Model Question Paper 4

⇒ Since l ∥ m,

∠BOD = ∠AOC = 65° (Vertically opposite angles are equal)

Now, ∠BOD and x are co-interior angles, so they must be supplementary.

⇒ ∠BOD + x = 180°

⇒ 65° + x = 180°

⇒ x = 180° − 65° = 115°.

Hence, option 3 is the correct option.

⇒ The third angle = 180° − (40° + 100°) = 40°.

Two angles are equal (40° each), so the triangle is isosceles.

Hence, option 2 is the correct option.

Since AB = AC, the base angles are equal, so ∠B = ∠C = x.

By the angle sum property of △ABC,

⇒ ∠A + ∠B + ∠C = 180°

⇒ 70° + x + x = 180°

⇒ 2x = 110°

⇒ x = 55°

Hence, x = 55°.

Since AB = BC = AC, △ABC is equilateral, so each of its angles is 60°.

∴ ∠ACB = 60°.

Now ∠ACB and ∠ACD form a linear pair, so

⇒ ∠ACD + ∠ACB = 180°

⇒ ∠ACD + 60° = 180°

⇒ ∠ACD = 180° − 60° = 120°

In △ACD, AC = CD, so the base angles are equal: ∠CAD = ∠CDA = x.

By the angle sum property of △ACD,

⇒ ∠CAD + ∠CDA + ∠ACD = 180°

⇒ x + x + 120° = 180°

⇒ 2x = 60°

⇒ x = 30°

Hence, x = 30°.

Under the congruence △PQR ≅ △EFD, the vertices correspond as P ↔ E, Q ↔ F and R ↔ D.

(i) ∠Q corresponds to ∠F.

(ii) $\overline{PR}$ corresponds to $\bm{\overline{ED}}$.

(iii) ∠P corresponds to ∠E.

(iv) $\overline{QR}$ corresponds to $\bm{\overline{FD}}$.

Since l ∥ m, the angle x corresponds to the given 75°:

⇒ x = 75° (corresponding angles are equal)

∠COD = ∠AOB = 75° (vertically opposite angles are equal)

Now, l is a transversal to the parallel lines p and q. ∠COD and y lie between p and q, on the same side of l, so they are co-interior angles.

Co-interior angles are supplementary, i.e. their sum is 180°.

⇒ ∠COD + y = 180°

⇒ 75° + y = 180°

⇒ y = 180° − 75° = 105°

As, ∠CED = z (vertically opposite angles are equal)

Since l ∥ m, ∠CED and y are co-interior angles.

⇒ ∠CED + y = 180°

⇒ z + 105° = 180°

⇒ z = 75°

Hence, x = 75°, y = 105° and z = 75°.

Since AB = AC, the base angles are equal, so ∠ABC = ∠ACB = x.

From figure,

∠BAC = 80° (Vertically opposite angles are equal)

By the angle sum property of △ABC,

⇒ ∠BAC + ∠ABC + ∠ACB = 180°

⇒ 80° + x + x = 180°

⇒ 2x = 100°

⇒ x = 50°

Now ∠ACB and y form a linear pair,

⇒ ∠ACB + y = 180°

⇒ x + y = 180°

⇒ y = 180° − 50° = 130°

Hence, x = 50° and y = 130°.

The third side must be less than the sum and greater than the difference of the other two sides.

Sum = 7.5 + 5 = 12.5 cm and difference = 7.5 − 5 = 2.5 cm.

Hence, the third side is greater than 2.5 cm and less than 12.5 cm.

Let ABCD be the rectangular orchard with length AB = 60 m, diagonal AC = 75 m and breadth BC = b m.

Given: AB = 60 m, AC = 75 m and ∠ABC = 90° (each angle of a rectangle is 90°).

(i) In right-angled △ABC, AC is the hypotenuse.

By Pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ 75² = 60² + b²

⇒ 5625 = 3600 + b²

⇒ b² = 5625 − 3600

⇒ b² = 2025

⇒ b = $\sqrt{2025}$

⇒ b = 45 m

Hence, the breadth of the orchard is 45 m.

(ii) By formula,

Perimeter of rectangle = 2(length + breadth)

⇒ Perimeter = 2(60 + 45)

⇒ Perimeter = 2 × 105

⇒ Perimeter = 210 m

Hence, the perimeter of the orchard is 210 m.

(i) The three pairs of equal parts in △ABD and △ACD are:

AB = AC (3.6 cm each)

AD = AD (common)

∠ADB = ∠ADC (90° each)

(ii) From part (i), the right angle, hypotenuse and one side of △ABD are equal to those of △ACD.

Hence, △ABD ≅ △ACD by the RHS rule of congruency.

(iii) Since △ABD ≅ △ACD, the corresponding sides BD and DC are equal, so D divides BC into two equal parts.

Hence, D is the mid-point of BC because BD = DC as corresponding parts of congruent triangles.