Question

please answer this urgently... its choose the correct answer. The area of a triangle with given two sides 18 cm and 10 cm respectively and perimeter equal to 42 cm is: * 1 point 20√11 square cm 19√11 square cm 22√11 square cm 21√11 square cm

Answer 1

Answer:

this can be done by using heron's formula which is $\sqrt{s(s-a)(s-b)(s-c)}$

the answer is 21$\sqrt{11}$

Step-by-step explanation:

Area of the triangle,

= √[s (s-a) (s-b) (s-c)]

= √[21(21 – 18) (21 – 10) (21 – 14)] cm^2

= √[21 × 3 × 11 × 7] cm^2

= 21√11 cm^2

[ hoping it helps you :-) ]

Answer 2

$\sf\green{Given:-}$

● Perimeter of triangle is 42 cm.

● Measures of two sides of triangle are 18 and 10 cm.

$\sf\green{To \ Find:-}$

● Area of triangle

$\sf\green{Solution:-}$

Let the third side of triangle be x

⇒ Perimeter = sum of all sides

⇒ 18 + 10 + x = 42

⇒ 28 + x = 40

⇒ x = 42 - 28

⇒ x = 14 cm

⇒ s = $\frac{sum \: of \: all \: sides }{2}$

⇒ s = $\frac{42}{2}$

⇒ s = 21

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

= $\sqrt{21(21 - 18)(21 - 10)(21 - 14)}$

= $\sqrt{21(3)(11)(7)}$

= $\sqrt{3 \times 7 \times 3 \times 11 \times 7}$

= $3 \times 7\sqrt{11}$

= 21√11 cm²

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