ICSE Class 7 Mathematics · Selina Solutions Concise Mathematics Class 7 ICSE · 28 solved questions
Triangle ABC is right-angled at vertex A. Calculate the length of BC, if AB = 18 cm and AC = 24 cm.
The given triangle is right-angled at vertex A means angle A = 90° and so BC is hypotenuse.
According to Pythagoras theorem :
⇒ Hypotenuse2 = Base2 + Perpendicular2
⇒ BC2 = AB2 + AC2
⇒ BC2 = 182 + 242
⇒ BC2 = 324 + 576
⇒ BC2 = 900
⇒ BC = $\sqrt{900}$ cm = 30 cm
Hence, BC = 30 cm.
Triangle XYZ is right-angled at vertex Z. Calculate the length of YZ, if XY = 13 cm and XZ = 12 cm.
The given triangle is right-angled at vertex Z means angle Z = 90° and so XY is hypotenuse.
According to Pythagoras theorem :
⇒ XY2 = XZ2 + YZ2
⇒ 132 = 122 + YZ2
⇒ 169 = 144 + YZ2
⇒ YZ2 = 169 - 144
⇒ YZ2 = 25
⇒ YZ = $\sqrt{25}$ cm = 5 cm
Hence, YZ = 5 cm.
Triangle PQR is right-angled at vertex R. Calculate the length of PR, if:
PQ = 2.6 cm and QR = 2.4 cm.
The given triangle is right-angled at vertex R means angle R = 90° and so PQ is hypotenuse.
According to Pythagoras theorem :
⇒ PQ2 = PR2 + QR2
⇒ (2.6)2 = PR2 + (2.4)2
⇒ 6.76 = PR2 + 5.76
⇒ PR2 = 6.76 - 5.76
⇒ PR2 = 1
⇒ PR = $\sqrt{1}$ cm = 1 cm
Hence, PR = 1 cm.
The sides of certain triangles are given below. Find whether they are right angled triangles or not.
(i) 16 cm, 20 cm and 12 cm
(ii) 6 m, 9 m and 13 m
A triangle is a right-angled triangle if the square of its longest side is equal to the sum of the squares of the other two sides.
(i) The longest side = 20 cm
⇒ 202 = 400
⇒ 162 + 122 = 256 + 144 = 400
Since, 202 = 162 + 122
Hence, the triangle with sides 16 cm, 20 cm and 12 cm is a right angled triangle.
(ii) The longest side = 13 m
⇒ 132 = 169
⇒ 62 + 92 = 36 + 81 = 117
Since, 132 ≠ 62 + 92
Hence, the triangle with sides 6 m, 9 m and 13 m is not a right angled triangle.
In the adjoining figure, angle BAC = 90°, AC = 400 m and AB = 300 m. Find the length of BC
Since angle BAC = 90°, BC is the hypotenuse.
According to Pythagoras theorem :
⇒ BC2 = AB2 + AC2
⇒ BC2 = 3002 + 4002
⇒ BC2 = 90000 + 160000
⇒ BC2 = 250000
⇒ BC = $\sqrt{250000}$ m = 500 m
Hence, BC = 500 m.
In the given figures, angle ACP = ∠BDP = 90°, AC = 12 m, BD = 9 m and PA = PB = 15 m. Find:
(i) CP
(ii) PD
(iii) CD
(i) In right-angled triangle ACP, angle ACP = 90°, so PA is the hypotenuse.
Applying Pythagoras theorem :
⇒ PA2 = AC2 + CP2
⇒ 152 = 122 + CP2
⇒ 225 = 144 + CP2
⇒ CP2 = 225 - 144
⇒ CP2 = 81
⇒ CP = $\sqrt{81}$ m = 9 m
Hence, CP = 9 m.
(ii) In right-angled triangle BDP, angle BDP = 90°, so PB is the hypotenuse.
Applying Pythagoras theorem :
⇒ PB2 = BD2 + PD2
⇒ 152 = 92 + PD2
⇒ 225 = 81 + PD2
⇒ PD2 = 225 - 81
⇒ PD2 = 144
⇒ PD = $\sqrt{144}$ m = 12 m
Hence, PD = 12 m.
(iii) Since C, P and D lie on the same line :
CD = CP + PD = 9 m + 12 m = 21 m
Hence, CD = 21 m.
In triangle PQR, angle Q = 90°, find:
(i) PR, if PQ = 8 cm and QR = 6 cm
(ii) PQ, if PR = 34 cm and QR = 30 cm
Since angle Q = 90°, PR is the hypotenuse.
(i) Applying Pythagoras theorem :
⇒ PR2 = PQ2 + QR2
⇒ PR2 = 82 + 62
⇒ PR2 = 64 + 36
⇒ PR2 = 100
⇒ PR = $\sqrt{100}$ cm = 10 cm
Hence, PR = 10 cm.
(ii) Applying Pythagoras theorem :
⇒ PR2 = PQ2 + QR2
⇒ 342 = PQ2 + 302
⇒ 1156 = PQ2 + 900
⇒ PQ2 = 1156 - 900
⇒ PQ2 = 256
⇒ PQ = $\sqrt{256}$ cm = 16 cm
Hence, PQ = 16 cm.
Show that the triangle ABC is a right-angled triangle; if:
AB = 9 cm, BC = 40 cm and AC = 41 cm.
A triangle is a right-angled triangle if the square of its longest side is equal to the sum of the squares of the other two sides.
The longest side = AC = 41 cm
AC2 = 412 = 1681
AB2 + BC2 = 92 + 402 = 81 + 1600 = 1681
Since, AC2 = AB2 + BC2
The angle opposite the longest side AC, i.e. angle B = 90°.
Hence, triangle ABC is a right-angled triangle, right-angled at B.
In the given figure, angle ACB = 90° = angle ACD. If AB = 10 cm, BC = 6 cm and AD = 17 cm, find:
(i) AC
(ii) CD
(i) In right-angled triangle ACB, angle ACB = 90°, so AB is the hypotenuse.
Applying Pythagoras theorem :
⇒ AB2 = AC2 + BC2
⇒ 102 = AC2 + 62
⇒ 100 = AC2 + 36
⇒ AC2 = 100 - 36
⇒ AC2 = 64
⇒ AC = $\sqrt{64}$ cm = 8 cm
Hence, AC = 8 cm.
(ii) In right-angled triangle ACD, angle ACD = 90°, so AD is the hypotenuse.
Applying Pythagoras theorem :
⇒ AD2 = AC2 + CD2
⇒ 172 = 82 + CD2
⇒ 289 = 64 + CD2
⇒ CD2 = 289 - 64
⇒ CD2 = 225
⇒ CD = $\sqrt{225}$ cm = 15 cm
Hence, CD = 15 cm.
In the given figure, angle ADB = 90°, AC = AB = 26 cm and BD = DC. If the length of AD = 24 cm; find the length of BC.
In right-angled triangle ADB, angle ADB = 90°, so AB is the hypotenuse.
Applying Pythagoras theorem :
⇒ AB2 = AD2 + BD2
⇒ 262 = 242 + BD2
⇒ 676 = 576 + BD2
⇒ BD2 = 676 - 576
⇒ BD2 = 100
⇒ BD = $\sqrt{100}$ cm = 10 cm
Since BD = DC,
BC = BD + DC = 10 cm + 10 cm = 20 cm
Hence, BC = 20 cm.
In the given figure, AD = 13 cm, BC = 12 cm, AB = 3 cm and angle ACD = angle ABC = 90°. Find the length of DC.
In right-angled triangle ABC, angle ABC = 90°, so AC is the hypotenuse.
Applying Pythagoras theorem :
⇒ AC2 = AB2 + BC2
⇒ AC2 = 32 + 122
⇒ AC2 = 9 + 144
⇒ AC2 = 153
In right-angled triangle ACD, angle ACD = 90°, so AD is the hypotenuse.
Applying Pythagoras theorem :
⇒ AD2 = AC2 + CD2
⇒ 132 = 153 + CD2
⇒ 169 = 153 + CD2
⇒ CD2 = 169 - 153
⇒ CD2 = 16
⇒ CD = $\sqrt{16}$ cm = 4 cm
Hence, DC = 4 cm.
A ladder, 6.5 m long, rests against a vertical wall. If the foot of the ladder is 2.5 m from the foot of the wall, find upto how much height does the ladder reach?
Let AC be the ladder = 6.5 m (hypotenuse), BC be the distance of its foot from the wall = 2.5 m and AB be the height reached on the wall. The wall is vertical, so angle B = 90°.
According to Pythagoras theorem :
⇒ Hypotenuse2 = Perpendicular2 + Base2
⇒ AC2 = AB2 + BC2
⇒ (6.5)2 = AB2 + (2.5)2
⇒ 42.25 = AB2 + 6.25
⇒ AB2 = 42.25 - 6.25
⇒ AB2 = 36
⇒ AB = $\sqrt{36}$ m = 6 m
Hence, the ladder reaches a height of 6 m.
A boy first goes 5 m due north and then 12 m due east. Find the distance between the initial and the final positions of the boy.
Let the boy start at O, go 5 m due north to A and then 12 m due east to B.
⇒ OA = 5 m and AB = 12 m
OB is the distance between the initial and the final positions of the boy.
Since north and east are perpendicular directions, angle OAB = 90°.
According to Pythagoras theorem in triangle OAB:
⇒ Hypotenuse2 = Perpendicular2 + Base2
⇒ OB2 = OA2 + AB2
⇒ OB2 = 52 + 122
⇒ OB2 = 25 + 144
⇒ OB2 = 169
⇒ OB = $\sqrt{169}$ m = 13 m
Hence, the distance between the initial and final positions of the boy is 13 m.
Use the information given in the adjoining figure to find the length of AD.
From the figure,
DO = CB = 24 cm, OB = DC = 10 cm and angle AOD = 90°.
AO = AB - OB = 20 cm - 10 cm = 10 cm
In right-angled triangle AOD, AD is the hypotenuse.
Applying Pythagoras theorem :
⇒ AD2 = AO2 + OD2
⇒ AD2 = 102 + 242
⇒ AD2 = 100 + 576
⇒ AD2 = 676
⇒ AD = $\sqrt{676}$ cm = 26 cm
Hence, AD = 26 cm.
The value of x is:
In the given figure,
Base = x, Perpendicular = 12 m, Hypotenuse = 15 m
According to Pythagoras theorem :
⇒ Hypotenuse2 = Base2 + Perpendicular2
⇒ Base2 = Hypotenuse2 - Perpendicular2
⇒ x2 = 152 - 122
⇒ x2 = 225 - 144
⇒ x2 = 81
⇒ x = $\sqrt{81}$ = 9 m
Hence, option 2 is the correct option.
A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. The original height of the tree is:
The standing part of the tree is 5 m and the broken part (from the break to the top) touches the ground 12 m from the base, forming the hypotenuse of a right-angled triangle.
Let the broken part = h.
According to Pythagoras theorem :
⇒ h2 = 52 + 122
⇒ h2 = 25 + 144
⇒ h2 = 169
⇒ h = $\sqrt{169}$ = 13 m
Original height of the tree = standing part + broken part = 5 m + 13 m = 18 m
Hence, option 3 is the correct option.
From a point P, a boy travels 12 km due east and then travels 9 km due north to point Q. The shortest distance between points P and Q is:
Since east and north are perpendicular directions, the shortest distance PQ is the hypotenuse of a right-angled triangle with legs 12 km and 9 km.
According to Pythagoras theorem :
⇒ PQ2 = 122 + 92
⇒ PQ2 = 144 + 81
⇒ PQ2 = 225
⇒ PQ = $\sqrt{225}$ = 15 km
Hence, option 1 is the correct option.
The value of x is:
The given triangle is an isosceles right-angled triangle in which the two equal sides (legs) are each x and the hypotenuse is $10\sqrt{2}$.
According to Pythagoras theorem :
⇒ Hypotenuse2 = x2 + x2
⇒ $\left(10\sqrt{2}\right)$2 = 2x2
⇒ 200 = 2x2
⇒ x2 = 100
⇒ x = $\sqrt{100}$ = 10
Hence, option 3 is the correct option.
The value of x is:
The altitude from the top vertex divides the base into two parts, 6 and 15, and is common to both right-angled triangles.
First, find the altitude h using the left right-angled triangle :
⇒ Hypotenuse2 = Perpendicular2 + Base2
⇒ 102 = 62 + h2
⇒ 100 = 36 + h2
⇒ h2 = 100 - 36
⇒ h2 = 64
⇒ h = $\sqrt{64}$ = 8
Now, in the right-angled triangle on the right, x is the hypotenuse with base 15 and height (h = 8) :
⇒ x2 = 152 + 82
⇒ x2 = 225 + 64
⇒ x2 = 289
⇒ x = $\sqrt{289}$ = 17
Hence, option 4 is the correct option.
The value of x is:
The altitude from the top vertex is 8 and is common to both right-angled triangles. AB = 10 and AD = 17. Base BD = x
Let BC be a units and CD be b units.
In right-angled triangle ACB,
⇒ AB2 = AC2 + BC2
⇒ 102 = 82 + a2
⇒ 100 = 64 + a2
⇒ a2 = 36
⇒ a = 6
In right-angled triangle ACD,
⇒ AD2 = AC2 + CD2
⇒ 172 = 82 + b2
⇒ 289 = 64 + b2
⇒ b2 = 225
⇒ b = $\sqrt{225}$
⇒ b = 15
x = a + b = 6 + 15 = 21
Hence, option 1 is the correct option.
The perimeter of quadrilateral (rhombus) ABCD is:
In a rhombus, the diagonals bisect each other at right angles at O.
AO = 8 cm and OB = 6 cm
In right-angled triangle AOB, AB is the hypotenuse :
⇒ AB2 = AO2 + OB2
⇒ AB2 = 82 + 62
⇒ AB2 = 64 + 36
⇒ AB2 = 100
⇒ AB = $\sqrt{100}$ = 10 cm
Since all sides of a rhombus are equal,
Perimeter = 4 × 10 = 40 cm
Hence, option 2 is the correct option.
The perimeter of given rectangle ABCD is:
In rectangle ABCD, the diagonal AC = 10 cm and the side AB = 8 cm.
In right-angled triangle ABC, AC is the hypotenuse :
⇒ AC2 = AB2 + BC2
⇒ 102 = 82 + BC2
⇒ 100 = 64 + BC2
⇒ BC2 = 100 - 64
⇒ BC2 = 36
⇒ BC = $\sqrt{36}$ = 6 cm
Perimeter = 2 × (AB + BC) = 2 × (8 + 6) = 2 × 14 = 28 cm
Hence, option 3 is the correct option.
Length of BC is:
In triangle ABC, angle B = 40° and angle C = 50°.
Angle A = 180° - (40° + 50°) = 90°
So the triangle is right-angled at A and BC is the hypotenuse.
According to Pythagoras theorem :
⇒ BC2 = AB2 + AC2
⇒ BC2 = 162 + 122
⇒ BC2 = 256 + 144
⇒ BC2 = 400
⇒ BC = $\sqrt{400}$ = 20 cm
Hence, option 4 is the correct option.
Statement 1: The sides of a triangle are of length 6 cm, 4.5 cm and 7.5 cm. The angle opposite to the side 7.5 cm is 90°.
Statement 2: A triangle having one right angle is called a right-angled triangle.
Which of the following options is correct?
Here the longest side is 7.5 cm.
(7.5)2 = 7.5 × 7.5 = 56.25
(4.5)2 + 62 = 20.25 + 36 = 56.25
Since (7.5)2 = (4.5)2 + 62, Pythagoras theorem holds, so the angle opposite the longest side 7.5 cm is 90°. Thus, statement 1 is true.
A triangle having one right angle is called a right-angled triangle. Thus, statement 2 is true.
Hence, option 1 is the correct option.
Statement 1: A right angled triangle may be an isosceles triangle.
Statement 2: Vertex angle of a right angled isosceles triangle is 90°.
Which of the following options is correct?
A right-angled triangle may be an isosceles triangle, because a triangle can have a right angle and, at the same time, two equal sides (an isosceles right-angled triangle).
Thus, statement 1 is true.
In a right-angled isosceles triangle, the two equal sides contain the right angle, so the vertex angle (the angle between the two equal sides) is 90°.
Thus, statement 2 is true.
Hence, option 1 is the correct option.
Assertion (A): A man goes 15 m due West and then 20 m due North, then he is 25 m from the starting point.
Reason (R): According to Pythagoras theorem, in a triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
West and North are perpendicular directions, so the distance from the starting point is the hypotenuse of a right-angled triangle with legs 15 m and 20 m.
⇒ Distance2 = 152 + 202
⇒ Distance2 = 225 + 400
⇒ Distance2 = 625
⇒ Distance = $\sqrt{625}$ = 25 m
So the man is 25 m from the starting point.
So, Assertion (A) is true.
According to Pythagoras theorem, in a triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
So, Reason (R) is also true.
Hence, option 3 is the correct option.
Assertion (A): Angles B and C of ∆ABC are 69° and 21°, then AC2 - AB2 = BC2.
Reason (R): In a triangle if sum of any two angles is 90° then the 3rd angle is right angle.
Angle B + Angle C = 69° + 21° = 90°, so Angle A = 180° - 90° = 90°.
Since the right angle is at A, the hypotenuse is BC and by Pythagoras theorem BC2 = AB2 + AC2.
The Assertion states AC2 - AB2 = BC2, which is incorrect (the correct relation is AC2 + AB2 = BC2). So, Assertion (A) is false.
If the sum of any two angles is 90°, the third angle = 180° - 90° = 90°, a right angle. So, Reason (R) is true.
Hence, option 2 is the correct option.
Assertion (A): If you walk diagonally across a rectangular field 60 m long and 80 m wide then you will walk 40 m shorter.
Reason (R): Line joining any two non-consecutive vertices of a rectangle is a diagonal of the rectangle.
Length of the diagonal :
⇒ Diagonal2 = 602 + 802
⇒ Diagonal2 = 3600 + 6400
⇒ Diagonal2 = 10000
⇒ Diagonal = $\sqrt{10000}$ = 100 m
Walking along the two sides = 60 + 80 = 140 m
Distance saved = 140 - 100 = 40 m
So walking diagonally is 40 m shorter. So, Assertion (A) is true.
The line joining any two non-consecutive (opposite) vertices of a rectangle is a diagonal.
So, Reason (R) is also true.
Hence, option 3 is the correct option.
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