ICSE Class 7 Mathematics · ML Aggarwal Solutions Understanding ICSE Mathematics Class 7 · 10 solved questions
If the area of a triangle is 48 cm2 and the length of its base is 12 cm, then its height is
Given,
Area of the triangle = 48 cm2
Base = 12 cm
Let the height be h cm.
As we know,
Area of a triangle = $\frac{1}{2}$ x base x height
⇒ $\frac{1}{2}$ x 12 x h = 48
⇒ 6h = 48
⇒ h = $\frac{48}{6}$ = 8 cm
Hence, option 2 is the correct option.
The letter which has both line and rotational symmetry is
A letter has line symmetry if it can be folded along a line so that its two parts coincide exactly. It has rotational symmetry if it looks exactly the same after a rotation of less than 360° about its centre.
Let us consider the given letters one-by-one.
W has only one line of symmetry (vertical) and no rotational symmetry.
N has rotational symmetry of order 2 but no line of symmetry.
X has two lines of symmetry (vertical and horizontal) and rotational symmetry of order 2.
E has only one line of symmetry (horizontal) and no rotational symmetry.
The lines of symmetry are shown by dotted lines and the centre of rotation by a dot in the figure below:
We note that only the letter X has both line and rotational symmetry.
Hence, option 3 is the correct option.
If the circumference of a circular sheet is 44 cm, then find its area. Take π = $\frac{22}{7}$
Given,
Circumference of the circular sheet = 44 cm
Let the radius of the sheet be $r$ cm.
Circumference of the circular sheet = $2\pi r$
$$\begin{aligned}\Rightarrow 2\pi r = 44 \\[1em] \Rightarrow 2 \times \frac{22}{7} \times r = 44 \\[1em] \Rightarrow r = 44 \times \frac{7}{44} \\[1em] \Rightarrow r = 7 \text{ cm}\end{aligned}$$
Area of the circular sheet = $\pi r^2$
= $\frac{22}{7}$ x 7 x 7 cm2
= 154 cm2
Hence, the area of the circular sheet is 154 cm2.
Copy the adjoining figure and draw all lines of symmetry.
A line of symmetry divides a figure into two parts which coincide exactly when the figure is folded along that line.
The given figure is a square with two equal shaded quarter circles at two opposite corners, and two parallel line segments joining the end points of these two arcs. These line segments are parallel to the diagonal of the square which passes through the two shaded corners.
(i) On folding the square along the diagonal passing through the two shaded corners, each shaded quarter circle falls on itself and the two parallel line segments fall on each other. So, this diagonal is a line of symmetry.
(ii) On folding the square along the other diagonal, the two shaded quarter circles fall on each other and each line segment falls on itself. So, this diagonal is also a line of symmetry.
The lines of symmetry are shown by dotted lines in the figure below:
Hence, the figure has two lines of symmetry, along the two diagonals of the square.
Fill in the blanks:
(i) A triangular prism has ...... faces, ...... edges and ...... vertices.
(ii) Out of ...... faces of a square pyramid, ...... are triangular and ...... is/are square (s).
(i) A triangular prism has 2 triangular faces (top and bottom) and 3 rectangular faces.
∴ Number of faces = 2 + 3 = 5
A triangular prism has 2 triangular faces and 3 rectangular faces.
Each triangular face has 3 edges.
∴ Number of edges of the two triangular faces = 2 × 3 = 6
There are 3 additional edges joining the corresponding vertices of the two triangular faces.
∴ Total number of edges = 6 + 3 = 9
Each triangular face has 3 vertices.
∴ Number of vertices = 3 + 3 = 6
Hence, a triangular prism has 5 faces, 9 edges and 6 vertices.
(ii) In a square pyramid, the base is a square and the vertices of the base are joined to a common vertex, so all the other faces are triangular. As the square base has 4 sides, there are 4 triangular faces.
∴ Number of faces = 1 + 4 = 5
Hence, out of 5 faces of a square pyramid, 4 are triangular and 1 is square.
A figure looks exactly the same at its original figure after a rotation of 72°. At what other angles will this figure appear the same?
What can you say if the angle of rotation is (i) 60° (ii) 75°?
The smallest angle through which a figure can be rotated to look exactly the same as its original position is called the smallest angle of rotational symmetry.
For this question, we consider the given angle of rotation as the smallest angle of rotational symmetry.
As we know,
Order of rotational symmetry = $\frac{360^{\circ}}{\text{Smallest angle of rotational symmetry}}$
Given: Angle of rotation = 72°
∴ Order of rotational symmetry = $\frac{360^{\circ}}{72^{\circ}}$ = 5
The figure will look the same at every multiple of 72° up to 360°.
∴ Other angles = 2 × 72°, 3 × 72°, 4 × 72°, 5 × 72°
= 144°, 216°, 288°, 360°
Hence, the figure will appear the same at 144°, 216°, 288° and 360°.
(i) If the smallest angle of rotation is 60°:
Order of rotational symmetry = $\frac{360^{\circ}}{60^{\circ}}$ = 6
Since 60° divides 360° exactly, the figure will look the same at every multiple of 60° up to 360°.
∴ Other angles = 2 × 60°, 3 × 60°, 4 × 60°, 5 × 60°, 6 × 60°
= 120°, 180°, 240°, 300°, 360°
Hence, the figure will appear the same at 120°, 180°, 240°, 300° and 360°.
(ii) If the smallest angle of rotation is 75°:
Order of rotational symmetry = $\frac{360^{\circ}}{75^{\circ}}$ = 4.8
The order of rotational symmetry must be a whole number.
Since 75° does not divide 360° exactly, the order of rotational symmetry is not a whole number.
Hence, 75° is not possible as the smallest angle of rotational symmetry.
In the adjoining figure, a circle is inscribed in a semicircle of radius 7 cm. Find the area of the shaded region. Take π = $\frac{22}{7}$
Given,
Radius of the semicircle (R) = 7 cm
From the figure, the inscribed circle touches the diameter of the semicircle at its centre and touches the arc of the semicircle at its lowest point. So, the diameter of the inscribed circle is equal to the radius of the semicircle.
∴ Diameter of the circle = 7 cm
⇒ Radius of the circle (r) = $\frac{7}{2}$ cm = 3.5 cm
Area of the semicircle = $\frac{1}{2}$ x πR2
= $\frac{1}{2} \times \frac{22}{7}$ x 7 x 7 cm2
= 77 cm2
Area of the circle = πr2
= $\frac{22}{7}$ x 3.5 x 3.5 cm2
= 38.5 cm2
∴ Area of the shaded region = Area of the semicircle - area of the circle
= 77 cm2 - 38.5 cm2
= 38.5 cm2
Hence, the area of the shaded region is 38.5 cm2.
Draw an isometric sketch of a cuboid of length 5 units, breadth 4 units and height 3 units.
Take an isometric dot paper of side 1 unit.
Steps:
Draw the base lines of two front faces, one of length 5 units (length) and the other of length 4 units (breadth), meeting at a common vertex.
Draw three vertical line segments each of length 3 units (height), one at the common vertex and two at the free end points of the base lines.
For the top face, draw a parallelogram of sides 5 units and 4 units whose two adjacent sides are formed by joining the top free ends of the three vertical line segments.
The required isometric sketch of the cuboid is shown below:
In the adjoining figure, ABCD is a parallelogram with AB = 6 cm and AD = 5 cm. If DM is perpendicular to AB and area of parallelogram ABCD is 24 cm2, find the length of AM.
Given,
AB = 6 cm, AD = 5 cm
Area of parallelogram ABCD = 24 cm2
Taking AB as the base of the parallelogram, its height is DM.
As we know,
Area of a parallelogram = base x height
⇒ AB x DM = 24
⇒ 6 x DM = 24
⇒ DM = $\frac{24}{6}$ = 4 cm
As ΔAMD is right angled at M, by Pythagoras property,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
AD2 = AM2 + DM2
⇒ 52 = AM2 + 42
⇒ AM2 = 25 - 16 = 9
⇒ AM = $\sqrt{9}$ cm = 3 cm
Hence, the length of AM is 3 cm.
By using ruler and compass, construct a triangle ABC with BC = 6.5 cm, AC = 5 cm and ∠C = 60°.
Steps of construction:
Draw a line segment BC of length 6.5 cm.
With C as centre and any convenient radius, draw an arc cutting CB at P.
With P as centre and the same radius, draw another arc cutting the first arc at Q.
Join CQ and produce it to form ray CX. Then ∠BCX = 60°.
With C as centre and radius 5 cm, draw an arc cutting ray CX at A.
Join AB, then ABC is the required triangle.
Hence, ΔABC is the required triangle.
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