Question

II. The base of an isosceles triangular lawn is 16 cm and equal sides are 17 cm each. Find
the cost of paving it at the rate of ₹2.50 per square metre.​

Answer 1

$\sf\large\underline{Given:}$

$\rm{\implies The\:base\:_{(isosceles\triangle)}=16cm}$

$\rm{\implies The\: equal\: sides\:_{(isosceles\triangle)}=17cm}$

$\rm{\implies Cost\:_{(paving\:rate)}=Rs.2.50\:per\:m^2}$

$\sf\large\underline{To\: Find:}$

$\rm{\implies Cost\:_{(paving\:rate)}=?}$

$\sf\large\underline{Solution:}$

• At first calculate semiperimeter of the triangle then calculate it's area by applying heron's formula. At last the area which we get by calculating from that we have to multiply by the given rate]

$\sf\large\underline{Formula\:used:}$

$\tt{\implies Semiperimeter=\dfrac{a+b+c}{2}}$

• As we have given in the question in the base of an isosceles triangular lawn is 16cm and it's equal sides are 17cm. It means here a=16 , b=17 , c=17 here]

$\tt{\implies Semiperimeter=\dfrac{16+17+17}{2}}$

$\tt{\implies Semiperimeter=\dfrac{50}{2}=25cm}$

• Now calculate the area of isosceles triangular lawn by applying heron's formula here:]

$\rm{\implies Area\:_{(isosceles\triangle)}=\sqrt{s(s-a)(s-b)(s-b)}}$

$\rm{\implies Area\:_{(isosceles\triangle)}=\sqrt{25(25-16)(25-17)(25-17)}}$

$\rm{\implies Area\:_{(isosceles\triangle)}=\sqrt{25*9*8*8}}$

$\rm{\implies Area\:_{(isosceles\triangle)}=\sqrt{14400}}$

$\rm{\implies Area\:_{(isosceles\triangle)}=120cm^2}$

• Now calculate the cost of paving rate of isosceles triangular lawn but in question rate is given m^2 so,we have to covert the area into m^2]

we know that,

• I metre=100cm
• 1 metre square=10000cm²
• so, 120 cm²=120/10000=0.0120m²

$\rm{\implies The\: cost\:_{(paving\: rate)}=Rate*Area}$

$\rm{\implies The\: cost\:_{(paving\: rate)}=2.50*0.120}$

$\rm{\implies The\: cost\:_{(paving\: rate)}=Rs.0.3}$

$\sf\large{Hence,}$

$\rm{\implies The\: cost\:_{(paving\: rate)}=Rs.0.3}$

Answer 2

${\huge Given}$:-

If height of the triangle is h, then its area

=$\frac{2} {1} ×40$

=20h cm

Cost of this triangular land = Rs. 125×20h

∴125×20h=50,000

∴h=20 cm

By Pythagoras theorem, we have

AB= AD +2BD = $\frac{4} {h} +20$

=$\frac{20} {2} cm$

=$\color{red} {} coat .of . paving$

=$\color{red} {} 125000rupees$