Q 1 – Find the value of the unknown x in the following diagrams:
(i) By angle sum property of a triangle, we have
∠x + 50° + 60° = 180°
⇒ ∠x + 110° = 180°
∴ ∠x = 180° – 110° = 70°
(ii) By angle sum property of a triangle, we have
∠x + 90° + 30 = 180° [∆ is right angled triangle]
⇒ ∠x + 120° = 180°
∴ ∠x – 180° – 120° = 60°
(iii) By angle sum property of a triangle, we have
∠x + 30° + 110° – 180°
⇒ ∠x + 140° = 180°
∴ ∠x = 180° – 140° = 40°
(iv) By angle sum property of a triangle, we have
∠x + ∠x + 50° = 180°
⇒ 2x + 50° = 180°
⇒ 2x = 180° – 50°
⇒ 2x = 130°
∴ x = 130°/2 = 65°
(v) By angle sum property of a triangle, we have
∠x + ∠x +∠x =180°
⇒ 3 ∠x = 180°
∴ ∠x = 180° = 60°
(vi) By angle sum property of a triangle, we have
x + 2 x + 90° = 180° (∆ is right angled triangle)
⇒ 3x + 90° = 180°
⇒ 3x = 180° – 90°
⇒ 3x = 90°
∴ x = 90º/3 =30°
Q 2 – AD is the median of a ΔABC, prove that AB + BC + CA > 2AD.
Q 3 – Find the values of the unknowns x and y in the following diagrams:
(i) ∠x + 50° = 120° (Exterior angle of a triangle)
∴ ∠x = 120°- 50° = 70°
∠x + ∠y + 50° = 180° (Angle sum property of a triangle)
70° + ∠y + 50° = 180°
∠y + 120° = 180°
∠y = 180° – 120°
∴ ∠y = 60°
Thus ∠x = 70 and ∠y – 60°
(ii) ∠y = 80° (Vertically opposite angles are same)
∠x + ∠y + 50° = 180° (Angle sum property of a triangle)
⇒ ∠x + 80° + 50° = 180°
⇒ ∠x + 130° = 180°
∴ ∠x = 180° – 130° = 50°
Thus, ∠x = 50° and ∠y = 80°
(iii) ∠y + 50° + 60° = 180° (Angle sum property of a triangle)
∠y + 110° = 180°
∴ ∠y = 180°- 110° = 70°
∠x + ∠y = 180° (Linear pairs)
⇒ ∠x + 70° = 180°
∴ ∠x = 180° – 70° = 110°
Thus, ∠x = 110° and y = 70°
(iv) ∠x = 60° (Vertically opposite angles)
∠x + ∠y + 30° = 180° (Angle sum property of a triangle)
⇒ 60° + ∠y + 30° = 180°
⇒ ∠y + 90° = 180°
⇒ ∠y = 180° – 90° = 90°
Thus, ∠x = 60° and ∠y = 90°
(v) ∠y = 90° (Vertically opposite angles)
∠x + ∠x + ∠y = 180° (Angle sum property of a triangle)
⇒ 2 ∠x + 90° = 180°
⇒ 2∠x = 180° – 90°
⇒ 2∠x = 90°
∴ ∠x= =45∘
Thus, ∠x = 45° and ∠y = 90°
(vi) From the given figure, we have
Adding both sides, we have
∠y + ∠1 + ∠2 = 3∠x
⇒ 180° = 3∠x (Angle sum property of a triangle)
∴ ∠x= 180º/3=60∘
∠x = 60°, ∠y = 60°
Q 4 – In the following figure, find the unknown angles a and b, if l || m.
Q 5 – Verify by drawing a diagram if the median and altitude of an isosceles triangle can be same.
∆ ABC is an isosceles triangle in which AB = AC
Draw AD as the median of the triangle.
Measure the angle ADC with the help of protractor we find, ∠ADC = 90°
Thus, AD is the median as well as the altitude of the ∆ABC.
Hence Verified.
Q 6 – Draw rough sketches for the following:
(a) In ∆ABC, BE is a median.
(b) In ∆PQR, PQ and PR are altitudes of the triangle.
(c) In ∆XYZ, YL is an altitude in the exterior of the triangle.
Q 7 – Find the value of the unknown exterior angle x in the following diagrams:
(i) ∠x = 50° + 70° = 120° (Exterior angle is equal to sum of its interior opposite angles)
(ii) ∠x = 65°+ 45° = 110° (Exterior angle is equal to sum of its interior opposite angles)
(iii) ∠x = 30° + 40° = 70° (Exterior angle is equal to sum of its interior opposite angles)
(iv) ∠x = 60° + 60° = 120° (Exterior angle is equal to sum of its interior opposite angles)
(v) ∠x = 50° + 50° =100° (Exterior angle is equal to sum of its interior opposite angles)
(vi) ∠x = 30° + 60° = 90° (Exterior angle is equal to sum of its interior opposite angle)
Q 8
– Find the value of the unknown interior angle x
in the following figures:
(i) ∠x + 50° = 115° (Exterior angle of a triangle)
∴ ∠x = 115°- 50° = 65°
(ii) ∠x + 70° = 110° (Exterior angle of a triangle)
∴ ∠x = 110° – 70° = 40°
(iii) ∠ x + 90° = 125° (Exterior angle of a right triangle)
∴ ∠x = 125° – 90° = 35°
(iv) ∠x + 60° = 120° (Exterior angle of a triangle)
∴ ∠x = 120° – 60° = 60°
(v)∠ X + 30° = 80° (Exterior angle of a triangle)
∴ ∠x = 80° – 30° = 50°
(vi) ∠ x + 35° = 75° (Exterior angle of a triangle)
∴ ∠ x = 75° – 35° = 40°
Q 9 – Is it possible to have a triangle with the following sides?
(i) 2 cm, 3 cm, 5 cm
(ii) 3 cm, 6 cm, 7 cm
(iii) 6 cm, 3 cm, 2 cm
We know that for a triangle, the sum of any two sides must be greater than the third side.
(i) Given sides are 2 cm, 3 cm, 5 cm
Sum of the two sides = 2 cm + 3 cm = 5 cm Third side = 5 cm
We have, Sum of any two sides = the third side i.e. 5 cm = 5 cm
Hence, the triangle is not possible.
(ii) Given sides are 3 cm, 6 cm, 7 cm
Sum of the two sides = 3 cm + 6 cm = 9 cm Third side = 7 cm
We have sum of any two sides > the third
side. i.e. 9 cm > 7 cm
Hence, the triangle is possible.
(iii) Given sides are 6 cm, 3 cm, 2 cm
Sum of the two sides = 3 cm + 2 cm – 5 cm Third side = 6 cm
We have, sum of any two sides < the third sid6 i.e. 5 cm < 6 cm Hence, the triangle is not possible.
Q 10 – Take any point O in the interior of a triangle PQR . Is
(i) OP + OQ > PQ?
(ii) OQ + OR > QR?
(iii) OR + OP > RP?
(i) Yes, In ∆OPQ, we haveOP + OQ > PQ
[Sum of any two sides of a triangle is greater than the third side]
(ii) Yes, In ∆OQR, we have OQ + OR > QR
[Sum of any two sides of a triangle is greater than the third side]
(iii) Yes, In ∆OPR, we have OR + OP > RP
[Sum of any two sides of a triangle is greater than the third side]
Q 11 – The length of two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall?
Sum of two sides
= 12 cm + 15 cm = 27 cm
Difference of the two sides
= 15 cm – 12 cm = 3 cm
∴ The measure of third side should fall between 3 cm and 27 cm.
Q 12 – PQR is a triangle, right angled at P. If PQ = 10 cm and PR = 24 cm, find QR.
In right angled triangle PQR, we have
QR2 = PQ2 + PR2 From Pythagoras property)
= (10)2 + (24)2
= 100 + 576 = 676
∴ QR = √676 = 26 cm
The, the required length of QR = 26 cm.
Q 13 – Which of the following can be the sides of a right triangle?
(i) 2.5 cm, 6.5 cm, 6 cm.
(ii) 2 cm, 2 cm, 5 cm
(iii) 1.5 cm, 2 cm, 2.5 cm
(i) Given sides are 2.5 cm, 6.5 cm, 6 cm.
Square of the longer side = (6.5)2 = 42.25 cm.
Sum of the square of other two sides
= (2.5)2 + (6)2 = 6.25 + 36
= 42.25 cm.
Since, the square of the longer side in a triangle is equal to the sum of the squares of other two sides.
∴ The given sides form a right triangle.
(ii) Given sides are 2 cm, 2 cm, 5 cm .
Square of the longer side = (5)2 = 25 cm Sum of the square of other two sides
= (2)2 + (2)2 =4 + 4 = 8 cm
Since 25 cm ≠ 8 cm
∴ The given sides do not form a right triangle.
(iii) Given sides are 1.5 cm, 2 cm, 2.5 cm
Square of the longer side = (2.5)2 = 6.25 cm Sum of the square of other two sides
= (1.5)2 + (2)2 = 2.25 + 4
Since 6.25 cm = 6.25 cm = 6.25 cm
Since the square of longer side in a triangle is equal to the sum of square of other two sides.
∴ The given sides form a right triangle.
Q 14 – 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. Find the original height of the tree.
Let AB be the original height of the tree and broken at C touching the ground at D such that
AC = 5 m
and AD = 12 m
In right triangle ∆CAD,
AD2 + AC2 = CD2 (By Pythagoras property)
⇒ (12)2 + (5)2 = CD2
⇒ 144 + 25 = CD2
⇒ 169 = CD2
∴ CD = √169 = 13 m
But CD = BC
AC + CB = AB
5 m + 13 m = AB
∴ AB = 18 m .
Thus, the original height of the tree = 18 m.
Q 15 – Angles Q and R of a Δ PQR are 25° and 65°. Write which of the following is true.
(i) PQ2 + QR2 = RP2
(ii) PQ2 + RP2 = QR2
(iii) RP2 + QR2 = PQ2
We know that
∠P + ∠Q + ∠R = 180° (Angle sum property)
∠P + 25° + 65° = 180°
∠P + 90° = 180°
∠P = 180° – 90° – 90°
∆PQR is a right triangle, right angled at P
(i) Not True
∴ PQ2 + QR2 ≠ RP2 (By Pythagoras property)
(ii) True
∴ PQ2 + RP2 = QP2 (By Pythagoras property)
(iii) Not True
∴ RP2 + QR2 ≠ PQ2 (By Pythagoras property)
Q 16 – The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.
Let ABCD be a rhombus whose diagonals intersect each other at O such that AC = 16 cm and BD = 30 cm
Since, the diagonals of a rhombus bisect each other at 90°.
∴ OA = OC = 8 cm and OB = OD = 15 cm
In right ∆OAB,
AB2 = OA2 + OB2 (By Pythagoras property)
= (8)2+ (15)2 = 64 + 225
= 289
∴ AB = √289 = 17 cm
Since AB = BC = CD = DA (Property of rhombus)
∴ Required perimeter of rhombus
= 4 × side = 4 × 17 = 68 cm.
Q 17 – How many elements are there in a triangle?
(a) 3
(b) 6
(c) 4
(d) None of these
(b) 6
Q 18 – How many vertices does a triangle have?
(a) 1
(b) 2
(c) 3
(d) 4
(c) 3
Q 19 – How many sides are there in a triangle?
(a) 1
(b) 2
(c) 3
(d) 4
(c) 3
Q 20 – How many angles are there in a triangle?
(a) 1
(b) 2
(c) 3
(d) 4
(c) 3
Q 21 – If two sides of a triangle are not equal, the triangle is called
(a) scalene
(b) isosceles
(c) equilateral
(d) right-angled
(a) scalene
Q 22 – If two sides of a triangle are equal, the triangle is called
(a) isosceles
(b) equilateral
(c) scalene
(d) right-angled
(a) isosceles
Q 23 – If all the three sides of a triangle are equal, the triangle is called
(a) equilateral
(b) right-angled
(c) isosceles
(d) scalene
(a) equilateral
Q 24 – If all the angles of a triangle are acute, the triangle is called
(a) obtuse-angled
(b) acute-angled
(c) right-angled
(d) none of these
(b) acute-angled
Q 25 – If one angle of a triangle measures 90°, the triangle is called
(a) acute-angled
(b) obtuse-angled
(c) right-angled
(d) none of these
(c) right-angled
Q 26 – If one angle of a triangle is obtuse, the triangle is called
(a) acute-angled
(b) obtuse-angled
(c) right-angled
(d) none of these
(c) right-angled
Q 27 – How many medians can a triangle have?
(a) 1
(b) 2
(c) 3
(d) 4
(c) 3
Q 28 – How many altitudes can a triangle have?
(a) 1
(b) 2
(c) 3
(d) 4
(c) 3
Q 29 – The total measure of the three angles of a triangle is
(a) 360°
(b) 90°
(c) 180°
(d) none of these
(c) 180°
Q 30 – The measure of each angle of an equilateral triangle is
(a) 30°
(b) 45°
(c) 90°
(d) 60°
(d) 60°
Q 31 – Which of the following statements is true?
(a) A triangle can have two right angles
(b) A triangle can have two obtuse angles
(c) A triangle can have two acute angles
(d) A triangle can have all the three angles less than 60°
(c) A triangle can have two acute angles
Q 32 – Which of the following statements is true?
(a) A triangle can have all the three angles equal to 60°.
(b) A triangle can have all the three angles greater than 60°.
(c) The sum of any two angles of a triangle is always greater than the third angle.
(d) The difference between the lengths of any two sides of a triangle is greater than the length of the third side
(a) A triangle can have all the three angles equal to 60°.
Q 33 – Which of the following statement is false?
(a) The sum of the lengths of any two sides of a triangle is less than the third side.
(b) In a right-angled triangle, the square on the hypotenuse = sum of the squares on the legs.
(c) If the Pythagorean property holds, the triangle must be right-angled.
(d) The diagonal of a rectangle produce ‘by itself the same area as produced by its length and breadth.
(a) The sum of the lengths of any two sides of a triangle is less than the third side.
Q 34 – Two angles of a triangle measure 90° and 30°. The measure of the third angle is
(a) 90°
(b) 30°
(c) 60°
(d) 120°
(c) 60°
Q 35 – The ratio of the measures of the three angles of a triangle is 2 : 3 : 4. The measure of the largest angle is
(a) 80°
(b) 60°
(c) 40°
(d) 180°
(a) 80°
Q 36 – In the following figure, the side BC of ∆ ABC is extended up to the point D. If ∠A = 55° and ∠B = 60°, then the measure of ∠ACD is
(a) 120°
(b) 110°
(c) 115°
(d) 125°
(c) 115°
Q 37 – In the following figure, the measure of ∠A is
(a) 30°
(b) 45°
(c) 90°
(d) 30°
(d) 30°
Q 38 – In the following figure, m || QR. Then, the measure of ∠QPR is
(a) 80°
(b) 85°
(c) 75°
(d) 70°
(b) 85°
Q 39 – In the following figure, find ∠ x and ∠ y, if ∠x – ∠y – 10°
(a) 65°, 55°
(b) 55°, 45°
(c) 45°, 35°
(d) 60°, 60°
(a) 65°, 55°
Q 40 – In the following figure, find ∠ B.
(a) 30°
(b) 45°
(c) 40°
(d) 60°
(d) 60°
Q 41 – In the following figure, ∆ ABC is an equilateral triangle. Find ∠x.
(a) 30°
(b) 45°
(c) 60°
(d) 90°
(a) 30°
Q 42 – In the following figure, one angle of triangle ABC is 40°. If the difference of the other two angles is 30°, find the larger of the other two angles.
(a) 85°
(b) 80°
(c) 75°
(d) 70°
(a) 85°
Q 43 – In the following figure, find
(a) 60°
(b) 70°
(c) 80°
(d) 75°
(b) 70°
Q 44 – In the following figure, find x if BA || CE.
(a) 60°
(b) 40°
(c) 45°
(d) 65°
(d) 65°
Q 45 – Find the value of the unknown interior angle x in the following figure:
(a) 30°
(b) 35°
(c) 40°
(d) 45°
(c) 40°
Q 46 – Find the value of unknown x in the following figure:
(a) 40°
(b) 50°
(c) 45°
(d) 55°
(b) 50°
Q 47 – Find the value of unknown x in the following figure:
(a) 10°
(b) 15°
(c) 20°
(d) 25°
(b) 15°
Q 48 – Find angle x in the following figure:
(a) 90°
(b) 80°
(c) 95°
(d) 100°
(a) 90°
Q 49 – Find angle x in the following figure:
(a) 40°
(b) 50°
(c) 45°
(d) 60°
(b) 50°
Q 50 – Find angle x in the following figure:
(a) 40°
(b) 30°
(c) 25°
(d) 35°
(b) 30°
Q 51 – Find angle x in the following figure:
(a) 40°
(b) 45°
(c) 35°
(d) 50°
(a) 40°
Q 52 – Find angle x in the following figure:
(a) 58°
(b) 59°
(c) 57°
(d) 56°
(a) 58°
Q 53 – Find angle x in the following figure:
(a) 45°
(b) 40°
(c) 35°
(d) 50°
(a) 45°
Q 54 – In which case of the following lengths of sides of a triangle, is it possible to draw a triangle?
(а) 3 cm, 4 cm, 7 cm
(b) 2 cm, 3 cm, 7 cm
(c) 3 cm, 4 cm, 5 cm
(d) 3 cm, 3 cm, 7 cm
(c) 3 cm, 4 cm, 5 cm
Q 55 – Which of the following cannot be the sides of a right triangle?
(а) 2 cm, 2 cm, 4 cm
(b) 5 cm, 12 cm, 13 cm
(c) 6 cm, 8 cm, 10 cm
(d) 3 cm, 4 cm, 5 cm
(а) 2 cm, 2 cm, 4 cm