Question

Type-1 Very Short Answer Type Questions
1. The sides of a triangle are 7,24 and 25 cm. Find its area is (in cm).​

Answer 1

Answer:

$\large \bold\red{Area = 84 \: {cm}^{2} }$

Step-by-step explanation:

Given,

Sides of a triangle are,

• a = 25 cm
• a = 25 cmb = 24 cm
• a = 25 cmb = 24 cmc = 7 cm

Now,

Semi Perimeter,

• $s = \frac{25 + 24 + 7}{2} = \frac{56}{2} = 28 \: cm$

Now,

We know that,

Area of triangle is given by,

• $\large \boxed{ \bold\purple{\sqrt{ s(s - a)(s - b)(s - c)} }}$

Putting the respective values,

We get,

$= > A = \sqrt{28(28 - 25)(28 - 24)(28 - 7)} \\ \\ = > A = \sqrt{28 \times 3 \times 4 \times 21} \\ \\ = > A = \sqrt{7 \times 4 \times 3 \times 4 \times 7 \times 3} \\ \\ = > A = \sqrt{ {7}^{2} \times {4}^{2} \times {3}^{2} } \\ \\ = > A = \sqrt{ {(7 \times 4 \times 3)}^{2} } \\ \\ = > A = 7 \times 4 \times 3 \\ \\ = > \large \boxed{\large \bold{A = 84 \: {cm}^{2} }}$

Answer 2

$\LARGE\underline{\underline{\sf \red{T}\blue{o}\:\green{F}\orange{i}\pink{n}\red{d}:}}$ Area of ∆ .

$\LARGE\underline{\underline{\sf \red{G}\blue{i}\green{v}\orange{e}\red{n}:}}$

• sides are 7 , 24 and 25 cm .

Lets Try to solve it by 2-3 methods...

______________________________

$\pink{\bold{\underline{\underline{Solution(1)}}}}$

Given its sides are 7, 24, and 25 cm .

so, Semiperimeter of ∆ = perimeter /2

S = 7+24+25/2 = 28cm

Now,

By heron' s Formula we know that area of ∆ =

$\large\red{\boxed{\sf \sqrt{s(s - a)(s - b)(s - c)}}}$

Here s = semi-perimeter of ∆ , and a,b and C are its sides.

Putting values now we get,

$\sqrt{28 \times (28 - 7)(28 - 24)(28 - 25)} \\ \\ \sqrt{28 \times 21 \times 4 \times 3} \\ \\ \sqrt{7 \times 4 \times 7 \times 3 \times 4 \times 3} \\ \\ = 7 \times 4 \times 3 \\ \\ = 84 \: cm ^{2}$

______________________________

$\red{\bold{\underline{\underline{Solution(2)}}}}$

Formula to be used :----

• Pythagoras Theoram = H²= P²+B² or sum of square two sides is equal to sum of square of third side .
• area of right angled ∆ = 1/2 × perpendicular Distance × Base of ∆

If we Prove 7,24 and 25 are Triplets of ∆, than it will be a right ∆. and we can easily find its area..

let check ,

(7)² + (24)² = (25)²

49 + 576 = 625

625 = 625 (PRoved)

So ,

it is a Right angled ∆ with prerpendicular and base as 7 or 24 , and its Hypotenuse is 25 .

So, Area of Right angled ∆ will be =

$\large\red{\boxed{\sf </strong><strong>\frac{1}{2} \times 24 \times 7 = 84 \: cm ^{2}</strong><strong>}}$

______________________________

$\huge\underline\mathfrak\green{Hope\:it\:Helps\:You}$