5 Marks Questions

Question 15 Marks

The length and the breadth of a rectangular field are in the ratio $7 : 4$. The cost of fencing the field at $Rs. 25$ per metre is $Rs. 3300.$ Find the dimensions of the field.

Answer

Total cost of fencing a rectangular field $= Rs. 3300$
Rate of fencing $= Rs 25$ per m
$\therefore$ Perimeter $=\frac{\text{Total cost}}{\text{Rate}}$
$=\frac{3300}{25}$
$=132\text{m}$ And length + breadth $=\frac{\text{Perimeter}}{2}$
$=\frac{132}{2}$
$=66\text{m}$ Now ratio in length and breadth $= 7 : 4$
Let length $= 7x$
Then breadth $= 4x$
$\therefore 7x + 4x = 66$
$\Rightarrow 11x = 66$
$\therefore \text{x}=\frac{66}{11} = 6$
$\therefore$ Length of the field $= 7x = 7 \times 6 = 42m$ And
breadth $= 4x = 4 \times 6 = 24m$

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Question 25 Marks

Calculate the area one of the shaded region given below (all measures are in cm):

Answer

Area of the square BCDE $= (side)^2$
$= (CD)^2$
$= (3)^2\ cm^2$
$= 9\ cm^2$
Area of the rectangle ABFK = Length $\times$ Breadth
$= AB \times AK {(AB = AC – BC) and (AK = AL + LK)}$
$= (5 \times 1)cm^2$
$= 5\ cm^2$
Area of the rectangle $MLKG =$ Length $\times $ Breadth
$= ML \times MG$
$= (2 \times 3)cm^2$
$= 6\ cm^2$
Area of the rectangle $JHGF =$ Length $\times $ Breadth
$= JH \times HG$
$= (2 \times 4)cm^2$
$= 8\ cm^2$
Area of the complete figure $=$ Area of the rectangle $ABFK +$ Area of the rectangle $MLKG +$ Area of the rectangle $JHGF +$ Area of the square $BCDE$
$= (9 + 5 + 6 + 8)cm^2$
$= 28cm^2$

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Question 35 Marks

Calculate the area one of the shaded region given below (all measures are in cm):

Answer

Area of the rectangle $CEFG =$ Length $\times $ Breadth
$= EF \times CE$
$=(1 \times 5) \mathrm{cm}^2(\mathrm{CE}=\mathrm{EA}-\mathrm{AC})$
$ =5 \mathrm{~cm}^2$
Area of the rectangle $ABDC =$ Length $\times $ Breadth
$= HI \times IJ$
$=(1 \times 2) \mathrm{cm}^2$
$ =2 \mathrm{~cm}^2$
Area of the rectangle $HIJG =$ Length $\times $ Breadth
$= HI \times IJ$
$=(1 \times 2) \mathrm{cm}^2 $
$ =2 \mathrm{~cm}^2 $
$ =8 \mathrm{~cm}^2$
Area of the complete figure = Area of the rectangle $CEFG +$ Area of the rectangle $HIJG +$ Area of the rectangle $ABDC$
$=(5+2+2) \mathrm{cm}^2$
$ =9 \mathrm{~cm}^2$

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Question 45 Marks

Match the following:

$(a)$	Area of a rectangle	$(i)$	$\pi\text{r}^2$
$(b)$	Area of a square	$(ii)$	$4 \times $ side
$(c)$	Perimeter of a rectangle	$(iii)$	$l \times b$
$(d)$	Perimeter of a square	$(iv)$	$(side)^2$
$(e)$	Area of a circle	$(v)$	$2(l + b)$

Answer

$(a)$	Area of a rectangle	$(iii)$	$l \times b$
$(b)$	Area of a square	$(iv)$	$(side)^2$
$(c)$	Perimeter of a rectangle	$(v)$	$2(l + b)$
$(d)$	Perimeter of a square	$(ii)$	$4 \times $ side
$(e)$	Area of a circle	$(i)$	$\pi\text{r}^2$

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MATHS 5 Marks Questions

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