Class 8 Mathematics

Chapter 14 Solutions — Area

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Solutions

Overview

Step-by-step NCERT solutions for Area (Chapter 14, NCERT Class 8 Mathematics) — every question and answer worked out in full, not just the final result. You can also read the Area textbook chapter.

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What these solutions cover

All 18 questions in Area are solved in the PDF. Here's what's inside, exercise by exercise:

Rectangle and Squares

Identify the missing sidelengths. (i) A staircase figure made of 4 rectangular strips, each 7 in wide. Heights (top to bottom): 4 in, 3 in, 2 in, and ? in. The two given arrows at the right show the top strip height = 4 in and the total height on the right = 14 in. The '?' at the top labels the unknown width of the top strip, and the '? in' at the bottom labels the unknown height of the bottom…
Find the area of the spiral tube shown in the figure. The tube has the same width throughout. The rectangular spiral has outer dimensions 20 × 20 (units), with inner concentric rectangles at 15, 10, and 5 (units). The tube width is 1 unit throughout. What should be the length of the straight tube if it is to have the same area as the bent (spiral) tube?
In this figure, if the sidelength of the square is doubled, what is the increase in the areas of the regions 1, 2 and 3? Give reasons.

Triangles

1.
Find the areas of the following triangles:
- (i) Triangle with base = 4 cm and corresponding altitude = 3 cm (the perpendicular height drawn to the base).
- (ii) Triangle DEF with base EF = 5 cm and the perpendicular DN = 3.2 cm drawn from vertex D to the base EF (N lies on EF).
- (iii) Triangle NAT: right angle at A, leg AT = 3 cm, leg NA = 4 cm.
Find the length of the altitude BY. Triangle ABC has AX = 4 units (altitude from A to BC, meeting BC at X), BC = 6 units, and AC = 8 units. BY is the altitude from B to AC.
Find the area of △SUB, given that it is isosceles, SE is perpendicular to UB, and the area of △SEB is 24 sq. units.
4.
ABCD, BCEF, and BFGH are identical squares.
- (i) If the area of the red region is 49 sq. units, then what is the area of the blue region?
- (ii) In another version of this figure, if the total area enclosed by the blue and red regions is 180 sq. units, then what is the area of each square?

Area of any Polygon

Find the area of the quadrilateral ABCD given that AC = 22 cm, BM = 3 cm, DN = 3 cm, BM is perpendicular to AC, and DN is perpendicular to AC.
Find the area of the shaded region given that ABCD is a rectangle. In the figure: AE = 10 cm and EB = 8 cm (so AB = 18 cm); AF = 6 cm and FD = 4 cm (so AD = 10 cm); DC = 18 cm; BC = 10 cm. E is on AB and F is on AD. The segments FE and EC are drawn, and the shaded region is the quadrilateral DFEC.

Parallelogram

1.
Observe the parallelograms in the figure below (parallelograms a through g, all drawn on a grid with the same base and the same height between the same pair of parallel lines).
- (i) What can we say about the areas of all these parallelograms?
- (ii) What can we say about their perimeters? Which figure appears to have the maximum perimeter, and which has the minimum perimeter?
2.
Find the areas of the following parallelograms:
- (i) base = 7 cm, height = 4 cm
- (ii) base = 5 cm, height = 3 cm
- (iii) base = 4.8 cm, height = 5 cm
- (iv) base = 4.4 cm, height = 2 cm
Find QN. In parallelogram PQSR: SR = 12 cm (base), QM = 6 cm (perpendicular height from Q to base SR, meeting SR at M), and PS = 7.6 cm (the slant side). QN is the perpendicular from Q to side PS.
Consider a rectangle and a parallelogram of the same sidelengths: 5 cm and 4 cm. Which has the greater area? [Hint: Imagine constructing them on the same base.]
Which has greater area — an equilateral triangle or a square of the same sidelength as the triangle? Which has greater area — two identical equilateral triangles together or a square of the same sidelength as the triangle? Give reasons.

Rhombus and Trapezium

Find the area of a rhombus whose diagonals are 20 cm and 15 cm.
2.
Find the areas of the following figures:
- (i) A rhombus with side = 10 ft, and one diagonal = 16 ft.
- (ii) A trapezium with parallel sides 24 m and 36 m, and height = 14 m.
- (iii) A right trapezoid with the two parallel sides (vertical) = 14 in and 6 in, and the horizontal distance between them = 10 in.
- (iv) A trapezium with parallel sides 12 ft and 18 ft, and height = 8 ft.
A regular hexagon is divided into a trapezium, an equilateral triangle, and a rhombus, as shown. Find the ratio of their areas.
ZYXW is a trapezium with ZY ∥ WX. A is the midpoint of XY. Show that the area of the trapezium ZYXW is equal to the area of △ZWB, where B is the point where line ZA (extended) meets line WX.

Sample

Worked examples

3 questions from Area, solved step by step.

Triangles

Find the areas of the following triangles: (i) Triangle with base = 4 cm and corresponding altitude = 3 cm (the perpendicular height drawn to the base). (ii) Triangle DEF with base EF = 5 cm and the perpendicular DN = 3.2 cm drawn from vertex D to the base EF (N lies on EF). (iii) Triangle NAT: right angle at A, leg AT = 3 cm, leg NA = 4 cm.

Part (i):

Base = 4 cm, height = 3 cm (perpendicular to the base).

Area = (1/2) × base × height = (1/2) × 4 × 3 = 6 cm²

Part (ii): Area of triangle DEF

Base = EF = 5 cm, and the perpendicular height from D to EF is DN = 3.2 cm.

Area = (1/2) × base × height = (1/2) × 5 × 3.2 = 8 cm²

Part (iii): Area of triangle NAT

Right angle at A, so NA and AT are the two legs (perpendicular to each other).

Base = AT = 3 cm, height = NA = 4 cm.

Area = (1/2) × 3 × 4 = 6 cm²

Answer: (i) 6 cm² (ii) 8 cm² (iii) 6 cm²

Area of any Polygon

Find the area of the quadrilateral ABCD given that AC = 22 cm, BM = 3 cm, DN = 3 cm, BM is perpendicular to AC, and DN is perpendicular to AC.

Diagonal AC divides quadrilateral ABCD into two triangles: △ABC and △ACD.

BM ⊥ AC with BM = 3 cm (altitude of △ABC to base AC)

DN ⊥ AC with DN = 3 cm (altitude of △ACD to base AC)

Area(△ABC) = (1/2) × AC × BM = (1/2) × 22 × 3 = 33 cm²

Area(△ACD) = (1/2) × AC × DN = (1/2) × 22 × 3 = 33 cm²

Area(ABCD) = Area(△ABC) + Area(△ACD) = 33 + 33 = 66 cm²

Answer: Area of quadrilateral ABCD = 66 cm²

Parallelogram

Find the areas of the following parallelograms: (i) base = 7 cm, height = 4 cm (ii) base = 5 cm, height = 3 cm (iii) base = 4.8 cm, height = 5 cm (iv) base = 4.4 cm, height = 2 cm

Area of a parallelogram = base × height

(i) Area = 7 × 4 = 28 cm²

(ii) Area = 5 × 3 = 15 cm²

(iii) Area = 4.8 × 5 = 24 cm²

(iv) Area = 4.4 × 2 = 8.8 cm²

Answer: (i) 28 cm² (ii) 15 cm² (iii) 24 cm² (iv) 8.8 cm²

All 18 questions in Area are solved step by step in the Solutions PDF above.

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Chapter 14 Solutions — Area

Overview

What these solutions cover

Rectangle and Squares

Triangles

Area of any Polygon

Parallelogram

Rhombus and Trapezium

Worked examples

Find the area of the quadrilateral ABCD given that AC = 22 cm, BM = 3 cm, DN = 3 cm, BM is perpendicular to AC, and DN is perpendicular to AC.

Find the areas of the following parallelograms: (i) base = 7 cm, height = 4 cm (ii) base = 5 cm, height = 3 cm (iii) base = 4.8 cm, height = 5 cm (iv) base = 4.4 cm, height = 2 cm

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