C-16-Q020mediumsingle_mcq

The area of a triangle with sides $25, 27, 32$ cm (semi-perimeter $42$ ) equals —

a $\sqrt{42\cdot 17\cdot 15\cdot 10}$ sq cm
b $\sqrt{42\cdot 25\cdot 27\cdot 32}$ sq cm
c $42\cdot 17\cdot 15\cdot 10$ sq cm
d $\dfrac{1}{2}\cdot 25\cdot 27$ sq cm

ব্যাখ্যা

With $s = 42$ , Heron's formula gives differences $s-a = 42-25 = 17$ , $s-b = 42-27 = 15$ , and $s-c = 42-32 = 10$ . So the area is $\sqrt{42 \cdot 17 \cdot 15 \cdot 10}$ sq cm.

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The area of a triangle with sides $25, 27, 32$ cm (semi-perimeter $42$ ) equals —

\sqrt{42\cdot 17\cdot 15\cdot 10}

sq cm

\sqrt{42\cdot 25\cdot 27\cdot 32}

sq cm

42\cdot 17\cdot 15\cdot 10

sq cm

\dfrac{1}{2}\cdot 25\cdot 27

sq cm

ব্যাখ্যা

With $s = 42$ , Heron's formula gives differences $s-a = 42-25 = 17$ , $s-b = 42-27 = 15$ , and $s-c = 42-32 = 10$ . So the area is $\sqrt{42 \cdot 17 \cdot 15 \cdot 10}$ sq cm.

The area of a triangle with sides 25,27,3225, 27, 3225,27,32 cm (semi-perimeter 424242) equals —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

The area of a triangle with sides 25,27,3225, 27, 3225,27,32 cm (semi-perimeter 424242) equals —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

The area of a triangle with sides $25, 27, 32$ cm (semi-perimeter $42$ ) equals —

The area of a triangle with sides $25, 27, 32$ cm (semi-perimeter $42$ ) equals —