Question

Calculate the following in a triangle ABC, D is the midpoint of AB and E is the midpoint of AC. $$$$
i)DE if BC=8cm$$$$
ii)$\mathrm{\angle }ADE$ if $\mathrm{\angle }DBC={125}^{\circ }$$$$$
b)If ${\displaystyle \frac{9^n{.3}^2\cdot 3^n{27}^n}{2^{3m}}}={\displaystyle \frac{3}{8}},$Find the value of $m-n$ where $m,n$ are integers. $$$$

Answer 1

Hint: We prove the similarity of triangles ADE and ABC in part(a). We use the ratio of sides to get DE. We use the equality of corresponding angles to get $\mathrm{\angle }ADE$. We prime factorize 9 and 27 in the second part(b) and use identities like ${\left(a^m\right)}^n=a^{mn}$,${\left(a^m\right)}^n=a^{mn}$ to get an expression. We check for which integral values of $x$ the equation satisfies.

Complete step-by-step solution:
(a) We have the triangle ABC , D is the midpoint o f AB and E is the midpoint of AC , $AD=DB,AE=EB$ So we have
$$\begin{array}{rl}& AB=AD+DB=AD+AD=2AD\\ & AC=AE+EC=AE+AE=2AE\end{array}$$
 

We have joined DE. . We know that lie joining the midpoints two sides will be parallel to the other side. So $DE||BC$. $$$$
 We observe the triangles ADE and ABC. We have corresponding angles formed by the line AB cutting the parallel lines DE and BC . They will be equal. So we have $\mathrm{\angle }ABC=\mathrm{\angle }ADE$ . Similarly AC cuts the parallel lines DE and BC and makes equal corresponding angles$\mathrm{\angle }AED=\mathrm{\angle }ACB$. The angle $\mathrm{\angle }BAC=\mathrm{\angle }DAE$ is the common angle to both the triangles. So use angle-angle-angle similarity to conclude triangles $\mathrm{\Delta }ADE\sim \mathrm{\Delta }ABC$
So the sides will be in equal ratio which means
$$\begin{array}{rl}& {\displaystyle \frac{AD}{AB}}={\displaystyle \frac{AE}{AC}}={\displaystyle \frac{DE}{BC}}\\ & \Rightarrow {\displaystyle \frac{AD}{2AD}}={\displaystyle \frac{AE}{2AE}}={\displaystyle \frac{DE}{BC}}\\ & \Rightarrow {\displaystyle \frac{1}{2}}={\displaystyle \frac{1}{2}}={\displaystyle \frac{DE}{BC}}\\ & \Rightarrow DE={\displaystyle \frac{1}{2}}BC\end{array}$$
 (i)We are given that BC=8cm. So we find $DE={\displaystyle \frac{1}{2}}BC={\displaystyle \frac{1}{2}}\times 8=4$cm.
(ii) We are given $\mathrm{\angle }DBC={125}^{\circ }$. So we have by corresponding angles $\mathrm{\angle }ADE=\mathrm{\angle }DBC={125}^{\circ }$ $$$$
(b) We are given
${\displaystyle \frac{9^n{.3}^2\cdot 3^n-{27}^n}{2^{3m}}}={\displaystyle \frac{1}{8}},$
We proceed by replacing the composite numbers 9 and 27 by their prime factorization.
$$\begin{array}{rl}& {\displaystyle \frac{{(3\times 3)}^n{.3}^2{.3}^n-{(3\times 3\times 3)}^n}{2^{3m}}}={\displaystyle \frac{1}{8}}\\ & \Rightarrow {\displaystyle \frac{{\left(3^2\right)}^n{.3}^2{.3}^n-{\left(3^3\right)}^n}{2^{3m}}}={\displaystyle \frac{1}{8}}\end{array}$$
We use the formula ${\left(a^m\right)}^n=a^{mn}$ where $a,m,n$ are real numbers and get
$$\Rightarrow {\displaystyle \frac{3^{2n}{.3}^2{.3}^n-3^{3n}}{2^{3m}}}={\displaystyle \frac{1}{8}}$$
We use the formula $a^m\cdot a^n=a^{m+n}$ and get
$$\begin{array}{rl}& \Rightarrow {\displaystyle \frac{3^{3n+2}-3^{3n}}{2^{3m}}}={\displaystyle \frac{1}{8}}\\ & \Rightarrow {\displaystyle \frac{3^{3n}{.3}^2-3^{3n}}{2^{3m}}}={\displaystyle \frac{1}{8}}\\ & \Rightarrow {\displaystyle \frac{3^{3n}(9-1)}{2^{3m}}}={\displaystyle \frac{1}{8}}\\ & \Rightarrow {\displaystyle \frac{3^{3n}}{2^{2m}}}={\displaystyle \frac{1}{8\times 8}}={\displaystyle \frac{1}{64}}\end{array}$$
The above result is true when for integral values of $m,n$. The above is result true when $3^{3n}=1=3^0$. Equating exponent we get $n=0$. Similarly we have $2^{2m}=64=2^6$ and by equating exponents we get $m=3$. So the asked result is $m-n=3-0=3$$$$$

Note: We need to take care of confusion of similarity from congruence which is the equality of angles and sides of two different triangles. We can only find a non-integral solution when the base of the exponents $m,n$ are the same. We take care of the fact when we use the formula ${\left(a^m\right)}^n=a^{mn}$ that both $a$ and $m$ are not zero at the same time.