ICSE Class 8 Mathematics · Solutions for Class 8 RS Aggarwal ICSE Foundation Mathematics · 21 solved questions
Construct a quadrilateral PQRS in which PQ = 4 cm, ∠P = 90°, QR = 4.3 cm, RS = 3.6 cm and SP = 3.2 cm.
Steps of construction :
Draw a line segment PQ = 4 cm.
At P, construct ∠QPX = 90°.
From PX, cut off PS = 3.2 cm.
With Q as centre and radius 4.3 cm, draw an arc.
With S as centre and radius 3.6 cm, draw another arc cutting the previous arc at R.
Join QR and RS.
Hence, PQRS is the required quadrilateral.
Construct a quadrilateral ABCD in which AB = 4.5 cm, BC = 5.2 cm, CD = 5 cm, DA = 4.7 cm and ∠ABC = 75°.
Steps of construction :
Draw a line segment AB = 4.5 cm.
At B, construct ∠ABX = 75°.
From BX, cut off BC = 5.2 cm.
With A as centre and radius 4.7 cm, draw an arc.
With C as centre and radius 5 cm, draw another arc cutting the previous arc at D.
Join AD and CD.
Hence, ABCD is the required quadrilateral.
Construct a quadrilateral ABCD in which AB = CD = 5.1 cm, BC = 4.7 cm, DA = 4.2 cm and ∠BCD = 60°.
Steps of construction :
Draw a line segment BC = 4.7 cm.
At C, construct ∠BCX = 60°.
From CX, cut off CD = 5.1 cm.
With B as centre and radius 5.1 cm, draw an arc.
With D as centre and radius 4.2 cm, draw another arc cutting the previous arc at A.
Join AB and AD.
Hence, ABCD is the required quadrilateral.
Construct a quadrilateral EFGH in which EF = 6 cm, FG = 5.3 cm, EH = 5 cm, ∠E = 60° and ∠F = 75°.
Steps of construction :
Draw a line segment EF = 6 cm.
At E, construct ∠FEX = 60°.
From EX, cut off EH = 5 cm.
At F, construct ∠EFY = 75° on the same side of EF as that of H.
From FY, cut off FG = 5.3 cm.
Join GH.
Hence, EFGH is the required quadrilateral.
Construct a quadrilateral PQRS in which PQ = 4.5 cm, QR = 5.6 cm, RS = 4.1 cm, ∠Q = 60° and ∠R = 120°.
Steps of construction :
Draw a line segment QR = 5.6 cm.
At Q, construct ∠RQX = 60°.
From QX, cut off QP = 4.5 cm.
At R, construct ∠QRY = 120° on the same side of QR as that of P.
From RY, cut off RS = 4.1 cm.
Join PS.
Hence, PQRS is the required quadrilateral.
Construct a quadrilateral ABCD in which AB = AD = 5.3 cm, BC = CD = 5 cm and diagonal BD = 6.4 cm.
The diagonal BD divides the quadrilateral ABCD into two triangles, ΔABD and ΔBCD, and all the three sides of each of them are known.
Steps of construction :
Draw a line segment AB = 5.3 cm.
With A as centre and radius 5.3 cm, draw an arc.
With B as centre and radius 6.4 cm, draw another arc cutting the previous arc at D.
Join AD and BD.
With B as centre and radius 5 cm, draw an arc on the side of BD opposite to that of A.
With D as centre and radius 5 cm, draw another arc cutting the previous arc at C.
Join BC and CD.
Hence, ABCD is the required quadrilateral.
Construct a quadrilateral PQRS in which PQ = 3 cm, QR = 2.4 cm, PS = 3.7 cm, PR = 4 cm and QS = 5.5 cm.
Steps of construction :
Draw a line segment PQ = 3 cm.
With P as centre and radius 4 cm, draw an arc.
With Q as centre and radius 2.4 cm, draw another arc cutting the previous arc at R.
Join PR and QR.
With P as centre and radius 3.7 cm, draw an arc on the side of PR opposite to that of Q.
With Q as centre and radius 5.5 cm, draw another arc cutting the previous arc at S.
Join PS, RS and QS.
Hence, PQRS is the required quadrilateral.
Construct a quadrilateral ABCD in which AB = 5.2 cm, BC = 3.6 cm, ∠A = 60°, ∠B = 90° and ∠C = 120°.
Steps of construction :
Draw a line segment AB = 5.2 cm.
At B, construct ∠ABX = 90°.
From BX, cut off BC = 3.6 cm.
At A, construct ∠BAY = 60° on the same side of AB as that of C.
At C, construct ∠BCZ = 120° such that CZ meets AY at D.
Hence, ABCD is the required quadrilateral.
Construct a parallelogram ABCD in which BC = 6 cm, CD = 4 cm and ∠C = 60°.
Property used: The opposite sides of a parallelogram are equal.
Thus, in parallelogram ABCD,
AB = CD = 4 cm and AD = BC = 6 cm
Steps of construction :
Draw a line segment BC = 6 cm.
At C, construct ∠BCX = 60°.
From CX, cut off CD = 4 cm.
With B as centre and radius 4 cm, draw an arc.
With D as centre and radius 6 cm, draw another arc cutting the previous arc at A.
Join AB and AD.
Hence, ABCD is the required parallelogram.
Construct a parallelogram PQRS in which QR = 4.7 cm, ∠Q = 120° and PQ = 2.8 cm.
Property used: The opposite sides of a parallelogram are equal.
Thus, in parallelogram PQRS,
PS = QR = 4.7 cm and RS = PQ = 2.8 cm
Steps of construction :
Draw a line segment PQ = 2.8 cm.
At Q, construct ∠PQX = 120°.
From QX, cut off QR = 4.7 cm.
With P as centre and radius 4.7 cm, draw an arc.
With R as centre and radius 2.8 cm, draw another arc cutting the previous arc at S.
Join PS and RS.
Hence, PQRS is the required parallelogram.
Construct a parallelogram ABCD in which BC = 5 cm, CD = 3 cm and diagonal BD = 6 cm.
Property used: The opposite sides of a parallelogram are equal.
Thus, in parallelogram ABCD,
AB = CD = 3 cm and AD = BC = 5 cm
Steps of construction :
Draw a line segment BC = 5 cm.
With B as centre and radius 6 cm, draw an arc.
With C as centre and radius 3 cm, draw another arc cutting the previous arc at D.
Join BD and CD.
With B as centre and radius 3 cm, draw an arc on the side of BD opposite to that of C.
With D as centre and radius 5 cm, draw another arc cutting the previous arc at A.
Join AB and AD.
Hence, ABCD is the required parallelogram.
Construct a parallelogram ABCD in which AB = 4.8 cm, BC = 3.5 cm and diagonal AC = 5.4 cm.
Property used: The opposite sides of a parallelogram are equal.
Thus, in parallelogram ABCD,
AD = BC = 3.5 cm and CD = AB = 4.8 cm
Steps of construction :
Draw a line segment AB = 4.8 cm.
With A as centre and radius 5.4 cm, draw an arc.
With B as centre and radius 3.5 cm, draw another arc cutting the previous arc at C.
Join AC and BC.
With A as centre and radius 3.5 cm, draw an arc on the side of AC opposite to that of B.
With C as centre and radius 4.8 cm, draw another arc cutting the previous arc at D.
Join AD and CD.
Hence, ABCD is the required parallelogram.
Construct a parallelogram ABCD in which diagonal AC = 5.6 cm, diagonal BD = 6.2 cm and angle between them is 60°.
Property used: The diagonals of a parallelogram bisect each other.
Thus, in parallelogram ABCD, if the diagonals AC and BD intersect at O, then
$$\begin{aligned}\text{OA} = \text{OC} = \frac{\text{AC}}{2} = \frac{5.6}{2} = 2.8 \text{ cm} \\[1em] \text{OB} = \text{OD} = \frac{\text{BD}}{2} = \frac{6.2}{2} = 3.1 \text{ cm}\end{aligned}$$
Steps of construction :
Draw a line segment AC = 5.6 cm.
Find the mid-point O of AC, so that OA = OC = 2.8 cm.
At O, construct ∠AOX = 60° and produce XO to Y.
From OX, cut off OD = 3.1 cm and from OY, cut off OB = 3.1 cm.
Join AB, BC, CD and AD.
Hence, ABCD is the required parallelogram.
Construct a rectangle whose adjacent sides are 4.7 cm and 3.2 cm.
Property used: In a rectangle, the opposite sides are equal and each angle is a right angle.
Let ABCD be the rectangle with AB = 4.7 cm and BC = 3.2 cm.
Thus, in rectangle ABCD,
AD = BC = 3.2 cm, CD = AB = 4.7 cm and ∠A = ∠B = ∠C = ∠D = 90°
Steps of construction :
Draw a line segment AB = 4.7 cm.
At B, construct ∠ABX = 90°.
From BX, cut off BC = 3.2 cm.
With A as centre and radius 3.2 cm, draw an arc.
With C as centre and radius 4.7 cm, draw another arc cutting the previous arc at D.
Join AD and CD.
Hence, ABCD is the required rectangle.
Construct a rectangle ABCD in which AB = 4.5 cm and diagonal AC = 6 cm.
Property used: In a rectangle, the opposite sides are equal and each angle is a right angle.
Thus, in rectangle ABCD,
CD = AB = 4.5 cm, AD = BC and ∠A = ∠B = ∠C = ∠D = 90°
Steps of construction :
Draw a line segment AB = 4.5 cm.
At B, construct ∠ABX = 90°.
With A as centre and radius 6 cm, draw an arc cutting BX at C.
Join AC.
With C as centre and radius 4.5 cm, draw an arc.
With A as centre and radius equal to BC, draw another arc cutting the previous arc at D.
Join AD and CD.
Hence, ABCD is the required rectangle.
Construct a rectangle ABCD in which diagonal AC = 6.3 cm and the angle between the two diagonals is 45°.
Property used: The diagonals of a rectangle are equal and they bisect each other.
Thus, in rectangle ABCD, if the diagonals AC and BD intersect at O, then
$$\begin{aligned}\text{BD} = \text{AC} = 6.3 \text{ cm} \\[1em] \text{OA} = \text{OC} = \text{OB} = \text{OD} = \frac{6.3}{2} = 3.15 \text{ cm}\end{aligned}$$
Steps of construction :
Draw a line segment AC = 6.3 cm.
Find the mid-point O of AC, so that OA = OC = 3.15 cm.
At O, construct ∠AOX = 45° and produce XO to Y.
From OX, cut off OD = 3.15 cm and from OY, cut off OB = 3.15 cm.
Join AB, BC, CD and AD.
Hence, ABCD is the required rectangle.
Construct a square of side 4.8 cm.
Property used: In a square, all the four sides are equal and each angle is a right angle.
Let ABCD be the square of side 4.8 cm.
Thus, in square ABCD,
AB = BC = CD = DA = 4.8 cm and ∠A = ∠B = ∠C = ∠D = 90°
Steps of construction :
Draw a line segment AB = 4.8 cm.
At B, construct ∠ABX = 90°.
From BX, cut off BC = 4.8 cm.
With C as centre and radius 4.8 cm, draw an arc.
With A as centre and radius 4.8 cm, draw another arc cutting the previous arc at D.
Join AD and CD.
Hence, ABCD is the required square.
Construct a square one of whose diagonals is 5.6 cm.
Property used: The diagonals of a square are equal and they bisect each other at right angles.
Let ABCD be the square with diagonal AC = 5.6 cm.
Thus, in square ABCD, if the diagonals AC and BD intersect at O, then
$$\begin{aligned}\text{BD} = \text{AC} = 5.6 \text{ cm} \\[1em] \text{OA} = \text{OC} = \text{OB} = \text{OD} = \frac{5.6}{2} = 2.8 \text{ cm}\end{aligned}$$
and, ∠AOD = 90°, i.e. AC ⊥ BD
Steps of construction :
Draw a line segment AC = 5.6 cm.
Draw the perpendicular bisector XY of AC, meeting AC at its mid-point O.
From OX, cut off OD = 2.8 cm.
Join AD and CD.
From OY, cut off OB = 2.8 cm.
Join AB and BC.
Hence, ABCD is the required square.
Construct a rhombus ABCD in which AB = 4.3 cm and ∠A = 45°.
Property used: All the sides of a rhombus are equal.
Thus, in rhombus ABCD,
AD = BC = CD = AB = 4.3 cm
Steps of construction :
Draw a line segment AB = 4.3 cm.
At A, construct ∠BAX = 45°.
From AX, cut off AD = 4.3 cm.
With B as centre and radius 4.3 cm, draw an arc.
With D as centre and radius 4.3 cm, draw another arc cutting the previous arc at C.
Join BC and CD.
Hence, ABCD is the required rhombus.
Construct a rhombus ABCD in which AB = 5 cm and diagonal AC = 6.8 cm.
Property used: All the sides of a rhombus are equal.
Thus, in rhombus ABCD,
AD = BC = CD = AB = 5 cm
Steps of construction :
Draw a line segment AB = 5 cm.
With A as centre and radius 6.8 cm, draw an arc.
With B as centre and radius 5 cm, draw another arc cutting the previous arc at C.
Join AC and BC.
With A as centre and radius 5 cm, draw an arc on the side of AC opposite to that of B.
With C as centre and radius 5 cm, draw another arc cutting the previous arc at D.
Join AD and CD.
Hence, ABCD is the required rhombus.
Construct a rhombus ABCD in which diagonal AC = 5.8 cm and diagonal BD = 6.4 cm.
Property used: The diagonals of a rhombus bisect each other at right angles.
Thus, in rhombus ABCD, if the diagonals AC and BD intersect at O, then
$$\begin{aligned}\text{OA} = \text{OC} = \frac{\text{AC}}{2} = \frac{5.8}{2} = 2.9 \text{ cm} \\[1em] \text{OB} = \text{OD} = \frac{\text{BD}}{2} = \frac{6.4}{2} = 3.2 \text{ cm}\end{aligned}$$
and, ∠AOD = 90°, i.e. AC ⊥ BD
Steps of construction :
Draw a line segment AC = 5.8 cm.
Draw the perpendicular bisector XOY of AC, meeting AC at its mid-point O.
From OX, cut off OD = 3.2 cm and from OY, cut off OB = 3.2 cm.
Join AB, BC, CD and AD.
Hence, ABCD is the required rhombus.
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