myhelper: RS Aggarwal solution class 8 chapter 18 Area of Trapezium and a Polygon Exercise 18B

Exercise 18B

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Q1 | Ex-18B | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon

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Question 1:

In the given figure, ABCD is a quadrilateral in which AC = 24 cm, BL ⊥ AC and DM ⊥ AC such that BL = 8 cm and DM = 7 cm. Find the area of quad. ABCD.

Answer 1:

$Area of quadrilateral ABCD = (Area of ∆ ADC) + (Area of ∆ ACB)$
$= (\frac{1}{2} \times AC \times DM) + (\frac{1}{2} \times AC \times BL) = [(\frac{1}{2} \times 24 \times 7) + (\frac{1}{2} \times 24 \times 8)] {cm}^{2} = (84 + 96) {cm}^{2} = 180 {cm}^{2}$

$Hence, the area of the quadrilateral is 180 {cm}^{2} .$

Q2 | Ex-18B | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon

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Question 2:

In the given figure, ABCD is a quadrilateral-shaped field in which diagonal BD is 36 m, AL ⊥ BD and CM ⊥ BD such that AL = 19 m and CM = 11 m. Find the area of the field.

Answer 2:

$Area of quadrilateral ABCD = (Area of ∆ ABD) + (Area of ∆ BCD)$
$= (\frac{1}{2} \times BD \times AL) + (\frac{1}{2} \times BD \times CM)$
$= [(\frac{1}{2} \times 36 \times 19) + (\frac{1}{2} \times 36 \times 11)] m^{2} = (342 + 198) m^{2} = 540 m^{2}$

$Hence, the area of the field is 540 m^{2} .$

Q3 | Ex-18B | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon

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Question 3:

Find the area of pentagon ABCDE in which BL ⊥ AC, DM ⊥ AC and EN ⊥ AC such that AC = 18 cm, AM = 14 cm, AN = 6 cm, BL = 4 cm, DM = 12 cm and EN = 9 cm.

Answer 3:

$Area of pentagon ABCDE = (Area of ∆ AEN) + (Area of trapezium EDMN) + (Area of ∆ DMC) + (Area of ∆ ACB)$
$= (\frac{1}{2} \times AN \times EN) + (\frac{1}{2} \times (EN + DM) \times NM) + (\frac{1}{2} \times MC \times DM) + (\frac{1}{2} \times AC \times BL) = (\frac{1}{2} \times AN \times EN) + (\frac{1}{2} \times (EN + DM) \times (AM - AN)) + (\frac{1}{2} \times (AC - AM) \times DM) + (\frac{1}{2} \times AC \times BL) = [(\frac{1}{2} \times 6 \times 9) + (\frac{1}{2} \times (9 + 12) \times (14 - 6)) + (\frac{1}{2} \times (18 - 14) \times 12) + (\frac{1}{2} \times 18 \times 4)] {cm}^{2} = (27 + 84 + 24 + 36) {cm}^{2} = 171 {cm}^{2}$

$Hence, the area of the given pentagon is 171 {cm}^{2} .$

Q4 | Ex-18B | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon

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Question 4:

Find the area of hexagon ABCDEF in which BL ⊥ AD, CM ⊥ AD, EN ⊥ AD and FP ⊥ AD such that AP = 6 cm, PL = 2 cm, LN = 8 cm, NM = 2 cm, MD = 3 cm, FP = 8 cm, EN = 12 cm, BL = 8 cm and CM = 6 cm.

Answer 4:

$Area of hexagon ABCDEF = (Area of ∆ AFP) + (Area of trapezium FENP) + (Area of ∆ ALB)$
$= (\frac{1}{2} \times AP \times FP) + (\frac{1}{2} \times (FP + EN) \times PN) + (\frac{1}{2} \times ND \times EN) + (\frac{1}{2} \times MD \times CM) + (\frac{1}{2} \times (CM + BL) \times LM) + (\frac{1}{2} \times AL \times BL) = (\frac{1}{2} \times AP \times FP) + (\frac{1}{2} \times (FP + EN) \times (PL + LN)) + (\frac{1}{2} \times (NM + MD) \times CM) + (\frac{1}{2} \times MD \times CM) + (\frac{1}{2} \times (CM + BL) \times (LN + NM)) + (\frac{1}{2} \times (AP + PL) \times BL) = [(\frac{1}{2} \times 6 \times 8) + (\frac{1}{2} \times (8 + 12) \times (2 + 8)) + (\frac{1}{2} \times (2 + 3) \times 12) + (\frac{1}{2} \times 3 \times 6) + (\frac{1}{2} \times (6 + 8) \times (8 + 2)) + (\frac{1}{2} \times (6 + 2) \times 8)] {cm}^{2} = (24 + 100 + 30 + 9 + 70 + 32) {cm}^{2} = 265 {cm}^{2}$

$Hence, the area of the hexagon is 265 {cm}^{2} .$

Q5 | Ex-18B | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon

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Question 5:

Find the area of pentagon ABCDE in which BL ⊥ AC, CM ⊥ AD and EN ⊥ AD such that AC = 10 cm, AD = 12 cm, BL = 3 cm, CM = 7 cm and EN = 5 cm.

Answer 5:

$Area of pentagon ABCDE = (Area of ∆ ABC) + (Area of ∆ ACD) + (Area of ∆ ADE)$
$= (\frac{1}{2} \times AC \times BL) + (\frac{1}{2} \times AD \times CM) + (\frac{1}{2} \times AD \times EM) = [(\frac{1}{2} \times 10 \times 3) + (\frac{1}{2} \times 12 \times 7) + (\frac{1}{2} \times 12 \times 5)] {cm}^{2} = (15 + 42 + 30) {cm}^{2} = 87 {cm}^{2}$

$Hence, the area of the pentagon is 87 {cm}^{2} .$

Q6 | Ex-18B | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon

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Question 6:

Find the area enclosed by the given figure ABCDEF as per dimensions given herewith.

Answer 6:

$Area enclosed by the given figure = (Area of trapezium FEDC) + (Area of square ABCF)$
$= [\{\frac{1}{2} \times (6 + 20) \times 8\} + (20 \times 20)] {cm}^{2} = (104 + 400) {cm}^{2} = 504 {cm}^{2}$

$Hence, the area enclosed by the figure is 504 {cm}^{2} .$

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Q7 | Ex-18B | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon

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Question 7:

Find the area of given figure ABCDEFGH as per dimensions given in it.

Answer 7:

$We will find the length of AC . From the right triangles ABC and HGF, we have : {AC}^{2} = {HF}^{2} = \{{(5)}^{2} - {(4)}^{2}\} cm$
$= (25 - 16) c m = 9 c m$

$AC = HF = \sqrt{9} c m = 3 c m$
$Area of the given figure ABCDEFGH = (Area of rectangle ADEH) + (Area of ∆ ABC) + (Area of ∆ HGF)$
$= (Area of rectangle ADEH) + 2 (Area of ∆ ABC) = (AD \times DE) + 2 (Area of ∆ ABC) = \{(AC + CD) \times DE\} + 2 (\frac{1}{2} \times BC \times AC) = \{(3 + 4) \times 8\} + 2 (\frac{1}{2} \times 4 \times 3) {cm}^{2} = (56 + 12) cm = 68 {cm}^{2}$
$Hence, the area of the given figure is 68 {cm}^{2}$ .

Q8 | Ex-18B | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon

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Question 8:

Find the area of a regular hexagon ABCDEF in which each side measures 13 cm and whose height is 23 cm, as shown in the given figure.

Answer 8:

$Let AL = DM = x cm LM = BC = 13 cm ∴ x + 13 + x = 23 \Rightarrow 2 x + 13 = 23 \Rightarrow 2 x = (23 - 13) \Rightarrow 2 x = 10 \Rightarrow x = 5 ∴ AL = 5 cm From the right ∆ AFL, we have : F L^{2} = A F^{2} - A L^{2} \Rightarrow F L^{2} = \{(13^{2}) - {(5)}^{2}\} \Rightarrow F L^{2} = (169 - 25) \Rightarrow F L^{2} = 144 \Rightarrow F L = \sqrt{144} \Rightarrow F L = 12 c m ∴ FL = BL = 12 cm Area of a regular hexagon = (A rea of t h e trapezium ADEF) + (A rea of t h e trapezium ABCD)$
$= 2 (A r e a o f t r a p e z i u m A D E F) = 2 \{\frac{1}{2} \times (A D + E F) \times F L\} = 2 \{\frac{1}{2} \times (23 + 13) \times 12\} c m^{2} = 2 (\frac{1}{2} \times 36 \times 12) c m^{2} = 432 c m^{2}$
$Hence, the area of the given regular hexagon is 432 {cm}^{2} .$

RS Aggarwal solution class 8 chapter 18 Area of Trapezium and a Polygon Exercise 18B

Exercise 18B

Page-210

Q1 | Ex-18B | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon

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Answer 1:

Q2 | Ex-18B | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon

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Q3 | Ex-18B | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon

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Q4 | Ex-18B | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon

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Q5 | Ex-18B | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon

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Q6 | Ex-18B | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon

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Q7 | Ex-18B | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon

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Q8 | Ex-18B | Class 8 | RS AGGARWAL | chapter 18 | Area of Trapezium and a Polygon

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