**Selina Concise Class 9 Maths Chapter 20 ****Area and Perimeter Of A Plane Figures**** Exercise 20B Solutions**

**Exercise 20B**

**(1) Find the area of a quadrilateral one of whose diagonals is 30 cm. long and the Perpendiculars from the other two vertices are 19 cm and 11 cm respectively.**

**Solution:**

Area (ABCD) = area (△ ABC) + area (△ ADC).

= 1/2 × AC × BM + 1/2 × AC × DN

= 1/2 × 30 × 19 + 1/2 × 30 × 11

= 30/2 × 19 + 30 / 2 × 11

= 15 (19 + 11)

= I5 (30)

= 450 cm^{2}

**∴ **The area of a quadrilateral is 450 cm^{2}

**(2) The diagonals of a quadrilateral are 16cm and 13cm. If they interest each other at right angles, find the area of the quadrilateral.**

**Solution:**

Area (ABCD) = area (△ABC) + area (△ADC)

= 1/2 × AC × BM + 1/2 × Ac × DM

= AC/2 [BM + DM]

= 16/2 [BD]

= 8 × BD

= 8 × 13

= 104 cm^{2}

**∴ **The area of the quadrilateral is 104 cm^{2}.

**(3) Calculate the area of quadrilateral ABCD, in which ****∠****ABD = 90****°****, triangle BCD is an equilateral triangle of side 24 cm and AD = 26 cm.**

**Solution:**

In △ABD,

By Pythagoras Theorem,

(hypotenuse)^{2} = (Base)^{2} + (height)^{2}

AD^{2} = BD^{2} + AB^{2}

(26)^{2} = (24)^{2} + AB^{2}

(26)^{2} – (24)^{2} = AB^{2}

By using formula,

( a^{2} – b^{2}) = (a + b) (a – b)

(26 + 24) (26 – 24) = AB^{2}

50 × 2 = AB^{2}

100 = AB^{2}

Taking Square root on both Sides

√100 = √AB^{2}

10 = AB

area (ABCD) = area (△ABD) + area (△BCD)

= 1/2 × BD × AB + √3/4 (24)^{2}

= 1/2 × 24 × 10 + √3/4 × 24 × 24

= 120 + √3 × 6 × 24

= 120 + 6 × 24 √3

= 120 + 144 × 1.732

= 120 + 244. 41

= 369. 41 cm^{2}

**(4) Calculate the area of ****quadrilateral ABCD in which AB = 32 cm, AD = 24cm, ****∠****A** **=** **90****° ****and BC = CD = 52 cm.**

**Solution:**

In △ABD,

By Pythagoras Theorem,

BD^{2} = AD^{2} + AB^{2}

= (24)^{2} + (32)^{2}

= 576 + 1024

BD^{2} = 1600

Taking Square root on both sides.

√BD^{2} = √1600

BD = 40 Cm

Area (ABCD) = area (△ABD) + area (△BCD)

= 1/2 × Base × height + b/4 √4a^{2} – b^{2}

= 1/2 × AB × AD + 40 √4 × (52)^{2} – (40)^{2}

= 1/2 × 32 × 24 + 10 √4 × 52 × 52 – 40 × 40

= 16 × 24 + 10 √4 × 4 × 13 × 52 – 4 × 10 × 10 × 4

= 384 + 10 √4 × 4 (13 × 52 – 100)

= 384 + 10 × 4 √676 – 100

= 384 + 40 √576

= 384 + 40 × 24

= 384 + 960

= 1344 cm^{2}

**∴ **The area of quadrilateral ABCD is 1344 Cm^{2}

**(5) The Perimeter of a rectangular field is 315 Km. If the length of the field is twice it’s width, find the area of the rectangle in Square meters.**

**Solution:**

Let breath of rectangle is x α m and the length of rectangle is 2xm.

Perimeter = 3/5 Km

= 3/5 × 1000 m

= 3 × 200

Perimeter = 600 m.

2 (l + b) = 600

2 (2x + x) = 600

2 (3x) = 600

3x = 600/2

3x = 200

X = 300/3

X = 100

**∴ **breadth = 100 m

**∴ **l = 2x = 2 × 100 = 200m

**∴ **Area of rectangle = l × b

= 200 × 100

= 20,000 m^{2} or Sq. m.

**Here is your solution of Selina Concise Class 9 Maths Chapter 20 Area and Perimeter Of A Plane Figures Exercise 20B**

**Dear Student, I appreciate your efforts and hard work that you all had put in. Thank you for being concerned with us and I wish you for your continued success.**

**For more solutions, See below ⇓**