Question

solve 9th and 10th questions ​

Answer 1

Answer:

9)32m

Step-by-step explanation:

area of trapezium =1440

area of trapezium=1/2×distance (sum of parallel sides)

1440=1/2×distance(54.6+35.4)

1440=1/2×distance×90

1440×2=distance×90

2880=distance×90

distance =2880/90=32

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Answer 2

$\large\underline{\sf{Solution-9}}$

Given that,

Area of trapezium = 1440 sq. cm.

Length of parallel sides are 54.6 cm and 35.4 cm

Let assume that length of parallel sides be denoted by a and b

So, a = 54.6 cm and b = 35.4 cm

Let assume that the distance between two parallel lines be h cm

We know,

Area of trapezium whose length of parallel sides are a units and b units respectively and distance between parallel lines is h units is given by

$\boxed{\sf{  \: \: \rm \: \: Area_{(trapezium)} \: = \: \frac{1}{2}(a + b) \times h \: \: }}$

So, on substituting the values, we get

$\rm \: 1440 = \dfrac{1}{2}(54.6 + 35.4) \times h$

$\rm \: 1440 = \dfrac{1}{2} \times 90 \times h$

$\rm \: 1440 = 45 \times h$

$\rm\implies \:h \: = \: 32 \: cm$

Hence,

Distance between the parallel lines is 32 cm

$\large\underline{\sf{Solution-10}}$

Given that,

The diagonal of the quadrilateral is 24 m and length of perpendiculars drawn from the opposite vertices on this diagonal is 8 m and 13 m.

Let assume that the perpendiculars be represented as

$\rm \: h_1 = 8 \: m$

$\rm \: h_2 = 13 \: m$

Now, we know that

$\boxed{\sf{ \rm \: \: Area_{(quadrilateral)} \: = \: \frac{1}{2}(h_1 + h_2) \times diagonal \: \: }}$

So, on substituting the values, we get

$\rm \: \: Area_{(quadrilateral)} \: = \: \frac{1}{2} \times (8 + 13) \times 24 \: \:$

$\rm \: \: Area_{(quadrilateral)} \: = \: 21 \times 12 \: \:$

$\rm \: \: Area_{(quadrilateral)} \: = \: 252 \: {m}^{2} \: \:$

Hence,

$\rm\implies \: \boxed{\sf{  \:\: \: \rm \: \: Area_{(quadrilateral)} \: = \: 252 \: {m}^{2} \: \: }}$

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$\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$