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 cm2
∴ The area of a quadrilateral is 450 cm2
(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 cm2
∴ The area of the quadrilateral is 104 cm2.
(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
AD2 = BD2 + AB2
(26)2 = (24)2 + AB2
(26)2 – (24)2 = AB2
By using formula,
( a2 – b2) = (a + b) (a – b)
(26 + 24) (26 – 24) = AB2
50 × 2 = AB2
100 = AB2
Taking Square root on both Sides
√100 = √AB2
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 cm2
(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,
BD2 = AD2 + AB2
= (24)2 + (32)2
= 576 + 1024
BD2 = 1600
Taking Square root on both sides.
√BD2 = √1600
BD = 40 Cm
Area (ABCD) = area (△ABD) + area (△BCD)
= 1/2 × Base × height + b/4 √4a2 – b2
= 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 cm2
∴ The area of quadrilateral ABCD is 1344 Cm2
(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 m2 or Sq. m.
Here is your solution of Selina Concise Class 9 Maths Chapter 20 Area and Perimeter Of A Plane Figures Exercise 20B
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