RBSE Solutions for Class 8 Maths Chapter 14 Area Additional Questions is part of RBSE Solutions for Class 8 Maths. Here we have given Rajasthan Board RBSE Class 8 Maths Chapter 14 Area Additional Questions.
| Board | RBSE |
| Textbook | SIERT, Rajasthan |
| Class | Class 8 |
| Subject | Maths |
| Chapter | Chapter 14 |
| Chapter Name | Area |
| Exercise | Additional Questions |
| Number of Questions | 34 |
| Category | RBSE Solutions |
Rajasthan Board RBSE Class 8 Maths Chapter 14 Area Additional Questions
I. Objective Type Questions
Question 1.
Area of a right angle triangle is 36 cm² and the base is 9 cm. Find the length of perpendicular is
(a) 8 cm.
(b) 4 cm.
(c) 16 cm.
(d) 32 cm.
Question 2.
The parallel sides of a trapezium are 32 m. and 20 m. respectively and distance between them is 15 m., then area of trapezium is
(a) 290 m²
(b) 390 m²
(c) 190 m²
(d) 400 m²
Question 3.
The diagonals of a rhombus(RBSESolutions.com)are 10 cm. and 8 cm. respectively then its area is
(a) 80 cm²
(b) 40 cm²
(c) 9 cm²
(d) 36 cm²
Question 4.
The area of a trapezium is 143 cm². Parallel sides are 15 cm. and 11 cm. respectively. The perpendicular distance between them is
(a) 11 cm.
(b) 15 cm.
(c) 13 cm.
(d) 14 cm.
Question 5.
The area of a triangular plot is 2.5 m². If the base is 100 meter then find the height is
(a) 10 cm.
(b) 5 cm.
(c) 10 meter
(d) 5 meter
Question 6.
A triangle and a rectangle are situated(RBSESolutions.com)between same base and parallel lines. The base and height of triangle are 9 cm. and 6 cm. respectively. Area of rectangle is
Question 7.
Formula of finding area of parallelogram is
(a) \(\frac { 1 }{ 2 }\) x product of both adjacent sides of parallelogram
(b) \(\frac { 1 }{ 2 }\) x Base x height
(c) Product of both adjacent sides
(d) Base x height
Question 8.
If area of a rhombus is 506 cm² and length of diagonal is 23 cm. then find the length of other diagonal
(a) 11 cm.
(b) 22 cm.
(c) 44 cm.
(d) 33 cm.
Answers
1. (a)
2. (b)
3. (b)
4. (a)
5. (d)
6. (c)
7. (d)
8. (c)
II. Fill in the blanks
Question 1.
Area of triangle =___x base x height.
Question 2.
Area of trapezium = \(\frac { 1 }{ 2 }\) x sum of parallel sides x ___
Question 3.
Diagonals of rhombus are___to each other.
Question 4.
Area of(RBSESolutions.com)rhombus = \(\frac { 1 }{ 2 }\) x ____
Question 5.
Area of rectangle = Length x ___
Answers
1. \(\frac { 1 }{ 2 }\),
2. height,
3. perpendicular,
4. product of diagonals,
5. breadth
III. True/False Type Questions
Question 1.
Area of a parallelogram, whose base is 6.5 cm. and height is 4 cm. is 26 cm².
Question 2.
1 meter = 100 square cm.
Question 3.
1 hectare = 10000 square meter.
Question 4.
If the diagonals of a(RBSESolutions.com)rhombus are 24 cm. and 7 cm. then area of rhombus is 168 cm².
Answers
1. True
2. True
3. True
4. False.
IV. Matching Type Questions
| Part 1 | Part 2 |
| 1. Area of trapezium | (a) \(\frac { 1 }{ 2 }\) x product of diagonals |
| 2. Area of rhombus | (b) \(\frac { 1 }{ 2 }\) x (sum of parallel sides) x height |
| 3. Side of a square whose area is 1 hectare | (c) 20 meter |
| 4. Length of rectangle whose breadth is 5 meter and area is 1 m² | (d) 100 meter |
Answers
1. ↔ (b)
2. ↔ (a)
3. ↔ (d)
4. ↔ (c)
V. Very Short Answer Type Questions
Question 1.
In a parallelogram ABCD, AB = 35 cm. and height is 12 cm. Corresponding to this side. BC = 25 cm. then find the corresponding height.
Solution:
Area of(RBSESolutions.com)parallelogram ABCD 35 x 12 = 25 x Height corresponding to BC
⇒ Height corresponding to side BC
= \(\frac { 35X12 }{ 25 }\)
= 16.8 cm
Question 2.
BC = 30 cm. in given figure. Calculate the area of ∆ABC.
Solution:
In ∆ ABD,
∵ ∠BAD = ∠ABD
∴BD = AD ….(1)
In ∆ ADC,
∵ ∠DAC = ∠DCA
∴ DC = AD
From (1) and (2)
BD = DC = AD
⇒ AD = \(\frac { 1 }{ 2 }\) BC = \(\frac { 1 }{ 2 }\) x 30 = 15 cm.
∴ Area of ∆ABC = \(\frac { 1 }{ 2 }\) x Base x Height
= \(\frac { 1 }{ 2 }\) x BC x AD
= \(\frac { 1 }{ 2 }\) x 30 x 15
= 225 cm²
Question 3.
In the given figure, ABCD is a trapezium shaped field. AB || DC, DC = 25 cm., CE = 12 cm. and EB = 10 cm. Find the area of quadrilateral.
Solution:
Area of trapezium ABCD = Area of rectangle AECD + Area of right angle triangle CEB
= 25 x 12 + \(\frac { 10X12 }{ 2 }\)
= 300 + 60
= 360 cm²
Question 4.
The perimeter of a rhombus is 52 cm. Distance between two parallel sides is 12 cm. Find the area of rhombus.
Solution:
Perimeter of rhombus = 52 cm.
∴ One side of rhombus = \(\frac { 52 }{ 4 }\) cm. = 13 cm.
Height of(RBSESolutions.com)rhombus = 12 cm.
∴ Area of rhombus = base x height
= 13 x 12
= 156 cm²
VI. Short Answer Type Questions
Question 1.
Area of trapezium is 140 cm.² Its height is 10 cm. Difference between parallel sides is 4 cm. Find the lengths of both parallel sides.
Solution:
Let a cm. and b cm. be the lengths of parallel sides where a > b.
Area of trapezium = \(\frac { 1 }{ 2 }\) x (Sum of parallel sides) x height
⇒ 140 = \(\frac { 1 }{ 2 }\)(a + b) 10
⇒ 140 x 2 = (a + b) 10
⇒ 280 = (a + b) 10
⇒ a + b = \(\frac { 280 }{ 10 }\)
⇒ a + b = 28 …(1)
Again, a – b = 4 …(2)
Adding (1) and (2)
2 a = 32
⇒ a = \(\frac { 32 }{ 2 }\) = 16 cm.
Putting a = 16 in (1)
16 + b = 28
⇒ b = 28 – 16
⇒ b = 12 cm.
Hence, the required lengths of parallel sides are 16 cm. and 13 cm. respectively.
Question 2.
A trapezium has area 60 m². Its parallel sides being 12 cm and 8 cm. Find its height
Solution:
Let the height of trapezium = h cm.
Then, Area of trapezium = \(\frac { 1 }{ 2 }\) x (sum of parallel sides) x height
or 60 = \(\frac { 1 }{ 2 }\) x (12 + 8) x h
or h = \(\frac { 60X2 }{ 20 }\) = 6 cm.
Hence height = 6 cm.
Question 3.
In a given figure ABCD is a trapezium in which AB || DC and DA ⊥ AB. If AB = 13 cm., AD = 8 cm. and CD = 7 cm., then find area of trapezium.
Solution:
Area of trapezium = \(\frac { 1 }{ 2 }\) x (sum of parallel sides) x height
= \(\frac { 1 }{ 2 }\) x (7 + 13) x 8
= \(\frac { 1 }{ 2 }\) x (7 + 13) x 8
= \(\frac { 1 }{ 2 }\) x 20 x 8 = 80 square cm.
Question 4.
Polygon ABCDE is divided in different parts as shown in figure. If AD = 8 cm., AH = 6 cm., AG = 4 cm., AF = 3 cm. and BF = 2 cm., CH = 3 cm., EG = 2.5 cm. Then And the area of polygon.
Solution:
Area of polygon ABCDE = Area of right ∆AFB + Area of trapezium BFHC + Area of right ∆CHD + Area of right ∆EGD + Area of right ∆AGE
FH = AH – AF = 6 – 3 = 3 cm.
HD = AD – AH = 8 – 6 = 2 cm.
GD = AD – AG = 8 – 4 = 4 cm.
= 3 + 7.5+ 3 + 5 + 5
= 23.5 cm²
Question 5.
Find the area of polygon MNOPQR if MP = 9 cm., MD = 7 cm., MC = 6 cm., MB = 4 cm., MA = 2 cm., NA, OC, QD and RB are perpendiculars on diagonal MP.
Solution:
Area of polygon MNOPQR
= Area of ∆MAN + Area of trapezium ACON + Area of ∆OCP + Area of ∆QDP + Area of trapezium DBRQ + Area of ∆RBM
AC = MC – MA = 6 – 2 = 4 cm.
CP = MP – MC = 9 – 6 = 3 cm.
PD = MP – MD = 9 – 7 = 2 cm.
BD = MD – MB = 7 – 4 = 3 cm.
= 2.5 + 11 + 4.5 + 2 + 6.75 + 5
= 31.75 cm²
Question 6.
The transverse cut of a canal is trapezium shaped. Canal is 10 meter wide at top and 6 meter wide at bottom. The area of transverse cut is 72 meter². Then find its depth.
Solution
Area of trapezium = \(\frac { 1 }{ 2 }\) x (Sum of parallel sides) x height
⇒ 72 = \(\frac { 1 }{ 2 }\) x (10 + 6) x height
⇒ 72 = 8 x height
⇒ height = \(\frac { 72 }{ 8 }\) = 9 meter
Hence, depth is 9 meter.
Question 7.
The diagonals of a rhombus are 15 cm. and 36 cm. Find its perimeter and area.
Solution
Area of rhombus
= \(\frac { 1 }{ 2 }\) x Product of diagonals
= \(\frac { 1 }{ 2 }\) x 15 x 36
= 270 cm²
We know that(RBSESolutions.com)the diagonal of rhombus bisect each other at right angle.
∴ Perimeter of rhombus = 4 x AB
= 4 x 19.5
= 78 cm.
Question 8.
As shown in the given picture a farm is similar in the shape of polygon ABCDEF. All measurements are given in meters. Find the area of polygon ABCDEF.
Solution:
Area of polygon ABCDEF = Area of fig. 1 + Area of fig. 2 + Area of fig. 3 + Area of fig. 4 + Area of fig. 5
= 300 + 4400 + 1750 + 2600 + 2800
= 11850 square meter
Question 9.
Find the area of the field shown below. AH measurements are in meters.
Solution
Area of polygon ABCDEA = Area of ∆ADE + Area of ∆DFC + Area of trapezium ABCF
= 45 x 40 + 30 x 30 + 45 x 15
= 1800 + 900 + 675
= 3375 meters.
Question 10.
Area of a rhombus ABCD is 264 cm². If length of its one diagonal AC = 24 cm then find length of diagonal BD.
Solution
We know that,
Area of rhombus = \(\frac { 1 }{ 2 }\) x product of diagonals
Given: Area of rhombus = 264 cm²,
diagonal AC = 24 cm
To find = diagonal BD ,
∴ 264 = \(\frac { 1 }{ 2 }\) x 24 x BD
BD = \(\frac { 264X2 }{ 24 }\) = 22 cm.
Question 11.
The area of rhombus is 360 cm². One of its diagonal is 20 cm, then find the length of other diagonal.
Solution
Area of the rhombus = \(\frac { 1 }{ 2 }\) x Product of the diagonals
⇒ 360 sq. cm. = \(\frac { 1 }{ 2 }\) x 20 cm, x other diagonal
⇒ other diagonal = \(\frac { 360X2 }{ 20 }\) cm
⇒ 18 x 2 cm.
⇒ other diagonal = 36 cm.
Question 12.
Find the area of the given figure (All measurements are in meters).
Solution:
Total area of figure = Area of ∆PEJ + area of trapezium PRIJ + area of ∆IRH + area of trapezium HGFQ + area of ∆QFE
Area of ∆PEJ = \(\frac { 1 }{ 2 }\) x Base x Height
= \(\frac { 1 }{ 2 }\) x 10 x 20
= 100 m²
Area of trapezium JPRI,
= \(\frac { 1 }{ 2 }\) (Sum of parallel sides) x (distance between parallel sides)
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