Question

8. A ground is in the shape of parallelogram. The height of the
parallelogram is 14 metres and the corresponding base is 8 metres longe a
than its height. Find the cost of levelling the ground at the rate of
rupees 20 per sq. m.
la
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Answer 1

Given:

• Height of the parallelogram (ground) = ${\bold{14\:m}}$
• Base of the parallelogram (ground) = $\bold{(8+Height)\:m}$
• Cost of levelling per m² = ${\bold {20/-}}$

To find:

• The cost of levelling the ground.

Solution:

To find the cost of levelling, first we've to find the area of the ground. And it is given that the ground is in the shape of a parallelogram, it's area will be the product of its base and its height. Once we find the area, we can find the cost of levelling by multiplying the area with cost per $\sf{{m}^{2}}$. So, now let's do it !!

$\normalsize{\boxed{\sf{Area\:of\: parallelogram=Base \times Height\:sq.units}}}$

Now substitute the measures given in the formula.

$\implies{\sf{(8+Height)\times 14}}$

$\implies{\sf{(8 + 14)\times 14}}$

$\implies{\sf{22 \times 14}}$

$\implies{\sf{308}}$

$\large\underline{\boxed{\rightarrow{\tt{Area\:of\:the\:ground=308\:{m}^{2}}}}}$

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Now, let's find the cost of levelling the ground.

$\normalsize{\boxed{\sf{Cost\:of\: levelling=Cost\:per\:{m}^{2}\times Area\:of\:the\:ground}}}$

$\implies{\sf{20\times308}}$

$\implies\underline{\sf{6160\:\:rupees}}$

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Answer:

• The cost of levelling the ground will be $\large\underline{\boxed{\tt{\red{6160/-}}}}$.

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Some important formulae for Area:

$\normalsize{\sf{1)\:Triangle=\frac{1}{2}\times Base\times Height\:sq.units}}$

$\normalsize{\sf{2)\:Right-angled\: triangle=\frac{1}{2}ab\:sq.units}}$

$\normalsize{\sf{3)\:Quadrilateral=\frac{1}{2}\times d\times(h1+h2)\:sq.units}}$

$\normalsize{\sf{4)\:Trapezium=\frac{1}{2}h\times(a+b)\:sq.units}}$

$\normalsize{\sf{5)\:Rhombus=\frac{1}{2}\times d1\times d2\:sq.units}}$

$\normalsize{\sf{6)\:Rectangle=Length\times Breadth\:sq.units}}$

$\normalsize{\sf{7)\:Square={Side}^{2}\:sq.units}}$

$\normalsize{\sf{8)\:Circle=\pi{r}^{2}\:sq.units\:or\:\frac{1}{4}\pi{d}^{2}\:sq.units}}$

$\normalsize{\sf{9)\:Semi-circle=\frac{1}{2}\pi{r}^{2}\:sq.units}}$

$\normalsize{\sf{10)\:Quadrant\:of\:circle=\frac{1}{4}\pi{r}^{2}\:sq.units\:or\:\frac{\theta}{360°}\pi{r}^{2}\:sq.units}}$

$\normalsize{\sf{11)\:Scalene\: triangle=\sqrt{s(s - a)(s - b)(s - c)}\:sq.units}}$

$\normalsize{\sf{12)\:Equilateral\: triangle=\frac{\sqrt{3}}{4}{a}^{2}=\frac{1}{2}ah\:sq.units}}$

$\normalsize{\sf{13)\:Isosceles\: triangle=\frac{1}{2}\times b \times AD\:sq.units}}$

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