Solution

Given: An isosceles right triangle has area $8 ~cm^{2}$.

Area of an isosceles right triangle $=\dfrac{1}{2} \times$ Base $\times$ Height

So, $8=\dfrac{1}{2} \times$ Base $\times$ Height

$(\text{ Base })^{2}=16 \quad$ [Base $=$ height, as triangle is an isosceles]

Base $=\sqrt{16}$

Base $=4 ~cm$

See the triangle $ABC$, using Pythagoras theorem:

$ \begin{aligned} A C^{2}+A B^{2}+B C^{2} & =4^{2}+4^{2} \\ & =16+16 \\ A C^{2} & =32 \\ A C & =\sqrt{32} \end{aligned} $

Therefore, the length of its hypotenuse is $\sqrt{32}$.

Hence, the correct option is (A).

Solution

Given: The perimeter of an equilateral triangle is $60 m$.

Suppose that each side of an equilateral be a.

$a+a+a=60 m$

$3 a=60 m$

$a=20 m$

Area of an equilateral triangle $=\dfrac{\sqrt{3}}{4} \times(\text{ Side })^{2}$

$ \begin{aligned} & =\dfrac{\sqrt{3}}{4} \times(20)^{2} \\ & =100 \sqrt{3} m^{2} \end{aligned} $

Therefore, the area of the triangle is $100 \sqrt{3} m^{2}$.

Solution

The sides of a triangle are $a=56 ~cm, b=60 ~cm$ and $c=52 ~cm$.

So, semi-perimeter of a triangle will be:

$ \begin{aligned} s & =\dfrac{a+b+c}{2} \\ & =\dfrac{56+60+52}{2} \\ & =\dfrac{168}{2} \\ & =84 ~cm \end{aligned} $

Area of the triangle $=\sqrt{s(s-a)(s-b)(s-c)} \quad$ [By heron’s formula]

$ \begin{aligned} & =\sqrt{84(84-56)(84-60)(84-52)} \\ & =\sqrt{84 \times 28 \times 24 \times 32} \\ & =\sqrt{4 \times 7 \times 3 \times 4 \times 7 \times 4 \times 2 \times 3 \times 4 \times 4 \times 2} \\ & =\sqrt{4^{6} \times 7^{2} \times 3^{2}} \\ & =4^{3} \times 7 \times 3 \\ & =1344 ~cm^{2} \end{aligned} $

Hence, the correct option is (C).

Solution

Given: The side of an equilateral triangle is $2 \sqrt{3} ~cm$.

Now, area of an equilateral triangle $=\dfrac{\sqrt{3}}{4}(\text{ Side })^{2}$

$ \begin{aligned} & =\dfrac{\sqrt{3}}{4} \times(2 \sqrt{3})^{2} \\ & =\dfrac{\sqrt{3}}{4} \times 4 \times 3 \\ & =3 \sqrt{3} \\ & =3 \times 1.732 \\ & =5.196 ~cm^{2} \end{aligned} $

Hence, the area of an equilateral triangle is $5.196 ~cm^{2}$.

Hence, the correct option is (A).

Solution

Given: area of an equilateral triangle $=9 \sqrt{3} ~cm^{2}$

Area of an equilateral triangle $=\dfrac{\sqrt{3}}{4} \times(\text{ Side })^{2}$

$ \begin{aligned} \dfrac{\sqrt{3}}{4} \times(\text{ Side })^{2} & =9 \sqrt{3} \\ (\text{ Side })^{2} & =9 \times 4 \\ \text{ Side } & =\sqrt{9 \times 4} \\ \text{ Side } & =3 \times 2 \\ \text{ Side } & =6 ~cm \end{aligned} $

Therefore, the length of an equilateral triangle is $6 ~cm$.

Hence, the correct option is (D).

Solution

Given: The area of an equilateral triangle is $16 \sqrt{3} ~cm^{2}$.

Area of equilateral triangle $=\dfrac{\sqrt{3}}{4} \times(\text{ side })^{2}$

$ \begin{aligned} 16 \sqrt{3} & =\dfrac{\sqrt{3}}{4} \times(\text{ side })^{2} \\ (\text{ Side })^{2} & =\dfrac{16 \sqrt{3} \times 4}{\sqrt{3}} \\ & =64 \end{aligned} $

$ \begin{aligned} & \text{ Side }=\sqrt{64} \\ & \text{ Side }=8 ~cm \end{aligned} $

Therefore, the perimeter of triangle $8+8+8=24 ~cm$

Hence, the correct option is (B).

Solution

Given: The sides of a triangle are $a=35 ~cm, b=54 ~cm$ and $c=61 ~cm$, respectively. So, semiperimeter of a triangle is:

$s=\dfrac{a+b+c}{2}=\dfrac{35+54+61}{2}=\dfrac{150}{2}=75$

Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$

$ \begin{aligned} & =\sqrt{75(75-35)(75-54)(75-61)} \\ & =\sqrt{75 \times 40 \times 21 \times 14} \\ & =\sqrt{5 \times 5 \times 3 \times 2 \times 2 \times 2 \times 5 \times 3 \times 7 \times 7 \times 2} \\ & =5 \times 3 \times 2 \times 2 \times 7 \sqrt{5} \\ & =420 \sqrt{5} \end{aligned} $

As know that,

Area of triangle $ABC=\dfrac{1}{2} \times$ Base $\times$ Altitude

$ \begin{aligned} \dfrac{1}{2} \times 35 \times \text{ Altitude } & =420 \sqrt{5} \\ \text{ Altitude } & =\dfrac{420 \sqrt{5} \times 2}{35} \\ \text{ Altitude } & =24 \sqrt{5} \end{aligned} $

Therefore, the length of altitude is $24 \sqrt{5}$.

Hence, the correct option is $(C)$.

Solution

Given: The length of side be $a=2 ~cm$ and $b=4 ~cm$.

As we know that,

Area of an isosceles triangle $=\dfrac{a}{4} \sqrt{4 b^{2}-a^{2}}$

$ \begin{aligned} & =\dfrac{2 \sqrt{4 \times(4)^{2}-2^{2}}}{4} \\ & =\dfrac{\sqrt{64-4}}{2} \end{aligned} $

$ \begin{aligned} & =\dfrac{\sqrt{60}}{2} \\ & =\dfrac{2 \sqrt{15}}{2} \\ & =\sqrt{15} ~cm^{2} \end{aligned} $

Hence, the correct option is (A).

Solution

Given: The edges of a triangular board are $a=6 ~cm, b=8 ~cm$ and $c=10 ~cm$.

Now, semi-perimeter of a triangular board will be:

$ \begin{aligned} s & =\dfrac{a+b+c}{2} \\ & =\dfrac{6+8+10}{2} \\ & =\dfrac{24}{2} \\ & =12 ~cm \end{aligned} $

Now, by Heron’s formula:

Area of a triangle board $=\sqrt{s(s-a)(s-b)(s-c)}$

$ \begin{aligned} & =\sqrt{12(12-6)(12-8)(12-10)} \\ & =\sqrt{12 \times 6 \times 4 \times 2} \\ & =\sqrt{12^{2} \times 2^{2}} \\ & =12 \times 2 \\ & =24 ~cm^{2} \end{aligned} $

As, the cost of painting for area $1 ~cm^{2}=$ Rs. 0.09

So, Cost of paint for area $24 ~cm^{2}=0.09 \times 24=$ Rs. 2.16

Therefore, the cost of a triangular board is Rs. 2.16.

Hence, the correct option is (B).

Solution

Given: The base and height of a triangle are $4 ~cm$ and $6 ~cm$ respectively.

As we know that, area of a triangle $=\dfrac{1}{2} \times$ Base $\times$ Height

$ \begin{aligned} & =\dfrac{1}{2} \times 4 \times 6 \\ & =12 ~cm^{2} \end{aligned} $

Hence, the given statement is false.

Solution

Area of a triangle $=\dfrac{1}{2} \times$ Base $\times$ Height

$ \begin{aligned} & =\dfrac{1}{2} \times 4 \times 4 \\ & =8 ~cm^{2} \end{aligned} $

Hence, the given statement is true.

Solution

Suppose that side of isosceles triangle be a.

Now, perimeter of an isosceles triangle:

$2 s=5+a+a[2 s=a+b+c]$

$11=5+2 a$

$2 a=11-5$

$2 a=6$

$a=3$

Now, the formula of an area of isosceles triangle $=\dfrac{a}{4} \sqrt{4 b^{2}-a^{2}}$

So, area of an isosceles triangle $=\dfrac{5 \sqrt{4 \times(3)^{2}-(5)^{2}}}{4}$

$ \begin{aligned} & =\dfrac{5 \sqrt{4 \times 9-25}}{4} \\ & =5 \times \dfrac{\sqrt{36-25}}{4} \\ & =\dfrac{5 \sqrt{11}}{4} ~cm^{2} \end{aligned} $

Hence, the given statement is true.

Solution

Given, side of an equilateral triangle be $8 ~cm$.

Area of the equilateral triangle $=\dfrac{\sqrt{3}}{4}(\text{ Side })^{2}$

$ \begin{aligned} & =\dfrac{\sqrt{3}}{4} \times(8)^{2} \\ & =\dfrac{64}{3} \sqrt{ } 3[\therefore \text{ side }=8 ~cm] \\ & =16 \sqrt{3} ~cm^{2} \end{aligned} $

Hence, the given statement is false.

Solution

Let $PQRS$ be the rhombus whose one diagonal is $16 ~cm$, the area of the rhombus is $10 ~cm$.

As we know that diagonal of a rhombus bisect each other at right angles. So, $OA=OC=8 ~cm$ and $OB=OD$

Now, in triangle $AOB, \angle A O B=90^{\circ}$

So, $A B^{2}=O A^{2}+O B^{2}$

[By Pythagoras theorem]

$ \begin{aligned} A B^{2} & =O A^{2}+O B^{2} \\ O B^{2} & =A B^{2}-O A^{2} \\ & =(10)^{2}-8^{2} \\ & =100-64 \\ & =36 \end{aligned} $

So, $O B=\sqrt{36}=6$

Also,

$ \begin{aligned} O B & =2(O A)=2 \times 6 \\ & =12 ~cm \end{aligned} $

Therefore, area of rhombus $=\dfrac{1}{2} \times$ Products of diagonals

$ \begin{aligned} & =\dfrac{1}{2} \times 16 \times 12 \\ & =96 ~cm^{2} \end{aligned} $

Hence, the given statement is true.

Solution

Given, parallelogram in which base $=10 ~cm$ and altitude $=3.5 ~cm$

Area of a parallelogram $=$ Base $x$ Altitude

$ \begin{aligned} & =10 \times 3.5 \\ & =35 ~cm^{2} \end{aligned} $

Hence, the given statement is false.

Solution

Given: The side of a regular hexagon is ’ $a$ ‘.

As we know that the regular hexagon is divided into six equilateral triangles. So,

Area of regular hexagon $=$ Sum of area of the six equilateral triangles.

Hence, the given statement is false.

Solution

Given: The sides of the ground are $a=51 m, b=37 ~cm$, and $c=20 ~cm$. Now, the semiparameter(s) of ground is:

$2 s=a+b+c$

$2 s=51 m+37 m+20 m$

$2 s=108 m$

$s=\dfrac{108 m}{2}$

$s=54 m$

Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$

$ \begin{aligned} & =\sqrt{54(54-51)(54-37)(54-20)} \\ & =\sqrt{54 \times 3 \times 17 \times 34} \\ & =\sqrt{3 \times 3 \times 3 \times 2 \times 3 \times 17 \times 17 \times 2} \\ & =3 \times 3 \times 17 \times 2 \\ & =306 m^{2} \end{aligned} $

The cost of levelling of $1 m^{2}$ area is Rs. 3 .

So, cost of levelling the ground of $306 m^{2}$ area = Rs. $3 \times 306=$ Rs. 918

Hence, the given statement is true.

Solution

Given: The length of the altitude is 10.25 . And in a triangle, the sides are $a=11 ~cm, b=12 ~cm$ and $c=13 ~cm$.

So, semi-perimeter(s) will be:

$2 s=a+b+c$

$2 s=11 ~cm+12 ~cm+13 ~cm$

$2 s=36 ~cm$

$s=\dfrac{36}{2}$

$s=18 ~cm$

So, area of triangle $=\dfrac{2 \times \text{ Area of } \Delta}{\text{ Base }}$

$ \begin{aligned} & =\dfrac{2 \times 6 \sqrt{105}}{12} \\ & =\sqrt{105} \\ & =10.25 \end{aligned} $

Hence, the given statement is true.

Solution

Given: The sides of the ground are $a=50 m, b=65 m$, and $c=65 m$. Now, the semiparameter(s) of the cost of levelling is:

$2 s=a+b+c$

$2 s=50 m+65 m+65 m$

$2 s=180 m$

$s=\dfrac{180 m}{2}$

$s=90 m$

Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$

$ \begin{aligned} & =\sqrt{90(90-50)(90-65)(90-65)} \\ & =\sqrt{90 \times 40 \times 25 \times 25} \\ & =3 \times 2 \times 10 \times 25 \\ & =6 \times 250 \\ & =1500 m^{2} \end{aligned} $

The cost of laying grass $1 m^{2}$ area is Rs. 7 .

Therefore, the cost of levelling grass per $1500 m^{2}=$ Rs. $7 \times 1500=$ Rs. 10500

Solution

Let the sides of a triangular walls are $a=13 m, b=14 m$ and $c=15 m$.

Now, the semi-perimeter of triangular side wall,

$ \begin{aligned} s & =\dfrac{a+b+c}{2} \\ & =\dfrac{13+14+15}{2} \\ & =21 m \end{aligned} $

Now, area of triangular wall $=\sqrt{s(s-a)(s-b)(s-c)}$ [By Heron’s formula]

$ \begin{aligned} & =\sqrt{21(21-13)(21-14)(21-15)} \\ & =\sqrt{21 \times(21-13) \times(21-14) \times(21-15)} \\ & =\sqrt{21 \times 8 \times 7 \times 6} \\ & =\sqrt{21 \times 4 \times 2 \times 7 \times 3 \times 2} \\ & =\sqrt{21^{2} \times 4^{2}} \\ & =21 \times 4 \\ & =84 m^{2} \end{aligned} $

The advertisement yield earning per year for $1 m^{2}$ area is Rs. 2000 .

Therefore, advertisement yield earning per year on $84 m^{2}=2000 \times 84=$ Rs. 168000 .

According to the question, the company hired one of its walls for 6 months, therefor company pay the rent $=\dfrac{1}{2} \times 168000=$ Rs. 84000 .

Hence, the company paid rent Rs. 84000 .

Solution

Let $ABC$ be an equilateral triangle, $O$ be the interior point and $OP=14 ~cm, OQ=10 ~cm$ and $OR=6 ~cm$. Also, sides of an equilateral triangle be $a m$.

Area of triangle $OAB=\dfrac{1}{2} \times A B \times O P[.$ Area of a triangle $=\dfrac{1}{2} \times($ Base $\times$ Height $.)]$

$ \begin{aligned} & =\dfrac{1}{2} \times a \times 14 \\ & =7 a ~cm^{2} \end{aligned} $

Similarly, Area of triangle $OBC=\dfrac{1}{2} \times B C \times O Q$

$ \begin{aligned} & =\dfrac{1}{2} \times a \times 10 \\ & =5 a ~cm^{2} \end{aligned} $

Again, area of triangle $OAC=\dfrac{1}{2} \times A C \times O R$

$ \begin{aligned} & =\dfrac{1}{2} \times a \times 6 \\ & =3 a ~cm^{2} \end{aligned} $

See the given figure, area of equilateral triangle $ABC=$ Area of $(\triangle O A B+\triangle O B C+\triangle O A C)$

$=(7 a+5 a+3 a) ~cm^{2}$

$=15 a ~cm^{2}$

Now, semi-perimeter of triangle $ABC$ is:

$s=\dfrac{a+a+a}{2}$

$s=\dfrac{3 a}{2} ~cm$

As, area of equilateral triangle $ABC=\sqrt{s(s-a)(s-b)(s-c)} \quad$ [By Heron’s formula]

$ \begin{aligned} & =\sqrt{\dfrac{3 a}{2}(\dfrac{3 a}{2}-a)(\dfrac{3 a}{2}-a)(\dfrac{3 a}{2}-a)} \\ & =\sqrt{\dfrac{3 a}{2} \times \dfrac{a}{2} \times \dfrac{a}{2} \times \dfrac{a}{2}} \\ & =\dfrac{\sqrt{3}}{4} a^{2} \ldots \text{ (II) } \end{aligned} $

According to the equation (I) and (II), get:

$ \begin{aligned} \dfrac{\sqrt{3}}{4} a^{2} & =15 a \\ a & =\dfrac{15 \times 4}{\sqrt{3}} \\ a & =\dfrac{60}{\sqrt{3}} \\ a & =20 \sqrt{3} \end{aligned} $

Putting $a=20 \sqrt{3}$ in equation (II), get:

Area of triangle $ABC=\dfrac{\sqrt{3}}{4} \times(20 \sqrt{3})^{2}$

$ \begin{aligned} & =\dfrac{\sqrt{3}}{4} \times 400 \times 3 \\ & =300 \sqrt{3} ~cm^{2} \end{aligned} $

Hence, the area of an equilateral triangle is $300 \sqrt{3} ~cm^{2}$.

Solution

Given: Perimeter of triangle $=32 ~cm$

The ratio of the equal side to its base of an isosceles triangle is $3: 2$. Let sides of an isosceles triangle be $3 x, 3 x$ and $2 x$.

So, perimeter of the triangle $=3 x+3 x+2 x=8 x$

$32=8 x$

$x=\dfrac{32}{8}$

$x=4$

Since, the sides of the isosceles triangle are $3 \times 4=12,3 \times 4=12$ and $2 \times 4=8 ~cm$.

Now, semi-perimeter of triangle will be:

$ \begin{aligned} s & =\dfrac{12+12+8}{2} \\ & =\dfrac{32}{2} \\ & =16 ~cm \end{aligned} $

Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$

$ \begin{aligned} & =\sqrt{16(16-12)(16-12)(16-8)} \\ & =\sqrt{16 \times 4 \times 4 \times 8} \\ & =4 \times 4 \times 2 \sqrt{2} ~cm^{2} \\ & =32 \sqrt{2} \end{aligned} $

Therefore, the area of an isosceles triangle $ABC=\sqrt{s(s-a)(s-b)(s-c)}$

$ \begin{aligned} & =\sqrt{16(16-12)(16-12)(16-8)} \\ & =\sqrt{16 \times 4 \times 4 \times 8} \\ & =32 \sqrt{2} ~cm^{2} \end{aligned} $

Therefore, the area of an isosceles triangle is $32 \sqrt{2} ~cm^{2}$.

Solution

Let the sides of a triangle $BCD$ are $a=12 ~cm, b=17 ~cm$ and $c=25 ~cm$ and altitude of $a$ parallelogram is $h$.

Area of parallelogram, $ABCD=2$ (Area of triangle BCD)

Now, semi-perimeter(s) of triangle BCD will be:

$ \begin{aligned} s & =\dfrac{a+b+c}{2} \\ & =\dfrac{12+17+25}{2} \\ & =\dfrac{54}{2} \\ & =27 ~cm \end{aligned} $

Area of triangle $BCD=\sqrt{s(s-a)(s-b)(s-c)} \quad$ [By heron’s formula]

$ \begin{aligned} & =\sqrt{27(27-12)(27-17)(27-25)} \\ & =\sqrt{27 \times 15 \times 10 \times 2} \\ & =\sqrt{9 \times 3 \times 3 \times 5 \times 5 \times 2 \times 2} \\ & =3 \times 3 \times 5 \times 2 ~cm^{2} \\ & =90 ~cm^{2} \end{aligned} $

So, area of parallelogram $ABCD=2 \times$ Area of triangle $BCD$

$$ \begin{align*} & =2 \times 90 ~cm^{2} \\ & =180 ~cm^{2} \tag{II} \end{align*} $$

As, Area of parallelogram $ABCD=$ Base $\times$ Altitude

$ \begin{aligned} 180 & =DC \times h \\ 180 & =12 \times h \\ h & =\dfrac{180}{12} \\ h & =15 ~cm \end{aligned} $

Therefore, the area of parallelogram is $180 ~cm^{2}$ and the length of altitude is $15 ~cm$.

Solution

Given: Let a field in the form of a parallelogram $A B C D$ has sides $60 m$ and $40 m$ and one of its diagonals is $80 m$ long.

See the figure, in triangle $A B C$, let $a=40 m, b=60 m$ and $c=80 m$.

Now, semi perimeter(s) of triangle $ABC$ :

$ \begin{aligned} s & =\dfrac{a+b+c}{2} \\ & =\dfrac{40+60+80}{2} \\ & =\dfrac{180}{2} \\ & =90 m \end{aligned} $

So, area of triangle $ABC$ will be $=\sqrt{s(s-a)(s-b)(s-c)}$

$ \begin{aligned} & =\sqrt{90(90-40)(90-60)(90-80)} \\ & =\sqrt{90 \times 50 \times 30 \times 10} \\ & =\sqrt{3 \times 30 \times 5 \times 10 \times 30 \times 10} \\ & =300 \sqrt{15} \\ & =1161.895 m^{2} \end{aligned} $

Now, from equation (I),

Area of parallelogram $ABCD=2 \times 1161.895 m^{2}=2323.79 m^{2}$.

Therefore, the area of parallelogram ABCD is $2323.79 m^{2}$.

Solution

Given: The perimeter of a triangular field is $420 m$ and its sides are in the ratio $6: 7: 8$.

According to the question, Let the sides in meters are $a=6 x, b=7 x$ and $c=8 x$.

So, perimeter of the triangle $=6 x+7 x+8 x$

$420=21 x$ $x=\dfrac{420}{21}$

$x=20$

Since, the sides of the triangular field are $a=6 \times 20 ~cm=120 m, b=7 \times 20 m=140 m$ and $c=8 \times 20 m=160 m$.

Now, semi-perimeter(s) of triangle will be:

$ \begin{aligned} s & =\dfrac{1}{2} \times 420 m \\ & =210 m \end{aligned} $

Area of the triangle field $=\sqrt{s(s-a)(s-b)(s-c)}$ [Using Heron’s formula]

$ \begin{aligned} & =\sqrt{210(210-120)(210-140)(210-160)} \\ & =\sqrt{210 \times 90 \times 70 \times 50} \\ & =100 \sqrt{7 \times 3 \times 3^{2} \times 7 \times 5} \\ & =100 \times 7 \times 3 \times \sqrt{15} \\ & =2100 \sqrt{15} \end{aligned} $

Therefore, the area of the triangular field is $2100 \sqrt{15}$.

Solution

Given: The sides of a quadrilateral $ABCD$ are $AB=6 ~cm, BC=8 ~cm$, and $CD=12 ~cm$ and $DA=14 ~cm$.

Construction: Join AC.

In the right triangle $ABC$, whose angle $B$ is right angle. So, $A C^{2}=A B^{2}+B C^{2}$ [By Pythagoras theorem]

$A C^{2}=6^{2}+8^{2}$

$A C^{2}=36+64$

$A C=\sqrt{100}$

$A C=10$

Area of quadrilateral $ABCD=$ Area of triangle $ABC+$ Area of triangle $ACD$

Now, area of triangle $ABC=\dfrac{1}{2} \times A B \times A C$

$ \begin{aligned} & =\dfrac{1}{2} \times 6 \times 8 \\ & =24 ~cm^{2} \end{aligned} $

In triangle $A C D$, let $A C=a=10 ~cm, C D=b=12 ~cm$, and $D A=c=14 ~cm$.

Now, semi-perimeter of triangle ACD will be:

$ \begin{aligned} s & =\dfrac{a+b+c}{2} \\ & =\dfrac{10+12+14}{2} \\ & =\dfrac{36}{2} \\ & =18 ~cm \end{aligned} $

So, area of triangle $ACD=\sqrt{s(s-a)(s-b)(s-c)}$ [By heron’s formula]

$ \begin{aligned} & =\sqrt{18(18-10)(18-12)(18-14)} \\ & =\sqrt{18 \times 8 \times 6 \times 4} \\ & =\sqrt{(3)^{2} \times 2 \times 4 \times 2 \times 3 \times 2 \times 4} \\ & =3 \times 4 \times 2 \sqrt{3 \times 2} \\ & =24 \sqrt{6} ~cm^{2} \end{aligned} $

Hence, the area of the quadrilateral $ABCD$ is $24 \sqrt{6} ~cm^{2}$.

Solution

Given: One diagonal $=12 ~cm$, Perimeter of rhombus $=40 ~cm$

So,

$4 \times$ Side $=40$

side $=\dfrac{40}{4}$

Side $=10 ~cm$

In triangle $A B C$, let $a=10 ~cm, b=10 ~cm$, and $c=12 ~cm$.

As we know that rhombus is also a parallelogram, so its diagonal divide it into two congruent triangles of equal area. So,

Area of rhombus $=2($ Area of triangle ABC)

Now, Semi-perimeter of triangle $ABC$ will be:

$ \begin{aligned} s & =\dfrac{a+b+c}{2} \\ & =\dfrac{10+10+12}{2} \\ & =\dfrac{32}{2} \\ & =16 ~cm \end{aligned} $

So, area of triangle $ABC=\sqrt{s(s-a)(s-b)(s-c)}$

$ \begin{aligned} & =\sqrt{16(16-10)(16-10)(16-12)} \\ & =\sqrt{16 \times 6 \times 6 \times 4} \\ & =\sqrt{2304} \\ & =48 ~cm^{2} \end{aligned} $

Since, area of rhombus $=2$ (Area of triangle $ABC)$

$ \begin{aligned} & =2 \times 48 ~cm^{2} \\ & =96 ~cm^{2} \end{aligned} $

The cost of painting of the sheet is Rs. 5 per $m^{2}$.

Therefore, cost of painting both sides of rhombus shaped sheet $ABCD=$ Rs. $(2 \times 5 \times 96)=$ Rs. 960 .

Solution

Let PQRS is a trapezium, in which draw a line RT perpendicular to PS.

See the figure, $ST=PS-TP=12-7=5 ~cm$

So, in right triangle STR,

$(S R)^{2}=(S T)^{2}+(T R)^{2}$ [By using Pythagoras theorem]

$(13)^{2}=(5)^{2}+(T R)^{2}$

$(T R)^{2}=169-25$

$(T R)^{2}=144$

$T R=12 m$

Now, area of triangle STR $=\dfrac{1}{2} \times T R \times T S$ [area of triangle $=\dfrac{1}{2} \times$ Base $\times$ Height

$ \begin{aligned} & =\dfrac{1}{2} \times 12 \times 5 \\ & =30 m^{2} \end{aligned} $

As, area of rectangle $PQRT=P Q \times R Q=12 \times 7=84 m^{2}$

Now, area of trapezium $=$ Area of DSTR + Area of rectangle PQRT

$ \begin{aligned} & =30+84 \\ & =114 m^{2} \end{aligned} $

Therefore, the area of trapezium is $114 m^{2}$.

Solution

Given: Diagonal of square $ABCD=44 ~cm$

Also, $ABCD$ is a square. $So, AB=BC=CD=DA$

Now, in triangle $A C D$,

$ \begin{aligned} A C^{2} & =A D^{2}+D C^{2} \\ 44^{2} & =A D^{2}+A D^{2} \\ 2 A D^{2} & =44 \times 44 \\ 2 A D^{2} & =22 \times 2 \times 22 \times 2 \\ A D^{2} & =22 \times 2 \times 22 \\ A D & =\sqrt{22 \times 2 \times 22} \\ A D & =22 \sqrt{2} \end{aligned} $

Now, area of square $ABCD=$ Side $\times$ Side $=22 \sqrt{2} \times 22 \sqrt{2}=968 ~cm^{2}$

Since, area of square is divided into four parts.

Now, the area of paper of Red shade needed to make the kite is: $=\dfrac{1}{4} \times 968 ~cm^{2}=242 ~cm^{2}$

Also, area of green portion is:

$=\dfrac{1}{4} \times 968 ~cm^{2}$

$=242 ~cm^{2}$

Similarly, area of yellow portion is:

$ =\dfrac{1}{2} \times 968 ~cm^{2}=484 ~cm^{2} $

In triangle $P C Q$, Let $P C=a=20 ~cm, C Q=b=20 ~cm$, and $P Q=c=14 ~cm$.

Now, semi-perimeter of triangle PCQ will be:

$ \begin{aligned} s & =\dfrac{a+b+c}{2} \\ & =\dfrac{20+20+14}{2} \\ & =\dfrac{54}{2} \\ & =27 ~cm \end{aligned} $

So, area of triangle PCQ $=\sqrt{s(s-a)(s-b)(s-c)}$

$ \begin{aligned} & =\sqrt{27 \times(27-20) \times(27-20)(27-14)} \\ & =\sqrt{27 \times 7 \times 7 \times 13} \\ & =\sqrt{3 \times 3 \times 3 \times 7 \times 7 \times 13} \\ & =21 \sqrt{39} \\ & =21 \times 6.24 \\ & =131.04 ~cm^{2} \end{aligned} $

Since, the total area of green portion $=242 ~cm^{2}+131.04 ~cm^{2}=373.04 ~cm^{2}$

Therefore, the paper required for each shade to make a kite is red paper $=242 ~cm^{2}$, yellow paper $=484 ~cm^{2}$, and green paper $=373.04 ~cm^{2}$.

Solution

Given: the perimeter of a triangle is $50 ~cm$.

Now, semi-perimeter(s) of the triangle is $=\dfrac{\text{ Perimeter of triangle }}{2}=\dfrac{50}{2}=25$

Suppose that the smaller side of the triangle be $a=x~cm$. So, the second side will $be b=(x+4)$ $~cm$ and $3^{\text{rd }}$ side will be $c=(2 x-6) ~cm$.

Now, perimeter of triangle $=a+b+c=x+(x+4)+(2 x-6)$

$50 ~cm=(4 x-2) ~cm$

$50=4 x-2$

$4 x=50+2$

$4 x=52$

$x=\dfrac{52}{4}$

$x=13$

Since, the three side of the triangle are:

$a=x=13$,

$b=x+4=13+4=17$

$c=2 x-6=2 \times 13-6=26-6=20$.

So, area of the triangle $=\sqrt{s(s-a)(s-b)(s-c)}$

$ \begin{aligned} & =\sqrt{25 \times(25-13) \times(25-17) \times(25-20)} \\ & =\sqrt{25 \times 12 \times 8 \times 5} \\ & =\sqrt{5 \times 5 \times 4 \times 3 \times 4 \times 2 \times 5} \\ & =5 \times 4 \times 20 \sqrt{30} ~cm^{2} \\ & =20 \sqrt{30} ~cm^{2} \end{aligned} $

Therefore, the area of triangle is $20 \sqrt{30} ~cm^{2}$.

Solution

Given:

Area of a trapezium $=475 ~cm^{2}$ and Height $=19 ~cm$.

According to the question, let one sides of trapezium is $x$. So, another side will be $x+4$.

Now, Area of trapezium $=\dfrac{1}{2} \times($ Sum of the parallel sides $) \times$ Height

$ \begin{aligned} 475 & =\dfrac{1}{2} \times(x+x+4) \times 19 ~cm \\ 2 x+4 & =\dfrac{950}{19} \\ & =50 \\ 2 x & =50-4 \\ 2 x & =46 \\ x & =23 \end{aligned} $

Therefore, the length of the parallel side of trapezium are $x=23 ~cm$ and $x+4=23+4=27$ $~cm$.

Solution

Given: Let a rectangular plot $ABCD$ is constructing a house, having a measurement of $40 m$ long and $15 m$ in the front.

According to the question,

Length of inner-rectangle $(E F)=40-3-3=34 m$

And breadth of inner rectangle $(FG)=15-2-2=11 m$

Now, area of inner rectangle $(EFGH)$ will be $=$ Length $x$ Breadth

$ \begin{aligned} & =E F \times F G \\ & =34 \times 11 m^{2} \\ & =374 m^{2} \end{aligned} $

Therefore, the largest area where house can be constructed $=374 m^{2}$.

Solution

Given: In the trapezium $ABCD$, the two parallel sides are $AB=90 m, CD=30 m$, and $E C \perp A B$.

So, $EB=AB-EA=90 m-30 m=60 m$

Now, in triangle BEC,

$ \begin{aligned} (B C)^{2} & =(B E)^{2}+(E C)^{2} \\ 100^{2} & =60^{2}+(E C)^{2} \\ (E C)^{2} & =10000-3600 \\ (E C)^{2} & =6400 \\ E C & =\sqrt{6400} \\ E C & =80 m \end{aligned} $

Now, area of trapezium $A B C D=\dfrac{1}{2} \times($ Sum of parallel sides $) \times($ Distance between parallel sides $)$

$ \begin{aligned} & =\dfrac{1}{2} \times(A B+C D) \times E C \\ & =\dfrac{1}{2} \times(90+30) \times 80 \\ & =\dfrac{1}{2} \times 120 \times 80 \\ & =4800 m^{2} \end{aligned} $

The cost of ploughing the field of $1 m^{2}$ is Rs. 4.

Now, The cost of ploughing the field of $4800 m^{2}$ area $=4800 \times$ Rs. $4=$ Rs. 19200 .

Therefore, the total cost of plughing the field is Rs. 19200.

Solution

Given: in triangle $ABC$, the sides are $AB=a=7.5 ~cm, BC=b=7 ~cm$, and $CA=c=6.5 ~cm$. Now, semi-perimeter of a triangle will be:

$ \begin{aligned} s & =\dfrac{a+b+c}{2} \\ & =\dfrac{7.5+7+6.5}{2} \\ & =\dfrac{21}{2} \\ & =10.5 \end{aligned} $

So, area of triangle $ABC=\sqrt{s(s-a)(s-b)(s-c)}$ [By heron’s formula]

$ \begin{aligned} & =\sqrt{10.5 \times(10.5-7.5)(10.5-7)(10.5-6.5)} \\ & =\sqrt{10.5 \times 3 \times 3.5 \times 4} \\ & =\sqrt{441} \\ & =21 ~cm^{2} \end{aligned} $

Also, the area of parallelogram BCED will be $=$ Base $\times$ Height

$ \begin{aligned} & =B C \times D F \\ & =7 \times D F \end{aligned} $

Now, according to the question,

Area of triangle $ABC=$ Area of parallelogram BCED

$21=7 \times D F$

$D F=\dfrac{21}{4}$

$D F=3 ~cm$

Hence, the height of parallelogram BCED is $3 ~cm$.

Solution

Given: $A B C D$ is a rectangle, where $A B=51 ~cm$ and $B C=25 ~cm$.

The parallel sides $QC$ and $PD$ of the trapezium $PQCD$ are in the ratio of 9:8. Let $QC=9 x$ and $P D=8 x$.

Now, the area of trapezium PQCD:

$=\dfrac{1}{2} \times($ Sum of parallel sides $) \times($ Distance between parallel sides $)$

$=\dfrac{1}{2} \times(9 x+8 x) \times 25 ~cm^{2}$

$=\dfrac{1}{2} \times 17 x \times 25$

Again, area of rectangle $ABCD=B C \times C D=51 \times 25$

Now, according to the question,

Area of trapezium PQCD $=\dfrac{5}{6} \times$ Area of rectangle ABCD

$ \begin{aligned} \dfrac{1}{2} \times 17 x \times 25 & =\dfrac{5}{6} \times 51 \times 25 \\ x & =\dfrac{5}{6} \times 51 \times 25 \times 2 \times \dfrac{1}{17 \times 25} \\ x & =5 \end{aligned} $

Therefore, the length of the trapezium $PQCD, QC=9 x=9 \times 5=45 ~cm$ and, $PD=8 x=$ $8 \times 5=40 ~cm$.

Solution

Given: the dimension of the rectangular tile are $50 ~cm \times 70 ~cm$.

So, area of the rectangular tile $=50 ~cm \times 70 ~cm=3500 ~cm^{2}$.

See the given figure in the question, the sides of the triangle $A B C$ be:

$a=25 ~cm, b=17 ~cm$, and $c=26 ~cm$

Since, semi-parameter(s) of triangle be:

$ \begin{aligned} s & =\dfrac{a+b+c}{2} \\ & =\dfrac{25+17+26}{2} \\ & =\dfrac{68}{2} \\ & =34 \end{aligned} $

So, area of triangle $ABC=\sqrt{s(s-a)(s-b)(s-c)}$

Since, Total area of eight triangle $=8 \times$ Area of triangle $ABC$

$ \begin{aligned} & =204 \times 8 \\ & =1632 ~cm^{2} \end{aligned} $

The area of the design will be equal to the area of eight triangle that is $1632 ~cm^{2}$.

Now, remaining area of the tile $=$ Area of the rectangle - Area of the design $=$ $3500 ~cm^{2}-1632 ~cm^{2}=1868 ~cm^{2}$

Therefore, total area of the design is $1632 ~cm^{2}$ and the remaining area of the tile is $1868 ~cm^{2}$.