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Practice each question, then open the answer or worked solution.
The solid shapes are two-dimensional.
Ans: The solid shapes are three-dimensional.
Examples of solid shapes are cone, cube etc.
Therefore, the given statement is False.
Triangle with length of sides as ${\mathbf{5cm}}$, ${\mathbf{6cm}}$ and ${\mathbf{11cm}}$ can be constructed.
Ans: Condition for constructing a triangle is that Sum of any two sides of a triangle should be greater than the third side.
But here, $6{\text{cm}} + 5{\text{cm}} = 11{\text{cm}}$ and the condition is not satisfied.
Therefore, the given statement is False.
Draw the top, side and front views of the solids given below in Figures 12.21 and 12.22:
(i)
(ii)
Ans: For the given figure,
Ans:
Draw a solid using the top. side and front views as shown below. {Use Isometric dot paper}.
Ans:
Construct a right-angled triangle whose hypotenuse measures ${\mathbf{5 cm}}$ and one of the other sides measures ${\mathbf{3}}{\mathbf{.2 cm}}{\mathbf{.}}$
Ans: Steps of construction:
(1) Draw a line AB= 3.2 cm.
(2) Construct a right angle ${\mathbf{90}}^\circ $ at point B, that is $\angle {\text{ABY} = {90}^{\circ} }$.
(3) From point A, cut an arc 5 cm on BY at ${\text{C}}$
(4) Join ${\text{AC}}$.
Therefore $\Delta {\text{ABC}}$ is the required triangle, having hypotenuse ${\text{AC = 5 cm}}$ and ${\text{AB = 3}}{\text{.2 cm}}{\text{.}}$
Construct a right-angled isosceles triangle with one side (other than hypotenuse) of length ${\mathbf{4}}{\mathbf{.5 cm}}$.
Ans: Steps of construction:
(1) Draw a line${\text{AB = 4}}{\text{.5 cm}}$.
(2) Construct a right angle $\left( {90^\circ } \right)$ at point ${\text{B}}$, that is $\angle ABY = {90}^{\circ}.$
(3) From point ${\text{B}}$, cut an arc ${\text{4}}{\text{.5 cm}}$ on ${\text{BY}}$ at ${\text{C}}$.
(4) Join ${\text{AC}}$.
Therefore,$\Delta {\text{ABC}}$ is the required triangle with one side of length ${\text{4}}{\text{.5 cm}}$.
Draw two parallel lines at a distance of ${\mathbf{2}}{\mathbf{.2cm}}$ apart.
Ans: Steps of construction:
(1) Draw a line ${\text{l}}$ and mark a point ${\text{C}}$ outside it.
(2) Take a point ${\text{B}}$ on line ${\text{l}}$ and join ${\text{BC}}$.
(3) Draw a line parallel to the line ${\text{l}}$ passing through ${\text{C}}$.
(4) Mark a point ${\text{D}}$ on line ${\text{m}}$, at a distance of $2.2{\text{cm}}$ from ${\text{C}}$.
(5) Mark angle arc at point D which should be equal to angle arc at point C.
(5) Through ${\text{D}}$ draw a line at the angle arc which will be parallel to the BC line.
Draw an isosceles triangle with each of equal sides of length ${\mathbf{3cm}}$ and the angle between them as ${\mathbf{45^\circ }}$.
Ans: Steps of construction:
(1) Draw a line segment ${\text{AB}}$ of length $3{\text{cm}}$.
(2) Draw an angle of$45^\circ $ on point ${\text{B}}$ and produce it to ray ${\text{Y}}$.
(3) With ${\text{B}}$ as center, draw an arc of $3{\text{cm}}$ which intersects ray ${\text{BY}}$ at ${\text{C}}$.
(4) Join ${\text{AC}}$.
Thus,$\Delta {\text{ABC}}$ is the required isosceles triangle.
Draw a triangle whose sides are of lengths $4 cm, 5 cm$ and $7 cm$.
Ans: Let us assume that given sides are ${\text{BC = 7 cm, AB = 4 cm}}$ and ${\text{AC = 5 cm}}$.
Steps of construction:
(1) Draw a line ${\text{BC = 7 cm}}.$
(2) With center ${\text{B}}$ and radius ${\text{4 cm}}$ draw an arc.
(3) With center ${\text{C}}$ and radius ${\text{5 cm}},$ draw an arc which cuts the previous arc at ${\text{A}}$.
(4) Join ${\text{AB}} $ and ${\text{AC}}$.
Hence,$\Delta {\text{ABC}}$ is the required triangle in which ${\text{BC = 7 cm, AB = 4 cm}}$ and ${\text{AC = 5 cm}}.$
Construct an obtuse angled triangle which has a base of ${\mathbf{5}}{\mathbf{.5 cm}}$ and base angles of ${\mathbf{30^\circ }}$ and ${\mathbf{120^\circ }}$.
Ans: Steps of construction:
(1) Draw a line segment ${\text{BC}}$ of length ${\text{5}}{\text{.5 cm}}.$
(2) Draw an angle of $120^\circ $ on point ${\text{B}}$ and produce it to ray ${\text{Y}}$.
(3) Draw an angle of $30^\circ $ on point ${\text{C}}$ and produce it to ray ${\text{X}}$.
(4) Extend ${\text{BY}}$ and ${\text{CX}}$ to intersect at point ${\text{A}}$.
Hence, $\Delta {\text{ABC}}$ is the required triangle.