RS Aggarwal 2019,2020 solution class 7 chapter 20 Mensuration Exercise 20D

Exercise 20D

Page-242

Question 1:

Find the area of the triangle in which

(i) base = 42 cm and height = 25 cm,
(ii) base = 16.8 m and height = 75 cm,
(iii) base = 8 dm and height = 35 cm,

Answer 1:

We know:
Area of a triangle = $\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}$
(i) Base = 42 cm
Height = 25 cm
     ∴ Area of the triangle = $\left(\frac{1}{2}\times 42\times 25\right)$ cm² = 525 cm²
(ii) Base = 16.8 m    
     Height = 75 cm = 0.75 m      [since 100 cm = 1 m]
     ∴ Area of the triangle = $\left(\frac{1}{2}\times 16.8\times 0.75\right)$ m² = 6.3 m²
(iii) Base = 8 dm = (8 $\times$ 10) cm = 80 cm     [since 1 dm = 10 cm]
       Height = 35 cm
     ∴ Area of the triangle = $\left(\frac{1}{2}\times 80\times 35\right)$ cm² = 1400 cm²

Question 2:

Find the height of a triangle having an area of 72 cm² and base 16 cm.

Answer 2:

Height of a triangle = $\frac{2\times \mathrm{Area}}{\mathrm{Base}}$
Here, base = 16 cm and area = 72 cm²

∴ Height = $\left(\frac{2\times 72}{16}\right)$ cm = 9 cm

Question 3:

Find the height of a triangle region having an area of 224 m² and base 28 m.

Answer 3:

Height of a triangle = $\frac{2\times \mathrm{Area}}{\mathrm{Base}}$
Here, base = 28 m and area = 224 m²

∴ Height = $\left(\frac{2\times 224}{28}\right)$ m = 16 m

Question 4:

Find the base of a triangle whose are is 90 cm² and height 12 cm.

Answer 4:

Base of a triangle = $\frac{2\times \mathrm{Area}}{\mathrm{Height}}$
Here, height = 12 cm and area = 90 cm²

∴ Base = $\left(\frac{2\times 90}{12}\right)$ cm = 15 cm

Question 5:

The base of a triangular field is three times its height. If the cost of cultivating the field at Rs 1080 per hectare is Rs 14580, find its base and height.

Answer 5:

Total cost of cultivating the field = Rs. 14580

Rate of cultivating the field = Rs. 1080 per hectare

Area of the field = $\left(\frac{\mathrm{Total} \mathrm{cost}}{\mathrm{Rate} \mathrm{per} \mathrm{hectare}}\right)$ hectare
                           = $\left(\frac{14580}{1080}\right)$ hectare
                           =  13.5 hectare
                           = (13.5 $\times$ 10000) m² = 135000 m²       [since 1 hectare = 10000 m² ]
Let the height of the field be x m.
Then, its base will be 3x m.
Area of the field = $\left(\frac{1}{2}\times 3x\times x\right)$ m² = $\left(\frac{3x^2}{2}\right)$ m²
∴ $\left(\frac{3x^2}{2}\right)$ = 135000
⇒ $x^{2 }=\left(135000\times \frac{2}{3}\right)=90000$
⇒ x = $\sqrt{90000}$ = 300
∴ Base = (3 $\times$ 300) = 900 m
Height = 300 m

Question 6:

The area of right triangular region is 129.5 cm². If one of the sides containing the right angle is 14.8 cm, find the other one.

Answer 6:

Let the length of the other leg be h cm.
Then, area of the triangle = $\left(\frac{1}{2}\times 14.8\times h\right)$ cm² = (7.4 h) cm²
But it is given that the area of the triangle is 129.5 cm².
∴ 7.4h = 129.5
⇒ h = $\left(\frac{129.5}{7.4}\right)$ = 17.5 cm
∴ Length of the other leg = 17.5 cm

Question 7:

Find the area of a right triangle whose base is 1.2 m and hypotenuse 3.7 m.

Answer 7:

Here, base = 1.2 m and hypotenuse = 3.7 m

In the right angled triangle:

Perpendicular = $\sqrt{(H\mathrm{ypotenuse}{)}^2-(B\mathrm{ase}{)}^2}$

                        $$\begin{gathered} =\sqrt{{\left(3.7\right)}^2-{\left(1.2\right)}^2} \\ =\sqrt{13.69-1.44} \\ =\sqrt{12.25} \\ =3.5 \end{gathered}$$
 Area = $\left(\frac{1}{2}\times \mathrm{base}\times \mathrm{perpendicular}\right)$ sq. units
          = $\left(\frac{1}{2}\times 1.2\times 3.5\right)$ m²
∴ Area of the right angled triangle = 2.1 m²

Question 8:

The legs of a right triangle are in the ratio 3 : 4 and its area is 1014 cm². Find the lengths of its legs.

Answer 8:

In a right angled triangle, if one leg is the base, then the other leg is the height.
Let the given legs be 3x and 4x, respectively.
Area of the triangle = $\left(\frac{1}{2}\times 3x\times 4x\right)$ cm²
⇒ 1014 = (6x²)
⇒ 1014 = 6x²
⇒ x² = $\left(\frac{1014}{6}\right)$ = 169
⇒ x = $\sqrt{169}$ = 13
∴ Base = (3 $\times$ 13) = 39 cm
Height = (4 $\times$ 13) = 52 cm

Question 9:

One side of a right-angled triangular scarf is 80 cm and its longest side is 1 m. Find its cost at the rate of Rs 250 per m².

Answer 9:

Consider a right-angled triangular scarf (ABC).
Here, ∠B= 90°
BC = 80 cm
AC = 1 m = 100 cm

Now, AB² + BC² = AC²
⇒ AB² = AC² - BC² = (100)² - (80)²
            = (10000 - 6400) = 3600
 ⇒ AB = $\sqrt{3600}$  = 60 cm
Area of the scarf ABC = $\left(\frac{1}{2}\times BC\times AB\right)$ sq. units
                               = $\left(\frac{1}{2}\times 80\times 60\right)$ cm²
                               = 2400 cm² = 0.24 m²     [since 1 m² = 10000 cm²]
Rate of the cloth = Rs 250 per m²
∴ Total cost of the scarf = Rs (250 $\times$ 0.24) = Rs 60
Hence, cost of the right angled scarf is Rs 60.
                            

Question 10:

Find the area of an equilateral triangle each of whose sides measures (i) 18 cm,  (ii) 20 cm.
[Take $\sqrt{3}$ = 1.73]

Answer 10:

(i) Side of the equilateral triangle = 18 cm
     Area of the equilateral triangle = $\frac{\sqrt{3}}{4}{\left(\mathrm{Side}\right)}^2$  sq. units
                                                      =  $\frac{\sqrt{3}}{4}{\left(18\right)}^2$ cm² = $\left(\sqrt{3}\times 81\right)$ cm²
                                                      = (1.73 $\times$ 81) cm² = 140.13 cm²

(ii) Side of the equilateral triangle = 20 cm
     Area of the equilateral triangle = $\frac{\sqrt{3}}{4}{\left(\mathrm{Side}\right)}^2$  sq. units
                                                      =  $\frac{\sqrt{3}}{4}{\left(20\right)}^2$ cm² = $\left(\sqrt{3}\times 100\right)$ cm²
                                                      = (1.73 $\times$ 100) cm² = 173 cm²

Question 11:

The area of an equilateral triangle is $(16\times \sqrt{3}){\mathrm{cm}}^2$. Find the length of each side the triangle.

Answer 11:

It is given that the area of an equilateral triangle is $16\sqrt{3}$ cm².

We know:
Area of an equilateral triangle = $\frac{\sqrt{3}}{4}{\left(\mathrm{side}\right)}^2$ sq. units

∴ Side of the equilateral triangle = $\left[\sqrt{\left(\frac{4\times \mathrm{Area}}{\sqrt{3}}\right)}\right]$ cm
                                                      =   $\left[\sqrt{\left(\frac{4\times 16\sqrt{3}}{\sqrt{3}}\right)}\right]$cm = $\left(\sqrt{4\times 16}\right)$cm = $\left(\sqrt{64}\right)$cm = 8 cm

Hence, the length of the equilateral triangle is 8 cm.

Question 12:

Find the length of the height of an equilateral triangle of side 24 cm. (Take $\sqrt{3}$=1.73)

Answer 12:

Let the height of the triangle be h cm.
Area of the triangle = $\left(\frac{1}{2}\times \mathrm{Base} \times \mathrm{Height}\right)$ sq. units
                          = $\left(\frac{1}{2}\times 24\times h\right)$ cm²

Let the side of the equilateral triangle be a cm.
Area of the equilateral triangle = $\left(\frac{\sqrt{3}}{4}{a}^2\right)$ sq. units
                                            = $\left(\frac{\sqrt{3}}{4}\times 24\times 24\right)$ cm² = $\left(\sqrt{3}\times 144\right)$ cm²
∴ $\left(\frac{1}{2}\times 24\times h\right)$ = $\left(\sqrt{3}\times 144\right)$
⇒ 12 h = $\left(\sqrt{3}\times 144\right)$
⇒ h = $\left(\frac{\sqrt{3}\times 144}{12}\right)=\left(\sqrt{3}\times 12\right)=\left(1.73\times 12\right)=20.76$ cm
∴ Height of the equilateral triangle = 20.76 cm

Question 13:

Find the area of the triangle in which

(i) a = 13 m, b = 14 m, c = 15m:
(ii) a = 52 m, b = 56 cm, c = 60 cm:
(iii) a = 91 m, b = 98 m, c = 105 m.

Answer 13:

(i) Let a = 13 m, b = 14 m and c = 15 m
     s = $\left(\frac{a+b+c}{2}\right)$ = $\left(\frac{13+14+15}{2}\right)=\left(\frac{42}{2}\right)$m = 21 m
∴  Area of the triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ sq. units
                               = $\sqrt{21\left(21-13\right)\left(21-14\right)\left(21-15\right)}$m²
                               =   $\sqrt{21\times 8\times 7\times 6}$ m²
                               = $\sqrt{3\times 7\times 2\times 2\times 2\times 7\times 2\times 3}$ m²
                               = (2 $\times$ 2 $\times$ 3 $\times$ 7) m²
                               = 84 m²

(ii) Let a = 52 cm, b = 56 cm and c = 60 cm
      s = $\left(\frac{a+b+c}{2}\right)$ = $\left(\frac{52+56+60}{2}\right)=\left(\frac{168}{2}\right)$ cm = 84 cm
∴  Area of the triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ sq. units
                               = $\sqrt{84\left(84-52\right)\left(84-56\right)\left(84-60\right)}$cm²
                               =   $\sqrt{84\times 32\times 28\times 24}$ cm²
                               = $\sqrt{12\times 7\times 4\times 8\times 4\times 7\times 3\times 8}$ cm²
                               = $\sqrt{2\times 2\times 3\times 7\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 7\times 3\times 2\times 2\times 2}$ cm²
                               = (2 $\times$ 2 $\times$ 2 $\times$ 2 $\times$ 2 $\times$ 2 $\times$ 3 $\times$ 3 $\times$ 7) m²
                               = 1344 cm²

(iii) Let a = 91 m, b = 98 m and c = 105 m
      s = $\left(\frac{a+b+c}{2}\right)$ = $\left(\frac{91+98+105}{2}\right)=\left(\frac{294}{2}\right)$ m = 147 m
∴  Area of the triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ sq. units
                               = $\sqrt{147\left(147-91\right)\left(147-98\right)\left(147-105\right)}$m²
                               =  $\sqrt{147\times 56\times 49\times 42}$ m²
                               = $\sqrt{3\times 49\times 8\times 7\times 49\times 6\times 7}$ m²
                               = $\sqrt{3\times 7\times 7\times 2\times 2\times 2\times 7\times 7\times 7\times 2\times 3\times 7}$ m²
                               = ( 2 $\times$ 2 $\times$ 3 $\times$ 7 $\times$ 7 $\times$ 7) m²
                               = 4116 m²

Question 14:

The lengths of the sides of a triangle are 33 cm, 44 cm and 55 respectively. Find the area of the triangle and hence find the height corresponding to the side measuring 44 cm.

Answer 14:

Let a = 33 cm, b = 44 cm and c = 55 cm
Then, s = $\frac{a+b+c}{2}$ = $\left(\frac{33+44+55}{2}\right)$ cm = $\left(\frac{132}{2}\right)$ cm = 66 cm
∴  Area of the triangle = $\sqrt{\mathrm{s}\left(\mathrm{s}-\mathrm{a}\right)\left(\mathrm{s}-\mathrm{b}\right)\left(\mathrm{s}-\mathrm{c}\right)}$ sq. units
                                      = $\sqrt{66\left(66-33\right)\left(66-44\right)\left(66-55\right)}$ cm²
                                      =  $\sqrt{66\times 33\times 22\times 11}$ cm²
                                      = $\sqrt{6\times 11\times 3\times 11\times 2\times 11\times 11}$ cm²
                                      = (6 $\times$ 11 $\times$ 11) cm² = 726 cm²

Let the height on the side measuring 44 cm be h cm.
Then, Area  = $\frac{1}{2}\times b\times h$
⇒ 726 cm² = $\frac{1}{2}\times 44\times h$
⇒ h = $\left(\frac{2\times 726}{44}\right)$ cm = 33 cm.
∴  Area of the triangle = 726 cm²
Height corresponding to the side measuring 44 cm = 33 cm

Question 15:

The sides of a triangle are in the ratio 13 : 14 : 15 and its perimeter is 84 cm. Find the area of the triangle.

Answer 15:

Let a = 13x cm, b = 14x cm and c = 15x cm
Perimeter of the triangle = 13x + 14x + 15x = 84 (given)
⇒ 42x = 84
⇒ x = $\frac{84}{42}=2$
∴ a = 26 cm , b = 28 cm and c = 30 cm
               
s = $\frac{a+b+c}{2}$ = $\left(\frac{26+28+30}{2}\right)$cm = $\left(\frac{84}{2}\right)$cm = 42 cm
∴  Area of the triangle = $\sqrt{\mathrm{s}\left(\mathrm{s}-\mathrm{a}\right)\left(\mathrm{s}-\mathrm{b}\right)\left(\mathrm{s}-\mathrm{c}\right)}$ sq. units
                               = $\sqrt{42\left(42-26\right)\left(42-28\right)\left(42-30\right)}$ cm²
                               =  $\sqrt{42\times 16\times 14\times 12}$ cm²
                               = $\sqrt{6\times 7\times 4\times 4\times 2\times 7\times 6\times 2}$ cm²
                               = (2 $\times$4 $\times$ 6 $\times$ 7) cm² = 336 cm²
Hence, area of the given triangle is 336 cm².

Question 16:

The sides of a triangle are 42 cm, 34 cm and 20 cm. Calculate its area and the length of the height on the longest side.

Answer 16:

Let a = 42 cm, b = 34 cm and c = 20 cm
Then, s = $\frac{a+b+c}{2}$ = $\left(\frac{42+34+20}{2}\right)$ cm = $\left(\frac{96}{2}\right)$ cm = 48 cm
∴  Area of the triangle = $\sqrt{\mathrm{s}\left(\mathrm{s}-\mathrm{a}\right)\left(\mathrm{s}-\mathrm{b}\right)\left(\mathrm{s}-\mathrm{c}\right)}$ sq. units
                               = $\sqrt{48\left(48-42\right)\left(48-34\right)\left(48-20\right)}$ cm²
                               =  $\sqrt{48\times 6\times 14\times 28}$ cm²
                               = $\sqrt{6\times 2\times 2\times 2\times 6\times 14\times 2\times 14}$ cm²
                               = (2 $\times$ 2 $\times$ 6 $\times$ 14) cm² = 336 cm²

Let the height on the side measuring 42 cm be h cm.
Then, Area  = $\frac{1}{2}\times b\times h$
⇒ 336 cm² = $\frac{1}{2}\times 42\times h$
⇒ h = $\left(\frac{2\times 336}{42}\right)$ cm = 16 cm
∴ Area of the triangle = 336 cm²
Height corresponding to the side measuring 42 cm = 16 cm

Question 17:

The base of an isosceles triangle is 48 cm and one of its equal sides is 30 cm. Find the area of the triangle.

Answer 17:

Let each of the equal sides be a cm.
b = 48 cm
a = 30 cm
Area of the triangle = $\left\{\frac{1}{2}\times b\times \sqrt{a^2-\frac{b^2}{4}}\right\}$ sq. units
                                = $\left\{\frac{1}{2}\times 48\times \sqrt{{\left(30\right)}^2-\frac{{\left(48\right)}^2}{4}}\right\}$ cm² = $\left(24\times \sqrt{900-\frac{2304}{4}}\right)$ cm²
                                = $\left(24\times \sqrt{900-576}\right)$ cm² = $\left(24\times \sqrt{324}\right)$ cm² = (24 $\times$ 18) cm² = 432 cm²
∴ Area of the triangle = 432 cm²

Question 18:

The base of an isosceles triangle is 12 cm and its perimeter is 32 cm. Find its area.

Answer 18:

Let each of the equal sides be a cm.
a + a + 12 = 32 ⇒ 2a = 20  ⇒ a = 10
∴ b = 12 cm and a = 10 cm
Area of the triangle = $\left\{\frac{1}{2}\times b\times \sqrt{a^2-\frac{b^2}{4}}\right\}$ sq. units
                                = $\left\{\frac{1}{2}\times 12\times \sqrt{100-\frac{144}{4}}\right\}$ cm² = $\left(6-\sqrt{100-36}\right)$ cm²
                                = $\left(6\times \sqrt{64}\right)$ cm² = (6 $\times$ 8) cm²
                                = 48 cm²

Question 19:

A diagonal of a quadrilateral is 26 cm and the perpendiculars drawn to it from the opposite vertices are 12.8 cm and 11.2 cm. Find the area of the quadrilateral.

Answer 19:

We have:
AC = 26 cm, DL = 12.8 cm and BM = 11.2 cm

Area of ΔADC = $\frac{1}{2}$ $\times$ AC $\times$ DL
                          = $\frac{1}{2}$ $\times$ 26 cm $\times$ 12.8 cm = 166.4 cm²
Area of ΔABC = $\frac{1}{2}$ $\times$ AC $\times$ BM
                         = $\frac{1}{2}$ $\times$ 26 cm $\times$ 11.2 cm = 145.6 cm²

∴ Area of the quadrilateral ABCD = Area of ΔADC  + Area of ΔABC
                                              = (166.4 + 145.6) cm²
                                                        = 312 cm²

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Question 20:

In a quadrilateral ABCD, AB = 28 cm, BC = 26 cm, CD = 50 cm, DA = 40 cm and diagonal AC = 30 cm. Find the area of the quadrilateral.

Answer 20:

First, we have to find the area of ΔABC and ΔACD.
For ΔACD:
Let a = 30 cm, b = 40 cm and c = 50 cm
     s = $\left(\frac{a+b+c}{2}\right)=\left(\frac{30+40+50}{2}\right)=\left(\frac{120}{2}\right)=60$ cm
∴  Area of triangle ACD = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$  sq. units
                                        = $\sqrt{60\left(60-30\right)\left(60-40\right)\left(60-50\right)}$ cm²
                                        =  $\sqrt{60\times 30\times 20\times 10}$ cm²
                                        = $\sqrt{360000}$ cm²
                                        = 600 cm²
For ΔABC:
Let a = 26 cm, b = 28 cm and c = 30 cm
     s = $\left(\frac{a+b+c}{2}\right)=\left(\frac{26+28+30}{2}\right)=\left(\frac{84}{2}\right)=42$ cm
∴  Area of triangle ABC = $\sqrt{s\left(s-a\right)\left(s-b\left(s-c\right)\right)}$  sq. units
                                        = $\sqrt{42\left(42-26\right)\left(42-28\right)\left(42-30\right)}$ cm²
                                        =  $\sqrt{42\times 16\times 14\times 12}$ cm²
                                        = $\sqrt{2\times 3\times 7\times 2\times 2\times 2\times 2\times 2\times 7\times 3\times 2\times 2}$ cm²
                                        = (2 $\times$2 $\times$ 2 $\times$ 2 $\times$ 3 $\times$ 7) cm²
                                        = 336 cm²
∴ Area of the given quadrilateral ABCD =  Area of ΔACD + Area of ΔABC
                                                                           = (600 + 336) cm² = 936 cm²

Question 21:

In the given figure, ABCD is a rectangle with length = 36 m and breadth = 24 m. In ∆ADE, EF ⊥ AD and EF = 15 m. Calculate the area of the shaded region.

Answer 21:

Area of the rectangle = AB $\times$ BC
                                   = 36 m $\times$ 24 m
                                   = 864 m²
Area of the triangle  = $\frac{1}{2}$ $\times$ AD $\times$ FE
                                 = $\frac{1}{2}$ $\times$ BC $\times$ FE       [since AD = BC]
                                 = $\frac{1}{2}$ $\times$ 24 m $\times$ 15 m
                                 = 12 m $\times$15 m = 180 m²
∴ Area of the shaded region = Area of the rectangle − Area of the triangle
                                              = (864 − 180) m²
                                                         =  684 m²

Question 22:

In the given figure, ABCD is a rectangle in which AB = 40 cm and BC = 25 cm. If P,Q,R.S be the midpoints of AB, BC, CD and DA respectively, find the area of the shaded region.

Answer 22:

Join points PR and SQ.
These two lines bisect each other at point O.

Here, AB = DC = SQ = 40 cm
AD = BC =RP = 25 cm

Also, OP = OR = $\frac{RP}{2}=\frac{25}{2}$ = 12.5 cm
From the figure we observe:
Area of ΔSPQ = Area of ΔSRQ
∴ Area of the shaded region   = 2 $\times$ (Area of ΔSPQ)
                                                       = 2 $\times$ ($\frac{1}{2}$$\times$ SQ $\times$OP)
                                                       = 2 $\times$ ($\frac{1}{2}$ $\times$ 40 cm $\times$ 12.5 cm)
                                                       = 500 cm²

Question 23:

In the following figures, find the area of the shaded region.

Answer 23:

(i) Area of rectangle ABCD = (10 cm x 18 cm) = 180 cm²
    
    Area of triangle I = $\left(\frac{1}{2}\times 6\times 10\right)$ cm² = 30 cm²
    Area of triangle II = $\left(\frac{1}{2}\times 8\times 10\right)$ cm² = 40 cm²
    ∴ Area of the shaded region = {180 - ( 30 + 40)} cm² = { 180 - 70}cm² = 110 cm²

(ii) Area of square ABCD = (Side)² = (20 cm)² = 400 cm²
     
     Area of triangle I = $\left(\frac{1}{2}\times 10\times 20\right)$ cm² = 100 cm²
    Area of triangle II = $\left(\frac{1}{2}\times 10\times 10\right)$ cm² = 50 cm²
    Area of triangle III = $\left(\frac{1}{2}\times 10\times 20\right)$ cm² = 100 cm²
   ∴ Area of the shaded region = {400 - ( 100 + 50 + 100)} cm² = { 400 - 250}cm² = 150 cm²

Question 24:

Find the area of quadrilateral ABCD in which diagonal BD = 24 cm. AL ⊥ BD and CM ⊥ BD such that AL = 5 cm and CM = 8 cm.

Answer 24:

Let ABCD be the given quadrilateral and let BD be the diagonal such that BD is of the length 24 cm.
Let AL ⊥ BD and CM ⊥ BD
Then, AL = 5 cm and CM = 8 cm
Area of the quadrilateral ABCD = (Area of ΔABD + Area of ΔCBD)
                                              =  $\left[\left(\frac{1}{2}\times BD\times AL\right)+\left(\frac{1}{2}\times BD\times CM\right)\right]$ sq. units
                                              =  $\left[\left(\frac{1}{2}\times 24\times 5\right)+\left(\frac{1}{2}\times 24\times 8\right)\right]$ cm²
                                              = ( 60 + 96) cm² = 156 cm²

∴ Area of the given quadrilateral = 156 cm