Question 14 Marks
Using ruler and compasses only, construct the quadrilateral $\text{ABCD}$, having given $AB = 5 \ cm, BC = 2.5 \ cm CD = 6 \ cm, \angle BAD = 90^o$ and diagonal $BD = 5.5 \ cm.$
Answer
Steps:
$1.$ Draw $A B=5 \ cm$.
$2.$ From A draw a line $A Y$ such that $\angle A=90^{\circ}$.
$3.$ Taking $B$ as a center with radius $5.5 \ cm$ draw an arc at $D$ on $A Y$.
$4.$ With $D$ and $B$ as center and radii, $6 \ cm$ and $2.5 \ cm$ draw arcs cutting each other at $C$.
$5.$ Join $DC$ and $BC$.
$\text{ABCD}$ is the required quadrilateral. View full question & answer→Question 24 Marks
Using ruler and compasses only, construct the quadrilateral $\text{ABCD}$, having given $A B=5 \ cm, B C=2.5 \ cm, C D$
$=6 \ cm$. $\angle B A D=90^{\circ}$ and the diagonal $A C=5.5 \ cm$.
Answer
Steps:
$1$. draw $A B=5 \ cm$.
$2$. Now draw $\angle X A B$ such that it is $90^{\circ}$.
$3$. Taking $A$ and $B$ as a center and radius $2.5 \ cm$ and $5.5 \ cm$ draw arcs cut off at $C$.
$4$. Now join $B C$ and $A C$.
$5$. Taking $C$ as a center and radius $6 \ cm$ draw arcs at $D$ on $A X$.
$\text{ABCD}$ is the required quadrilateral. View full question & answer→Question 34 Marks
Construct a rhombus, having given one side $= 4.8 \ cm$ and one angle $= 75^o.$
Answer
Steps:
$1$.draw a line $A B=4.8 \ cm$
$2$. At $A$ Draw $A X$ such that $\angle B A X=75^{\circ}$.
$3$. With $A$ as a center and measurement equal to $A B$ cut off an arc at $D$ on $AX.$
$4.$ Using the same radius taking $D$ and $B$ as centers cut off arcs, which will intersect at $C$.
$5$. Join $C D$ and $C B$.
$\text{ABCD}$ is the required rhombus. View full question & answer→Question 44 Marks
Construct a square $\text{ABCD}$, when: One diagonal $= 5.4 \ cm.$
AnswerWe know that the diagonals of a square are equal and bisect each other at right angles.
Steps:
$1.$ draw $A C=5.4\ cm$.
$2.$ Draw the right bisector $X Y$ of $A C$, meeting $A C$ at $O$.
$3.$ From $O$, set off $O B=\frac{1}{2}(5.4)=2.7\ cm$ along $O Y$ and along $O X$.
$4$. Join $A B, B C, C D$, and $D A$.
$\text{ABCD}$ is the required square. View full question & answer→Question 54 Marks
Construct a square $\text{ABCD},$ when: One side $= 4.5 \ cm.$
Answer
Steps:
$1.$ Draw a line segment $A B=4.5 \ cm$
$2.$ Draw $AP \perp AB$.
$3.$ From $A P$ cut off $A D=4.5 \ cm$.
$4.$ With $B$ as a center and radius $4.5 \ cm$ draw an arc.
$5.$ With $D$ as center and radius $4.5 \ cm$ draw another arc cutting the former arc at $C.$
$6.$ Join $B C$ and $C D$.
$\text{ABCD}$ is the required square. View full question & answer→Question 64 Marks
Construct a rhombus $\text{ABCD},$ when: Diagonal $AC = 6.0 \ cm$ and height $= 3.5 \ cm.$
Answer
Steps:
$1.$ draw a line$ AP.$
$2.$ now draw $A C=6\ cm$ and $C P=3.5\ cm$.
$3.$ Now draw a line $B C$ such that $A B=B C$.
$4.$ Now at $C$ draw a line $C Y$ parallel to $AP.$
$5.$ At point $C$ and $A$, taking radius same as $A B$ draw arcs cutting each other at $D.$
$6.$ Now join $AD.$
$\text{ABCD}$ is the required rhombus. View full question & answer→Question 74 Marks
Construct a rhombus $\text{ABCD},$ when: $A = 60^\circ$ and height $= 3.0 \ cm.$
Answer
Steps:
$1.$ Draw a line $AP.$
$2.$ Now draw a line $A F$ such that $\angle A=60^{\circ}$
$3.$ At $S$ draw a perpendicular $S E$ of length $3\ cm$ such that it cut at $A F$ at $D$.
$4.$Through $D$ draw a line $QR$ parallel to $AP.$
$5.$ Now taking the radius same as $AD$ draws an arc at $B$ on $AP.$
$6.$ Now through and $B$ taking radius same as $A D$ and $A B$ draw arcs cutting each other at $C$.
$7.$ Now join $BC.$
$\text{ABCD}$ is the required rhombus. View full question & answer→Question 84 Marks
Construct a rhombus $\text{ABCD},$ when:One side $= 5.0 \ cm$ and height $= 2.6 \ cm.$
Answer
Steps:
$1.$ Draw $AB =5 \ cm$.
$2.$ At $B$, draw $B P \perp A B$.
$3.$ From $B P$, cut $B E=2.6 \ cm =$ height.
$4.$ Through $E$ draw a perpendicular to $CP$ to get $QR$ parallel to $A B$.
With $A$ and $B$ as a centre and radii $5 \ cm$ draw arcs cutting $QR$ at $D$ and $C$.
$\text{ABCD}$ is the required rhombus. View full question & answer→Question 94 Marks
Construct a rhombus $\text{ABCD}$, when:One side $= 5.4 \ cm$ and one diagonal is $7.0 \ cm.$
Answer
Steps:
$1.$ We construct the segment $A C=7\ cm$.
$2.$ With $A$ as a centre and radius $5.4 \ cm$, we draw an arc extending on both sides of $A C$.
$3.$ With $C$ as a centre and same radius as in step 2,
we draw an arc extending on both sides of $A C$ to cut the first arc at $B$ and $D$.
$4$. Join $\text{AB, BC, CD}$ and $D A$.
$\text{ABCD}$ is the required rhombus. View full question & answer→Question 104 Marks
Construct a rectangle $\text{ABCD,}$ when:Its sides are $6.0 \ cm$ and $7.2 \ cm.$
AnswerSince each angle of a rectangle is $90^{\circ}$ and the opposite sides are equal.
Therefore,
Steps:
$1.$ Draw $BC =7.2 \ cm$.
$2.$ With $B$ as a centre draw a line $BX$ taking as a $90^{\circ}$.
$3. $Now taking radius $6 \ cm$ to draw an arc at $A$.
$4.$ From the point, $A$ draw a line $A Y$ parallel to $B C$.
$5.$ With $A$ as a center taking radius $7.2 \ cm$ to draw an arc at $D$.
$6.$ Now join $C D$.
$\text{ABCD}$ is the required rectangle. View full question & answer→Question 114 Marks
Construct a parallelogram $\text{ABCD},$ when:Base $BC = 5.6 \ cm,$ diagonal $BD = 6.5 \ cm$ and altitude $= 3.2 \ cm.$
Answer
Steps:
$1$. Draw $BC =5.6 \ cm$.
$2$. At $C$, draw $CX$ perpendicular to $BC$.
$3.$ with $C$ as a centre and taking radius $3.2 \ cm$ to draw an arc to cut $C X$ at $Y.$
$4.$ Through $Y$ draw a straight line $PQ$ parallel to $BC.$
$5.$ With $B$ as a centre and radius, $6.5 \ cm$ draw an arc to meet $PQ$ at $D$.
$6.$ With $D$ as a centre and radius equal to $5.6 \ cm$, draw an arc to meet $PQ$ at $A.$
$7$.Join $BA, BD$ and $CD.$
$\text{ABCD}$ is the required parallelogram. View full question & answer→Question 124 Marks
Construct a parallelogram $\text{ABCD},$ when:Base $AB = 6.5 \ cm, BC = 4 \ cm$ and the altitude corresponding to $AB = 3.1 \ cm.$
Answer
Steps :
$1.$ Draw $A B=6.5 \ cm$.
$2. $At $B$, draw $BP \perp AB$.
$3.$ From $B P$ cut $B E=3.1 \ cm$.
$4.$ Through $E$ draw a perpendicular to $BP$ to get $QR$ parallel to $AB$.
$5.$ With $B$ as a center and radius $=A C=4 \ cm$, draw an arc that cuts $Q R$ at $C.$
$6.$ With $A$ as a center and radius $= AD =4 \ cm$, draw an arc that cuts $QR$ at $D.$
$\text{ABCD}$ is the required parallelogram. View full question & answer→Question 134 Marks
Construct a parallelogram $\text{ABCD},$ when:$AB = 5.8 \ cm,$ diagonal $AC = 8.2 \ cm$ and diagonal $BD = 6.2 \ cm.$
Answer
Steps:
$1.$ Since diagonal of a parallelogram bisect each other, construct $O A B$
such that;
$ OA =\frac{1}{2} AC =\frac{1}{2} \times 8.2 \ cm =4.1 \ cm$
$ OB =\frac{1}{2} BD =\frac{1}{2} \times 6.2 \ cm =3.1 \ cm $
and $A B=5.8 \ cm$.
$2.$ Produce $A O$ up to $C$, such that $O C=O A=4.1 \ cm$ and $B O$ up to $D$,
such that $D O=O B=3.1 \ cm$.
$3.$ Join $AD , DC$, and $CB$.
$\text{ABCD}$ is the required parallelogram. View full question & answer→Question 144 Marks
Draw parallelogram $\text{ABCD}$ with the following data:$A B=6 \ cm , A D=5 \ cm$ and $\angle D A B=45^{\circ}$.Let $A C$ and $D B$ meet in $O$ and let $E$ be the mid$-$point of $B C$. Join $O E$.Prove that:$(i) OE \| AB;(ii) OE =\frac{1}{2} AB$.
AnswerTo draw the parallelogram follows the steps:
$1.$ First, draw a line $A B$ of measure $6 \ cm$. Then draw an angle of measure $45^{\circ}$ at point $A$ such that $\angle D A B=45^{\circ}$ and $A D=5 \ cm$.
$2.$ Now draw a line $C D$ parallel to the line $A B$ of measure $6 \ cm$. Then join $BC$ to construct the parallelogram as shown below:
$3.$ Now it is given that $E$ is the mid-point of $BC$. We join $OE$. Now we are to prove that $OE \| AB$ and $OE =\frac{1}{2} AB$.
$4.$ Since $O$ is the mid$-$point of $A C$ and $E$ is the midpoint of $B C$, therefore the line is parallel to $A B$ and $O E=\frac{1}{2} A B$ View full question & answer→Question 154 Marks
The perpendicular distance between the pair of opposite sides of a parallelogram are $3 \ cm$ and $4 \ cm,$ and one of its angles measures $60^o$. Using a ruler and compasses only,construct the parallelogram.
Answer
Steps:
$1.$ Draw a baseline $AQ$.
$2.$ A take some random distance in compass and draw one are below and above the line.
Now without changing the distance in compass draw one are below and above the line.
These arcs intersect each other above and below the line.
Draw the line passing through these intersecting points, you will get perpendicular to the line $A Q$.
$3.$ Take distance of $4 \ cm$ in compass and mark an arc on the perpendicular above the line.
Draw a line parallel to line $A Q$ passing through this arc.
$4.$ From point $A$ measure an angle of $60$ degrees and draw the line which intersects the above$-$drawn line at some point label it as $D$.
$5.$ Using the procedure given in step $2$ again draw a perpendicular to line $A D$.
$6.$ Take distance of $3 \ cm$ in compass and mark an arc on the perpendicular above the line.
Draw a line parallel to line $AD$ passing through this arc which intersects the line $AQ$ at some point label it as $B$ and to other lines at point $C$.
$\text{ABCD}$ is the required parallelogram. View full question & answer→Question 164 Marks
Using ruler and compasses only, construct a parallelogram $\text{ABCD}$ using the following data : $AB = 6 \ cm, AD = 3 \ cm$ and$\angle DAB = 45^\circ$. If the bisector of $\angle DAB$ meets $DC$ at $P,$prove that $\angle APB$ is a right angle.
Answer
Steps:
$1.$ draw $A B=6 \ cm$.
$2.$ With $A$ as a center draw a line $A X$ such that $\angle B A X=45^{\circ}$.
$3.$ With $A$ as a center and radii, $3 \ cm$ draw an arc on $AD$.
$4.$ now with $D$ and $B$ as a center and radii $6 \ cm$ and $3 \ cm$ draw arcs cutting each other at $C$.
$5.$ Join $DC$ and $BC$.
$\text{ABCD}$ is the required parallelogram. Here
$\angle PAB =\angle APD\dots ...[$ Alternate angles $]$
$\angle C P B=\angle PBA\dots ...[$ Alternate angles $]$
Now,
$\angle DPA +\angle APB +\angle CPB =180^{\circ} \ldots \ldots . \text { (i) }$
Also, considering APB,
$\angle PAB +\angle PBA +\angle APB =180^{\circ} \ldots \ldots . \text { (ii) }$
Therefore, from $(i)$ and $(ii)$
$\angle APB =90^{\circ}$
Hence proved. View full question & answer→Question 174 Marks
Construct a regular hexagon of side $3.2 \ cm$
AnswerThe length of the side of a regular hexagon is equal to the radius of its circumcircle.
Steps of construction:
$1.$ Draw a circle of radius $3.2 \ cm$
$2.$ Taking any point $A$ on the circumference of the circle as a centre, draw arcs of same radii $($i.e. $3.2 \ cm)$ which cut the circumference at $B$ and $F$.
$3$. With $B$ and $F$ as centres, again draw two arcs of same radii which cut the circumference at $C$ and $E$ respectively.
$4$. With $C$ or $E$ as a centre, draw one more arc of the same radius which cuts the circumference at point $D.$
In this way, the circumference of the circle is divided into six equal parts.
$5$. Join $A B, B C, C D, D E, E F$ and $F A$.
$\text{A B C D E F}$ is the required regular hexagon. View full question & answer→Question 184 Marks
Construct a regular hexagon of side $2.5 \ cm$
AnswerThe length of the side of a regular hexagon is equal to the radius of its circumcircle.
Steps of construction:
$1.$ Draw a circle of radius $2.5 \ cm$
$2.$ Taking any point $A$ on the circumference of the circle as the centre, draw arcs of same radii $($i.e. $2.5 \ cm)$ which cut the circumference at $B$ and $F.$
$3.$ With $B$ and $F$ as centres, again draw two arcs of same radii which cut the circumference at $C$ and $E$ respectively.
$4$. With $C$ or $E$ as the centre, draw one more arc of the same radius which cuts the circumference at point $D$.
In this way, the circumference of the circle is divided into six equal parts.
$5.$ Join $AB , BC , CD , DE , EF$ and $FA$.
$\text{A B C D E F}$ is the required regular hexagon.
View full question & answer→Question 194 Marks
Construct a square $\text{ABCD},$ when: Perimeter $= 24 \ cm.$
AnswerThe perimeter of a square
$P=4 a$
Where $a$ is the length of each side
We have Perimeter $=24 \ cm$
Therefore,
$24=4 a$
$ 4=6$
Therefore the sides of the squares are of length $6 \ cm$.
Steps:
$1$. Draw a line segment $A b=6 \ cm$.
$2$. Draw $A P \perp A B$.
$3$. From $A P$ cut off $A D=6 \ cm$.
$4$. With $B$ as a center and radius, $6 \ cm$ draw an arc.
$5$. With $D$ as center and radius, $6 \ cm$ draw another arc cutting the former arc at $C$.
$6$. Join $B C$ and $C D$.
$\text{ABCD}$ is the required square. View full question & answer→Question 204 Marks
Construct a trapezium $\text{ABCD},$ when:$AB= CD = 3.2 \ cm, BC = 6.0 \ cm, AD = 4.4 \ cm$ and $AD \| BC.$
AnswerRough diagram $-$
Steps:
$1$. Draw $BC =6 \ cm$.
$2$. From $B C$ cut $B E=A D=4.4 \ cm$.
$3$. Draw triangle $D E C$ such that $D E=A B=2 \ cm$ and $C D=3.2 \ cm$.
$4$. Taking $B$ and $D$ as a center and radii $3.2 \ cm$ and $4.4 \ cm$ respectively, draw arcs cutting each other at $A$.
$5$. Join $A B$ and $A D$.
$\text{ABCD}$ is the required trapezium.
Construction $-$
View full question & answer→Question 214 Marks
Construct a rectangle $\text{ABCD},$ when :Area $= 36 \ cm^2$ and height $= 4.5 \ cm$.
AnswerGiven that the height $=4.5 \ cm$ and Area $=36 \ cm ^2$
We know that the area of rectangle $=$ base $\times$ Height
Therefore,
$36=$ base $\times \ 4.5$
Base $=8 \ cm$
With height $=4.5$ and base $=8 \ cm$, the rectangle is shown below :
Steps :
$1.$ Draw base $A B=8 \ cm$.
$2.$ With $A$ and $B$ as a center draw arc taking radius $=4.5 \ cm$ at $D$ and $C$
$3.$ Now join $AD , BC$, and $DC$.
$\text{ABCD}$ is the required rectangle. View full question & answer→Question 224 Marks
Construct a rectangle $\text{ABCD},$ when:Area $= 24 \ cm^2$ and base $= 4.8 \ cm^2$.
AnswerGiven that the base $=4.8 \ cm ^2$ and Area $=24 \ cm ^2$
We know that the area of the rectangle $=$ base $\times$ Height.
Therefore,
$24=4.8 \ \times$ height
Height $=5$
With base $=4.8 \ cm ^2$ and height $=5 \ cm ^2,$
the rectangle is shown below:
Steps :
$1.$ Draw base $A B=4.8 \ cm ^2$.
$2.$ With $A$ and $B$ as a center draw an arc taking radius $5 \ cm ^2$ at $D$ and $C$.
$3.$ Now join $A D, B C$, and $D C$.
$\text{ABCD}$ is the required rectangle. View full question & answer→