Question

EXERCISE 16.2
1. Find the perimeter and area of an isosceles triangle ABC, with base BC = 6 cm and
side AB = side AC = 5 cm. Area prob.​

Answer 1

Answer:

perimeter =16 cm

area =12 cm^2

Answer 2

✧ Given :-

• The sides is an Isoceles ∆ i.e., AB = BC = 5cm. and BC (base) = 6cm.

✧ To find :-

• The perimeter and area of the Isoceles ∆.

✧ Solution :-

$\sf {\underline{✯ Perimeter \: of an \: Isoceles \: ∆ :- (2a + b).}}$

• $\bf \: Here \: 2a \: refers \: to \: the \: equal \: sides \: of \: the \: ∆ \\ \sf i.e., \: (2 \times 5 + 6)</li><li> \\ \sf \: = 16cm$
• $\sf \:{\boxed{\boxed{ \sf perimeter \: of \: the \: \triangle \: is \: 16cm.}}}$

$\sf {\underline{✯ Area\: of \:the \: Isoceles \:∆\: :- .}}$

• $\sf{Although \: height \: is \: not \: given \: we \: \: can }$$\sf{ \: find \: it \: by \: the \: { \blue{ \underline{HERON'S \: formula}}}}$.

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

• Here, s refresh to the semi-circle.
• To find s (semi-circle) the formula is :-

$= \huge \bf\frac{a + b + c}{2}$

$\sf \implies \: { \: \frac{5 + 5 + 6}{2} } = \frac{16}{2} = 8cm.$

✯ Area of the Isoceles ∆ :-

(using HERON'S formula)

$= \bf \sqrt{8( 8- 6)( 8- 5)(8 - 5)} \\ \bf \: = \sqrt{8 \times 2 \times 3 \times 3} \\ \bf \: = \sqrt{144} = 12cm^2$

$\rm{\boxed{\boxed{\mathfrak{\blue✧ <u>\</u>purple A \pink n \red s \orange w \green \purple \blue e \red r \pink s :- }}}}$

• $\rm\bold{ <strong> </strong><strong> </strong><strong>Area</strong><strong> </strong><strong>=</strong><strong> </strong><strong>1</strong><strong>2</strong><strong>c</strong><strong>m</strong><strong>^</strong><strong>2</strong><strong> </strong>}$
• Perimeter = 16cm.