Question

please solve 4th and 9th questions

Answer 1

4) Area of rectangle= length× breadth

= 22 × 12

= 264 m^2

Area of 4 quadrants
$= 4 \times \frac{1}{4} \times \pi {r}^{2}$


Now, r(radius) = 2.5 m

area of 4 quadrants =
$\cancel4 \times \frac{1}{ \cancel4} \times \pi {r}^{2}$


$= \frac{22}{7} \times 2.5 \times 2.5$



$= \frac{137.5}{7} {m}^{2}$


= 19.64 m^2 (approx)


Area remaining= Area of rectangle- Area of 4 quadrants

= 264 - 19.64

$\boxed{ = \bf244.36 \: {m}^{2} }$


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9) Refer to the attachment for figure.


$\bf{Given :-}$ The adjacent sides of parallelogram is 10 m and 8 m.

So, AB = CD = 10 m and

AD= BC= 8 m

Given, distance between AB and CD (height AE) = 4 m.


$\bf{To\: find :}$ Distance between AD and BC (height BF) = ?


$\bf{Solution:- }$

Area of parallelogram= Base × height

$\bf{Case\:1}$

Base = CD = 10 m

Height = AE = 4 m

Area = 10× 4

=40 m^2 ------(1)

$\bf{Case\:2}$

Base = AD= 8 m

Height= BF=?

$\textbf{But area must be equal as the figure is same.}$

So,

40= AD×BF

40= 8 ×BF

40/8=BF

5m = BF

$\textbf{So,distance between smaller sides is 5m}$


$\huge{ \bf{hope \: it \: helps}}$