Worksheet on Triangles for Class 7

Explore three-sided shapes with our printable grade 7 triangles worksheet pdf. This geometric learning sheet teaches students about triangle properties, classifications and angle relationships through construction activities and practical applications. Kids will understand triangle theorems and solve geometric problems confidently for advanced geometry. Download free Class 7 triangles practice questions, and master geometric concepts.

Solved Questions on Triangles

1. In the given quadrilateral PQRS, PQ = PR = PS, PR bisects ∠QRS and ∠RPS = 72°. What is the value of q?

a) 44°
b) 54°
c) 64°
d) 74°

Answer: b) 54°

Explanation: In the given quadrilateral PQRS,

PQ = PR = PS

PR bisects ∠QRS. Therefore, ∠PRQ = ∠PRS

∠RPS = 72°

In △PSR,

∠PRS + ∠PSR + ∠RPS = 180°            (Angle Sum Property of a Triangle)

∠PRS + ∠PRS + 72° = 180°                (PR = PS, ∠PSR = ∠PRS)

2∠PRS = 180° − 72°

2∠PRS = 108°

∠PRS = 54°

In △PQR,

∠PRQ = ∠PRS = 54°                             (PR bisects ∠    QRS)

q = ∠PRQ = 54°                                     (PQ = PR)

2. Which of the following options expresses x in terms of y?

a) x = 180° + $\frac{\frac{y}{}}{\frac{2}{}}$
b) x = 180° − $\frac{\frac{y}{}}{\frac{2}{}}$
c) x = 180° + 2y
d) x = 180° − 2y

Answer: b) x  = 180° − $\frac{\frac{y}{}}{\frac{2}{}}$

Explanation: ∠ACB +  ∠ACE = 180° [Linear Pair]

∠ACB +  x = 180°
∠ACB = 180° − x

∠BAC = ∠ACB = 180° − x (Angle opposite to equal sides are equal, AB = BC)

∠ABC +  ∠CBD = 180° [Linear Pair]
∠ABC +  y = 180°
∠ABC = 180° − y

In △ABC,
∠ACB + ∠BAC  + ∠ABC  = 180°     (Angle Sum Property of a Triangle)
180° − x + 180° − x  + 180° − y  = 180°
540° − 2x − y  = 180°
2x + y  = 540° − 180°
2x + y  = 360°
2x  = 360° − y
x  = $\frac{\frac{\mathrm{360°\; -\; y}}{}}{\frac{2}{}}$
x  = ^360°⁄₂ − ^y⁄₂
x  = 180° − ^y⁄₂

3. What is the length of PR using the information given in the figure?

a) 22 cm
b) 24 cm
c) 26 cm
d) 28 cm

Answer: c) 26 cm

Explanation: In △PQR,

RQ = TS = 24 cm
PQ = PS − QS = PS − RT = 20 − 10 = 10 cm
PR² = OD² + OA² [Pythagoras’ Theorem]
PR = √(RQ² + PQ²)
PR = √(24² + 10²)
PR = √(576 + 100)
PR = √676
PR = √(26 × 26)
PR = 26 cm

4. In △XYZ, OZ bisects ∠Z and ∠OXZ = ∠OYZ. Which of the following is true?

a) ∠XOZ = ∠YOZ; OX = OY; XY = YZ
b) ∠XOZ = ∠YZO; OX = OY; XZ = YZ
c) ∠XOZ = ∠YOZ; OX = OZ; XZ = YZ
d) ∠XOZ = ∠YOZ; OX = OY; XZ = YZ

Answer: d) ∠XOZ = ∠YOZ; OX = OY; XZ = YZ

Explanation: In △XOZ and △YOZ,

∠OXZ = ∠OYZ (Given)
OZ = OZ (Common)
∠OZX = ∠OZY  (OZ bisects ∠Z)
△XOZ ≅ △YOZ (A.S.A)
OX = OY  (Corresponding Parts of Congruent Triangles are Equal)
XZ = YZ  (Corresponding Parts of Congruent Triangles are Equal)
∠XOZ = ∠YOZ  (Corresponding Parts of Congruent Triangles are Equal)

5. The ratio between a base angle and the vertical angle of an isosceles triangle is 1 : 2. What is the value of each angle of the triangle?

a) Base Angle = 75° and Vertical Angle = 30°
b) Base Angle = 60° and Vertical Angle = 60°
c) Base Angle = 45° and Vertical Angle = 90°
d) Base Angle = 30° and Vertical Angle = 120°

Answer: c) Base Angle = 45° and Vertical Angle = 90°

Explanation: In △ABC,

Base Angle : Vertical Angle =1 : 2

Let each base angle be x° and vertical angle be 2x°.
∠ABC = ∠ACB = x°   (Base Angles)
∠CAB = 2x°  (Vertical Angle)
∠ABC + ∠ACB + ∠CAB = 180° (Angle Sum Property of a Triangle)
x° + x° + 2x° = 180°
4x° = 180°
x° = $\frac{\frac{\mathrm{180°}}{}}{\frac{4}{}}$
x° = 45°

Each Base Angle = ∠ABC = ∠ACB = x° = 45°

Vertical Angle = ∠CAB = 2x° = 2 × 45° = 90°