Herons Formula Test

Result

Herons Formula Test - 29

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

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