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

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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 = $\sqrt{150(150-60)(150-100)(150-140) }$
= $\sqrt{150 \times 90 \times 50 \times 10}$
= $1500 \sqrt{3} \mathrm{~m}^2$

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Question 33 Marks

A triangular park $ABC$ 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 $3$ m wide for a gate on one side.

Answer

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

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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 $\mathrm{a}=8 \mathrm{~cm}, \mathrm{~b}=11 \mathrm{~cm}$ and $2 \mathrm{~s}=32 \mathrm{~cm}$ i.e. $\mathrm{s}=16 \mathrm{~cm}$
Now,
$a+b+c=2 s$
$\Rightarrow 8+11+c=32$
$\Rightarrow c=13$
$\therefore s-a=16-8=8, s-b=16-11=5 \text { and } s-c=16-13=3$
$\text { Hence, Area of given triangle }=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{16 \times 8 \times 5 \times 3}=8 \sqrt{30} \mathrm{~cm}^2$

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Maths 3 Marks Question

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