## AP State Syllabus 8th Class Maths Solutions 9th Lesson Area of Plane Figures InText Questions

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 9 Area of Plane Figures InText Questions and Answers.

### 8th Class Maths 9th Lesson Area of Plane Figures InText Questions and Answers

Do this

Question 1.

Find the area of the following figures: [Page No. 200]

i)

Answer:

Area of a parallelogram = b × h = 7 × 4 = 28 sq.cm.

ii)

Answer:

Area of a triangle = \(\frac{1}{2}\) bh = \(\frac{1}{2}\) × 7 × 4

= 14 sq.cm.

iii)

Answer:

Area of a triangle = \(\frac{1}{2}\) bh = \(\frac{1}{2}\) × 5 × 4

= 10 sq.cm.

iv)

Answer:

Area of rhombus = \(\frac{1}{2}\) d_{1}d_{2}

= \(\frac{1}{2}\) × (4+4) × (3+3)

[∴ d_{1} = 4 + 4 = 8, d_{2} = 3 + 3 = 6]

= \(\frac{1}{2}\) × 8 × 6

= 24 cm^{2}

v)

Answer:

Area of a rectangle = l × b

= 20 × 14 = 280 sq.cm

vi)

Answer:

Area of a square = s^{2}

= s × s

= 5 × 5 = 25 cm^{2}

Question 2.

The measurements of some plane figures are given in the table below. However, they are incomplete. Find the missing information. [Page No. 200]

Answer:

Question 3.

Find the area of the following trapezium. [Page No. 204]

fig (i)

Answer:

Area of a trapezium

fig (ii)

Answer:

Area of a trapezium

Question 4.

Area of a trapezium is 16 cm^{2}. Length of one parallel side is 5 cm and distance between two parallel sides is 4 cm. Find the length of the other parallel side. Try to draw this trapezium on a graph paper and check the area.

[Page No. 204]

Answer:

Given that

Area of a trapezium = 16 sq.cm

Length of one of the parallel sides is a = 5 cm; h = 4 cm

Length of 2nd parallel side (b) = ?

A = \(\frac{1}{2}\)h(a + b)

Graph Sheet:

Area of parallelogram ABCD = 12 sq.cm + (S + P) + (Q + R) + (W + T) + (V + U)

= 12 + 1 + 1 + 1 + 1

= 12 + 4

= 16 sq.cm

Question 5.

ABCD is a parallelogram whose area is 100 sq.cm. P is any point insile the parallelogram (see fig.) find tie area of △APB + △CPD. [Page No. 204]

Answer:

Area of parallelogram ABCD = 100 sq.cm

From the given figure,

ar (△APB) + ar (△CPD) = ar (△PD) + ar (△BPC)

Question 6.

The following details are noted in meters in the field book of a surveyor. Find the area of the fields. [Page No. 213]

i)

Answer:

From the above figure

i) A, B, C, D, E are the vertices of pentagonal field,

ii) AD is the diagonal.

iii) Now the area of the field = Areas of 4 triangles and a trapezium.

PQ = AQ – AP = 50 – 30 = 20

QD = AD – AQ = 140 – 50 = 90

RD = AD – AR = 140 – 80 = 60

Area of △APB:

Area of trapezium PBCQ:

Area of △QCD:

Area of △DER:

Area of △ERA:

∴ Area of the field = ar △APB + ar trapezium PBCQ + ar △QCD + ar △DER + ar △ERA

= 450 + 800 + 2250 + 1500 + 2000 = 7000 sq. units

ii)

Answer:

From the above figure

i) A, B, C, D, E are the vertices of a pentagonal field.

ii) AC is the diagonal.

iii) The area of a field is equal to areas of 4 triangles and a trapezium.

QC = AC – AQ = 160 – 90 = 70

RC = AC – AR = 160 – 130 = 30

PR = AR – AP = 130 – 60 = 70

Area of △AQB:

Area of △QBC :

Area of △DRC :

Area of trapezium EPRD:

Area of △EPA :

∴ Area of the field = ar △AQB + ar △QBC + ar △DRC + ar trapezium EPRD + ar △EPA

= 2700 + 2100 + 450 + 2450 + 1200 = 8900 sq. units

Try these

Question 1.

We know that parallelogram is also a quadrilateral. Let us split such a quadrilateral into two triangles. Find their areas and subsequently that of the parallelogram. Does this process in turn with the formula that you already know? [Page No. 209]

Answer:

Area of a parallelogram ABCD

Area of parallelogram ABCD

= base x height

= bh sq. units

(OR)

Area of parallelogram ABCD

= ar △ABC + ar △ACD

= \(\frac{1}{2}\) BC × h_{1} + \(\frac{1}{2}\) AD × h_{2}

= \(\frac{1}{2}\) bh + \(\frac{1}{2}\) bh [∵ h_{1} = h_{2}]

= bh sq. units.

∴ This process in turn with already known formula.

Question 2.

Find the area of following quadrilaterals. [Page No. 213]

i)

Answer:

d = 6 cm, h_{1} = 3 cm, h_{2} = 5 cm

Area of a quadrilateral

= \(\frac{1}{2}\)d(h_{1} + h_{2})

= \(\frac{1}{2}\) × 6 (3 + 5) = 3(8) = 24 cm^{2}

ii)

Answer:

d_{1} = 7 cm; d_{2} = 6 cm

Area of a rhombus A = \(\frac{1}{2}\) d_{1}d_{2}

= \(\frac{1}{2}\) × 7 × 6

= 7 × 3 = 21 cm^{2}

iii)

Answer:

Area of a parallelogram (A) = bh

(∵ The given fig. is a parallelogram in which two opposite sides are parallel)

Area of a parallelogram = 2 ar AADC

= 2 × \(\frac{1}{2}\) × 8 × 2 = 16 Sq. cm.

[∵ Area of a parallelogram = ar △ADC + ar △ABC. But ar △ABC = ar △ADC]

Question 3.

i) Divide the following polygon into parts (triangles and trapezium) to find out its area. [Page No. 214]

Answer:

FI is a diagonal of polygon EFGHI.

If perpendiculars GA, HB are drawn on the diagonal FI, then the given figure pentagon is divided into 4 parts.

∴ Area of a pentagon EFGHI = ar △AFG + ar AGHB + ar △BHI + ar △EFI.

NQ is a diagonal of polygon MNOPQR. Here the polygon is divided into two parts.

∴ Area of a hexagon MNOPQR = ar NOPQ + ar MNQR.

ii) Polygon ABCDE is divided into parts as shown in the figure. Find the area. [Page No. 215].

If AD = 8 cm, AH = 6 cm, AF = 3 cm and perpendiculars BF = 2 cm, GH = 3 cm and EG = 2.5 cm.

Answer:

Area of polygon ABCDE = ar △AFB + ar FBCH + ar △HCD + ar △AED

So, the area of polygon ABCDE = 3 + 7.5 + 3 + 10 = 23.5 sq.cm

iii) Find the area of polygon MNOPQR if MP = 9 cm, MD = 7 cm, MC = 6 cm, MB = 4 cm, MA = 2 cm. [Page No. 215].

NA, OD, QC and RB are perpendiculars to diagonal MP.

Answer:

Area of MNOPQR

= ar △MAN + ar ADON + ar △DOP + ar △CQP + ar BCQR + ar △MBR

Hence CP = MP – MC = 9 – 6 = 3 cm

BC = MC – MB = 6 – 4 = 2 cm

AB = MB – MA = 4 – 2 = 2 cm

DP = MP – MD = 9 – 7 = 2 cm

AD = MD – MA = 7 – 2 = 5 cm

= 2.5 + (2.5 × 5.5) + 3 + 3 + 4.5 + (2 × 2.5)

= 2.5 + 13.75 + 3 + 3 + 4.5 + 5

= 31.75 sq.cms

Think, discuss and write

Question 1.

A parallelogram is divided into two congruent triangles by drawing a diagonal across it. Can we divide a trapezium into two congruent triangles? [Page No. 213]

Answer:

No, we cannot divide a trapezium into two congruent triangles.

∵ From the adjacent figure,

△ABC ≆ △ADC