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.

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

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

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

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

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

Correct Answer: 4 cm

Area of equilateral triangle = \( \Large \frac{\sqrt{3}}{4}a^{2} \)

=> \( \Large 4\sqrt{3}=\frac{\sqrt{3}}{4}a^{2}\) => \( \Large a^{2} \)=16

a=\( \Large \sqrt{16} \)=4cm

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

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

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

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

 

Correct Answer: 3 cm

According to the question,

\( \Large \frac{\sqrt{3}}{4}a^{2}=\frac{\sqrt{243}}{4} \)

=> \( \Large a^{2}=\frac{\sqrt{81\times 3}}{\sqrt{3}}=\frac{9\sqrt{3}}{\sqrt{3}} \)

a=\( \Large \sqrt{9} \)=3cm