online user counter

MATHMAA

Answers Of Sample Paper contain all the solutions of the question in sample paper .  It has answers with detail explanation.

1Q)  If a, b , c are measures which form a triangle. For all n=2,3,4...etc, Prove  that $$\sqrt[n]{a}, \sqrt[n]{b}, \sqrt[n]{c}$$
also form a triangle .

Solution:

If a, b , c  are the measures of triangle then sum of two sides is greater than the third side.

a+b > c

For all n=2,3,4,...etc, to prove $$\sqrt[n]{a}, \sqrt[n]{b}, \sqrt[n]{c}$$ forms a triangle , it is enough to show that $$\sqrt[n]{a}+\sqrt[n]{b}> \sqrt[n]{c}$$ .

Using Binomial Theorem we get

$$( \sqrt[n]{a}+\sqrt[n]{b} )^{n}$$ = $$a+\binom{n}{1}(\sqrt[n]{a})^{n-1}\sqrt[n]{b}+\binom{n}{2}(\sqrt[n]{a})^{n-2}(\sqrt[n]{b})^{2}+\cdots +b$$.

.

$$(\sqrt[n]{a}+\sqrt[n]{b})^{n} > a+b > c= (\sqrt[n]{c})^{n}$$

$$(\sqrt[n]{a}+\sqrt[n]{b}) > \sqrt[n]{c}$$.

Hence for all values of n=2, 3, 4 ....etc, $$\sqrt[n]{a}, \sqrt[n]{b}, \sqrt[n]{c}$$ also form a triangle .

2Q) A square sheet of paper PQRS is so folded that the point Q fall on the mid point M of RS. Prove that the crease will divide QR in the ratio 5:3 .

Solution :

Let side of square = x .

Q falls on the mid point M of RS.

Let crease be XY and  YR=a then QY=x-a

We know that QY=YM as Q falls on M .

YR = a , YM=x-a  and triangle YMR is right angled triangle.

Using Pythagoras Theorem we get

$$MR^{2}+RY^{2}=YM^{2}$$

$$\left ( \frac{x}{2} \right )^{2}$$ +$$a^{2}$$=$$\left ( x-a \right )^{2}$$

$$\frac{x^{2}}{4} + a^{2} =x^{2}-2ax + a^{2}$$

$$2ax=\frac{3x^2}{4}$$

$$a =\frac{3x}{8}$$

$$\frac{x}{a}=\frac{8}{3}$$

$$\frac{x-a}{a}=\frac{8-3}{3}$$

$$\frac{x-a}{a} =\frac{5}{3}$$

$$\frac{QY}{YR} =\frac{5}{3}$$

QY:YR =5:3

Hence it is proved .

3Q) Prove that in any triangle ABC, if one angle is $$120^{\circ}$$, the triangle formed by the feet of the angle bisectors is a right  angled.