Rd Sharma 2021 for Class 9 Maths Chapter 17 - Heron

Answer:

The quadrilateral ABCD having sides AB, BC, CD, DA and diagonal AC=5 cm is given, where AC divides quadrilateralABCD into two triangles ΔABC and ΔADC. We will find the area of the two triangles separately and them to find the area of quadrilateral ABCD.

In triangle ΔABC, observe that,

So the triangle ΔABC is right angled triangle.

Area of right angled triangle ΔABC, is given by

In ΔACD all the sides are known, so just use Heron’s formula to find out Area of triangle ΔACD, 

$$\begin{gathered} s=\frac{AD+DC+AC}{2} \\ =\frac{5+4+5}{2} \\ =7 \mathrm{cm} \end{gathered}$$

The area of the ΔACD is:

Area of quadrilateral ABCD will be,

Area = Area of triangle ABC + Area of triangle ADC

Answer:

We assume quadrilateral ABCD be the quadrangular field having sides AB, BC, CD, DA and.

We take a diagonal AC, where AC divides quadrilateral ABCD into two triangles ΔABC and ΔADC

We will find the area of two triangles ΔABC and ΔADC separately and add them to find the area of the quadrangular ABCD.

In triangle ΔADC, we have

AD = 24 m; DC = 7 m

We use Pythagoras theorem to find side AC, 

AC² = AD² + DC²

Area of right angled triangle ΔADC, say A₁ is given by

Where, Base = DA = 24 m; Height = DC = 7 m

Area of triangle ΔABC, say A₂ having sides a, b, c and s as semi-perimeter is given by

Where, a = AC = 25 m; b = AB = 26 m; c = BC = 27 m 

Area of quadrilateral ABCD, say A

A = Area of triangle ΔADC + Area of triangle ΔABC

Answer:

We assume ABCD be the quadrilateral having sides AB, BC, CD, DA and angle.

We take a diagonal AC, where AC divides quadrilateral ABCD into two triangles ΔABC and ΔADC. We will find the area of these two triangles and add them to find the area of the quadrilateral ABCD

In triangle ΔABC, we have

AB = 5 m; BC = 12 m

We will use Pythagoras theorem to calculate AC 

AC² = AB² + BC²

Area of right angled triangle ΔABC, say A₁ is given by

Where, Base = AB = 5 m; Height = BC = 12 m

Area of triangle ΔADC, say A₂ having sides a, b, c and s as semi-perimeter is given by

Where, a = AC = 13 m; b = DC = 14 m; c = AD = 15 m 

Area of quadrilateral ABCD, say A

A = Area of triangle ΔABC + Area of triangle ΔADC

Answer:

We assume ABCD be the quadrilateral having sides AB, BC, CD, DA and.

We take a diagonal DB, where DB divides ABCD into two triangles ΔBCD and ΔABD

In ΔBCD, we have

DC = 5 m; BC = 12 m

Use Pythagoras theorem 

BD² = DC² + BC²

Area of right angled triangle ΔBCD, say A₁ is given by

Where, Base = DC = 5 m; Height = BC = 12 m

Area of triangle ΔABD, say A₂ having sides a, b, c and s as semi-perimeter is given by

Where, a = AD = 8 m; b = AB = 9 m; c = BD = 13 m 

Area of quadrilateral ABCD, say A

A = Area of triangle DCB + Area of triangle ABD

Answer:

We assume ABCD be the given trapezium where AB is parallel to DC.

We draw CE parallel to AD from point C.

Therefore, a parallelogram ADCE is formed having AD parallel to CE and DC parallel to AE.

AE = 60 cm; CE = 25 cm; BE = 

Basically we will find the area of the triangle BCE and area of the parallelogram AECD and add them to find the area of the trapezium ABCD.

Area of triangle ECB, say A₁ having sides a, b, c and s as semi-perimeter is given by

Where, a = EB = 17 cm; b = EC = 25 cm; c = BC = 26 cm 

Here we need to find the height of the parallelogram AECD which is CM to calculate area of AECD.

Where, BE = Base = 17 cm ; Height = CM = h

Thus area of parallelogram will be,
A₂= $b\times h$
$$\begin{gathered} =60\times 24 \\ =1440 c{m}^2 \end{gathered}$$

Total area of the trapezium will be
A = A₁ + A₂
= 204 + 1440
= 1644 cm²

Answer:

We assume ABCD be the given rhombus having 

AB = BC = CD = DA 

BD and AC be the diagonals of rhombus

We need to find cost of painting both sides

Perimeter of rhombus ABCD, say P is 32 m

We know that BD and AC diagonals of rhombus and.So,

(Diagonals in rhombus intersect at right angle)

BD = 24 m; AC = h; side = AB = 8m

Taking square of both sides, we get

Area of rhombus, say A₁

Area of both sides of rhombus;

Answer:

We assume ABCD be the quadrilateral having sides AB, BC, CD, DA, diagonal BD and angle where BCD forms an equilateral triangle having equal sides.

We need to find area of ABCD

In triangle BAD, we have

BD² = BA² + AD² .So

Area of right angled triangle ABD, say A₁ is given by

Where,

Base = BA = 10 cm; Height = AD = 24 cm

Area of equilateral triangle BCD, say A₂ having sides a, b, c is given by

, where

a = BC = CD = BD = 26 cm

Area of quadrilateral ABCD, say A

A = Area of triangle BAD + Area of triangle BCD

Answer:

The quadrilateral ABCD having sides AB,BC,CD,DA and diagonal BD is given, where BD divides ABCD into two triangles 

ΔDBC and ΔDAB

In triangle DBC, we can observe that

DC² = DB² + BC²

Therefore, it is a right angled triangle. 

Area of right angled triangle DBC, say A₁ is given by

Where,

Base = BC = 21 cm; Height = BD = 20 cm

Area of triangle DAB, say A₂ having sides a, b, c and s as semi-perimeter is given by

, where

a = DB = 20 cm; b = AD = 34 cm; c = AB = 42 cm

Area of quadrilateral ABCD, say A

A = Area of triangle DBC + Area of triangle DAB

Answer:

We are given the measure of adjacent sides of a parallelogram AB and BC that is the sides having same point of origin and the diagonal AC which divides parallelogram ABCD into two congruent triangles ABC and ADC.

Area of triangle ABC is equal to Area of triangle ADC as they are congruent triangles.

Area of parallelogram ABCD, say A is given by

A =

The area of a triangle having sides a, b, c and s as semi-perimeter is given by,

, where

Therefore the area of a triangle, say A₁ having sides 20 cm, 34 cm and 42 cm is given by

a = 20 cm ; b = 34 cm ; c = 42 cm

Area of parallelogram ABCD, say A is given by

Answer:

The blades of the magnetic compass are forming a rhombus having all equal sides measuring 5 cm each. A diagonal measuring 1 cm is given which is forming the triangular shape of the blades of the magnetic compass and diving the rhombus into two congruent triangles, say triangle ABC and triangle DBC having equal dimensions.

The area of a triangle having sides a, b, c and s as semi-perimeter is given by,

, where

Therefore the area of a triangle ABC, say A₁ having sides 5 cm, 5 cm and 1 cm is given by:

a = 5 cm ; b = 5 cm ; c = 1 cm

Area of blades of magnetic compass, say A is given by

A = Area of one of triangle ABC

Answer:

It is given that the area of triangle and parallelogram are equal.

We will calculate the area of triangle with the given values and it will also give us the area of parallelogram as both are equal.

The area of a triangle having sides a, b, c and s as semi-perimeter is given by,

, where

Therefore the area of a triangle; say A, having sides 15 cm, 13 cm and 14 cm is given by

a = 15 cm ; b = 13 cm ; c = 14 cm

We have to find the height of the parallelogram, say h

Area of parallelogram AECD say A₁ is given by

Base = 60 cm; Height = h cm; A =A₁ = 84 cm²

Answer:

Consider a trapezium ABCD as shown below.

Draw a line BF parallel to AD such that ABFD becomes a parallelogram. Also, draw a perpendicular BE on CD as shown in the figure.


In ΔBCF, the three sides are given as $a=\mathrm{BF}=25 \mathrm{cm}, b=\mathrm{BC}=26 \mathrm{cm}, c=\mathrm{CF}=\mathrm{CD}-\mathrm{FD}=17\mathrm{cm}$.
The semi-perimeter of ΔBCF = $\frac{25+26+17}{2}=34$.
The area of ΔBCF can be calculated using Heron's formula as
$$\begin{gathered} \sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{34\left(34-25\right)\left(34-26\right)\left(34-17\right)} \\ =\sqrt{34\times 9\times 8\times 17} \\ =204 {\mathrm{cm}}^2 \end{gathered}$$

Also, the area of ΔBCF = $\frac{1}{2}\times \mathrm{base}\times \mathrm{height}$
$$\begin{gathered} \Rightarrow 204=\frac{1}{2}\times 17\times \mathrm{BE} \\ \Rightarrow 204=\frac{1}{2}\times 17\times \mathrm{BE} \\ \Rightarrow \mathrm{BE}=24 \mathrm{cm} \end{gathered}$$

Therefore, Area of Trapezium =$\frac{1}{2}\left(\mathrm{AB}+\mathrm{CD}\right)\times \mathrm{BE}=\frac{1}{2}\left(60+77\right)\times 24=1644 {\mathrm{cm}}^2$.

Answer:

We assume ABCD be the quadrilateral having sides AB, BC, CD, DA and ∠ACB = 90^∘.

We take a diagonal AC, where AC divides ABCD into two triangles ΔACB and ΔADC

Since ∆ACB is right angled at C, we have

AC = 15 cm; AB = 17 cm

AB² = AC² + BC²

Area of right angled triangle ABC, say A₁ is given by

, where,

Base = BC = 8 cm; Height = AC = 15 cm

Area of triangle ADC, say A₂ having sides a, b, c and s as semi-perimeter is given by

, where

a = AD = 9 cm; b = DC = 12 cm; c = AC = 15 cm 

Area of quadrilateral ABCD, say A

A = Area of ∆ACB + Area of ∆ADC

Perimeter of quadrilateral ABCD, say P

Answer:

We have to find the area of each type of triangular strips needed for the fan.

There are 5 strips of each type having equal dimensions, so we will calculate the area of a single strip and then multiply it by 5 to ascertain the area of each type of strip needed. 

The area of a triangle having sides a, b, c and s as semi-perimeter is given by,

, where

Therefore the area of a triangular strip, say A₁ having sides 25 cm, 25 cm and 14 cm is given by: 

a = 25 cm ; b = 25 cm ; c = 14 cm

Area of each type of strip needed, say A.

Answer:

The area of a triangle having sides a, b, c and s as semi-perimeter is given by,

, where

Therefore the area of a triangle say A, having sides 16 cm, 30 cm and 34 cm is given by

a = 16 cm ; b = 30 cm ; c = 34 cm

Therefore the area of the triangle is 

Hence, the correct option is (b).

Answer:

$$\begin{gathered} \mathrm{Let} \mathrm{ABC} \mathrm{be} \mathrm{the} \mathrm{right} \mathrm{triangle} \mathrm{in} \mathrm{which} \angle \mathrm{B}=90°. \\ \mathrm{Now}, \mathrm{base}=\mathrm{BC}; \mathrm{perpendicular}=\mathrm{AB}; \mathrm{Hypotenuse}=\mathrm{AC} \\ \mathrm{Now}, \mathrm{BC}=30 \mathrm{cm} \left(\mathrm{given}\right) \\ \mathrm{Now}, ∆\mathrm{ABC} \mathrm{is} \mathrm{an} \mathrm{isosceles} \mathrm{right} \mathrm{angled} ∆ \mathrm{and} \mathrm{we} \mathrm{know} \mathrm{that} \mathrm{hypotenuse} \mathrm{is} \mathrm{the} \mathrm{longest} \mathrm{side} \mathrm{of} \mathrm{the} \mathrm{right} ∆. \\ \mathrm{So}, \mathrm{AB}=\mathrm{BC}=30 \mathrm{cm} \\ \mathrm{area} \mathrm{of} ∆\mathrm{ABC}=\frac{1}{2}\times \mathrm{base}\times \mathrm{height} \\ =\frac{1}{2}\times \mathrm{BC}\times \mathrm{AB} \\ =\frac{1}{2}\times 30\times 30 \\ =450 {\mathrm{cm}}^2 \end{gathered}$$


Hence, the correct option is (d).

Answer:

The area of a triangle having sides a, b, c and s as semi-perimeter is given by,

, where

Therefore the area of a triangle having sides 7 cm, 9 cm and 14 cm is given by

a = 7 cm ; b = 9 cm ; c = 14 cm

Therefore the answer is (a).

Answer:

The area of a triangle having sides a, b, c and s as semi-perimeter is given by,

, where

Therefore the area of a triangular field, say A having sides 325 m, 300 m and 125 m is given by

a = 325 m ; b = 300 m ; c = 125 m

Therefore, the correct answer is (a).

Answer:

The area of a triangle having sides a, b, c and s as semi-perimeter is given by,

, where

Therefore the area of a triangle, say A having sides 50 cm, 78 cm and 112 cm is given by

The area of a triangle, having p as the altitude will be,

$\mathrm{Area} = \frac{1}{2}\times \mathrm{base}\times \mathrm{height}$

Where, A = 1680

We have to find the smallest altitude, so will substitute the value of the base AC with the length of each side one by one and find the smallest altitude distance i.e. p

Case 1 

Case 2

Case 3

Therefore, the answer is (b).

Answer:

The area of a triangle having sides a, b, c and s as semi-perimeter is given by,

, where

We need to find the altitude to the smallest side
 

Therefore the area of a triangle having sides 11 m, 60 m and 61 m is given by

a = 11 m ; b = 60 m ; c = 61 m

The area of a triangle having base AC and height p is given by

We have to find the height p corresponding to the smallest side of the triangle. Here smallest side is 11 m 

AC = 11 m

Therefore, the answer is (d).

Answer:

The area of a triangle having sides a, b, c and s as semi-perimeter is given by,

, where

We need to find the altitude corresponding to the longest side
 

Therefore the area of a triangle having sides 11 cm, 15 cm and 16 cm is given by

a = 11 m ; b = 15 cm ; c = 16 cm

The area of a triangle having base AC and height p is given by

We have to find the height p corresponding to the longest side of the triangle.Here longest side is 16 cm, that is AC=16 cm 

Therefore, the answer is (c).

Answer:

In right angled triangle ABC having base 5 cm and hypotenuse 13 cm we are asked to find its area

Using Pythagorean Theorem

Where, AB = hypotenuse = 13 cm, AC = Base = 5 cm, BC = Height

Area of a triangle, say A having base 5 cm and altitude 12 cm is given by

Where, Base = 5 cm; Height = 12 cm

Therefore, the answer is (c).

Answer:

Area of an equilateral triangle say A, having each side a cm is given by 

We are asked to find the side of the triangle

Therefore, the side of the equilateral triangle says a, having area is given by

Therefore, the correct answer is (a). 

Answer:

Given: Area of an isosceles right triangle is 8 cm²

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{an} \mathrm{isosceles} \mathrm{right} \mathrm{triangle}=\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height} \\ =\frac{1}{2}\times \mathrm{Base}\times \mathrm{Base} \left(\because \mathrm{Base}=\mathrm{Height}\right) \\ =\frac{1}{2}\times {\left(\mathrm{Base}\right)}^2 \\ \Rightarrow 8=\frac{1}{2}\times {\left(\mathrm{Base}\right)}^2 \\ \Rightarrow 16={\left(\mathrm{Base}\right)}^2 \\ \Rightarrow \mathrm{Base}=4 \mathrm{cm} =\mathrm{Perpendicular} \\ \mathrm{In} \mathrm{a} \mathrm{right} \mathrm{angled} \mathrm{triangle}, \mathrm{using} \mathrm{pythagoras} \mathrm{theorem} \\ {\left(\mathrm{Hypotenuse}\right)}^2={\left(\mathrm{Base}\right)}^2+{\left(\mathrm{Perpendicular}\right)}^2 \\ ={\left(4\right)}^2+{\left(4\right)}^2 \\ =16+16 \\ =32 \\ \Rightarrow \mathrm{Hypotenuse}=\sqrt{32} \mathrm{cm} \end{gathered}$$

Hence, the correct option is (a).
 

Answer:

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

Let the length of the side of an equilateral triangle be x m.

$$\begin{gathered} \mathrm{Perimeter}=60 \mathrm{m} \\ \Rightarrow 3x=60 \\ \Rightarrow x=\frac{60}{3} \\ \Rightarrow x=20 \mathrm{m} \\ \mathrm{Area} \mathrm{of} \mathrm{an} \mathrm{equilateral} \mathrm{triangl}e=\frac{\sqrt{3}}{4}\times x^2 \\ =\frac{\sqrt{3}}{4}{\left(20\right)}^2 \\ =\frac{\sqrt{3}}{4}\left(400\right) \\ =100\sqrt{3} {\mathrm{m}}^2 \end{gathered}$$

Hence, the correct option is (d).

Answer:

Given: 
The base of an isosceles triangle is 2 cm.
The length of one of the equal sides 4 cm.

Using Heron's formula:
If a, b and c are three sides of a triangle, then
area of the triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ , where $s=\frac{a+b+c}{2}$.


Here, $s=\frac{2+4+4}{2}=\frac{10}{2}=5$

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle}=\sqrt{5\left(5-2\right)\left(5-4\right)\left(5-4\right)} \\ =\sqrt{5\left(3\right)\left(1\right)\left(1\right)} \\ =\sqrt{15} {\mathrm{cm}}^2 \end{gathered}$$

Hence, the correct option is (a).

Answer:

Given: The area of an equilateral triangle is $9\sqrt{3} {\mathrm{cm}}^2$.

Let the length of the side of an equilateral triangle be x cm.

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{an} \mathrm{equilateral} \mathrm{triangle}=\frac{\sqrt{3}}{4}\times x^2 \\ \Rightarrow 9\sqrt{3}=\frac{\sqrt{3}}{4}{\left(x\right)}^2 \\ \Rightarrow \frac{9\sqrt{3}\times 4}{\sqrt{3}}={\left(x\right)}^2 \\ \Rightarrow x^2=36 \\ \Rightarrow x=6 \mathrm{cm} \end{gathered}$$

Hence, the correct option is (d).

Answer:

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

Let the length of the side of an equilateral triangle be x cm.

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{an} \mathrm{equilateral} \mathrm{triangle}=\frac{\sqrt{3}}{4}\times x^2 \\ \Rightarrow 16\sqrt{3}=\frac{\sqrt{3}}{4}{\left(x\right)}^2 \\ \Rightarrow \frac{16\sqrt{3}\times 4}{\sqrt{3}}={\left(x\right)}^2 \\ \Rightarrow x^2=64 \\ \Rightarrow x=8 \mathrm{cm} \\ \mathrm{Thus}, \mathrm{Perimeter} \mathrm{of} \mathrm{triangle}=3x \\ =3\left(8\right) \\ =24 \mathrm{cm} \end{gathered}$$

Hence, the correct option is (b).

Answer:

Given: 
The sides of a triangle are 35 cm, 54 cm and 61 cm respectively.



Let CD be the longest altitude of the triangle.

Using Heron's formula:
If a, b and c are three sides of a triangle, then
area of the triangle =  , where .


Here, 




We know,


Thus, the length of its longest altitude is .

Hence, the correct option is (c).

Answer:

Given: 
The sides of a triangle are 56 cm, 60 cm and 52 cm.

Using Heron's formula:
If a, b and c are three sides of a triangle, then
area of the triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ , where $s=\frac{a+b+c}{2}$.


Here, $s=\frac{56+60+52}{2}=\frac{168}{2}=84$

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle}=\sqrt{84\left(84-56\right)\left(84-60\right)\left(84-52\right)} \\ =\sqrt{84\left(28\right)\left(24\right)\left(32\right)} \\ =\sqrt{3\times 7\times 2\times 2\left(7\times 2\times 2\right)\left(3\times 2\times 2\times 2\right)\left(2\times 2\times 2\times 2\times 2\right)} \\ =\left(3\times 7\times 2\times 2\times 2\times 2\times 2\times 2\right) \\ =1344 {\mathrm{cm}}^2 \end{gathered}$$

Hence, the correct option is (c).

Answer:

Given: 
The edges of a triangular board are 6 cm, 8 cm and 10 cm long.
The cost of painting per cm² is 9 paise.


Using Heron's formula:
If a, b and c are three sides of a triangle, then
area of the triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ , where $s=\frac{a+b+c}{2}$.


Here, $s=\frac{6+8+10}{2}=\frac{24}{2}=12$

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle}=\sqrt{12\left(12-6\right)\left(12-8\right)\left(12-10\right)} \\ =\sqrt{12\left(6\right)\left(4\right)\left(2\right)} \\ =\sqrt{\left(3\times 2\times 2\right)\left(3\times 2\right)\left(2\times 2\right)\left(2\right)} \\ =\left(3\times 2\times 2\times 2\right) \\ =24 {\mathrm{cm}}^2 \end{gathered}$$

The cost of painting per cm² = 9 paise.
The cost of painting 24 cm² = 24 × 9 paise
                                             = 216 paise
                                             = ₹ 2.16


Hence, the correct option is (b).

Answer:

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

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{an} \mathrm{equilateral} \mathrm{triangl}e=\frac{\sqrt{3}}{4}\times x^2 \\ =\frac{\sqrt{3}}{4}{\left(2\sqrt{3}\right)}^2 \\ =\frac{\sqrt{3}}{4}\left(12\right) \\ =3\sqrt{3} \\ =5.196 {\mathrm{cm}}^2 \end{gathered}$$

Hence, the correct option is (a).

Answer:

Given: The area of a regular hexagon is $54\sqrt{3} {\mathrm{cm}}^2$.

We know, a hexagon is formed by joining 6 equilateral triangles together.
Therefore, Area of 6 equilateral triangles = $54\sqrt{3} {\mathrm{cm}}^2$

$$\begin{gathered} \mathrm{Area} \mathrm{of} 6 \mathrm{equilateral} \mathrm{triangles}=6\times \frac{\sqrt{3}}{4}\times x^2 \\ \Rightarrow 54\sqrt{3}=6\times \frac{\sqrt{3}}{4}\times x^2 \\ \Rightarrow \frac{54\sqrt{3}\times 4}{6\times \sqrt{3}}=x^2 \\ \Rightarrow 9\times 4=x^2 \\ \Rightarrow 36=x^2 \\ \Rightarrow x=6 \mathrm{cm} \end{gathered}$$

Hence, the correct option is (c).

Answer:

Given: The length of each edge of a regular tetrahedron is 'a' units.

We know, the surface area of tetrahedron = area of 4 equilateral triangles

$$\begin{gathered} \mathrm{Thus}, \\ \mathrm{Surface} \mathrm{Area}=4\times \frac{\sqrt{3}}{4}\times a^2 \\ =\sqrt{3} a^2 \mathrm{sq}. \mathrm{units} \end{gathered}$$

Hence, the correct option is (a).

Answer:

We are given the area of an isosceles right triangle and we have to find its perimeter. 

Two sides of isosceles right triangle are equal and we assume the equal sides to be the base and height of the triangle. We are asked to find the perimeter of the triangle

Let us take the base and height of the triangle be x cm. 

Area of a isosceles right triangle, say A having base x cm and height x cm is given by

A = 8 cm²; Base = Height = x cm

Using Pythagorean Theorem we have;

Let ABC be the given triangle

Perimeter of triangle ABC, say P is given by

AB = 4 cm; BC = 4 cm; AC =

Therefore, the answer is (b). 

Answer:

We are given that triangle ABC has equal perimeter as to the perimeter of an equilateral triangle having side 9 cm. The sides of triangle ABC are consecutive integers. We are asked to find the smallest side of the triangle ABC 

Perimeter of an equilateral triangle, say P having side 9 cm is given by

Let us assume the three sides of triangle ABC be x, x+1, x−1

Perimeter of triangle ABC, say P₁ is given by

P₁ = AB + BC + AC

AB = x; BC = x +1; AC = x−1. Since P₁ = P. So

By using the value of x, we get the sides of triangle as 8 cm, 9 cm and 10 cm

Therefore, the answer is (c).

Answer:

Area of triangle ABC is given 40 cm².

Also

We are asked to find the area of the triangle BDC

Let us take BE perpendicular to base AC in triangle ABC.

We assume AC equal to y and BE equal to x in triangle ABC

Area of triangle ABC, say A is given by

We are given the ratio between AD to DC equal to 3:2

So,

In triangle BDC, we take BE as the height of the triangle

Area of triangle BDC, say A₁ is given by

Therefore, the answer is (a).

Answer:

We are given the length of median of an equilateral triangle by which we can calculate its side. We are asked to find area of triangle in terms of x

Altitude of an equilateral triangle say L, having equal sides of a cm is given by, where, L = x cm

Area of an equilateral triangle, say A₁ having each side a cm is given by 

Since .So

Therefore, the answer is (c).

Answer:

The area of a triangle having sides a, b, c and s as semi-perimeter is given by,

, where

We take the sides of a new triangle as 2a, 2b, 2c that is twice the sides of previous one

Now, the area of a triangle having sides 2a, 2b, and 2c and as semi-perimeter is given by,

Where,

Now,

Therefore, increase in the area of the triangle

Percentage increase in area 

Therefore, the answer is (c).

Answer:

It is given the perimeter of a square ABCD is equal to the perimeter of triangle PQR.

The measure of the diagonal of the square is given .We are asked to find the area of the triangle

In square ABCD, we assume that the adjacent sides of square be a.

Since, it is a square then

By using Pythagorean Theorem

Therefore, side of the square is 12 cm.

Perimeter of the square ABCD say P is given by

Side = 12 cm

Perimeter of the equilateral triangle PQR say P₁ is given by

The side of equilateral triangle PQR is equal to 16 cm.

Area of an equilateral triangle say A, having each side a cm is given by 

Area of the given equilateral triangle having each equal side equal to 4 cm is given by

a = 16 cm

Therefore, the answer is (d).

Answer:

Let ABC be an equilateral triangle of side 'a' units and AD be the altitude of the triangle.

In ∆ABD,
AB = a units
BD = $\frac{a}{2}$ units

Using pythagoras theorem,
$$\begin{gathered} A{B}^2=B{D}^2+A{D}^2 \\ \Rightarrow a^2={\left(\frac{a}{2}\right)}^2+A{D}^2 \\ \Rightarrow A{D}^2=a^2-\frac{a^2}{4} \\ \Rightarrow A{D}^2=\frac{4a^2-a^2}{4} \\ \Rightarrow A{D}^2=\frac{3a^2}{4} \\ \Rightarrow AD=\frac{\sqrt{3}}{2}a ...\left(1\right) \end{gathered}$$

Now,
$$\begin{gathered} \frac{\mathrm{Side}}{\mathrm{Altitude}}=\frac{AB}{AD} \\ =\frac{a}{{\displaystyle \frac{\sqrt{3}}{2}}a} \\ =\frac{2}{\sqrt{3}} \end{gathered}$$

Hence, the side and altitude of an equilateral triangle are in the ratio $\underline{2 : \sqrt{3}}$.

Answer:

Given: Area of an isosceles right-angled triangle is 72 cm²

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{an} \mathrm{isosceles} \mathrm{right} \mathrm{triangle}=\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height} \\ =\frac{1}{2}\times \mathrm{Base}\times \mathrm{Base} \left(\because \mathrm{Base}=\mathrm{Height}\right) \\ =\frac{1}{2}\times {\left(\mathrm{Base}\right)}^2 \\ \Rightarrow 72=\frac{1}{2}\times {\left(\mathrm{Base}\right)}^2 \\ \Rightarrow 144={\left(\mathrm{Base}\right)}^2 \\ \Rightarrow \mathrm{Base}=12 \mathrm{cm} =\mathrm{Perpendicular} \\ \mathrm{In} \mathrm{a} \mathrm{right}-\mathrm{angled} \mathrm{triangle}, \mathrm{using} \mathrm{pythagoras} \mathrm{theorem} \\ {\left(\mathrm{Hypotenuse}\right)}^2={\left(\mathrm{Base}\right)}^2+{\left(\mathrm{Perpendicular}\right)}^2 \\ ={\left(12\right)}^2+{\left(12\right)}^2 \\ =144+144 \\ =288 \\ \Rightarrow \mathrm{Hypotenuse}=\sqrt{288} \mathrm{cm} \\ \Rightarrow \mathrm{Hypotenuse}=12\sqrt{2} \mathrm{cm} \\ \mathrm{Thus}, \\ \mathrm{Perimeter}=12+12+12\sqrt{2} =12\left(2+\sqrt{2}\right) \mathrm{cm} \end{gathered}$$

Hence, its perimeter is $\underline{12\left(2+\sqrt{2}\right) \mathrm{cm}}.$

Answer:

Given: The height of an equilateral triangle is $\sqrt{3}$a units.

Let ABC be an equilateral triangle of side 'x' units and AD be the altitude of the triangle of length $\sqrt{3}$a units.

In ∆ABD,
AB = x units
BD = $\frac{x}{2}$ units

Using pythagoras theorem,
$$\begin{gathered} A{B}^2=B{D}^2+A{D}^2 \\ \Rightarrow x^2={\left(\frac{x}{2}\right)}^2+{\left(\sqrt{3}a\right)}^2 \\ \Rightarrow 3a^2=x^2-\frac{x^2}{4} \\ \Rightarrow 3a^2=\frac{4x^2-x^2}{4} \\ \Rightarrow 3a^2=\frac{3x^2}{4} \\ \Rightarrow a^2=\frac{x^2}{4} \\ \Rightarrow a=\frac{x}{2} \\ \Rightarrow x=2a ...\left(1\right) \end{gathered}$$

Now,
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle}=\frac{\sqrt{3}}{4}\times {\left(\mathrm{side}\right)}^2 \\ =\frac{\sqrt{3}}{4}\times {\left(2a\right)}^2 \\ =\frac{\sqrt{3}}{4}\times 4a^2 \\ =\sqrt{3}{a}^2 \end{gathered}$$


Hence, its area is $\underline{\sqrt{3}{a}^2 \mathrm{sq}. \mathrm{units}}$.

Answer:

Given: 
Base of the triangle = 4 cm
Height = 6 cm

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle}=\frac{1}{2}\times \mathrm{base}\times \mathrm{height} \\ =\frac{1}{2}\times 4\times 6 \\ =\frac{24}{2} \\ =12 {\mathrm{cm}}^2 \end{gathered}$$


Hence, the area is $\underline{12 {\mathrm{cm}}^2}$.

Answer:

Given: 
ΔABC is an isosceles right triangle right-angled at A
AB = 4 cm

Since, ΔABC is an isosceles right triangle right-angled at A
Therefore, AB = AC = 4 cm

In ∆ABC,
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle}=\frac{1}{2}\times \mathrm{base}\times \mathrm{height} \\ =\frac{1}{2}\times AB\times AC \\ =\frac{1}{2}\times 4\times 4 \\ =8 {\mathrm{cm}}^2 \end{gathered}$$


Hence, its area is $\underline{8 {\mathrm{cm}}^2}$.

Answer:

Given: 
The side of a rhombus is 10 cm
One diagonal is 16 cm

Let ABCD be a rhombus.
AB = 10 cm     ...(1)
AC = 16 cm     ...(2)

We know, the diagonals of a rhombus bisects each other at right angles.
Let the point of intersection of the diagonals be O.

Then,
AO = OC = $\frac{16}{2}=8$ cm     ...(3)

$$\begin{gathered} \mathrm{In} ∆AOB, \\ \mathrm{Using} \mathrm{pythagoras} \mathrm{theorem}, \\ A{B}^2=A{O}^2+B{O}^2 \\ \Rightarrow {10}^2=8^2+B{O}^2 \\ \Rightarrow 100=64+B{O}^2 \\ \Rightarrow B{O}^2=100-64 \\ \Rightarrow B{O}^2=36 \\ \Rightarrow BO=6 \\ \mathrm{Thus}, BD=2\times BO \\ =2\times 6 \\ =12 \mathrm{cm} ...\left(4\right) \end{gathered}$$



$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{rhombus}=\frac{1}{2}\times \left(\mathrm{product} \mathrm{of} \mathrm{diagonals}\right) \\ =\frac{1}{2}\times 12\times 16 \left(\mathrm{from} \left(2\right) \mathrm{and} \left(4\right)\right) \\ =6\times 16 \\ =96 {\mathrm{cm}}^2 \end{gathered}$$


Hence, the area is $\underline{96 {\mathrm{cm}}^2}$.

Answer:

Given: The side of a regular hexagon is 6 cm.

We know, a hexagon is formed by joining 6 equilateral triangles together.
Therefore,
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{hexagon}=\mathrm{Area} \mathrm{of} 6 \mathrm{equilateral} \mathrm{triangles} \\ =6\times \frac{\sqrt{3}}{4}\times x^2 \\ =6\times \frac{\sqrt{3}}{4}\times {\left(6\right)}^2 \\ =6\times \frac{\sqrt{3}}{4}\times 36 \\ =6\times 9\times \sqrt{3} \\ =54\sqrt{3} {\mathrm{cm}}^2 \end{gathered}$$

Hence, the area of a regular hexagon of side 6 cm is $\underline{54\sqrt{3} {\mathrm{cm}}^2}$.

Answer:

Given: 
The base of a right triangle is 8 cm
hypotenuse is 10 cm


$$\begin{gathered} \mathrm{Using} \mathrm{pythagoras} \mathrm{theorem}, \\ {\left(\mathrm{Hypotenuse}\right)}^2={\left(\mathrm{Base}\right)}^2+{\left(\mathrm{Height}\right)}^2 \\ \Rightarrow {10}^2=8^2+{\left(\mathrm{Height}\right)}^2 \\ \Rightarrow 100=64+{\left(\mathrm{Height}\right)}^2 \\ \Rightarrow {\left(\mathrm{Height}\right)}^2=100-64 \\ \Rightarrow {\left(\mathrm{Height}\right)}^2=36 \\ \Rightarrow \mathrm{Height}=6 \end{gathered}$$


$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle}=\frac{1}{2}\times \left(\mathrm{base}\right)\times \left(\mathrm{height}\right) \\ =\frac{1}{2}\times 8\times 6 \\ =4\times 6 \\ =24 {\mathrm{cm}}^2 \end{gathered}$$


Hence, the area is $\underline{24 {\mathrm{cm}}^2}$.

Answer:

Let the sides of the triangle be a, b and c.

According to the question,
$$\begin{gathered} a\left(b^2+c^2\right)+b\left(a^2+c^2\right)+c\left(b^2+a^2\right)=6abc \\ \Rightarrow a{b}^2+a{c}^2+b{a}^2+b{c}^2+c{b}^2+c{a}^2=6abc \\ \Rightarrow a{b}^2+a{c}^2+b{a}^2+b{c}^2+c{b}^2+c{a}^2-6abc=0 \\ \Rightarrow a{b}^2+a{c}^2-2abc+b{a}^2+b{c}^2-2abc+c{b}^2+c{a}^2-2abc=0 \\ \Rightarrow a\left(b^2+c^2-2bc\right)+b\left(a^2+c^2-2ac\right)+c\left(b^2+a^2-2ab\right)=0 \\ \Rightarrow a{\left(b-c\right)}^2+b{\left(a-c\right)}^2+c{\left(b-a\right)}^2=0 \\ \Rightarrow {\left(b-c\right)}^2=0 \mathrm{and} {\left(a-c\right)}^2=0 \mathrm{and} {\left(b-a\right)}^2=0 \\ \Rightarrow \left(b-c\right)=0 \mathrm{and} \left(a-c\right)=0 \mathrm{and} \left(b-a\right)=0 \\ \Rightarrow b=c \mathrm{and} a=c \mathrm{and} b=a \\ \Rightarrow a=b=c \end{gathered}$$


Hence, the triangle must be equilateral.

Answer:

Given: The perimeter of the triangle is 36 cm.

Let the sides of the triangle be a, b and c.

According to the question,
b = a − 4 and c = ​a − 5


Perimeter of the triangle = a + b + c
$$\begin{gathered} \Rightarrow 36=a+a-4+a-5 \\ \Rightarrow 36=3a-9 \\ \Rightarrow 3a=36+9 \\ \Rightarrow 3a=45 \\ \Rightarrow a=15 \mathrm{cm} \\ \Rightarrow b=a-4=15-4=11 \mathrm{cm} \\ \Rightarrow c=a-5=15-5=10 \mathrm{cm} \\ \mathrm{Therefore}, \mathrm{the} \mathrm{sides} \mathrm{are} 15 \mathrm{cm}, 11 \mathrm{cm} \mathrm{and} 10 \mathrm{cm}. \end{gathered}$$

Now, using Heron's formula:
If a, b and c are three sides of a triangle, then
area of the triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ , where $s=\frac{a+b+c}{2}$.


Here, $s=\frac{15+11+10}{2}=\frac{36}{2}=18$

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle}=\sqrt{18\left(18-15\right)\left(18-11\right)\left(18-10\right)} \\ =\sqrt{18\left(3\right)\left(7\right)\left(8\right)} \\ =\sqrt{\left(3\times 3\times 2\right)\left(3\right)\left(7\right)\left(2\times 2\times 2\right)} \\ =\left(3\times 2\times 2\right)\sqrt{21} \\ =12\sqrt{21} {\mathrm{cm}}^2 \end{gathered}$$


Hence, the area of the triangle is $\underline{12\sqrt{21} {\mathrm{cm}}^2}$.

Answer:

We know, when the triangles have same perimeter, then the equilateral triangle have the greatest area.

Hence, among an equilateral triangle, an isosceles triangle and a scalene triangle, an equilateral triangle has the maximum area if the perimeter of each triangle is same.

Answer:

Given: 
Base  of isosceles triangle is 6 units.
equal sides of isosceles triangle is 5 units.


Now, using Heron's formula:
If a, b and c are three sides of a triangle, then
area of the triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ , where $s=\frac{a+b+c}{2}$.


Here, $s=\frac{6+5+5}{2}=\frac{16}{2}=8$

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle}=\sqrt{8\left(8-5\right)\left(8-5\right)\left(8-6\right)} \\ =\sqrt{8\left(3\right)\left(3\right)\left(2\right)} \\ =\sqrt{\left(2\times 2\times 2\right)\left(3\right)\left(3\right)\left(2\right)} \\ =\left(3\times 2\times 2\right) \\ =12 {\mathrm{cm}}^2 \end{gathered}$$


Hence, the area of the triangle is $\underline{12 {\mathrm{cm}}^2}$.

Answer:

Given: 
The sides of a triangle are 11 cm, 12 cm and 13 cm respectively.

Let ABC be a triangle with sides
AB = 11 cm
BC = 12 cm
AC = 13 cm

Let AD be the altitude corresponding to the side having length 12 cm.

Using Heron's formula:
If a, b and c are three sides of a triangle, then
area of the triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ , where $s=\frac{a+b+c}{2}$.


Here, $s=\frac{11+12+13}{2}=\frac{36}{2}=18$

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle}=\sqrt{18\left(18-11\right)\left(18-12\right)\left(18-13\right)} \\ =\sqrt{18\left(7\right)\left(6\right)\left(5\right)} \\ =\sqrt{\left(2\times 3\times 3\right)\left(7\right)\left(3\times 2\right)\left(5\right)} \\ =\left(2\times 3\right)\sqrt{105} \\ =6\sqrt{105} {\mathrm{cm}}^2 \end{gathered}$$


We know,
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle}=\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height} \\ \Rightarrow 6\sqrt{105}=\frac{1}{2}\times 12\times AD \\ \Rightarrow \sqrt{105}=AD \\ \Rightarrow AD=\sqrt{105} \mathrm{cm} \end{gathered}$$

Hence, the length of the altitude corresponding to the side having length 12 cm is $\underline{\sqrt{105} \mathrm{cm}}$.

Answer:

Given: 
A ground is in the form of a triangle having sides 51 m, 37 m and 20 m. 
The cost of levelling the ground per m² is ₹ 3.


Using Heron's formula:
If a, b and c are three sides of a triangle, then
area of the triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ , where $s=\frac{a+b+c}{2}$.


Here, $s=\frac{51+37+20}{2}=\frac{108}{2}=54$

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle}=\sqrt{54\left(54-51\right)\left(54-37\right)\left(54-20\right)} \\ =\sqrt{54\left(3\right)\left(17\right)\left(34\right)} \\ =\sqrt{\left(2\times 3\times 3\times 3\right)\left(3\right)\left(17\right)\left(2\times 17\right)} \\ =\left(3\times 3\times 2\times 17\right) \\ =306 {\mathrm{m}}^2 \end{gathered}$$

The cost of painting per m² = ₹ 3
The cost of painting 306 m² = ₹ 306 × 3
                                             = ₹ 918

Hence, the cost of levelling the ground at the rate of ₹ 3 per m² is ₹ 918.