ICSE Class 6 Mathematics · Selina Solutions Concise Mathematics Class 6 ICSE · 36 solved questions
In each of the following, find the marked unknown angles :
From the figure, the three angles of the triangle are 70°, 72° and z.
By the angle sum property of a triangle,
⇒ 70° + 72° + z = 180°
⇒ 142° + z = 180°
⇒ z = 180° − 142°
⇒ z = 38°
Hence, z = 38°.
In each of the following, find the marked unknown angles :
From the figure, the upper triangle has angles 80°, 50° and b.
By the angle sum property of a triangle,
⇒ 80° + 50° + b = 180°
⇒ 130° + b = 180°
⇒ b = 180° − 130°
⇒ b = 50°
In the lower triangle, the angles are 40°, 45° and a.
By the angle sum property of a triangle,
⇒ 40° + 45° + a = 180°
⇒ 85° + a = 180°
⇒ a = 180° − 85°
⇒ a = 95°
Hence, a = 95° and b = 50°.
In each of the following, find the marked unknown angles :
From the figure,
The angle at the top vertex is made up of 20° and 45°, so the top angle is (20° + 45°) = 65°. The other two angles of the triangle are x and 60°.
By the angle sum property of a triangle,
⇒ 65° + x + 60° = 180°
⇒ 125° + x = 180°
⇒ x = 180° − 125°
⇒ x = 55°
Hence, x = 55°.
Can a triangle have the following angles?
55°, 55° and 80°
A triangle is possible only if the sum of its three angles is 180°.
Solving,
⇒ 55° + 55° + 80° = 190° ≠ 180°.
Since the sum is not 180°, such a triangle is not possible.
Hence, a triangle with angles 55°, 55° and 80° cannot be formed.
Can a triangle have the following angles?
33°, 74° and 73°
A triangle is possible only if the sum of its three angles is 180°.
Solving,
⇒ 33° + 74° + 73° = 180°.
Since the sum is 180°, such a triangle is possible.
Hence, a triangle with angles 33°, 74° and 73° can be formed.
Can a triangle have the following angles?
85°, 95° and 22°
A triangle is possible only if the sum of its three angles is 180°.
Solving,
⇒ 85° + 95° + 22° = 202° ≠ 180°.
Since the sum is not 180°, such a triangle is not possible.
Hence, a triangle with angles 85°, 95° and 22° cannot be formed.
Find x, if the angles of a triangle are :
x°, x°, x°
By the angle sum property of a triangle,
⇒ x° + x° + x° = 180°
⇒ 3x° = 180°
⇒ x° = $\frac{180^{\circ}}{3}$
⇒ x° = 60°
⇒ x = 60
Hence, x = 60.
Find x, if the angles of a triangle are :
x°, 2x°, 2x°
By the angle sum property of a triangle,
⇒ x° + 2x° + 2x° = 180°
⇒ 5x° = 180°
⇒ x° = $\frac{180^{\circ}}{5}$
⇒ x° = 36°
⇒ x = 36
Hence, x = 36.
Find x, if the angles of a triangle are :
2x°, 4x°, 6x°
By the angle sum property of a triangle,
⇒ 2x° + 4x° + 6x° = 180°
⇒ 12x° = 180°
⇒ x° = $\frac{180^{\circ}}{12}$
⇒ x° = 15°
⇒ x = 15
Hence, x = 15.
One angle of a right-angled triangle is 70°. Find the other acute angle.
In a right-angled triangle, one angle is 90°. Let the other acute angle be x.
By the angle sum property of a triangle,
⇒ 90° + 70° + x = 180°
⇒ 160° + x = 180°
⇒ x = 180° − 160°
⇒ x = 20°
Hence, the other acute angle is 20°.
In △ABC, ∠A = ∠B = 62°; find ∠C.
By the angle sum property of a triangle,
⇒ ∠A + ∠B + ∠C = 180°
⇒ 62° + 62° + ∠C = 180°
⇒ 124° + ∠C = 180°
⇒ ∠C = 180° − 124°
⇒ ∠C = 56°
Hence, ∠C = 56°.
In △ABC, ∠B = ∠C and ∠A = 100°; find ∠B.
Given,
∠A = 100° and ∠B = ∠C. Let ∠B = ∠C = x.
By the angle sum property of a triangle,
⇒ ∠A + ∠B + ∠C = 180°
⇒ 100° + x + x = 180°
⇒ 100° + 2x = 180°
⇒ 2x = 180° − 100°
⇒ 2x = 80°
⇒ x = $\frac{80^{\circ}}{2}$
⇒ x = 40°
⇒ ∠B = x = 40°
Hence, ∠B = 40°.
Find, giving reasons, the unknown marked angles in the triangle drawn below :
From the figure, in triangle ABC, ∠B = 30° and the side BC is produced to D, so ∠ACD = 110° is the exterior angle at C.
By the exterior angle property, an exterior angle of a triangle is equal to the sum of its two interior opposite angles.
⇒ ∠ACD = ∠A + ∠B
⇒ 110° = x + 30°
⇒ x = 110° − 30°
⇒ x = 80°
Hence, x = 80°.
Find, giving reasons, the unknown marked angles in the triangle drawn below :
From the figure, SQ is a straight line and ∠PQS = 115° is the exterior angle at Q, while the base angles ∠PQR and ∠PRQ are each equal to x.
Since ∠PQS and ∠PQR form a linear pair,
⇒ ∠PQS + ∠PQR = 180°
⇒ 115° + x = 180°
⇒ x = 180° − 115°
⇒ x = 65°
So the base angles ∠PQR = ∠PRQ = x = 65°.
By the angle sum property of a triangle,
In triangle PQR,
⇒ y + x + x = 180°
⇒ y + 65° + 65° = 180°
⇒ y + 130° = 180°
⇒ y = 180° − 130°
⇒ y = 50°
Hence, x = 65° and y = 50°.
Find, giving reasons, the unknown marked angles in the given figure :
From the figure,
∠XYM is an exterior angle of triangle and the two opposite interior angles are ∠YMZ and ∠YZM
Thus, 110° is the exterior angle of the triangle and the two interior opposite angles are 2x and 3x.
By the exterior angle property, an exterior angle of a triangle is equal to the sum of its two interior opposite angles.
⇒ 110° = 2x + 3x
⇒ 110° = 5x
⇒ x = $\frac{110^{\circ}}{5}$
⇒ x = 22°
So, 2x = 44° and 3x = 66°.
Hence, the marked angles are 2x = 44° and 3x = 66°.
Classify the following triangle according to the measures of its angles :
From the figure, the angles of the triangle are 28°, 120° and 32°.
Since one of the angles, 120°, is more than 90°, it is an obtuse angle.
Hence, the given triangle is an obtuse-angled triangle.
Classify the following triangle according to the measures of its angles :
From the figure, the angles of the triangle are 80°, 70° and 30°.
Since each of the three angles is less than 90°, all the angles are acute.
Hence, the given triangle is an acute-angled triangle.
Classify the following triangle according to the measures of its angles :
From the figure, the angles of the triangle are 30°, 90° and 60°.
Since one of the angles is exactly 90°, it is a right angle.
Hence, the given triangle is a right-angled triangle.
Classify the following triangle according to the lengths of its sides :
From the figure, the sides of the triangle are 2.5 cm, 3 cm and 3 cm.
Two sides AC and BC are equal.
Hence, the given triangle is an isosceles triangle.
Classify the following triangle according to the lengths of its sides :
From the figure, the sides of the triangle are 3 cm, 4 cm and 2.5 cm.
Since all three sides are of different lengths, the given triangle is a scalene triangle.
Hence, the given triangle is a scalene triangle.
Classify the following triangle according to the lengths of its sides :
From the figure, the sides of the triangle are 3.5 cm, 3 cm and 2 cm.
Since all three sides are of different lengths, the given triangle is a scalene triangle.
Hence, the given triangle is a scalene triangle.
Classify the following triangle according to the lengths of its sides :
From the figure, the sides of the triangle are 3 cm, 3 cm and 3 cm.
Since all three sides are equal, the given triangle is an equilateral triangle.
Hence, the given triangle is an equilateral triangle.
Exterior angle of a triangle is equal to:
An exterior angle of a triangle is always equal to the sum of its two opposite interior angles.
Hence, option 2 is the correct option.
In a right angled triangle, the other two angles are always:
In a right-angled triangle, one angle is 90°, so the other two angles add up to 90°. Therefore each of them must be less than 90°, i.e. acute. They need not be equal.
Hence, option 3 is the correct option.
The angles of a triangle are in the ratio 2 : 3 : 4. The largest angle is:
Let the angles be 2x, 3x and 4x.
By the angle sum property of a triangle,
⇒ 2x + 3x + 4x = 180°
⇒ 9x = 180°
⇒ x = $\frac{180^{\circ}}{9}$
⇒ x = 20°
So the largest angle = 4x = 4 × 20° = 80°.
Hence, option 3 is the correct option.
The angles of a triangle are 50°, 70° and x°. The value of x° is:
By the angle sum property of a triangle,
⇒ 50° + 70° + x° = 180°
⇒ 120° + x° = 180°
⇒ x° = 180° − 120°
⇒ x° = 60°
Hence, option 1 is the correct option.
The value of x is:
From the figure,
The two base angles of the triangle are each equal to x and the side BC is produced to D, so 110° is the exterior angle.
By the exterior angle property,
⇒ 110° = x + x
⇒ 110° = 2x
⇒ x = $\frac{110^{\circ}}{2}$
⇒ x = 55°
Hence, option 2 is the correct option.
On extending each side of a triangle, on both the sides, the number of exterior angles formed is:
When the sides of a triangle are extended on both the sides, two exterior angles are formed at each of the three vertices, giving 2 × 3 = 6 exterior angles in all.
Hence, option 2 is the correct option.
The value of x is:
From the figure, the three angles of the triangle are 3x, 60° and x.
By the angle sum property of a triangle,
⇒ 3x + 60° + x = 180°
⇒ 4x + 60° = 180°
⇒ 4x = 180° − 60°
⇒ 4x = 120°
⇒ x = $\frac{120^{\circ}}{4}$
⇒ x = 30°
Hence, option 1 is the correct option.
The given triangle is:
From the figure, ∠ABC is an obtuse angle, as it is greater than 90°.
Thus, the given triangle is an obtuse angled triangle.
Hence, option 3 is the correct option.
The measure of angle BAC is:
From the figure,
In the triangle ABC, AB = BC = CA, thus it is an equilateral triangle.
Each angle of an equilateral triangle = 60°.
Thus, ∠BAC = 60°.
Hence, option 2 is the correct option.
In the given figure, AP is:
From the figure, AP is drawn from the vertex A perpendicular to the opposite side BC, i.e. AP ⊥ BC.
The perpendicular drawn from a vertex of a triangle to the opposite side is called an altitude.
Hence, option 4 is the correct option.
Statement 1: In the given figure, P is mid-point of side BC, then AP is a median of triangle ABC.
Statement 2: Every triangle has three sides so three medians. For this reason, BQ and CR are also medians of △ABC.
Which of the following options is correct?
A median of a triangle is the line joining a vertex to the mid-point of the opposite side. Since P is the mid-point of BC, AP joins vertex A to the mid-point of the opposite side, so AP is a median. Thus, statement 1 is true.
Every triangle has three vertices and three sides, so it has three medians. As Q and R are the mid-points of AC and AB respectively, BQ and CR are also medians of △ABC. Thus, statement 2 is true.
Hence, option 1 is the correct option.
Statement 1: In the given figure, AP is altitude corresponding to side BC and CR is altitude corresponding to side AB then BQ is altitude corresponding to side AC.
Statement 2: Altitudes of a triangle intersect each other at the same point.
Which of the following options is correct?
From the right-angle marks in the figure,
AP ⊥ BC, CR ⊥ AB and BQ ⊥ AC.
Therefore, AP, CR and BQ are the three altitudes of △ABC. Thus, statement 1 is true.
The three altitudes of a triangle are concurrent, i.e. they pass through the same point. Thus, statement 2 is true.
Hence, option 1 is the correct option.
Assertion (A): As shown in the given figure, sides AB, BC and CA are extended to get the exterior angles x, y and z then ∠x + ∠y + ∠z = 360°
Reason (R): ∠x + ∠y + ∠z = (a + c) + (a + b) + (b + c) = 2(a + b + c) = 2 × 180°
Let the interior angles of the triangle be a, b and c, so a + b + c = 180°.
By the exterior angle property, each exterior angle equals the sum of the two interior opposite angles:
⇒ ∠x + ∠y + ∠z = (b + c) + (a + c) + (a + b)
⇒ ∠x + ∠y + ∠z = 2(a + b + c)
⇒ ∠x + ∠y + ∠z = 2 × 180° = 360°
So the sum of the three exterior angles is 360°, which makes the Assertion (A) true.
The Reason (R) gives the correct derivation of this result, so the Reason (R) is also true, and it correctly explains the Assertion.
Hence, option 3 is the correct option.
Assertion (A): Every isosceles triangle is equilateral but every equilateral triangle is not necessarily isosceles.
Reason (R): A triangle with atleast two sides equal is an isosceles triangle.
An isosceles triangle has at least two sides equal, whereas an equilateral triangle has all three sides equal. So every equilateral triangle is isosceles, but an isosceles triangle need not be equilateral. The Assertion states this the wrong way round, so the Assertion (A) is false.
A triangle with at least two sides equal is indeed called an isosceles triangle, so the Reason (R) is true.
Hence, option 2 is the correct option.
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