Question

Help me with these sums!!!???!??!?
Don't give wrong answers, don't even answer if You don't know...
Help and get points!!​

Answer 1

Step-by-step explanation:

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Answer 2

Required Answers:

$\bigstar \sf \: solution \: to \: question \: 1 : \:$

Given:

• Area of a rhombus is 216cm²
• diagonal 1 is 24

To Find:

• it's other diagonal and the sides of the rhombus

Solution:

we know,

$\leadsto \sf \: area \: of \: a \: rhombus = \dfrac{ d_{1} \times d_{2}}{2}$

let's substitute the values and find the other diagonal

$\longrightarrow \sf \: 216 {cm}^{2} = \frac{24 \times d_{2} }{2} \\ \\ \\ \longrightarrow \sf \: 24 \times d_{2} = 216 \times 2 \\ \\ \\ \longrightarrow \sf \: 24 \times d_{2} = 432 \: \: \: \: \: \: \: \\ \\ \\ \longrightarrow \sf \: d_{2} = \cancel \frac{432}{24} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \longrightarrow \sf \: d_{2} = 18cm \bigstar \: \: \: \: \: \: \: \: \:$

hence,

• the other diagonal is 18cm

Now, let's find the sides

Using pythagoras therom ,

$\longrightarrow \sf \: {24}^{2} + {18}^{2} = \: side^{2} \\ \\ \\ \longrightarrow \sf \: 576 + 324 = side ^{2} \\ \\ \\ \longrightarrow \sf \: 900 = {side}^{2} \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \longrightarrow \sf \: side \: = \sqrt{900} \: \: \: \: \: \: \: \\ \\ \\ \longrightarrow \sf \: side \: = 30 \bigstar \: \: \: \: \: \: \: \:$

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$\bigstar \sf \: solution \: to \: question \: 2 :$

Given:

• pqrs is a parallelogram
• diagonal qn is 9cm
• side rs measures 16cm
• side ps measures 24cm

To find:

• QN

Solution :

we know that,

• area of a parallelogram is base times height

${ : \implies} \sf \: QN \: \times RS = QM \times PS \\ \\ \\ { : \implies} \sf 9 \times 16 = QM \times 24 \: \: \: \: \: \: \\ \\ \\{ : \implies} \sf \: QM = \cancel\frac{9 \times 16}{24} \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ { : \implies} \sf \: QM = 6cm \bigstar \: \: \: \: \: \: \: \: \: \: \: \: \:$

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$\bigstar \sf \: solution \: to \: question \: 3 :$

• Answer given in the attachment

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$\bigstar \sf \: solution \: to \: question \: 4 :$

Given:

• diameter of a circle is 91cm

To Find:

• the distance travelled when made 1000 revelutions

Solution :

we know that,

$\leadsto \sf circumference \: of \: a \: circle \: = 2\pi \: r$

As diameter is 91cm

• radius equals 45.5cm

Distance travelled per one revolution:

$\longrightarrow \sf \: 2\pi \: r \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \longrightarrow \sf2 \times \frac{22} {\cancel{{7} }} \times \cancel{45.5} \\ \\ \\ \longrightarrow \sf \: \: 2 \times 22 \times 6.5 \: \\ \\ \\ \longrightarrow \sf \: 286cm \bigstar \: \: \: \: \: \: \: \: \: \: \:$

so, distance covered per 1000 revolutions;

➺ 286 × 1000

➺ 286000cm★

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$\bigstar \sf \: solution \: to \: question \: 5 :$

• refer to the attachment

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$\bigstar \sf \: solution \: to \: question \: 6 :$

Given :

• diameter is 56m
• a 2m road runs around it

To find :

• the cost of fencing the road

solution:

• Refer to the attachment

Check:

I have given 4 attachments don't missout