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# [Smart Math] Geometry Problem 13

Here’s and example of a SMART MATH problem for GEOMETRY. ### Problem

The equal sides of an isosceles triangle are even integers only. What is the area of the triangle if the perimeter is 7 units?

1. 7 sq. units
2. 3 sq. units
3. 8 sq. units
4. $\frac{3\sqrt{7}}{4}$sq. units
5. $3\sqrt{7}$sq. units

### The Usual Method

Even integers are 2, 4, 6 … Now to have the perimeter as 7 i.e. a + a + b = 7, where ‘1’ is the length of equal side and is an even integer.

Hence, 2a + b = 7

Now a can take the value of 2 only.

So 2(2) + b = 7 $\therefore$b = 3

Hence dimension of the triangle = 2, 2, and 3 units.

Hence, area of this triangle using semi-perimeter formula (Area = $\sqrt{s(s-a)(s-b)(s-c)}$, where $s=\frac{a+b+c}{2}$ = semi-perimeter)
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In this case, s = $\frac{2+2+3}{2}=\frac{7}{2}$
<br /> $\therefore$Area = $\sqrt{\frac{7}{2}\left( \frac{7}{2}-2 \right)\left( \frac{7}{2}-2 \right)\left( \frac{7}{2}-3 \right)}$
<br />

= $\sqrt{\frac{7}{2}\left( \frac{3}{2} \right)\left( \frac{3}{2} \right)\left( \frac{1}{2} \right)}$
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= $\sqrt{\frac{63}{16}}=\sqrt{\frac{7}{4}\times \frac{9}{4}}=\frac{3\sqrt{7}}{4}$sq. units

(Ans: 4)

Estimated Time to arrive at the answer = 45 seconds.

### Using Technique

Once you reach to a the level of identifying the dimensions of the triangle (2, 2 and 3), put the values as $\sqrt{\frac{7}{2}\left( \frac{7}{2}-2 \right)\left( \frac{7}{2}-2 \right)\left( \frac{7}{2}-3 \right)}$, to get the answer.
Now, do not solve this. Just by observing it, you will see that there are four 2s in the denominator with the $\sqrt{{}}$ sign. Hence, the denominator will be $\sqrt{2^{4}}$ = 4. Also 7 is a prime number so $\sqrt{7}\times$… will have $\sqrt{7}$(or one of its form) as the numerator. Hence the answer should have $\frac{n\sqrt{7}}{4}$. This format is only in option ‘4’.