Area and perimeter Questions and answers

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Elementary Mathematics

1). Find the area of a triangle whose sides measure 8 cm, 10 cm and 12 cm.

A). \( \Large 8\sqrt{63} \)sq cm

B). \( \Large 5\sqrt{63} \)sq cm

C). \( \Large 6\sqrt{53} \)sq cm

D). \( \Large 7\sqrt{93} \)sq cm

View Answer

Correct Answer: \( \Large 5\sqrt{63} \)sq cm

Given that,

a=8cm, b = 10 cm and c=12cm

We know, that

s = \( \Large \frac{a+b+c}{2} \) = \( \Large \frac{8+10+12}{2} \)

= \( \Large \frac{30}{2} \)=15cm

(s-a)=(15-8)=7cm

(s-b)=(15-10)=5cm

(s-c)=(15-12)=3cm

Area = \( \Large \sqrt{s(s-a)(s-b)(s-c)} \)

= \( \Large \sqrt{15\times 7\times 5\times 3} \)= \( \Large \sqrt{1575} \)= \( \Large \sqrt{25\times 63} \)

= \( \Large 5\sqrt{63} \)sq cm.

2). The area of a right angled triangle is 40 sq cm. If its base is equal to 28 cm. find its height

A). \( \Large 3\frac{6}{7} \)cm

B). \( \Large 4\frac{6}{7} \)cm

C). \( \Large 2\frac{6}{7} \)cm

D). \( \Large 5\frac{6}{7} \)cm

View Answer

Correct Answer: \( \Large 2\frac{6}{7} \)cm

Given that, area = 40 sq cm. base = 28 cm and height = perpendicular = ?

Area = \( \Large \frac{1}{2}\times Base\times Perpendicular \)

=> 40 = \( \Large \frac{1}{2}\times 28\times Perpendicular \)

Perpendicular =\( \Large \frac{40}{14}=\frac{20}{7} \)

=\( \Large 2\frac{6}{7} \)cm

3). The three sides of a triangle are 15, 25 and x units. Which one of the following is correct?

A). \( \Large 10 < x < 40 \)

B). \( \Large 10 \le x \le 40 \)

C). \( \Large 10 \le x < 40 \)

D). \( \Large 10 < x \le 40 \)

View Answer

Correct Answer: \( \Large 10 < x < 40 \)

In a triangle,

Sum of two sides is always greater than 3rd

side i.e., x < 25 + 15=40 ...(i)

Difference of two sides is always less than

3rd side i.e., 25 - 15 =10 < x ...(ii)

Form Eqs. (i) and (ii), we get

10 < x < 40

4). A triangle with three equal sides has its area equal to \( \Large \Large 3\sqrt{3} \)sq cm. Find its perimeter.

A). \( \Large 6\sqrt{3} \)cm

B). \( \Large 2\sqrt{3} \)cm

C). \( \Large 5\sqrt{3} \)cm

D). \( \Large 7\sqrt{3} \)cm

View Answer

Correct Answer: \( \Large 6\sqrt{3} \)cm

According to the question,

\( \Large \frac{\sqrt{3}}{4}a^{2}= 3\sqrt{3} \)

=> \( \Large \frac{1}{4}a^{2} \)=3 => \( \Large a^{2}=3\times 4 \)

a=\( \Large 2\sqrt{3} \)

Required perimeter=3a=\( \Large 3\times 2\sqrt{3} \)

=\( \Large 6\sqrt{3} \)

5). The sides of a triangle area in the ratio of \( \Large \frac{1}{3}:\frac{1}{4}:\frac{1}{5} \) and its perimeter is 94 Cm.Find the length of the smallest side of the triangle.

A). 18 cm

B). 22.5 cm

C). 24 m

D). 27 cm

View Answer

Correct Answer: 24 m

Given ratio = \( \Large \frac{1}{3}:\frac{1}{4}:\frac{1}{5} \)

Let lengths of the sides be 20x,15x and 12x.

Then, according to the question,

20x + 15x + 12x = 94 =>47x = 94

x = \( \Large \frac{94}{47} \)=2

Smallest side =12x = \( \Large 12\times 2 \) = 24 cm

6). The area of an equilateral triangle is \( \Large \Large 4\sqrt{3} \)sq cm. Find the length of each side of the triangle

A). 3 cm

B). \( \Large 2\sqrt{2} \)cm

C). \( \Large 2\sqrt{3} \)cm

D). 4 cm

View Answer

7). The sides of a right angled triangle are equal to three consecutive numbers expressed in centimeters. What can be the area of such a triangle?

A). \( \Large 6cm^{2} \)

B). \( \Large 8cm^{2} \)

C). \( \Large 10cm^{2} \)

D). \( \Large 12cm^{2} \)

View Answer

Correct Answer: \( \Large 6cm^{2} \)

Since, the triangle is right angled.

All the three consecutive sides must satisfy Pythagoras theorem 3, 4 and 5 are the sides of triangle which satisfy Pythagoras theorem.

\( \Large (5^{2}=4^{2}+3^{2}) \)

Area of triangle = \( \Large \frac{1}{2}\times 4\times 3=6cm^{2} \)

8). If the area of an equilateral triangle is x and its perimeter is y. then which one of the following is correct?

A). \( \Large y^{4}=432x^{2} \)

B). \( \Large y^{4}=216x^{2} \)

C). \( \Large y^{2}=432x^{2} \)

D). None of these

View Answer

Correct Answer: \( \Large y^{4}=432x^{2} \)

Area of equilateral triangle

\( \Large \frac{\sqrt{3} a^{2}}{4} \)=x ....(i)

and perimeter = 3a = y => a = \( \Large \frac{y}{3} \) ...(ii)

Now, putting the value of a from Eq. (ii) in Eq. (i), we get

\( \Large \frac{\sqrt{3}\left(\frac{y}{3}\right)^{2}}{4} \)=x => x=\( \Large \frac{\sqrt{3}\times y^{2}}{9\times 4} \)

=> x=\( \Large \frac{y^{2}}{3\sqrt{3}\times 4} \) => x=\( \Large \frac{y^{2}}{12\sqrt{3}} \)

=> \( \Large 12\sqrt{3}x=y^{2} \)

On squaring both sides, we get

\( \Large y^{4}=432x^{2} \)

9). The area of an isosceles \( \Large \triangle ABC \) with AB = AC and altitude AD = 3 cm is 12 sq cm. What is its perimeter?

A). 18 cm

B). 16 cm

C). 14 cm

D). 12 cm

View Answer

Correct Answer: 18 cm

Let AB = CA = a cm and base = b cm

Now, area of the \( \Large \triangle ABC \)= \( \Large \frac{1}{2}\times b\times h \)

=> 12= \( \Large \frac{1}{2}\times b\times 3 \)

b= \( \Large \frac{12\times 2}{3} \)=8cm

Here, BD=CD=\( \Large \frac{b}{2}=\frac{8}{2} \)=4cm

In right angled \( \Large \triangle ABD \) , by Pythagoras theorem,

AB=\( \Large \sqrt{BD^{2}+AD^{2}} \) => a=\( \Large \sqrt{4^{2}+3^{2}} \)

=\( \Large \sqrt{16+9}=\sqrt{25} \)=5cm

Now, perimeter of an isosceles triangle

= 2a + b = \( \Large 2\times 5+8 \)

= 10 + 8 = 18 cm

10). The area of an equilateral triangle is \( \Large \Large \frac{\sqrt{243}}{4} \)sq cm. Find the length of its side.

A). 3 cm

B). \( \Large 3\sqrt{3} \)cm

C). 9 cm

D). \( \Large \sqrt{3} \)cm

View Answer

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