Herons Formula Test

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Herons Formula Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

If the sides of triangle are 4, 5 and 6 cm. Then the area (in sq cm) of triangle is

A
$\dfrac{\pi}{4}$

B
$\dfrac{\pi}{4} \sqrt{7}$

C
$\dfrac{4}{15}$

D
$\dfrac{4}{15} \sqrt{7}$

E
$\dfrac{15}{4} \sqrt{7}$

Solution

Given, triangle of sides $a,b,c=4,5,6cm$
$\therefore S=\dfrac{a+b+c}{2}$

$=\dfrac{4+5+6}{2}=\dfrac{15}{2}$
Then, area of triangle
$A=\sqrt{S(S-a)(S-b)(S-c)}$

$=\sqrt{\dfrac{15}{2}\left(\dfrac{15}{2}-4\right)\left(\dfrac{15}{2}-5\right)\left(\dfrac{15}{2}-6\right)}$

$=\sqrt{\dfrac{15}{2}.\left(\dfrac{7}{2}\right)\left(\dfrac{5}{2}\right)\left(\dfrac{3}{2}\right)}$
$=\dfrac{15}{4}\sqrt{7}$
Question 2
1 / -0

Find the area of equilateral triangle each of whose sides measures 20 cm.

A
173 ${cm^2}$

B
150 ${cm^2}$

C
120 ${cm^2}$

D
153 ${cm^2}$

Solution

Area of Equilateral Triangle = $\dfrac{\sqrt{3}{b^2}}{4}$
where b is the Side of the Triangle.
Area= $\dfrac{\sqrt{3}\times {20^2}}{4}$
Area= $173.2{cm^2}$
Question 3
1 / -0

Find the area of equilateral triangle (in ${cm^2}$ )each of whose sides measure 18 cm.

A
150

B
140

C
180

D
190

Solution

Area of Equilateral Triangle = $\dfrac{\sqrt{3}{a^2}}{4}$
where $a$ is the Side of the Triangle.
Given $a= 18\ cm$

Area $=\dfrac{\sqrt{3}\times {18^2}}{4}$

Area $=140.29\ {cm^2}$
Question 4
1 / -0

The area of an equilateral triangle is $4 \sqrt{3} \,cm^2.$ The length of each of its sides is

A
$3 \,cm$

B
$4 \,cm$

C
$2 \sqrt{3} \,cm$

D
$\dfrac{1}{2} \sqrt{3} \,cm$

Solution

From the question is given that,
Area of an equilateral triangle is $(4 \sqrt{3})$
Area of the equilateral triangle $= ((\sqrt{3}) / 4) \times (sides)^2$ sq. units
$Area= ((\sqrt{3} / 4) \times (sides)^2$
$= (sides)^2 = (4 \sqrt{3}) / (4 / (\sqrt{3})$
$= (sides)^2 = (4 \sqrt{3} \times 4) / \sqrt{3}$
$= (sides)^2 = 4 \times 4$
$= Sides = \sqrt{} 16$
$= Sides = 4 \,cm$
Hence, the length of each side of the triangle is $4 \,cm$
Question 5
1 / -0

The sides of a triangle measures $13\,cm , 14\,cm$ and $15\,cm.$ Its area is

A
$84 \,cm^2$

B
$91\,cm^2$

C
$168 \,cm^2$

D
$182 \,cm^2$

Solution

Let $a, b, c$ be the sides of the triangle

Then, $s =\dfrac{a + b + c}{2}$

$= \dfrac{13 + 14 + 15}{2}$

$= \dfrac{42}{2}$

$= 21$

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

$= \sqrt{(21 \times (21 - 13) \times (21 - 14) \times (21 - 15)}$

$= \sqrt{(21 \times (8) \times (7) \times (6)}$

$= \sqrt{(3 \times 7 \times 2 \times 2 \times 2 \times 7 \times 2 \times 3)}$

$= \sqrt{(3^2 \times 7^2 \times 2^2 \times 2^2)}$
By cancelling square and square root we get,
$= (3 \times 7 \times 2 \times 2)$
$= 84\ cm^2$
Question 6
1 / -0

Each side of an equilateral triangle is $8 \,cm$ long. Its area is:

A
$32 \,cm^2$

B
$64 \,cm^2$

C
$16 \sqrt{3} \,cm^2$

D
$16 \sqrt{2} \,cm^2$

Solution

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

$= \dfrac{\sqrt 3} {4} \times (8)^2$

$= \dfrac{\sqrt 3}{ 4} \times 64$

$= \sqrt 3 \times 16$

$= 16 \sqrt 3 \,cm^2$
Question 7
1 / -0

Find the area of $\triangle ABD$

A
$12 \ cm^2$

B
$24 \ cm^2$

C
$28 \ cm^2$

D
none of these

Solution
Question 8
1 / -0

In an isosceles triangle $PQR$ , the altitude $PS = 6$ $cm$ and $PQ = PR = 10$ $cm$ . Find the area of $\triangle PSR$ .

A
$20$ $cm^2$

B
$22$ $cm^2$

C
$24$ $cm^2$

D
None of these

Solution
Question 9
1 / -0

If $AD = 4$ $cm$ and $AB = 5$ $cm$ in the given figure, then find the area of $\triangle ADB$ .

A
$4$ $cm^2$

B
$6$ $cm^2$

C
$8$ $cm^2$

D
$10$ $cm^2$

Solution
Question 10
1 / -0

In the given figure, if $PL = 2 QL$ , then what will be the area of $\triangle PLQ$ ?

A
$28.8$ square units

B
$26.8$ square units

C
$28.3$ square units

D
$26.3$ square units

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Herons Formula Test - 29

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Herons Formula Test - 29

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Question 1

π4\dfrac{\pi}{4}4π​

π47\dfrac{\pi}{4} \sqrt{7}4π​7​

415\dfrac{4}{15}154​

4157\dfrac{4}{15} \sqrt{7}154​7​

1547\dfrac{15}{4} \sqrt{7}415​7​

Solution

Question 2

173cm2{cm^2}cm2

150cm2{cm^2}cm2

120cm2{cm^2}cm2

153cm2{cm^2}cm2

Solution

Question 3

150

140

180

190

Solution

Question 4

3 cm3 \,cm3cm

4 cm4 \,cm4cm

23 cm2 \sqrt{3} \,cm23​cm

123 cm\dfrac{1}{2} \sqrt{3} \,cm21​3​cm

Solution

Question 5

84 cm284 \,cm^284cm2

91 cm291\,cm^291cm2

168 cm2168 \,cm^2168cm2

182 cm2182 \,cm^2182cm2

Solution

Question 6

32 cm232 \,cm^232cm2

64 cm264 \,cm^264cm2

163 cm216 \sqrt{3} \,cm^2163​cm2

162 cm216 \sqrt{2} \,cm^2162​cm2

Solution

Question 7

12 cm212 \ cm^212 cm2

24 cm224 \ cm^224 cm2

28 cm228 \ cm^228 cm2

none of these

Solution

Question 8

202020 cm2cm^2cm2

222222 cm2cm^2cm2

242424 cm2cm^2cm2

None of these

Solution

Question 9

444 cm2cm^2cm2

666 cm2cm^2cm2

888 cm2cm^2cm2

101010 cm2cm^2cm2

Solution

Question 10

28.828.828.8 square units

26.826.826.8 square units

28.328.328.3 square units

26.326.326.3 square units

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$\dfrac{\pi}{4}$

$\dfrac{\pi}{4} \sqrt{7}$

$\dfrac{4}{15}$

$\dfrac{4}{15} \sqrt{7}$

$\dfrac{15}{4} \sqrt{7}$

173 ${cm^2}$

150 ${cm^2}$

120 ${cm^2}$

153 ${cm^2}$

$3 \,cm$

$4 \,cm$

$2 \sqrt{3} \,cm$

$\dfrac{1}{2} \sqrt{3} \,cm$

$84 \,cm^2$

$91\,cm^2$

$168 \,cm^2$

$182 \,cm^2$

$32 \,cm^2$

$64 \,cm^2$

$16 \sqrt{3} \,cm^2$

$16 \sqrt{2} \,cm^2$

$12 \ cm^2$

$24 \ cm^2$

$28 \ cm^2$

$20$ $cm^2$

$22$ $cm^2$

$24$ $cm^2$

$4$ $cm^2$

$6$ $cm^2$

$8$ $cm^2$

$10$ $cm^2$

$28.8$ square units

$26.8$ square units

$28.3$ square units

$26.3$ square units