Students can Download Class 8 Maths Chapter 15 Quadrilaterals Ex 15.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka State Syllabus Class 8 Maths Chapter 15 Quadrilaterals Ex 15.4

Question 1.

The sides of a rectangle are in the ratio 2 : 1. The perimeter is 30 cm. Calculate the measure of all the sides.

Answer:

The ratio of the sides is 2 : 1

Let the sides be 2x and x

Perimeter = 30

2x + x + 2x + x=30

6x = 30

x = \(\frac { 30 }{ 6 }\) = 5 cm

2x = 2 x 5 = 10cm

∴The sides aer 10 cm, 5 cm, 10 cm, 5 cm.

Question 2.

In the adjacent rectangle ABCD ∠OCD = 30° Calculate ∠BOC. What type of triangle is BOC.

Answer:

∠BCD = 90° [Angle of a rectangle]

∠OCD + ∠OCB = 90°

30 + ∠QCB = 90°

∠OCB = 90 – 30

∠OCB = 60°

∠OCB = ∠OBC = 60°

OC = OB [Diagonals of a rectangle bisect each other]

∴ ∠BOC = 180 – (60 + 60) = 180 -120

∠BOC = 60°

In Δ BOC, ∠BOC = ∠OBC = ∠OCB = 60°

& equiangular triangle

∴ BOC is an equiangular triangle.

Question 3.

All the rectangles are parallelograms but all the parallelograms are not rectangles. Justify this statement.

Answer:

A rectangle has all the properties of a parallelogram, therefore, all rectangles are parallelograms. No angle of a Parallelogram is a right angle but all angles of a rectangle are right angles.

∴ All parallelograms are not rectangular.

Question 4.

The side of a rectangular park are in the ratio 4 : 3. If the area is 1728 m find the cost of fencing it at the rate of Rs 2.50/m

Answer:

The ratio of the sides = 4 : 3

Let the sides be 4x and 3x

Area = 1728

l × b = 1728

4x × 3x = 1728

12x² = 1728

x² = \(\frac { 12 }{ 1728 }\)

x² = 144

x = √144 = 12m

4x = 4 × 12=48m

3x = 3x ×12 = 36m

Perimeter =4x + 3x + 4x + 3x

= 48 + 36 + 48 + 36

= 168m

The cost of fencing 1m is Rs 2.50

The cost of fencing 168m = 168 × 250 = Rs.420.

Question 5.

A rectangular yard contains two flower beds in the shape of congruent isosceles right triangles. The remaining portion is a yard of a trapezoidal shape whose parallel sides have length 15m and 25m. What fraction of the yard is occupied by the flower bed?

Answer:

In the fig AB = 25 m and FE = 15 m

∴ DF + EC = 25 – 15= 10m

But DF = EC

∴DF = EC = 5m also AD = BC = 5m

Area of rectangle ∆BCD = AB x BC = 1 x b = 25 x 5 = 125. m²

Area of ∆ADF = \(\frac { 1 }{ 2 }\) base × height

= \(\frac { 1 }{ 2 }\) 5 × 5 = \(\frac { 25 }{ 2 }\) m²

∆ADF ≅ ABCE [data]

Area of ∆BCE = \(\frac { 25 }{ 2 }\) m²

Sum of Area of the two triangles

= \(\frac{25}{2}+\frac{25}{2}=\frac{50}{2}=25 \mathrm{m}^{2}\)

\(\frac{\text { Area of } 2 \text { triangles }}{\text { Area of rectangles }}=\frac{25}{125}=\frac{1}{5}\)

∴\(\frac { 1 }{ 5 }\)of the field is occupied by the flower bed.

Question 6.

In a rhombus ABCD ∠C = 70°. Find the other angles of the rhombus.

Answer:

∠A = ∠C = 70° [Opposite angles]

∠A +∠B = 180° [Adjacent angles of a rhombus]

70 + ∠B = 180°

∠B = 180 – 70

∠B = 110°

∠B = ∠D = 110° [Opposite angles are equal].

Angles are 70°, 110°, 70° & 110°

Question 7.

In a rhombus PQRS, if PQ = 3x – 7 and QR = x + 3 find PS.

Answer:

PQ = QR [Sides of a rhombus]

3x – 7 = x + 3

3x – x = 3 + 7

2x = 10

x = \(\frac { 10 }{ 2 }\)

x = 5

PQ = 3x – 7 = 3 × 5 – 7 =15 – 7

PQ = 8cm

∴ SP = 8

Question 8.

Rhombus is a parallelogram. Justify.

Answer:

In a rhombus opposite sides are equal and parallel to each other therefore it is a parallelogram.

Question 9.

In a given square ABCD, if the area of triangle ABD is 36cm^{2} Find

1. The area of the triangle BCD and

2. Area of the square ABCD

Answer:

Diagonal of a square divides the square into two congruent triangles.

∴ Area of ∆ABD = ∆BCD = 36 cm^{2}

Area of square ABCD = 36 + 36 = 72 cm^{2}.

Question 10.

The side of square ABCD is 5cm and another square PQRS has a perimeter equal to 40 cm. Find the ratio of perimeter ABCD to the perimeter of PQRS. Find the ratio of the area of ABCD to the area of PQRS.

Answer:

Side of ABCD = 5 cm

Perimeter of ABCD = 5 × 4 = 20 cm

Perimeter of PQRS = 40 cm 4 × side = 40

side = \(\frac{40}{4}\)

side = 10cm

Perimeter of ABCD : Perimeter PQRS

= 20 : 40 = 1 : 2

Area of ABCD : Area of PQRS

= 52 : 102 = 25 : 100 = 1 : 4

Question 11.

A square field has a side of 20m. Find the length of the wire required to fence it four times.

Answer:

Perimeter = 4a = 4 × 20 = 80m

Length of the wire required to fence 4 times. = 4 × 80 = 320 m.

Question 12.

List out the difference between square and rhombus

Answer: