NCERT Exemplar Class 9 Maths Solutions Chapter 12 Herons Formula

Question:1 An isosceles right triangle has area $8c{m}^{{}^2}$. The length of its hypotenuse is
(A) $\sqrt{32}cm$
(B) $\sqrt{16}cm$
(C) $\sqrt{48}cm$
(D) $\sqrt{24}cm$

Answer:

Question:2 The perimeter of an equilateral triangle is 60 m. The area is
(A) $10\sqrt{3}{m}^2$
(B) $15\sqrt{3}{m}^2$
(C) $20\sqrt{3}{m}^2$
(D) $100\sqrt{3}{m}^2$

Answer:

Question:3 The sides of a triangle are 56 cm, 60 cm and 52 cm long. Then the area of the triangle is
(A) 1322 cm²
(B) 1311 cm²
(C) 1344 cm²
(D) 1392 cm²

Answer:

Question:4 The area of an equilateral triangle with side 2$\sqrt{3}$ cm is
(A) 5.196 cm²
(B) 0.866 cm²
(C) 3.496 cm²
(D) 1.732 cm²

Answer:

Question:5 The length of each side of an equilateral triangle having an area of $9\sqrt{3}c{m}^2$ is
(A) 8 cm
(B) 36 cm
(C) 4 cm
(D) 6 cm

Answer:

Question:6 If the area of an equilateral triangle is $16\sqrt{3}c{m}^2$, then the perimeter of the triangle is:
(A) 48 cm
(B) 24 cm
(C) 12 cm
(D) 36 cm

Answer:

Question:7 The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. The length of its longest altitude
(A) $16\sqrt{5}cm$
(B) $10\sqrt{5}cm$
(C) $24\sqrt{5}cm$
(D) 28 $cm$

Answer:

Question:8 The area of an isosceles triangle having a base 2of cm and the length of one of the equal sides 4 cm, is
(A) $\sqrt{15}c{m}^2$
(B) $\sqrt{\frac{15}{2}}c{m}^2$
(C) $2\sqrt{15}c{m}^2$
(D) $4\sqrt{15}c{m}^2$

Answer:

^{$Weknowthat,$$Weknowthat,semi-perimeter$}

^{$S=\frac{a+b+c}{2}$}

^{$S=\frac{2+4+4}{2}$}

^{$S=\frac{10}{2}=2cm$}

^{Using Heron’s formula area of $\mathrm{\Delta }$ABC $=\sqrt{S(S-a)(S-b)(S-c)}$}

^{$=\sqrt{5(5-2)(5-4)(5-4)}$}

^{$=\sqrt{5\times 3\times 1\times 1}$}

^{$=\sqrt{15}c{m}^2$}

^{Hence, the area of the given triangle is $\sqrt{15}c{m}^2$.}

^{Hence option (A) is correct.}

Question:9 The edges of a triangular board are 6 cm, 8 cm and 10 cm. The cost of painting it at the rate of 9 paise per $c{m}^2$ is
(A) Rs 2.00
(B) Rs 2.16
(C) Rs 2.48
(D) Rs 3.00

Answer:

To find the cost of painting, we have to find the area of the triangular board
Let the sides be denoted as, a = 6 cm, b = 8 cm, c = 10 cm
$semi-perimeter,S=\frac{a+b+c}{2}$
$S=\frac{6+8+10}{2}=\frac{24}{2}=12cm$
We know that,
Using Heron’s formula, area of triangle $=\sqrt{S(S-a)(S-b)(S-c)}$
$=\sqrt{12(12-6)(12-8)(12-10)}$
$=\sqrt{12\times 6\times 4\times 2}$
$=\sqrt{6\times 2\times 6\times 2\times 2\times 2}$
$6\times 2\times 2=24c{m}^2$
$Nowcostofpainting1c{m}^2=9paise=\frac{9}{100}Rs=0.09Rs.$
$[1Rs=100paise,1paise=\frac{1}{100}Rs]$
$\therefore costofpainting24c{m}^2=24\times 0.09=24\times \frac{9}{100}=Rs2.16$
Hence the cost of painting is Rs. 2.16.
Hence option (B) is correct.

Question:1 Write True or False and justify your answer:
The area of a triangle with a base of 4 cm and height of 6 cm is 24 $c{m}^2$

Answer:

Question:2 Write True or False and justify your answer:
The area of $\mathrm{\Delta }$ABC is 8 $c{m}^{{}^2}$ in which AB = AC = 4 cm and $\mathrm{\angle }$A = ${90}^{\circ }$

Answer:

Question:3 Write True or False and justify your answer:
The area of the isosceles triangle is $\frac{5}{4}\sqrt{11}c{m}^2$ if the perimeter is 11 cm and the base is 5 cm.

Answer:

Question:4 Write True or False and justify your answer:
The area of the equilateral triangle is$20\sqrt{3}c{m}^2$ whose each side is 8 cm.

Answer:

Question:5 Write True or False and justify your answer:
If the side of a rhombus is 10 cm and one diagonal is 16 cm, the area of the rhombus is 96 $c{m}^2$.

Answer:

Question:6 Write True or False and justify your answer:
The base and the corresponding altitude of a parallelogram are 10 cm and 3.5 cm, respectively. The area of the parallelogram is 30 $c{m}^2$

Answer:

Question:7 Write True or False and justify your answer:
The area of a regular hexagon of side ‘a’ is the sum of the areas of the five equilateral triangles with side a.

Answer:

Question:8 Write True or False and justify your answer:
The cost of levelling the ground in the form of a triangle having the sides 51 m, 37 m and 20 m at the rate of Rs $3per{m}^2$ is Rs 918.

Answer:

Question:9 Write True or False and justify your answer:
In a triangle, the sides are given as 11 cm, 12 cm and 13 cm. The length of the altitude is 10.25 cm corresponding to the side having length 12 cm.

Answer:

Question:1 Find the cost of laying grass in a triangular field of sides 50 m, 65 m and 65 m at the rate of Rs$7per{m}^2$

Answer:

Question:2 The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 13 m, 14 m and 15 m. The advertisements yield an earning of Rs $2000per{m}^2$ a year. A company hired one of its walls for 6 months. How much rent did it pay?

Answer:

Question:3 From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. The lengths of the perpendiculars are 14 cm, 10 cm and 6 cm. Find the area of the triangle.

Answer:

Question:4 The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3: 2. Find the area of the triangle.

Answer:

Question:5 Find the area of a parallelogram given in the figure. Also, find the length of the altitude from vertex A on the side DC.

Answer:
$LengthofthealtitudefromvertexAonthesideDC=15cm$

We know $AB=CD$ and $AD=BC[\because ABCD$ is a parallelogram $]$

$Areaofparallelogram=2\times areaof\mathrm{\Delta }DBC$
$Sohere\mathrm{\Delta }DBCsideshavingDB=25cm,BC=17cmandCD=12cm$
Using Heron’s formula
$S=\frac{25+17+12}{2}=\frac{54}{2}=27cm$
$Areaoftriangle\mathrm{\Delta }DBC=\sqrt{S(S-a)(S-b)(S-c)}$
$=\sqrt{27(27-25)(27-17)(27-12)}$
$=\sqrt{27\times 2\times 10\times 15}$
$=\sqrt{9\times 3\times 2\times 5\times 2\times 5\times 3}$
$=\sqrt{3\times 3\times 3\times 3\times 5\times 5\times 2\times 2}$
$=2\times 3\times 3\times 5=90c{m}^2$
$AreaofparallelogramABCD=2\times Ar(\mathrm{\Delta }DBC)=2\times 90c{m}^2=180c{m}^2.$
$Weknowthatareaofparallelogram=base\times height$
$=baseCD\times lengthofthealtitudefromvertexAonthesideDC$
$180c{m}^2=12cm(lengthofthealtitudefromvertexAonthesideDC)$
$LengthofthealtitudefromvertexAonthesideDC=180/12=15cm.$
Hence, the length of the altitude from vertex A on the side DC = 15 cm