Question

1. The area of a rhombus is 60 cm². One diagonal is 10 cm. The other diagonal is _____cm *

a. 6
b. 12
c. 3
d. 24

2. The area of a trapezium is 480 cm², the distance between two parallel sides is 15 cm and one of the parallel side is 20 cm. The other parallel side is ____cm *

a. 20
b. 34
c. 44
d. 50

3. If the length of a diagonal of a quadrilateral is 4cm and the perpendiculars to the diagonal are 1cm and 2cm, the area is ____cm² *

a. 6
b. 12
c. 3
d. 8

4. The area of the trapezium with parallel bases 2cm and 4cm and the distance between them is 3cm is ____cm² *

a. 9
b. 6
c. 7
c. 24

5. The area of a parallelogram of base 6cm and height 2cm is ____cm².

a. 8
b. 6
c. 12
d. 16

6. A rectangular wire of length 40 cm and breadth 20 cm is bent in the shape of a square. The side of the square is _____cm *

a. 10
b. 20
c. 30
d. 40

7. The area of a rhombus whose diagonals are of lengths 10 cm and 8.2 cm is ____cm² *

a. 41
b. 82
c. 410
d. 820


8. The area of a trapezium is 40 cm². Its parallel sides are 12 cm and 8 cm. The distance between the parallel sides is _____cm *

a. 1
b. 2
d. 3
c. 4​

Answer 1

Question 1 :

Given :

• Area of rhombus = 60 cm²
• Length of one diagonal = 10 cm

To find :

• Length of the other diagonal

Concept :

→ Formula of area of rhombus :-

$\boxed{ \sf{ \pmb{ \blue{Area \: \: of \: \: rhombus \: \: = \dfrac{1}{2} \times d_1 \times d_2}}}}$

where,

• d₁ = First diagonal
• d₂ = Second diagonal

Solution :

Length of the other diagonal :-

$\\ \dashrightarrow \sf \quad Area \: \: of \: \: rhombus \: \: = \dfrac{1}{2} \times d_1 \times d_2$

$\\ \dashrightarrow \sf \quad 60 = \dfrac{1}{2} \times 10 \times d_2$

$\\ \dashrightarrow \sf \quad 60 = \dfrac{1}{ \not2} \times \not10 \times d_2$

$\\ \dashrightarrow \sf \quad 60 = 5 \times d_2$

Transpoing 5 to the other side.

$\\ \dashrightarrow \sf \quad \dfrac{60}{5} = \times d_2$

$\\ \dashrightarrow \sf \quad \dfrac{ \not60}{ \not5} = \times d_2$

$\\ \dashrightarrow \sf \quad 12 = \times d_2$

$\boxed{ \sf{ \pmb{ \blue{ Length \: \: of \: \: the \: \: other \: \: diagonal = 12 \: cm}}}}$

Answer 1. → Option (b)

• 1. The area of a rhombus is 60 cm². One diagonal is 10 cm. The other diagonal is 12 cm.

Question 2 :

Given :

• Area of trapezium = 480 cm²
• Distance between two parallel sides = 15 cm
• One of the parallel side = 20 cm

To find :

• Other parallel side

Concept :

→ Formula of area of trapezium :-

$\boxed{ \sf{ \pmb{ \blue{Area \: \: of \: \: trapezium \: = \dfrac{1}{2} \times Sum \: \: of \: \: parallel \: \: sides \times Distance \: \: between \: \: them}}}}$

Solution :

Other parallel side :-

Using formula,

$\\ \dashrightarrow\sf \quad Area \: \: of \: \: trapezium \: = \dfrac{1}{2} \times Sum \: \: of \: \: parallel \: \: sides \times Distance \: \: between \: \: them$

Let,

• First parallel side = a
• Second parallel side = b

$\\ \dashrightarrow\sf \quad 480= \dfrac{1}{2} \times (a + b)\: \: \times 15$

$\\ \dashrightarrow\sf \quad 480= \dfrac{1}{2} \times (20 + b) \: \: \times 15$

Transposing 2 and 15 to the other side. [2 is denominator here so, when we will shift it to he other side it will will become neumerator and similarly 15 will become denominator.]

$\\ \dashrightarrow\sf \quad \dfrac{480 \times 2}{15}= (20 + b)$

$\\ \dashrightarrow\sf \quad \dfrac{ \not480 \times 2}{ \not15}= (20 + b)$

$\\ \dashrightarrow\sf \quad 32 \times 2= (20 + b)$

$\\ \dashrightarrow\sf \quad 64= (20 + b)$

$\\ \dashrightarrow\sf \quad 64 - 20= b$

$\\ \dashrightarrow\sf \quad 44= b$

$\boxed{ \sf{ \pmb{ \blue{ Length \: \: of \: \: the \: \: other \: \: parallel \: \: side = 44\: cm}}}}$

Answer → Option (c)

• 2. The area of a trapezium is 480 cm², the distance between two parallel sides is 15 cm and one of the parallel side is 20 cm. The other parallel side is 44 cm.

Question 3 :

Given :

• Length of a diagonal of a quadrilateral = 4 cm
• Perpendiculars to the diagonal = 1 cm and 2 cm

To find :-

• Area of quadrilateral

Concept :-

→ Formula of area of quadrilateral :-

$\boxed{\sf{Area\:of\:quadrilateral=\dfrac{1}{2}\times\:{length}_{(Diagonals)}\times\:{Sum\:of\:the\:length}_{(Perpendiculars\:)}}}$

Solution :

Area of quadrilateral :-

$\\ \dashrightarrow\sf \quad Area \: \: of \: \: quadrilateral \: \: = \dfrac{1}{2} \times 4 \times (1 + 2)$

$\\ \dashrightarrow\sf \quad Area \: \: of \: \: quadrilateral \: \: = \dfrac{1}{2} \times 4 \times 3$

$\\ \dashrightarrow\sf \quad Area \: \: of \: \: quadrilateral \: \: = 2 \times 3$

$\\ \dashrightarrow\sf \quad Area \: \: of \: \: quadrilateral \: \: = 6$

$\boxed{ \sf{ \pmb{ \blue{Area \: \: of \: \: quadrilateral \: \: = 6 \: \: {cm}^{2} }}}}$

Answer → Option (a)

• 3. If the length of a diagonal of a quadrilateral is 4cm and the perpendiculars to the diagonal are 1cm and 2cm, the area is 6 cm².

Question 4 :

Given :

• Length of the parallel bases = 2 cm and 4 cm
• Distance between them = 3 cm

To find :

• Area of trapezium

Concept :-

→ Formula of area of trapezium :-

$\boxed{ \sf{ \pmb{\blue{Area \: \: of \: \: trapezium \: = \dfrac{1}{2} \times Sum \: \: of \: \: parallel \: \: sides \times Distance \: \: between \: \: them}}}}$

Solution :

Area of trapezium :-

$\\ \dashrightarrow \sf \quad Area \: \: of \: \: trapezium \: = \dfrac{1}{2} \times Sum \: \: of \: \: parallel \: \: sides \times Distance \: \: between \: \: them$

$\\ \dashrightarrow\sf \quad Area \: \: of \: \: trapezium \: = \dfrac{1}{2} \times (2 + 4) \times 3$

Answer 2

Answer- Continuity of @AestheticSoul's answer...

$\\ \dashrightarrow\sf  \quad Area  \:  \: of \:  \: trapezium \: =  \dfrac{1}{2} \times 6 \times 3$

$\\ \dashrightarrow\sf  \quad Area  \:  \: of \:  \: trapezium \: =  \dfrac{1}{ \not2} \times  \not6 \times 3$

$\\ \dashrightarrow\sf  \quad Area  \:  \: of \:  \: trapezium \: =  3 \times 3$

$\\ \dashrightarrow\sf  \quad Area  \:  \: of \:  \: trapezium \: =  9$

$\boxed{ \sf{ \pmb{  \blue{Area  \:  \: of \:  \: trapezium \: =  9 \:  \:  {cm}^{2} }}}}$

Answer → Option (a)

4. The area of the trapezium with parallel bases 2cm and 4cm and the distance between them is 3 cm is 9 cm² .

Question 5 :

Given :

• Base of parallelogram = 6 cm
• Height of parallelogram = 2 cm

To find :

• Area of parallelogram

Concept :-

→ Formula of area of parallelogram :-

$\boxed{ \sf{ \pmb{  \blue{Area  \:  \: of \:  \: parallelogram \: = \:  \: b \times h }}}}$

where b, h are the base, height of the parallelogram.

Solution :

Area of parallelogram :-

$\\  \dashrightarrow\sf \quad Area  \:  \: of \:  \: parallelogram \: = \:  \: b \times h$

$\\  \dashrightarrow\sf \quad Area  \:  \: of \:  \: parallelogram \: = \:  \: 6 \times 2$

$\\  \dashrightarrow\sf \quad Area  \:  \: of \:  \: parallelogram \: = 12$

$\boxed{ \sf{ \pmb{  \blue{Area  \:  \: of \:  \: parallelogram \: = \:  \:12 \:  \:  {cm}^{2}  }}}}$

Answer → Option (c)

5. The area of a parallelogram of base 6 cm and height 2 cm is 12 cm².

Question 6 :

Given :

• Length of the rectangular wire = 40 cm
• Breadth of the rectangular wire = 20 cm
• The same rectangular wire is bent in the shape of a square

To find:

• Side of the square

Concept :

When anything is intercasted from one shape to the other the perimeter remains the same. So, firstly we will find the perimeter of the rectangular wire and that will be equal to the perimeter of the square. So, by substituting the values in the formula of perimeter of square we will get our required answer.

→ Formula of perimeter of rectangle :-

$\boxed{ \sf{ \pmb{  \blue{ Perimeter \:  \:  of  \:  \: rectangle = 2(l + b)  }}}}$

where l = length, b = breadth

→ Formula of perimeter of square :-

$\boxed{ \sf{ \pmb{  \blue{ Perimeter \:  \:  of  \:  \: square = 4a }}}}$

where a = side of the square

Solution :

Perimeter of rectangular wire :-

$\\ \dashrightarrow\sf \quad Perimeter \:  \:  of  \:  \: rectangle = 2(l + b)$

$\\ \dashrightarrow\sf \quad Perimeter \:  \:  of  \:  \: rectangle = 2(40 + 20)$

$\\ \dashrightarrow\sf \quad Perimeter \:  \:  of  \:  \: rectangle = 2(60)$

$\\ \dashrightarrow\sf \quad Perimeter \:  \:  of  \:  \: rectangle = 120 \: cm$

Perimeter of rectangle = Perimeter of square

$\\ \dashrightarrow\sf \quad Perimeter \:  \:  of  \:  \: square =4a$

$\\ \dashrightarrow\sf \quad 120 =4a$

$\\ \dashrightarrow\sf \quad  \dfrac{120}{4}  =a$

$\\ \dashrightarrow\sf \quad  \dfrac{ \not120}{ \not4}  =a$

$\\ \dashrightarrow\sf \quad  30  =a$

$\boxed{ \sf{ \pmb{  \blue{Side \:  \:  of  \:  \: square = 30~~cm }}}}$

Answer → Option (c)

6. A rectangular wire of length 40 cm and breadth 20 cm is bent in the shape of a square. The side of the square is 30 cm.

Question 7 :

Given :

• First diagonal of rhombus = 10 cm
• Second diagonal of rhombus = 8.2 cm

To find :

• Area of rhombus

Concept :

→ Formula of area of rhombus :-

$\boxed{ \sf{ \pmb{  \blue{Area  \:  \: of  \:  \: rhombus \:  \:  =  \dfrac{1}{2} \times  d_1 \times d_2}}}}$

Solution :

$\\  \dashrightarrow \sf \quad Area  \:  \: of  \:  \: rhombus \:  \:  =  \dfrac{1}{2} \times  d_1 \times d_2$

$\\  \dashrightarrow \sf \quad Area  \:  \: of  \:  \: rhombus \:  \:  =  \dfrac{1}{2} \times  10 \times 8.2$

$\\  \dashrightarrow \sf \quad Area  \:  \: of  \:  \: rhombus \:  \:  =  5 \times 8.2$

$\\  \dashrightarrow \sf \quad Area  \:  \: of  \:  \: rhombus \:  \:  =  41$

$\boxed{ \sf{ \pmb{  \blue{Area  \:  \: of  \:  \: rhombus \:  \:  =  41  \:  \: {cm}^{2} }}}}$

Answer → Option (a)

7. The area of a rhombus whose diagonals are of lengths 10 cm and 8.2 cm is 41 cm².