myhelper: RD Sharma 2020 solution class 9 chapter 17 Heron's Formula FBQS

RD Sharma 2020 solution class 9 chapter 17 Heron's Formula FBQS

FBQS

Page-17.26

Question 1:

The side and altitude of an equilateral triangle are in the ratio __________.

Answer 1:

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,

$A B^{2} = B D^{2} + A D^{2} \Rightarrow a^{2} = {(\frac{a}{2})}^{2} + A D^{2} \Rightarrow A D^{2} = a^{2} - \frac{a^{2}}{4} \Rightarrow A D^{2} = \frac{4 a^{2} - a^{2}}{4} \Rightarrow A D^{2} = \frac{3 a^{2}}{4} \Rightarrow A D = \frac{\sqrt{3}}{2} a . . . (1)$

Now,

$\frac{Side}{Altitude} = \frac{A B}{A D} = \frac{a}{\frac{\sqrt{3}}{2} a} = \frac{2}{\sqrt{3}}$

Hence, the side and altitude of an equilateral triangle are in the ratio

$2 : \sqrt{3}$ .

Question 2:

If the area of an isosceles right-angled triangle is 72 cm², then its perimeter is _________.

Answer 2:

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

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

Hence, its perimeter is

$12 (2 + \sqrt{2}) cm .$

Question 3:

The height of an equilateral triangle is

$\sqrt{3}$ a units. Then its area is _________.

Answer 3:

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,

$A B^{2} = B D^{2} + A D^{2} \Rightarrow x^{2} = {(\frac{x}{2})}^{2} + {(\sqrt{3} a)}^{2} \Rightarrow 3 a^{2} = x^{2} - \frac{x^{2}}{4} \Rightarrow 3 a^{2} = \frac{4 x^{2} - x^{2}}{4} \Rightarrow 3 a^{2} = \frac{3 x^{2}}{4} \Rightarrow a^{2} = \frac{x^{2}}{4} \Rightarrow a = \frac{x}{2} \Rightarrow x = 2 a . . . (1)$

Now,

$Area of triangle = \frac{\sqrt{3}}{4} \times {(side)}^{2} = \frac{\sqrt{3}}{4} \times {(2 a)}^{2} = \frac{\sqrt{3}}{4} \times 4 a^{2} = \sqrt{3} a^{2}$

Hence, its area is

$\sqrt{3} a^{2} sq . units$ .

Question 4:

The area of a triangle with base 4 cm and height 6 cm is __________.

Answer 4:

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

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

Hence, the area is

$12 {cm}^{2}$ .

Question 5:

ΔABC is an isosceles right triangle right-angled at A. If AB = 4 cm, then its area is _________.

Answer 5:

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,

$Area of triangle = \frac{1}{2} \times base \times height = \frac{1}{2} \times A B \times A C = \frac{1}{2} \times 4 \times 4 = 8 {cm}^{2}$

Hence, its area is

$8 {cm}^{2}$ .

Question 6:

If the side of a rhombus is 10 cm and one diagonal is 16 cm, then its area is _________.

Answer 6:

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)

$In ∆ A O B, Using pythagoras theorem, A B^{2} = A O^{2} + B O^{2} \Rightarrow 10^{2} = 8^{2} + B O^{2} \Rightarrow 100 = 64 + B O^{2} \Rightarrow B O^{2} = 100 - 64 \Rightarrow B O^{2} = 36 \Rightarrow B O = 6 Thus, B D = 2 \times B O = 2 \times 6 = 12 cm . . . (4)$

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

Hence, the area is

$96 {cm}^{2}$ .

Question 7:

The area of a regular hexagon of side 6 cm is __________.

Answer 7:

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

We know, a hexagon is formed by joining 6 equilateral triangles together.
Therefore,

$Area of hexagon = Area of 6 equilateral triangles = 6 \times \frac{\sqrt{3}}{4} \times x^{2} = 6 \times \frac{\sqrt{3}}{4} \times {(6)}^{2} = 6 \times \frac{\sqrt{3}}{4} \times 36 = 6 \times 9 \times \sqrt{3} = 54 \sqrt{3} {cm}^{2}$

Hence, the area of a regular hexagon of side 6 cm is

$54 \sqrt{3} {cm}^{2}$ .

Question 8:

The base of a right triangle is 8 cm and hypotenuse is 10 cm. Then its area is _________.

Answer 8:

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

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

$Area of triangle = \frac{1}{2} \times (base) \times (height) = \frac{1}{2} \times 8 \times 6 = 4 \times 6 = 24 {cm}^{2}$

Hence, the area is

$24 {cm}^{2}$ .

Question 9:

Each side of a triangle is multiplied by with the sum of the squares of the other two sides. The sum of all such possible results is 6 times the product of sides. The triangle must be _________.

Answer 9:

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

According to the question,

$a (b^{2} + c^{2}) + b (a^{2} + c^{2}) + c (b^{2} + a^{2}) = 6 a b c \Rightarrow a b^{2} + a c^{2} + b a^{2} + b c^{2} + c b^{2} + c a^{2} = 6 a b c \Rightarrow a b^{2} + a c^{2} + b a^{2} + b c^{2} + c b^{2} + c a^{2} - 6 a b c = 0 \Rightarrow a b^{2} + a c^{2} - 2 a b c + b a^{2} + b c^{2} - 2 a b c + c b^{2} + c a^{2} - 2 a b c = 0 \Rightarrow a (b^{2} + c^{2} - 2 b c) + b (a^{2} + c^{2} - 2 a c) + c (b^{2} + a^{2} - 2 a b) = 0 \Rightarrow a {(b - c)}^{2} + b {(a - c)}^{2} + c {(b - a)}^{2} = 0 \Rightarrow {(b - c)}^{2} = 0 and {(a - c)}^{2} = 0 and {(b - a)}^{2} = 0 \Rightarrow (b - c) = 0 and (a - c) = 0 and (b - a) = 0 \Rightarrow b = c and a = c and b = a \Rightarrow a = b = c$

Hence, the triangle must be equilateral.

Question 10:

In a scalene triangle, one side exceeds the other two sides by 4 cm and 5 cm respectively and the perimeter of the triangle is 36 cm. The area of the triangle is _________.

Answer 10:

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

$\Rightarrow 36 = a + a - 4 + a - 5 \Rightarrow 36 = 3 a - 9 \Rightarrow 3 a = 36 + 9 \Rightarrow 3 a = 45 \Rightarrow a = 15 cm \Rightarrow b = a - 4 = 15 - 4 = 11 cm \Rightarrow c = a - 5 = 15 - 5 = 10 cm Therefore, the sides are 15 cm, 11 cm and 10 cm .$

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

$\sqrt{s (s - a) (s - b) (s - c)}$ , where

$s = \frac{a + b + c}{2}$ .

Here,

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

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

Hence, the area of the triangle is

$12 \sqrt{21} {cm}^{2}$ .

Question 11:

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

Answer 11:

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.

Question 12:

Area of an isosceles triangle, one of whose equal side is 5 units and base 6 units is __________.

Answer 12:

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 =

$\sqrt{s (s - a) (s - b) (s - c)}$ , where

$s = \frac{a + b + c}{2}$ .

Here,

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

$Area of triangle = \sqrt{8 (8 - 5) (8 - 5) (8 - 6)} = \sqrt{8 (3) (3) (2)} = \sqrt{(2 \times 2 \times 2) (3) (3) (2)} = (3 \times 2 \times 2) = 12 {cm}^{2}$

Hence, the area of the triangle is

$12 {cm}^{2}$ .

Question 13:

The sides of a scalene triangle are 11 cm, 12 cm and 13 cm. The length of the altitude corresponding to the side having length 12 cm is ________.

Answer 13:

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 =

$\sqrt{s (s - a) (s - b) (s - c)}$ , where

$s = \frac{a + b + c}{2}$ .

Here,

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

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

We know,

$Area of triangle = \frac{1}{2} \times Base \times Height \Rightarrow 6 \sqrt{105} = \frac{1}{2} \times 12 \times A D \Rightarrow \sqrt{105} = A D \Rightarrow A D = \sqrt{105} cm$

Hence, the length of the altitude corresponding to the side having length 12 cm is

$\sqrt{105} cm$ .

Question 14:

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

Answer 14:

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 =

$\sqrt{s (s - a) (s - b) (s - c)}$ , where

$s = \frac{a + b + c}{2}$ .

Here,

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

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

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.

Question 15:

If each side of a triangle is doubled, then its area is __________ times the area of the original triangle.

Answer 15:

Let the base of the original triangle be a units and height be b units.

Then, the area of original triangle is

$\frac{1}{2} a b$ ...(1)

According to the question,
Each side of a triangle is doubled

Thus, base of the new triangle is 2a units and height is 2b units.

Area of new triangle =

$\frac{1}{2} \times 2 a \times 2 b$
=

$\frac{1}{2} (4 a b)$
=

$4 \times \frac{1}{2} a b$
= 4 × area of original triangle (from (1))

Hence, if each side of a triangle is doubled, then its area is 4 times the area of the original triangle.

Question 16:

In a triangle, the sum of any two sides exceeds the third by 6 cm. The area of the triangle is __________.

Answer 16:

Given:
In a triangle, the sum of any two sides exceeds the third by 6 cm.

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

According to the question,
$a + b - c = 6 . . . (1) b + c - a = 6 . . . (2) a + c - b = 6 . . . (3) Adding (1), (2) and (3), we get 2 a + 2 b + 2 c - a - b - c = 18 \Rightarrow a + b + c = 18 . . . (4) Subtracting (1) from (4), we get a + b + c - a - b + c = 18 - 6 \Rightarrow 2 c = 12 \Rightarrow c = 6 . . . (5) Subtracting (2) from (4), we get a + b + c - b - c + a = 18 - 6 \Rightarrow 2 a = 12 \Rightarrow a = 6 . . . (6) Subtracting (3) from (4), we get a + b + c - a - c + b = 18 - 6 \Rightarrow 2 b = 12 \Rightarrow b = 6 . . . (7)$

From (5), (6) and (7),
a = b = c = 6 cm
⇒ triangle is equilateral

$Area = \frac{\sqrt{3}}{4} \times {(side)}^{2} = \frac{\sqrt{3}}{4} \times {(6)}^{2} = \frac{\sqrt{3}}{4} \times 36 = 9 \sqrt{3} {cm}^{2}$

Hence, the area of the triangle is $9 \sqrt{3} {cm}^{2} .$

Question 17:

If the circumradius of a right triangle is 10 cm and one of the two perpendicular sides is 12 cm, then the area of the triangle is ________.

Answer 17:

Given:
The circumradius of a right triangle is 10 cm
One of the two perpendicular sides is 12 cm

We know, the circumradius of the right-angled triangle is half of its hypotenuse.

Thus,
$Hypotenuse = 2 \times circumradius = 2 \times 10 = 20 cm$

Using Pythagoras theorem,
${(Hypotenuse)}^{2} = {(Base)}^{2} + {(Perpendicular)}^{2} \Rightarrow {(20)}^{2} = {(Base)}^{2} + {(12)}^{2} \Rightarrow 400 = {(Base)}^{2} + 144 \Rightarrow {(Base)}^{2} = 400 - 144 \Rightarrow {(Base)}^{2} = 256 \Rightarrow Base = 16 cm Area of right triangle = \frac{1}{2} \times Base \times Height = \frac{1}{2} \times 16 \times 12 = 8 \times 12 = 96 {cm}^{2}$

Hence, the area of the triangle is $96 {cm}^{2} .$

RD Sharma 2020 solution class 9 chapter 17 Heron's Formula FBQS

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