all sides and diagonals given, pick a diagonal to form the first triangle. This makes it easier to locate the remaining vertices. Ensure arcs intersect clearly.

 

Question 3. Construct quadrilateral PQRS with PQ = QR = 3.5 cm, RS = 5.2 cm, SP = 5.3 cm and \( \angle Q = 120^\circ \). Also, find its area.
Answer: First, we draw a rough diagram to plan the construction. Then, we follow these steps to construct the quadrilateral PQRS:

• Draw a line segment \(PQ = 3.5\) cm.
• At point Q, use a protractor to make an angle of \(120^\circ\) (\(\angle Q = 120^\circ\)). Draw a ray QX in that direction.
• With Q as the center, draw an arc with a radius of \(3.5\) cm along the ray QX. This arc will cut the ray at point R, since \(QR = 3.5\) cm.
• Now, with R as the center, draw an arc with a radius of \(5.2\) cm (\(RS = 5.2\) cm).
• With P as the center, draw another arc with a radius of \(5.3\) cm (\(SP = 5.3\) cm). These arcs will intersect at point S.
• Join PS and RS to complete the quadrilateral.
• PQRS is the required quadrilateral.

Next, we calculate the area. The diagonal PR divides the quadrilateral into two triangles. We use the formula for the area of a quadrilateral when a diagonal and perpendiculars to it are known. The calculation for the area is as follows (assuming measured heights from the diagram):
Area of the quadrilateral PQRS \( = \frac { 1 }{ 2 } \times d \times (h_1 + h_2) \) sq. units
\( = \frac { 1 }{ 2 } \times 6 \times (4.3 + 1.7)\) cm\(^2\)
\( = \frac { 1 }{ 2 } \times 6 \times 6\) cm\(^2\)
\( = 3 \times 6 = 18 \) cm\(^2\)
The area of the constructed quadrilateral is \(18\) cm\(^2\). Understanding how angles define the shape is key to accurate construction.In simple words: Start by drawing the side with the given angle, then use the angle to find the next point. After that, use compasses to find the last point. To find the area, measure the main diagonal and the heights from the other two corners to it, then use the formula.

🎯 Exam Tip: When an angle is given, use a protractor carefully to draw it. Remember that if two sides meeting at the angle are known, that triangle can be drawn first to fix the position of points.

 

Question 4. Construct quadrilateral MIND with MI = 3.6 cm, ND = 4 cm, MD = 4 cm, \( \angle M = 50^\circ \) and \( \angle D = 100^\circ \). Also, find its area.
Answer: First, we draw a rough diagram to plan the construction. Then, we follow these steps to construct the quadrilateral MIND:

• Draw a line segment \(MI = 3.6\) cm.
• At point M on MI, make an angle of \(50^\circ\) (\(\angle IMX = 50^\circ\)). Draw a ray MX.
• With M as the center, draw an arc with a radius of \(4\) cm along the ray MX. This arc will cut the ray at point D, since \(MD = 4\) cm.
• At point D on DM, make an angle of \(100^\circ\) (\(\angle MDY = 100^\circ\)). Draw a ray DY.
• With I as the center, draw an arc with a radius of \(4\) cm (\(IN = 4\) cm). This arc will cut the ray DY at point N.
• Join DN and IN to complete the quadrilateral.
• MIND is the required quadrilateral.

Next, we calculate the area. The diagonal IM divides the quadrilateral into two triangles. We use the formula for the area of a quadrilateral when a diagonal and perpendiculars to it are known. The calculation for the area is as follows (assuming measured heights from the diagram):
Area of the quadrilateral MIND \( = \frac { 1 }{ 2 } \times d \times (h_1 + h_2) \) sq. units
\( = \frac { 1 }{ 2 } \times 3.2 \times (2.7 + 3.3)\) cm\(^2\)
\( = \frac { 1 }{ 2 } \times 3.2 \times 6\) cm\(^2\)
\( = 1.6 \times 6 = 9.6 \) cm\(^2\)
The area of the constructed quadrilateral is \(9.6\) cm\(^2\). This construction shows how multiple angles can be used to define a polygon.In simple words: Draw the first side, then use angles at its ends to draw rays. Use side lengths to mark points along these rays. Connect the final points to complete the shape. For area, measure the diagonal and heights, then apply the formula.

🎯 Exam Tip: When two adjacent angles and their included sides are given, construct the base and then the angles at its endpoints first. This provides a strong framework for the rest of the construction.

 

Question 5. Construct quadrilateral AGRI with AG = 4.5 cm, GR = 3.8 cm, \( \angle A = 60^\circ \), \( \angle G = 110^\circ \) and \( \angle R = 90^\circ \). Also, find its area.
Answer: First, we draw a rough diagram to plan the construction. Then, we follow these steps to construct the quadrilateral AGRI:

• Draw a line segment \(AG = 4.5\) cm.
• At point A on AG, make an angle of \(60^\circ\) (\(\angle GAY = 60^\circ\)). Draw a ray AY.
• At point G on AG, make an angle of \(110^\circ\) (\(\angle AGX = 110^\circ\)). Draw a ray GX.
• With G as the center, draw an arc with a radius of \(3.8\) cm along the ray GX. This arc will cut the ray at point R, since \(GR = 3.8\) cm.
• At point R on GR, make an angle of \(90^\circ\) (\(\angle GRZ = 90^\circ\)). Draw a ray RZ.
• The rays AY and RZ will meet at point I.
• AGRI is the required quadrilateral.

Next, we calculate the area. The diagonal GR divides the quadrilateral into two triangles. We use the formula for the area of a quadrilateral when a diagonal and perpendiculars to it are known. The calculation for the area is as follows (assuming measured heights from the diagram):
Area of the quadrilateral AGRI \( = \frac { 1 }{ 2 } \times d \times (h_1 + h_2) \) sq. units
\( = \frac { 1 }{ 2 } \times 6.8 \times (2.9 + 2.4)\) cm\(^2\)
\( = \frac { 1 }{ 2 } \times 6.8 \times 5.3\) cm\(^2\)
\( = 3.4 \times 5.3 = 18.02 \) cm\(^2\)
The area of the constructed quadrilateral is \(18.02\) cm\(^2\). Using three angles and two sides allows for a precise construction.In simple words: Draw the first side, then use protractors to draw the angles at its ends. Mark the next point using its given length. Draw the third angle, and where the last two rays meet will be the final corner. Then, calculate the area using the diagonal and measured heights.

🎯 Exam Tip: When three angles and two adjacent sides are given, it's often best to construct the base and the angles at its ends first. The third angle helps determine the intersection point for the final vertex.

II. Construct the following trapeziums with the given measures and also find their area.

 

Question 1. Construct trapezium AIMS with \( \overline{\mathrm{AI}} || \overline{\mathrm{SM}} \), AI = 6cm, IM = 5cm, AM = 9cm and MS = 6.5cm. Also, find its area.
Answer: First, we draw a rough diagram to plan the construction. Then, we follow these steps to construct the trapezium AIMS:

• Draw a line segment \(AI = 6\) cm.
• With A as the center, draw an arc with a radius of \(9\) cm (\(AM = 9\) cm).
• With I as the center, draw another arc with a radius of \(5\) cm (\(IM = 5\) cm). These arcs will intersect at point M.
• Join AM and IM to form triangle AIM.
• Draw a line MX parallel to AI through point M. (Since AI is parallel to SM, this line will contain SM).
• With M as the center, draw an arc with a radius of \(6.5\) cm along the line MX. This arc will cut MX at point S, since \(MS = 6.5\) cm.
• Join AS to complete the trapezium.
• AIMS is the required trapezium.

Next, we calculate the area. The area of a trapezium is given by the formula: \( \text{Area} = \frac { 1 }{ 2 } \times h \times (a + b) \) where \(h\) is the height and \(a, b\) are the lengths of the parallel sides. The calculation for the area is as follows (assuming measured height from the diagram):
Area of the trapezium AIMS \( = \frac { 1 }{ 2 } \times h \times (a + b) \) sq. units
\( = \frac { 1 }{ 2 } \times 4.6 \times (6 + 6.5)\) cm\(^2\)
\( = \frac { 1 }{ 2 } \times 4.6 \times 12.5\) cm\(^2\)
\( = 2.3 \times 12.5 = 28.75 \) cm\(^2\)
The area of the constructed trapezium is \(28.75\) cm\(^2\). Knowing the parallel sides and the height is crucial for this calculation.In simple words: First, draw one of the non-parallel sides and complete a triangle using given lengths. Then, draw a line through one vertex parallel to the first side. Mark the last vertex on this parallel line. For area, measure the height between the parallel sides and use the formula with the lengths of the parallel sides.

🎯 Exam Tip: When constructing a trapezium, first draw a triangle using the non-parallel side and the diagonals (if given). Then, draw the parallel line and locate the fourth vertex. Remember to measure the perpendicular height accurately for area calculation.

 

Question 2. Construct trapezium CUTE with \( \overline{\mathrm{CD}} || \overline{\mathrm{ET}} \), CU = 7cm, \( \angle UCE = 80^\circ \), CE = 6cm, TE = 5cm and CD = 4cm. Also, find its area.
Answer: First, we draw a rough diagram to plan the construction. Then, we follow these steps to construct the trapezium CUTE:

• Draw a line segment \(CU = 7\) cm.
• At point C, make an angle of \(80^\circ\) (\(\angle UCE = 80^\circ\)). Draw a ray CY.
• With C as the center, draw an arc with a radius of \(6\) cm along the ray CY. This arc will cut the ray at point E, since \(CE = 6\) cm.
• Mark a point D on CU such that \(CD = 4\) cm. (This is implicitly derived from parallel construction.)
• Draw a line EX parallel to CU through point E.
• With E as the center, draw an arc with a radius of \(5\) cm along the line EX. This arc will cut EX at point T, since \(TE = 5\) cm.
• Join UT to complete the trapezium.
• CUTE is the required trapezium.

Next, we calculate the area. The area of a trapezium is given by the formula: \( \text{Area} = \frac { 1 }{ 2 } \times h \times (a + b) \). The calculation for the area is as follows (assuming measured height from the diagram):
Area of the trapezium CUTE \( = \frac { 1 }{ 2 } \times h \times (a + b) \) sq. units
\( = \frac { 1 }{ 2 } \times 5.9 \times (7 + 5)\) sq. units
\( = \frac { 1 }{ 2 } \times 5.9 \times 12\) sq. units
\( = 5.9 \times 6 = 35.4 \) sq. cm
The area of the constructed trapezium is \(35.4\) sq. cm. This construction demonstrates how to use an angle and parallel sides.In simple words: Draw the first base, then use the angle at one end to find the next point. Draw a parallel line from this point. Then, use the length of the other non-parallel side to find the final point on the parallel line. Connect the remaining ends. For area, measure the height between parallel sides and add the lengths of the parallel sides in the formula.

🎯 Exam Tip: When an angle is given, use it to fix one of the non-parallel sides. Drawing a parallel line from the known vertex helps locate the remaining point of the other parallel side.

 

Question 3. Construct trapezium ARMY with \( \overline{\mathrm{AR}} || \overline{\mathrm{YM}} \), AR = 7cm, RM = 6.5cm, \( \angle RAY = 100^\circ \) and \( \angle ARM = 60^\circ \). Also, find its area.
Answer: First, we draw a rough diagram to plan the construction. Then, we follow these steps to construct the trapezium ARMY:

• Draw a line segment \(AR = 7\) cm.
• At point A on AR, construct an angle of \(100^\circ\) (\(\angle RAX = 100^\circ\)). Draw a ray AX.
• At point R on AR, construct an angle of \(60^\circ\) (\(\angle ARN = 60^\circ\)). Draw a ray RN.
• With R as the center, draw an arc with a radius of \(6.5\) cm along the ray RN. This arc will cut RN at point M, since \(RM = 6.5\) cm.
• Draw a line MY parallel to AR through point M.
• The ray AX and the line MY will meet at point Y.
• ARMY is the required trapezium.

Next, we calculate the area. The area of a trapezium is given by the formula: \( \text{Area} = \frac { 1 }{ 2 } \times h \times (a + b) \). The calculation for the area is as follows (assuming measured height from the diagram):
Area of the trapezium ARMY \( = \frac { 1 }{ 2 } \times h \times (a + b) \) sq. units
\( = \frac { 1 }{ 2 } \times 5.6 \times (7 + 4.8)\) sq. units
\( = \frac { 1 }{ 2 } \times 5.6 \times 11.8\) sq. units
\( = 2.8 \times 11.8 = 33.04 \) sq. cm
The area of the constructed trapezium is \(33.04\) sq. cm. Constructing angles at the base first helps set up the shape correctly.In simple words: Draw the base line. Then, draw the angles at each end of the base. Mark the third point using a given side length from one of the base ends. Draw a line from this third point that is parallel to the base. Where this parallel line meets the other angle's ray, that's the final point. To find the area, measure the height between the parallel lines and sum their lengths for the formula.

🎯 Exam Tip: When two angles at the base and one non-parallel side are given, construct the base and angles first. Then, use the non-parallel side length and a parallel line to locate the remaining vertex.

 

Question 4. Construct trapezium CITY with \( \overline{\mathrm{CI}} || \overline{\mathrm{YT}} \), CI = 7cm, IT = 5.5cm, TY = 4cm and YC = 6cm. Also, find its area.
Answer: First, we draw a rough diagram to plan the construction. Then, we follow these steps to construct the trapezium CITY:

• Draw a line segment \(CI = 7\) cm.
• Mark a point D on CI such that \(CD = 4\) cm. (This point helps to construct a triangle that will provide the height of the trapezium).
• With D as the center, draw an arc with a radius of \(6\) cm (\(YC = 6\) cm, we will use this to form a triangle CDY or CTY indirectly).
• With I as the center, draw another arc with a radius of \(5.5\) cm (\(IT = 5.5\) cm). These arcs will intersect at point T.
• Join DT and IT.
• With C as the center, draw an arc with a radius of \(6\) cm (\(YC = 6\) cm).
• Draw a line TY parallel to CI through point T. This line will cut the arc drawn from C at point Y.
• Join CY.
• CITY is the required trapezium.

Next, we calculate the area. The area of a trapezium is given by the formula: \( \text{Area} = \frac { 1 }{ 2 } \times h \times (a + b) \). The calculation for the area is as follows (assuming measured height from the diagram):
Area of the trapezium CITY \( = \frac { 1 }{ 2 } \times h \times (a + b) \) sq. units
\( = \frac { 1 }{ 2 } \times 5.5 \times (7 + 4)\) sq. units
\( = \frac { 1 }{ 2 } \times 5.5 \times 11\) sq. units
\( = 2.75 \times 11 = 30.25 \) sq. cm
The area of the constructed trapezium is \(30.25\) sq. cm. Constructing a triangle by drawing a parallel line is a common technique for trapeziums.In simple words: Draw the longer parallel side. From one end, draw an arc for a non-parallel side. From a point on the longer side, draw an arc for the other non-parallel side. Where these meet helps form a triangle. Then, draw a line parallel to the base from one vertex, and mark the last point on it. To get the area, measure the distance between the parallel sides and use the formula with their lengths.

🎯 Exam Tip: When constructing a trapezium with all four sides given and parallel lines specified, creating an auxiliary triangle by drawing a line parallel to one of the non-parallel sides can simplify the construction.