SELINA Solution Class 9 Chapter 20 Area and perimeter of plane figures Exercise 20A

Question 1

Find the area of a triangle whose sides are 18 cm, 24 cm, and 30 cm. Also, find the length of altitude corresponding to the largest side of the triangle.

Sol:

Since the sides of the triangle are 18 cm, 24 cm and 30 cm respectively.

s = ${\displaystyle \frac{18+24+30}{2}}$
  = 36

Hence the area of the triangle is 

A  = ${\displaystyle \sqrt{s(s-a)(s-b)(s-c)}}$

    = ${\displaystyle \sqrt{36(36-18)(36-24)(36-30)}}$

    = ${\displaystyle \sqrt{36\times 18\times 12\times 6}}$

    = ${\displaystyle \sqrt{46656}}$

    = 216 sq.cm.

Again

A = ${\displaystyle \frac{1}{2}\text{base}\times \text{altitude}}$

Hence

216 = ${\displaystyle \frac{1}{2}\times 30\times h}$
h = 14.4cm

Question 2

The length of the sides of a triangle are in the ratio 3: 4: 5. Find the area of the triangle if its perimeter is 144 cm.

Sol:

Let the sides of the triangle are
a = 3x
b = 4x
c = 5x
Given that the perimeter is 144 cm.
hence
3x + 4x + 5x = 144
⇒ 12x = 144

⇒ x = ${\displaystyle \frac{144}{12}}$

⇒ x = 12

s = ${\displaystyle \frac{a+b+c}{2}=\frac{12x}{2}=6x=72}$

The sides are a = 36 cm, b = 48 cm and c = 60 cm.

Area of the triangle is

A = ${\displaystyle \sqrt{s(s-a)(s-b)(s-c)}}$

    = ${\displaystyle \sqrt{72(72-36)(72-48)(72-60)}}$

    = ${\displaystyle \sqrt{72\times 36\times 24\times 12}}$

     =  ${\displaystyle \sqrt{746496}}$

    =  864 cm²  

Question 3

ABC is a triangle in which AB = AC = 4 cm and ∠A = 90°. Calculate:
(i) The area of ΔABC,
(ii) The length of the perpendicular from A to BC.

Sol:

(i) Area of the triangle is given by

A = ${\displaystyle \frac{1}{2}\times \text{AB}\times \text{AC}}$

   = ${\displaystyle \frac{1}{2}\times 4\times 4}$

    = 8 sq .cm

(ii) Again area of the triangle

A = ${\displaystyle \frac{1}{2}\times \text{BC}\times h}$

8 = ${\displaystyle \frac{1}{2}\times \left(\sqrt{4^2+4^2}\right)\times h}$  

h = 2.83 cm

Question 4

The area of an equilateral triangle is 36${\displaystyle \sqrt{3}}$ sq. cm. Find its perimeter.

Sol:

Area of an equilateral triangle is given by

${\displaystyle \frac{\sqrt{3}}{4}\times {\left(\text{side}\right)}^2}$ = A

${\displaystyle \frac{\sqrt{3}}{4}\times {\left(\text{side}\right)}^2=36\sqrt{3}}$

            (side)² = 144 

              side = 12 cm

Hence
perimeter = 3 x ( its side )

                = 3 x 12

                = 36 cm

Question 5

Find the area of an isosceles triangle with perimeter is 36 cm and the base is 16 cm.

Sol:

Since the perimeter of the isosceles triangle is 36cm and the base is 16cm. hence the length of each of equal sides are  ${\displaystyle \frac{36-16}{2}}$ = 10 cm

Here

It is given that
a = equal sides = 10 cm
b = base = 16 cm

Let 'h' be the altitude of the isosceles triangle.
Since the altitude from the vertex bisects the base perpendicularly, we can apply Pythagoras Theorem,
Thus we have,

h = ${\displaystyle \sqrt{a^2-{\left(\frac{b}{2}\right)}^2}}$  = ${\displaystyle \frac{1}{2}\sqrt{4a^2-b^2}}$
We know that

Area of the triangle =${\displaystyle \frac{1}{2}\times \text{base}\times \text{altitude}}$

Area of the triangle
=${\displaystyle \frac{1}{4}\times b\times \sqrt{4a^2-b^2}}$

= ${\displaystyle \frac{1}{4}\times 16\times \sqrt{400-256}}$
= 48 sq. cm

Question 6

The base of an isosceles triangle is 24 cm and its area is 192 sq. cm. Find its perimeter.

Sol:

It is given that

Area = 192 sq .cm
base = 24 cm
Knowing the length of the equal side, a, and base, b, of an isosceles triangle, the area can be calculated using the formula,

A = ${\displaystyle \frac{1}{4}\times b\times \sqrt{4a^2-b^2}}$     

Let 'a' be the length of an equal side.

Area = ${\displaystyle \frac{1}{4}\times b\times \sqrt{4a^2-b^2}}$   

 192 = ${\displaystyle \frac{1}{4}\times 24\times \sqrt{4a^2-576}}$ 

192 = ${\displaystyle 6\sqrt{4a^2-576}}$ 

${\displaystyle \sqrt{4a^2-576}}$  = 32

4a² - 576 = 1024

         4a² = 1600

           a = 20 cm

Hence perimeter = 20 + 20 + 24 = 64 cm

Question 7

The given figure shows a right-angled triangle ABC and an equilateral triangle BCD. Find the area of the shaded portion.

Sol:

From ΔABC,

AB = ${\displaystyle \sqrt{{\text{AC}}^2-{\text{BC}}^2}}$

     = ${\displaystyle \sqrt{{16}^2-8^2}}$

     = ${\displaystyle \sqrt{192}}$

Area of ΔABC

ΔABC = ${\displaystyle \frac{1}{2}\times 8\times \sqrt{192}}$

        =  ${\displaystyle 4\sqrt{192}}$ 

Area of ΔBCD

ΔBCD = ${\displaystyle \frac{\sqrt{3}}{4}\times 8^2}$

          = ${\displaystyle 16\sqrt{3}}$

Now

ABD  = ABC - BDC
         = ${\displaystyle 4\sqrt{192}-16\sqrt{3}}$
        = ${\displaystyle 32\sqrt{3}-16\sqrt{3}}$
        = ${\displaystyle 16\sqrt{3}\text{sq . cm}}$

Question 8

Find the area and the perimeter of quadrilateral ABCD, given below; if AB = 8 cm, AD = 10 cm, BD = 12 cm, DC = 13 cm and ∠DBC = 90°.

Sol:

Given, AB = 8 cm, AD = 10 cm, BD = 12 cm, DC = 13 cm and ∠ DBC = 90^°

BC = ${\displaystyle \sqrt{{\text{DC}}^2-{\text{BD}}^2}}$

     = ${\displaystyle \sqrt{{13}^2-{12}^2}}$

     = 5 cm
Hence perimeter = 8 + 10 + 13 + 5 = 36 cm
Area of ΔABD,
ΔABD = ${\displaystyle \sqrt{15(15-8)(15-10)(15-12)}}$

= ${\displaystyle \sqrt{15\times 7\times 5\times 3}}$

= ${\displaystyle 15\sqrt{7}}$

= 39.7
Area of ΔBDC,
ΔBDC = ${\displaystyle \frac{1}{2}}$ x 12 x 5 = 30
Now,
Area of ABCD = area of ΔABD + area of ΔBDC
= 39.7 + 30
= 69.7 sq.cm

Question 9

The base of a triangular field is three times its height. If the cost of cultivating the field at ₹ 36.72 per 100 m² is ₹ 49,572; find its base and height.

Sol:

Since the cost of cultivating the field at the rate of Rs. 36.72 per 100 square metre is Rs. 49,572.

Area of the triangular field = ${\displaystyle \frac{49572\times 100}{36.72}}$ = 1,35,000 sq. m

Let the height of the triangle be h.

Now,

Area of the triangular field = ${\displaystyle \frac{1}{2}}$ × base × height

1,35,000 = ${\displaystyle \frac{1}{2}\times h\times 3h}$

⇒ h² = 90,000
⇒ h = 300

Height (h) = 300m
Base (3h) = 3 × 300 = 900 m.

Question 10

The sides of a triangular field are in the ratio 5: 3: 4 and its perimeter is 180 m. Find:
(i) its area.
(ii) the altitude of the triangle corresponding to its largest side.
(iii) the cost of leveling the field at the rate of Rs. 10 per square meter.

Sol:

(i) Given that the sides of a triangle are in the ratio 5: 3: 4.
Also, given that the perimeter of the triangle is 180.
Thus, we have, 5x + 4x + 3x = 180
⇒ 12x = 180

⇒ x = ${\displaystyle \frac{180}{12}}$

⇒ x = 15.
Thus, the sides are 5 x 15, 3 x 15 and q x 15
That is the sides are 75m, 45m and 60m.
Since the sides are in the ratio, 5: 3: 4, it is a Pythagorean triplet,
Therefore, the triangle is a right-angled triangle,

Area of a right-angled triangle = ${\displaystyle \frac{1}{2}}$ x base x altitude
⇒ ${\displaystyle \frac{1}{2}}$ x 45 x 60
⇒ 45 x 30 = 1350 m².

(ii) Consider the following figure.

In the above figure,
The largest side is AC = 75 m,
The altitude corresponding to AC is BD.
We need to find the value of BD.

Now consider the triangles ΔBCD and ΔBAD.
We have,
∠B = ∠B             .....[ common ]
BD = BD            .....[ common ]
∠D = ∠D = 90°
Thus, by Angle-Side-Angle criterion of congruence,

we have ΔBCD - ΔABD,
Similar triangles have similar proportionality.
Thus, we have,
${\displaystyle \frac{\text{CD}}{\text{BD}}=\frac{\text{BD}}{\text{AD}}}$
⇒ BD² = AD x CD             ....(1)

From subpart (i), the sides of the triangle are 
AC = 75 m, AB = 60 m and BC = 45 m
Let AD = x m ⇒ CD = ( 75 - x ) m

Thus applying Pythagoras Theorem,
from right triangle ΔBCD, we have
452 = ( 75 - x )² + BD2
⇒ BD² = 452 - ( 75 - x )²
⇒ BD² = 2025 - ( 5625 + x² - 150x )
⇒ BD² = 2025 - 5625 - x² + 150x
⇒ BD² = - 3600 - x² + 150x          ....(2)

Now applying Pythagoras Theorem,
from right triangle ΔABD, we have
602 = x² + BD²
⇒ BD² = 60² - x²
⇒ BD²  = 3600 - x²                 ......(3)
From equations (2) and (3), we have,
- 3600 - x² + 150x = 3600 -x²
⇒ 150x = 3600 + 3600
⇒ 150x = 7200

⇒ x = ${\displaystyle \frac{7200}{150}}$

⇒ x = 48

Thus, AD = 48 and CD = 75 - 48 = 27.

Substituting the values AD = 48 m
and CD = 27 m in equation (1), we have
BD² = 48 x 27
⇒ BD² = 1296
⇒ BD = 36 m

The altitude of the triangle corresponding to
its largest side is BD = 36 m

(iii)
The area of the triangular field
from subpart (i) is 1350 m²
The cost of levelling the field is Rs. 10 per square metre,
Thus the total cost of levelling the field
= 1350 x 10 = Rs. 13,500.

Question 11

Each of the equal sides of an isosceles triangle is 4 cm greater than its height. If the base of the triangle is 24 cm; calculate the perimeter and the area of the triangle.

Sol:

Let the congruent side of the isosceles triangle be a and the height be h.
Given,

a = h + 4
Since the height of the triangle, divides it into two right angled triangles,
${\displaystyle a^2=h^2+{(\frac{b}{2})}^2\Rightarrow {(h+4)}^2=h^2+{12}^2}$
${\displaystyle h^2+16+8h=h^2+144}$
8h = 128
h = 16 cm
a = 20 cm
Perimeter of the triangle = Sum of all sides = a + a + b = 20 + 20 + 24 = 64cm

Area of triangle = ${\displaystyle \frac{1}{2}}$ ​× base × height

= ${\displaystyle \frac{1}{2}}$ ​× 24 × 16 = 192 sq cm

Question 12

Calculate the area and the height of an equilateral triangle whose perimeter is 60 cm.

Sol:

Each side of the triangle is ${\displaystyle \frac{60}{3}}$ =  20 cm

Hence the area of the equilateral triangle is given by
A = ${\displaystyle \frac{\sqrt{3}}{4}}$ x 20²

   = 100√3
   = 173.2 sq.cm

The height h of the triangle is given by
${\displaystyle \frac{1}{2}}$ x 20 x h = 173.2
h = 17.32 cm.

Question 13

In triangle ABC; angle A = 90^o, side AB = x cm, AC = (x + 5) cm and area = 150 cm². Find the sides of the triangle.

Sol:

The area of the triangle is given as 150 sq.cm
${\displaystyle \frac{1}{2}\times x\times (x+5)=150}$

x² + 5x - 300 = 0
( x + 20 )( x - 15 ) = 0
x = 15

Hence AB = 15cm, AC = 20 cm and
BC = ${\displaystyle \sqrt{{15}^2+{20}^2}}$ = 25 cm.

Question 14

If the difference between the sides of a right-angled triangle is 3 cm and its area is 54 cm²; find its perimeter.

Sol:

Let the two sides be x cm and ( x - 3 ) cm.
Now,
${\displaystyle \frac{1}{2}\times x\times (x-3)}$ = 54

x² - 3x - 108 = 0
( x - 12 )( x + 9 ) = 0
x = 12 cm

Hence the sides are 12 cm, 9 cm and ${\displaystyle \sqrt{{12}^2+9^2}}$ = 15 cm.
The required perimeter is 12 + 9 + 15 = 36 cm.

Question 15

AD is altitude of an isosceles triangle ABC in which AB = AC = 30 cm and BC = 36 cm. A point O is marked on AD in such a way that ∠BOC = 90^o. Find the area of quadrilateral ABOC.

Sol:

Area of ΔABC = ${\displaystyle \frac{1}{4}\times 36\times \sqrt{4\times {30}^2-{36}^2}}$

                       = ${\displaystyle \frac{1}{4}\times 36\times \sqrt{2304}}$

                       = ${\displaystyle \frac{1}{4}}$ x 36 x 48

                       = 432
Since AB = AC and ∠BOC = 90°
∠ BOD = ∠ COD = 45°
hence ∠ OBD = 45° and OD = BD = 18 cm
Now,
Area of ΔBOC = ${\displaystyle \frac{1}{2}}$ x 36 x 18 = 324

Area of ABOC = Area of ΔABC - Area of ΔBOC 
                       = 432 - 324 
                       = 108 sq.cm