RS Aggarwal 2019,2020 solution class 7 chapter 20 Mensuration Test Paper 20

Test Paper 20

Page-257

Question 1:

Find the area of a rectangular plot on side of which is 48 m and its diagonal 50 m.

Answer 1:

We know that all the angles of a rectangle are 90° and the diagonal divides the rectangle into two right angled triangles.
So, one side of the triangle will be 48 m and the diagonal, which is 50 m, will be the hypotenuse.
According to Pythagoras theorem:
(Hypotenuse)² = (Base)² + (Perpendicular)²
Perpendicular = $\sqrt{{\left(\mathrm{Hypotenuse}\right)}^2-(\mathrm{Base}{)}^2}$
Perpendicular = $\sqrt{{\left(50\right)}^2-{\left(48\right)}^2}=\sqrt{2500-2304}=\sqrt{196}=14$ m
∴ Other side of the rectangular plot = 14 m

∴ Area of the rectangular plot = 48 m × 14 m = 672 m²
Hence, the area of a rectangular plot is 672 m².

Question 2:

A room is 9 m by 8 m by 6.5 m. It has one door of dimensions (2 m × 1.5 m) and four windows each of dimensions (1.5 m × 1 m). Find the cost of painting the walls at Rs 50 per m².

Answer 2:

Length = 9 m
Breadth = 8 m
Height = 6.5 m

Area of the four walls = {2(l + b) × h} sq. units
                          = {2(9 + 8) × 6.5} m² = {34 × 6.5} m² = 221 m²
Area of one door = (2 × 1.5) m² = 3 m²
Area of one window = (1.5 × 1) m² = 1.5 m²
 ∴ Area of four windows = (4 × 1.5) m² = 6 m²
Total area of one door and four windows = (3 + 6) m²
                                                           = 9 m²
Area to be painted = (221 - 9) m² = 212 m²
Rate of painting = Rs 50 per m²
Total cost of painting = Rs ( 212 × 50) = Rs 10600

Question 3:

Find the area of a square, the length of whose diagonal is 64 cm.

Answer 3:

Given that the diagonal of a square is 64 cm.

Area of the square = $\left\{\frac{1}{2}\times {\left(\mathrm{diagonal}\right)}^2\right\}$ sq. units.
                               = $\left\{\frac{1}{2}\times {\left(64\right)}^2\right\}$ cm² =$\left\{\frac{1}{2}\times 4096\right\}$ cm² = 2048 cm²
∴ Area of the square = 2048 cm²

Question 4:

A square lawn has a 2-m-wide path surrounding it. If the area of the path is 136 m², find the area of the lawn.

Answer 4:

Let ABCD be the square lawn and PQRS be the outer boundary of the square path.

Let one side of the lawn (AB) be x m.
Area of the square lawn = x²
Length PQ = (x m + 2 m + 2 m) = (x + 4) m
∴ Area of PQRS = (x + 4)² = (x² + 8x + 16) m²

Now, Area of the path = Area of PQRS − Area of the square lawn
⇒ 136 = x² + 8x + 16 − x²
⇒ 136 = 8x + 16
⇒ 136 − 16 = 8x
⇒ 120 = 8x
∴ x = 120 ÷ 8 = 15
∴ Side of the lawn = 15 m

∴ Area of the lawn = (Side)² = (15 m)² = 225 m²

Question 5:

A rectangular lawn is 30 m by 20 m. It has two roads each 2 m wide running in the middle of it, one parallel to the length and the other parallel to the breadth. Find the area of the roads.

Answer 5:

Let ABCD be the rectangular park.
EFGH and IJKL are the two rectangular roads with width 2 m.

Length of the rectangular park AD = 30 cm 
 Breadth of the rectangular park CD = 20 cm
Area of the road EFGH = 30 m × 2 m = 60 m²
Area of the road IJKL = 20 m × 2 m = 40 m²
Clearly, area of MNOP is common to the two roads.
∴ Area of MNOP = 2 m × 2 m = 4 m²

∴ Area of the roads = Area (EFGH) + Area (IJKL) − Area (MNOP)
                            = (60  + 40 ) m² − 4 m² = 96 m²

Question 6:

Find the area of a rhombus having each side equal to 13 cm and one of the diagonals equal to 24 cm.

Answer 6:

Let ABCD be the rhombus whose diagonals intersect at O.

Then, AB = 13 cm
        AC = 24 cm
The diagonals of a rhombus bisect each other at right angles.
Therefore, ΔAOB is a right-angled triangle, right angled at O, such that:
OA = $\frac{1}{2}$AC = 12 cm
AB = 13 cm

By Pythagoras theorem:
(AB)² = (OA)² + (OB)²
⇒ (13)² = (12)² + (OB)²
⇒ (OB)² = (13)² − (12)²
⇒ (OB)² = 169 − 144 = 25
⇒ (OB)² = (25)²
⇒ OB = 5 cm
∴ BD = 2 × OB = 2 × 5 cm = 10 cm

∴ Area of the rhombus ABCD = $\frac{1}{2}$ × AC × BD cm²
                                              = $\left(\frac{1}{2}\times 24\times 10\right)$ cm² = 120 cm²

Question 7:

The area of a parallelogram is 3385 m². If its altitude is twice the corresponding base, find the base and the altitude.

Answer 7:

Let the base of the parallelogram be x m.
Then, the altitude of the parallelogram will be 2x m.
It is given that the area of the parallelogram is 338 m².

Area of a parallelogram =  Base × Altitude
                        ∴ 338 m² =  x × 2x
                            338 m² = 2x²
⇒ x² = $\left(\frac{338}{2}\right)$ m² = 169 m²
⇒ x² = 169 m²
⇒ x = 13 m
∴ Base = x m = 13 m
Altitude = 2x m = (2 × 13)m = 26 m

Question 8:

Find the area of a right triangle having base = 24 cm and hypotenuse = 25 cm.

Answer 8:

Consider ΔABC.
Here, ∠B = 90°
          AB = 24 cm
          AC = 25 cm

Now, AB² + BC² = AC²
⇒ BC² = AC² - AB²
               = (25² - 24²)
               = (625 - 576)
               = 49
⇒ BC = $\left(\sqrt{49}\right)$cm = 7 cm
Area of ΔABC = $\left(\frac{1}{2}\times \mathrm{BC}\times \mathrm{AB}\right)$ sq. units
                         = $\left(\frac{1}{2}\times 7\times 24\right)$ cm² = 84 cm²
Hence, area of the right angled triangle is 84 cm².

Question 9:

The radius of the wheel of a car is 35 cm. How many revolutions will it make to travel 33 km?

Answer 9:

Radius of the wheel = 35 cm
Circumference of the wheel = $2\mathrm{\pi r}$
                                              = $\left(2\times \frac{22}{7}\times 35\right)$cm = (44 × 5) cm = 220 cm
                                                                                                       = $\left(\frac{220}{100}\right)$ m = $\left(\frac{11}{5}\right)$ m
Distance covered by the wheel in 1 revolution = $\left(\frac{11}{5}\right)$ m
Now, $\left(\frac{11}{5}\right)$ m is covered by the car in 1 revolution.
Thus, (33 × 1000) m will be covered by the car in $\left(1\times \frac{5}{11}\times 33\times 1000\right)$ revolutions, i.e. 15000 revolutions.
∴ Required number of revolutions = 15000

Question 10:

Find the radius of a circle whose area is 616 cm².

Answer 10:

Let the radius of the circle be r cm.
∴ Area = $\left({\mathrm{\pi r}}^2\right)$ cm²
∴ ${\mathrm{\pi r}}^2$ = 616
⇒ $\left(\frac{22}{7}\times r\times r\right)$ = 616
⇒ r² = $\left(\frac{616\times 7}{22}\right)$ = 196
⇒ r = $\sqrt{196}$ = 14 cm
Hence, the radius of the given circle is 14 cm.

Question 11:

Mark (✓) against the correct answer
The area of a circle is 154 cm². Its diameter is

(a) 14 cm
(b) 11 cm
(c) 7 cm
(d) 22 cm

Answer 11:

(a) 14 cm

Let the radius of the circle be r cm.
Then, its area will be $\left({\mathrm{\pi r}}^2\right)$ cm².
∴ ${\mathrm{\pi r}}^2$ = 154
⇒ $\left(\frac{22}{7}\times r\times r\right)$ = 154
⇒ r² = $\left(\frac{154\times 7}{22}\right)$ = 49
⇒ r = $\sqrt{49}$ = 7 cm
∴ Diameter of the circle = 2r = (2 × 7) cm = 14 cm

Question 12:

Mark (✓) against the correct answer
The circumference of a circle is 44 cm. Its area is

(a) 308 cm²
(b) 154 cm²
(c) 77 cm²
(d) 616 cm²

Answer 12:

(b) 154 cm²

Let the radius of the circle be r cm.
Circumference = $\left(2\mathrm{\pi r}\right)$cm
∴ $\left(2\mathrm{\pi r}\right)$ = 44
⇒ $\left(2\times \frac{22}{7}\times r\right)=44$
⇒ r = $\left(\frac{44\times 7}{2\times 22}\right)$ = 7 cm
∴ Area of the circle = ${\mathrm{\pi r}}^2$
                                 = $\left(\frac{22}{7}\times 7\times 7\right)$ cm² = 154 cm²

Question 13:

Mark (✓) against the correct answer
Each diagonal of a square is 14 cm long. Its area is

(a) 196 cm²
(b) 88 cm²
(c) 98 cm²
(d) 147 cm²

Answer 13:

(c) 98 cm²

Given that the diagonal of a square is 14 cm.

Area of a square = $\left\{\frac{1}{2}\times {\left(D\mathrm{iagonal}\right)}^2\right\}$sq. units.
                               = $\left\{\frac{1}{2}\times {\left(14\right)}^2\right\}$ cm² =$\left\{\frac{1}{2}\times 196\right\}$ cm² = 98 cm²
Hence, area of the square is 98 cm².

Question 14:

Mark (✓) against the correct answer
The area of a square is 50 cm². The length of its diagonal is

(a) $5\sqrt{2}$ cm
(b) 10 cm
(c) $10\sqrt{2}$ cm
(d) 8 cm

Answer 14:

(b) 10 cm

Given that the area of the square is 50 cm².
We know:
Area of a square = $\left\{\frac{1}{2}\times {\left(\mathrm{Diagonal}\right)}^2\right\}$sq. units
∴ Diagonal of the square = $\sqrt{2\times \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{square}}$ = $\left(\sqrt{2\times 50}\right)$cm = $\left(\sqrt{100}\right)$cm = 10 cm
Hence, the diagonal of the square is 10 cm.

Question 15:

Mark (✓) against the correct answer
The length and breadth of a rectangular park are in the ratio 4 : 3 and its perimeter is 56 m. The area of the field is

(a) 192 m²
(b) 300 m²
(c) 432 m²
(d) 228 m²

Answer 15:

(a) 192 m²

Let the length of the rectangular park be 4x.
∴ Breadth = 3x
Perimeter of the park = 2(l + b) = 56 m (given)
⇒ 56 = 2(4x + 3x)  
⇒ 56 = 14x
⇒ x = $\frac{56}{14}$= 4
Length = 4x = (4 × 4) = 16 m
Breadth = 3x = (3 × 4) = 12 m
∴ Area of the rectangular park = 16 m × 12 m = 192 m²

Question 16:

Mark (✓) against the correct answer
The sides of triangle are 13 cm, 14 cm and 15 cm. The area of the triangle is

(a) 84 cm²
(b) 91 cm²
(c) 105 cm²
(d) 97.5 cm²

Answer 16:

(a) 84 cm²

Let a = 13 cm, b = 14 cm and c = 15 cm
 s = $\frac{a+b+c}{2}$ = $\left(\frac{13+14+15}{2}\right)$cm = 21 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
                               = $\sqrt{21\left(21-13\right)\left(21-14\right)\left(21-15\right)}$ cm²
                               =  $\sqrt{21\times 8\times 7\times 6}$ cm²
                               = $\sqrt{3\times 7\times 2\times 2\times 2\times 7\times 2\times 3}$ cm²
                               = (2 × 2 × 3 × 7) cm² = 84 cm²

Question 17:

Mark (✓) against the correct answer
Each side of an equilateral triangle is 8 cm. Its area is

(a) $16\sqrt{3 }{\mathrm{cm}}^2$
(b) $32\sqrt{3 }{\mathrm{cm}}^2$
(c) $24\sqrt{3 }{\mathrm{cm}}^2$
(d) $8\sqrt{3 }{\mathrm{cm}}^2$

Answer 17:

(a) $16\sqrt{3 }{\mathrm{cm}}^2$

Given that each side of an equilateral triangle is 8 cm.
∴ Area of the equilateral triangle = $\frac{\sqrt{3}}{4}{\left(\mathrm{side}\right)}^2$ sq. units

                                                      =  $\frac{\sqrt{3}}{4}{\left(8\right)}^2$ cm²

                                                      = $\left(\frac{\sqrt{3}}{4}\times 64\right)$ cm² = $16\sqrt{3}$ cm²

Question 18:

Mark (✓) against the correct answer
One side of a parallelogram is 14 cm and the distance of this side from the opposite side is 6.5 cm. The area of the parallelogram is

(a) 45.5 cm²
(b) 91 cm²
(c) 182 cm²
(d) 190 cm²

Answer 18:

(b) 91 cm²

Base  = 14 cm
Height = 6.5 cm
∴ Area of the parallelogram = Base × Height
                                               = (14 × 6.5) cm²
                                               =  91 cm²

Question 19:

Mark (✓) against the correct answer
The lengths of the diagonals of a rhombus are 18 cm and 15 cm. The area of the rhombus is

(a) 270 cm²
(b) 135 cm²
(c) 90 cm²
(d) 180 cm²

Answer 19:

(b) 135 cm²

Area of the rhombus = $\frac{1}{2}$ ×  (Product of the diagonals)
                                  = $\left(\frac{1}{2}\times 18\times 15\right)$ cm² = 135 cm²
Hence, the area of the rhombus is 135 cm².

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Question 20:

Fill in the blanks.

(i) If d₁ and d₂ be the diagonals of a rhombus, then its are is (......) sq units.
(ii) If l,b and h be the length, breadth and height respectively of a room, then area of its 4 walls = (......) sq units.
(iii) 1 hectare = (......) m².
(iv) 1 are = ......m².
(v) If each side of a triangle is a cm, then its area = ...... cm².

Answer 20:

(i) If d₁ and d₂ be the diagonals of a rhombus, then its area is $\left(\frac{1}{2}{d}_1{d}_2\right)$ sq. units.
      Area of a rhombus = $\frac{1}{2}$× (Product of its diagonals)

(ii) If l, b and h are the length, breadth and height respectively of a room, then area of its 4 walls = 2h(l + b) sq. units.
     
(iii) 1 hectare = (10000) m²
      (since 1 hectometre = 100 m)
        ∴1 hectare = (100 × 100) m²
(iv) 1 acre = 100 m²
(v) If each side of a triangle is a cm, then its area = $\frac{\sqrt{3}}{4}{a}^2$ cm².
    Area of equilateral triangle with side a = $\left(\frac{\sqrt{3}}{4}{a}^2\right)$sq. units.

Question 21:

Write 'T' for true and 'F' for false

(i) Area of a triangle = (base × height)
(ii) Area of a || gm = (base × height)
(iii) Area of a circle = $2{\mathrm{\pi r}}^2$
(iv) Circumference of a circle = $2\mathrm{\pi r}$

Answer 21:

(i) F
   Area of a triangle = $\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}$

(ii) T
      Area of a parallelogram = Base × Height

(iii) F
        Area of a circle = ${\mathrm{\pi r}}^2$

(iv) T
       Circumference of a circle = $2\mathrm{\pi r}$2πr2