ICSE Class 7 Mathematics · ML Aggarwal Solutions Understanding ICSE Mathematics Class 7 · 29 solved questions
In the adjoining figure, if ∠AOC and ∠COB form a linear pair, then the value of x
Given,
∠AOC and ∠COB form a linear pair.
∴ ∠AOC + ∠COB = 180°
⇒ (2x)° + (x + 15)° = 180°
⇒ 2x° + x° + 15° = 180°
⇒ 3x° + 15° = 180°
⇒ 3x° = 180° − 15°
⇒ 3x° = 165°
⇒ x° = $\frac{165^{\circ}}{3}$
⇒ x° = 55°
⇒ x = 55
Hence, option 2 is the correct option.
An exterior angle of a triangle is 118°. If one of the two interior opposite angle is 54°, then the other interior opposite angle is
We know that,
An exterior angle of a triangle is equal to the sum of its two interior opposite angles.
Let the other interior opposite angle be x.
⇒ 118° = 54° + x
⇒ x = 118° − 54°
⇒ x = 64°
Hence, option 3 is the correct option.
In a right angled triangle, the lengths of two legs are 8 cm and 15 cm. The length of the hypotenuse is
In a right angled triangle, by Pythagoras theorem,
(Hypotenuse)2 = (Base)2 + (Perpendicular)2
Substituting the values, we get :
⇒ (Hypotenuse)2 = 82 + 152
⇒ (Hypotenuse)2 = 64 + 225
⇒ (Hypotenuse)2 = 289
⇒ Hypotenuse = $\sqrt{289}$ = 17 cm.
Hence, option 3 is the correct option.
If ΔABC ≅ ΔPRQ, then the correct statement is
Since, ΔABC ≅ ΔPRQ, the corresponding vertices are :
A ↔ P, B ↔ R and C ↔ Q
∴ The corresponding parts are :
AB = PR, BC = RQ, AC = PQ, ∠A = ∠P, ∠B = ∠R and ∠C = ∠Q.
So, AC = PQ is the correct statement.
Hence, option 4 is the correct option.
The number of lines that can be drawn parallel to a given line l through a point outside the line l is
Through a point outside a given line, one and only one line can be drawn parallel to the given line.
Hence, option 2 is the correct option.
If the area of a circle is numerically equal to its circumference, then the diameter of the circle is
Let the radius of the circle be $r$ units.
Given,
Area of circle = Circumference of circle
$$\begin{aligned}\Rightarrow \pi r^2 = 2\pi r \\[1em] \Rightarrow \frac{\pi r^2}{\pi r} = 2 \\[1em] \Rightarrow r = 2 \text{ units}.\end{aligned}$$
∴ Diameter = $2r$ = 2 × 2 = 4 units.
Hence, option 2 is the correct option.
A quadrilateral having exactly two lines of symmetry and a rotational symmetry of order 2 is a
A square has four lines of symmetry and rotational symmetry of order 4.
A parallelogram has no line of symmetry.
A kite has only one line of symmetry.
A rhombus has exactly two lines of symmetry (its diagonals) and rotational symmetry of order 2.
Hence, option 3 is the correct option.
Mode is the observation of the data
The observation which occurs maximum number of times in the given data is called mode of the data.
Hence, option 3 is the correct option.
An angle is 30° more than one-half of its complement. Find the angle.
Let the angle be $x^{\circ}$.
Then, its complement = $(90 − x)^{\circ}$.
Given,
Angle = $\frac{1}{2}$ × Complement + 30°
$$\begin{aligned}\Rightarrow x = \frac{1}{2}(90 - x) + 30 \\[1em] \Rightarrow x = 45 - \frac{x}{2} + 30 \\[1em] \Rightarrow x + \frac{x}{2} = 75 \\[1em] \Rightarrow \frac{3x}{2} = 75 \\[1em] \Rightarrow x = \frac{75 \times 2}{3} \\[1em] \Rightarrow x = 50\end{aligned}$$
Hence, the angle is 50°.
In the adjoining figure, find the value of x.
From figure,
The angle at the top vertex of the triangle and the angle of 70° are vertically opposite angles.
∴ Angle at the top vertex of the triangle = 70°.
We know that,
An exterior angle of a triangle is equal to the sum of its two interior opposite angles.
⇒ 125° = 70° + x
⇒ x = 125° − 70°
⇒ x = 55°
Hence, the value of x = 55°.
A die is thrown once. Find the probability of getting
(i) a prime number
(ii) a composite number.
When a die is thrown once, the possible outcomes are the numbers 1, 2, 3, 4, 5 and 6.
So, the total number of possible outcomes = 6
(i) The event is getting a prime number and the prime numbers (outcomes) are 2, 3 and 5.
So, the number of outcomes favourable to the event getting a prime number = 3
∴ P(getting a prime number) = $\frac{3}{6} = \frac{1}{2}$
Hence, the probability of getting a prime number = $\bm{\frac{1}{2}}$.
(ii) The event is getting a composite number and the composite numbers (outcomes) are 4 and 6.
So, the number of outcomes favourable to the event getting a composite number = 2
∴ P(getting a composite number) = $\frac{2}{6} = \frac{1}{3}$
Hence, the probability of getting a composite number = $\bm{\frac{1}{3}}$.
You want to show that ΔDEF ≅ ΔPQR by SAS congruence rule. It is given that ∠E = ∠Q, you need to have
(i) EF = ......
(ii) PQ = ......
Given,
∠E = ∠Q
For ΔDEF ≅ ΔPQR by S.A.S. congruence rule, the two sides including ∠E must be equal to the corresponding two sides including ∠Q.
In ΔDEF, the sides including ∠E are DE and EF.
In ΔPQR, the sides including ∠Q are PQ and QR.
Since, D ↔ P, E ↔ Q and F ↔ R,
∴ EF = QR and PQ = DE.
Hence, (i) EF = QR and (ii) PQ = DE.
In the adjoining figure, two adjacent sides of a parallelogram are 15 cm and 10 cm. If the distance between the longer sides is 8 cm, find the
(i) area of parallelogram
(ii) the distance between the shorter sides.
(i) Taking the longer side AB = 15 cm as base, the corresponding height = 8 cm.
By formula,
Area of parallelogram = Base × Height
Substituting the values, we get :
Area of parallelogram = 15 × 8
= 120 cm2
Hence, the area of parallelogram = 120 cm2.
(ii) Let the distance between the shorter sides be h cm.
Taking the shorter side BC = 10 cm as base, the corresponding height = h cm.
∴ Area of parallelogram = BC × h
⇒ 120 = 10 × h
⇒ h = $\frac{120}{10}$
⇒ h = 12 cm.
Hence, the distance between the shorter sides = 12 cm.
Find the mean age of six students whose ages (in years) are:
15, 13, 16, 13, 14, 16
Sum of all observations = (15 + 13 + 16 + 13 + 14 + 16) years
= 87 years
and, number of observations = 6
By formula,
Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$
Substituting the values, we get :
Mean age = $\frac{87}{6}$ years
= 14.5 years
= 14 years + 0.5 × 12 months
= 14 years 6 months
Hence, the mean age of the six students = 14 years 6 months.
In the adjoining figure, lines l and m are parallel. Find the values of x, y and z.
From figure,
Since, l ∥ m,
x = 65° [Alternate angles are equal]
The angles x, 45° and z are adjacent angles at the same point on line l, and together they make a straight angle.
⇒ x + 45° + z = 180° [Sum of angles on a straight line = 180°]
⇒ 65° + 45° + z = 180°
⇒ 110° + z = 180°
⇒ z = 180° − 110°
⇒ z = 70°
Since, l ∥ m,
y = z [Alternate angles are equal]
∴ y = 70°
Hence, x = 65°, y = 70° and z = 70°.
In the adjoining figure, BC = AC. Find the value of x.
From figure,
∠ABC + 155° = 180° [Linear pair]
⇒ ∠ABC = 180° − 155°
⇒ ∠ABC = 25°
In ΔABC,
Given, BC = AC
∴ ∠BAC = ∠ABC = 25° [Angles opposite to equal sides are equal]
By angle sum property of triangle,
∠BAC + ∠ABC + ∠ACB = 180°
⇒ 25° + 25° + x = 180°
⇒ 50° + x = 180°
⇒ x = 180° − 50°
⇒ x = 130°
Hence, the value of x = 130°.
If the lengths of the two sides of a triangle are 6 cm and 8.5 cm, then what can be the length of the third side?
We know that,
In a triangle, the sum of the lengths of any two sides is greater than the length of the third side, and the difference of the lengths of any two sides is less than the length of the third side.
Let the length of the third side be x cm.
Sum of the given two sides = 6 + 8.5 = 14.5 cm
∴ x < 14.5
Difference of the given two sides = 8.5 − 6 = 2.5 cm
∴ x > 2.5
∴ 2.5 < x < 14.5
Hence, the length of the third side can be any length greater than 2.5 cm but less than 14.5 cm.
In the adjoining figure, AD is perpendicular bisector of $\overline{BC}$.
(i) State three pairs of equal parts in ΔABD and ΔACD.
(ii) Is ΔABD ≅ ΔACD? Give reasons.
(iii) Is ABC an isosceles triangle? Give reasons.
(i) In ΔABD and ΔACD,
BD = CD [∵ AD bisects BC]
∠ADB = ∠ADC = 90° [∵ AD ⊥ BC]
AD = AD [Common side]
Hence, the three pairs of equal parts are BD = CD, ∠ADB = ∠ADC (each = 90°) and AD = AD.
(ii) From part (i),
In ΔABD and ΔACD,
BD = CD
∠ADB = ∠ADC
AD = AD
∴ ΔABD ≅ ΔACD [By S.A.S. congruence rule]
Hence, ΔABD ≅ ΔACD by S.A.S. congruence rule.
(iii) Since, ΔABD ≅ ΔACD,
∴ AB = AC [By C.P.C.T.]
So, two sides of ΔABC are equal.
Hence, ABC is an isosceles triangle as AB = AC.
Draw a line, say AB, take a point P outside line AB. Through P, draw a line parallel to line AB using ruler and compasses only.
Steps of construction :
Draw a line AB and take a point P outside it.
Take any point Q on AB and join P and Q.
With Q as centre and a convenient radius, draw an arc cutting QB at R and QP at S.
With P as centre and the same radius, draw an arc cutting PQ at T.
With the compasses, measure the length RS. With T as centre and radius equal to RS, draw an arc cutting the arc drawn in step 4 at U, on the side of PQ opposite to R.
Join PU and produce it on both sides.
By construction, ∠TPU = ∠SQR, i.e. ∠QPU = ∠PQB.
These are alternate angles formed by the transversal PQ with lines PU and AB. Since the alternate angles are equal, PU ∥ AB.
Hence, the line through P is the required line parallel to line AB.
In the adjoining figure, ABCD is a rectangle with AB = 20 cm and BC = 14 cm. Two semicircles are cut from each of two breadths as diameters. Find
(i) the area of the shaded region
(ii) the perimeter of the shaded region.
Take $\pi = \frac{22}{7}$
(i) Area of rectangle ABCD = AB × BC
= 20 × 14
= 280 cm2
Diameter of each semicircle = Breadth = 14 cm
∴ Radius of each semicircle (r) = $\frac{14}{2}$ = 7 cm
Area of two semicircles = $2 \times \frac{1}{2}\pi r^2$
= πr2
= $\frac{22}{7}$ × 7 × 7
= 22 × 7
= 154 cm2.
Area of shaded region = Area of rectangle − Area of two semicircles
= 280 − 154
= 126 cm2
Hence, the area of the shaded region = 126 cm2.
(ii) Length of each semicircular arc = πr
= $\frac{22}{7}$ × 7
= 22 cm.
Perimeter of shaded region = AB + DC + 2 × Length of semicircular arc
= 20 + 20 + 2 × 22
= 40 + 44
= 84 cm
Hence, the perimeter of the shaded region = 84 cm.
Draw the reflection of the letter E in the given mirror line shown dotted.
In a reflection, each point of the image is at the same distance behind the mirror line as the corresponding point of the object is in front of it. So, the image of letter E is laterally inverted.
The reflection of letter E in the given mirror line is shown below :
Hence, the above figure shows the reflection of letter E in the given mirror line.
Three cubes each with edge 2 units are placed side by side to form a cuboid. Find the dimensions of the cuboid so formed and draw an isometric sketch of this cuboid.
When three cubes each with edge 2 units are placed side by side, the edges along the length get added, while the breadth and height remain equal to the edge of one cube.
Length of cuboid = 2 + 2 + 2 = 6 units
Breadth of cuboid = 2 units
Height of cuboid = 2 units
Steps of construction :
Take an isometric dot paper of side 1 unit.
Draw the base lines of the two front faces of length 6 units and breadth 2 units.
Draw three vertical lines each of height 2 units, one on the middle vertex and the other two on the end points of these line segments.
For the top face, draw a parallelogram of length 6 units and breadth 2 units whose adjacent sides are formed by joining the top free ends of the three vertical line segments to get the required cuboid.
Hence, the dimensions of the cuboid are 6 units × 2 units × 2 units.
Draw two nets of a regular tetrahedron.
A regular tetrahedron has four faces, each of which is an equilateral triangle.
Two nets of a regular tetrahedron are shown below :
Hence, the above figures are two nets of a regular tetrahedron.
A boy scored the following marks in various class tests during a year, each test being marked out of 20:
15, 17, 16, 7, 9, 12, 14, 16, 3, 19, 12, 16
(i) Arrange the marks in ascending order.
(ii) What are his modal marks?
(iii) What are his median marks?
(iv) What are his mean marks?
(i) On arranging the given marks in ascending order of their magnitudes, we get :
3, 7, 9, 12, 12, 14, 15, 16, 16, 16, 17, 19
Hence, the marks in ascending order are 3, 7, 9, 12, 12, 14, 15, 16, 16, 16, 17, 19.
(ii) In the given data, the number 16 appears maximum times, i.e. 3 times.
∴ Mode = 16
Hence, the boy's modal marks = 16.
(iii) As the number of terms (n) is 12 and 12 is even :
$$\begin{aligned}\therefore \text{Median} = \frac{1}{2}\Big[\Big(\frac{n}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{n}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}\Big[\Big(\frac{12}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{12}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}[6^{\text{th}}\text{ term} + 7^{\text{th}}\text{ term}] \\[1em] = \frac{1}{2}(14 + 15) \\[1em] = \frac{1}{2} \times 29 \\[1em] = 14.5\end{aligned}$$
Hence, the boy's median marks = 14.5.
(iv) Sum of all observations = 3 + 7 + 9 + 12 + 12 + 14 + 15 + 16 + 16 + 16 + 17 + 19
= 156
and, number of observations = 12
By formula,
Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$
Substituting the values, we get :
Mean marks = $\frac{156}{12}$
= 13
Hence, the boy's mean marks = 13.
In the adjoining figure, all measurements are in centimeters. If AD is perpendicular to BC, find the length of AB.
From figure,
AD = 15 cm, AC = 25 cm and BC = 28 cm.
In right angled ΔADC, by Pythagoras theorem,
AC2 = AD2 + DC2
⇒ 252 = 152 + DC2
⇒ 625 = 225 + DC2
⇒ DC2 = 625 − 225
⇒ DC2 = 400
⇒ DC = $\sqrt{400}$
⇒ DC = 20 cm.
BD = BC − DC
= 28 − 20
= 8 cm
In right angled ΔADB, by Pythagoras theorem,
AB2 = AD2 + BD2
⇒ AB2 = 152 + 82
⇒ AB2 = 225 + 64
⇒ AB2 = 289
⇒ AB = $\sqrt{289}$
⇒ AB = 17 cm.
Hence, the length of AB = 17 cm.
In the adjoining figure, show that ΔABC ≅ ΔDBC. Hence, find the values of x and y.
In ΔABC and ΔDBC,
AB = DB [From figure]
AC = DC [From figure]
BC = BC [Common side]
∴ ΔABC ≅ ΔDBC [By S.S.S. congruence rule]
Since, ΔABC ≅ ΔDBC,
∠ABC = ∠DBC [By C.P.C.T.]
⇒ (2y + 6)° = 56°
⇒ 2y = 56 − 6
⇒ 2y = 50
⇒ y = $\frac{50}{2}$
⇒ y = 25
∠DCB = ∠ACB = 40° [By C.P.C.T.]
In ΔDBC, by angle sum property of triangle,
∠BDC + ∠DBC + ∠DCB = 180°
⇒ x + 56° + 40° = 180°
⇒ x + 96° = 180°
⇒ x = 180° − 96°
⇒ x = 84°
Hence, ΔABC ≅ ΔDBC, x = 84° and y = 25.
Using ruler and compass only, construct a triangle ABC with BC = 7.5 cm, ∠B = 60° and ∠A = 90°
In ΔABC, by angle sum property of triangle,
∠A + ∠B + ∠C = 180°
⇒ 90° + 60° + ∠C = 180°
⇒ 150° + ∠C = 180°
⇒ ∠C = 180° − 150°
⇒ ∠C = 30°
So, we construct ΔABC with BC = 7.5 cm, ∠B = 60° and ∠C = 30°.
Steps of construction :
Draw a line segment BC = 7.5 cm.
With B as centre and a convenient radius, draw an arc cutting BC at P. With P as centre and the same radius, draw an arc cutting the previous arc at Q. Join BQ and produce it to X. Then, ∠CBX = 60°.
With C as centre and a convenient radius, draw an arc cutting CB at R. With R as centre and the same radius, draw an arc cutting the previous arc at S, so that ∠BCS = 60°.
With R and S as centres and radius more than half of RS, draw two arcs intersecting each other at T. Join CT and produce it to Y. Then, ∠BCY = 30°.
BX and CY intersect each other at A.
On measuring, ∠BAC = 90°.
Hence, ABC is the required triangle.
Anjali took a wire of length 88 cm and bent it into the shape of a circle. Find the area enclosed by that circle. If the same wire is bent in the shape of a square, then find the area enclosed by that square. Which shape encloses more area and by how much? Take $\pi = \frac{22}{7}$
Length of wire = 88 cm
When the wire is bent into the shape of a circle,
Circumference of circle = Length of wire
Let the radius of the circle be $r$ cm.
$$\begin{aligned}\Rightarrow 2\pi r = 88 \\[1em] \Rightarrow 2 \times \frac{22}{7} \times r = 88 \\[1em] \Rightarrow \frac{44}{7}r = 88 \\[1em] \Rightarrow r = \frac{88 \times 7}{44} \\[1em] \Rightarrow r = 14 \text{ cm}.\end{aligned}$$
Area enclosed by the circle = $\pi r^2$
= $\frac{22}{7}$ × 14 × 14
= 22 × 2 × 14
= 616 cm2.
When the same wire is bent into the shape of a square,
Perimeter of square = Length of wire
⇒ 4 × Side = 88
⇒ Side = $\frac{88}{4}$ = 22 cm
Area enclosed by the square = (Side)2
= 22 × 22
= 484 cm2
Since, 616 > 484, the circle encloses more area.
Difference in areas = 616 − 484
= 132 cm2
Hence, the area enclosed by the circle = 616 cm2, the area enclosed by the square = 484 cm2 and the circle encloses more area by 132 cm2.
Given alongside is the data of school going students (boys and girls):
| Mode of transport | Boys | Girls |
|---|---|---|
| School bus | 75 | 135 |
| Walking | 120 | 60 |
| Bicycle | 240 | 180 |
| Other vehicles | 150 | 90 |
Draw a double bar graph to represent the above data and hence answer the following questions:
(i) Which mode of transport is used by more students?
(ii) Which mode of transport is used by more number of girls than the number of boys?
Take the modes of transport along X-axis and the number of students along Y-axis.
Scale : Take 1 unit height along Y-axis = 30 students.
The double bar graph representing the given data is shown below :
(i) Total number of students using each mode of transport :
School bus = 75 + 135 = 210
Walking = 120 + 60 = 180
Bicycle = 240 + 180 = 420
Other vehicles = 150 + 90 = 240
Since, 420 is the greatest, bicycle is used by the maximum number of students.
Hence, bicycle is used by more students.
(ii) From the bar graph, the bar for girls is higher than the bar for boys only for school bus.
Girls using school bus = 135 and boys using school bus = 75.
Hence, school bus is used by more number of girls than the number of boys.
Press ↓ to reveal a solution and move to the next question. Back to ML Aggarwal Solutions Understanding ICSE Mathematics Class 7.