3 Marks Question

Question 13 Marks

Sides of a triangle are in the ratio of $12: 17: 25$ and its perimeter is $540 \ cm.$ Find its area.

Answer

Let the sides of the triangle be $12x,17x$ and $25x$
Therefore, $12x+17x+25x = 540$
$\Rightarrow 54x=540$
$\Rightarrow x = 10$
$\therefore$ The sides are $120 \ cm, 170 \ cm$ and $250 \ cm.$
Semi$-$perimeter of triangle $s =\frac{120+170+250}{2} $
$= 270 \ cm$
Now, Area of triangle $= \sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$
$= \sqrt{270\left( 270-120 \right)\left( 270-170 \right)\left( 270-250 \right)}$
$= \sqrt{270\times 150\times 100\times 20}$
$= 9000\ cm^2$

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Question 23 Marks

The sides of a triangular plot are in the ratio of $3 : 5 : 7$ and its perimeter is $300\ m.$ Find its area.

Answer

Suppose that the sides in metres are $3x, 5x$ and $7x.$
Then, we know that $3x + 5x + 7x = 300 ($Perimeter of the triangle$)$
Therefore$, 15x = 300,$ which gives $x = 20.$
So the sides of the triangles are $3 \times 20\ m, 5 \times 20\ m$ and $7\times 20\ m$
$i.e., 60\ m, 100\ m$ and $140\ m.$
We have $s = \frac{60+100+140}{2} $
$= 150\ m$
and area will be $= \sqrt{150(150-60)(150-100)(150-140) }$
$= \sqrt{150 \times 90 \times 50 \times 10}$
$= 1500\sqrt{3} m^2$

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Question 33 Marks

A triangular park $\text{ABC}$ has sides $120\ m, 80\ m$ and $50\ m. ($in a given figure$)$. A gardener Dhania has to put a fence all around it and also plant grass inside. How much area does she need to plant? Find the cost of fencing it with barbed wire at the rate of $₹ 20$ per metre leaving a space 3m wide for a gate on one side.

Answer

Computation of area: Clearly, the park is trianglar with sides
$a = BC = 120\ m, b = CA = 80 \ m$ and $c = AB = 50 \ m$
Ifs denotes the semi$-$perimeter of the park, then
$2s = a + b + c$
$ \Rightarrow 2s = 120 + 80 + 50$
$\Rightarrow s = 125$
$\therefore s - a = 125 - 120 = 5, s - b = 125 - 80 = 45$ and $s - c = 125 - 50 = 75$
Hence, Area of the park $= \sqrt{s(s-a)(s-b)(s-c)}$
$= \sqrt{125 \times 5 \times 45 \times 75}m^2$
$= 375\sqrt{15} m^2$
Length of the wire needed for fencing $=$ perimeter of the park $-$ width of the gate
$= 250\ m - 3\ m$
$ = 247 \ m$
Cost of fencing $= Rs.(20 \times 247)$
$= Rs.4940$

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Question 43 Marks

Find the area of a triangle, two sides of which are $8 \ cm$ and $11 \ cm$ and the perimeter is $32 \ cm.$

Answer

Let $a, b, c$ be the sides of the given triangle and $2\ s$ be its perimeter such that
$a = 8 \ cm, b = 11 \ cm$ and $2s = 32 \ cm i.e. s = 16 \ cm$
Now,
$a + b + c = 2s$
$\Rightarrow 8 + 11 + c = 32$
$\Rightarrow c = 13$
$\therefore s - a = 16 - 8 = 8, s - b = 16 - 11 = 5$ and $s - c = 16 - 13 = 3$
Hence, Area of given triangle = $\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{16 \times 8 \times 5 \times 3} $
$= 8\sqrt{30} cm^2$

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