# Triangle MPT with bar(NR) || bar(MT) is shown below. The dimensions are in centimeters. 12210107121.jpg Which measurement is closest to the length of bar(RT) in centimeters? 1)1.4 2)1.8 3)3.4 4)4.3

Question
Similarity
Triangle MPT with $$\displaystyle\overline{{{N}{R}}}{\mid}{\mid}\overline{{{M}{T}}}$$ is shown below. The dimensions are in centimeters.

Which measurement is closest to the length of $$\displaystyle\overline{{{R}{T}}}$$ in centimeters?
1)1.4
2)1.8
3)3.4
4)4.3

2020-11-06
Step 1
Given,
$$\displaystyle\triangle{M}{P}{T}$$ with $$\displaystyle\overline{{{N}{R}}}{\mid}{\mid}\overline{{{M}{T}}}$$
PN = 4 cm
PR = 2.5 cm
NM = 2.9 cm
We have to find the length of $$\displaystyle\overline{{{R}{T}}}$$ in centimeters.
Step 2
We have
$$\displaystyle\overline{{{N}{R}}}{\mid}{\mid}\overline{{{M}{T}}}$$
Then, using the properties for two parallel lines cut by transversal lines
$$\displaystyle\angle{P}{N}{R}=\angle{P}{M}{T}$$ ..........(1)
$$\displaystyle\angle{P}{R}{N}=\angle{P}{T}{M}$$ ............(2)
Now,
In $$\displaystyle\triangle{N}{P}{R}{\quad\text{and}\quad}\triangle{M}{P}{T}$$
$$\displaystyle\angle{P}{N}{R}=\angle{P}{M}{T}$$ (from (1))
$$\displaystyle\angle{P}{R}{N}=\angle{P}{T}{M}$$ (from (2))
$$\displaystyle\angle{N}{P}{R}=\angle{M}{P}{T}$$ (common angle )
So,
From AAA (angle-angle-angle) rule of similarity
$$\displaystyle\triangle{N}{P}{R}\sim\triangle{M}{P}{T}$$
Step 3
Now,
In similar triangle, ratio of corresponding sides will be equal.
$$\displaystyle\frac{{{N}{P}}}{{{M}{P}}}=\frac{{{P}{R}}}{{{P}{T}}}=\frac{{{N}{R}}}{{{M}{T}}}$$
Taking
$$\displaystyle\frac{{{N}{P}}}{{{M}{P}}}=\frac{{{P}{R}}}{{{P}{T}}}$$
$$\displaystyle\Rightarrow\frac{{4}}{{{M}{N}+{N}{P}}}=\frac{{2.5}}{{{P}{R}+{R}{T}}}$$ (substituted the values)
$$\displaystyle\Rightarrow\frac{{4}}{{{2.9}+{4}}}=\frac{{2.5}}{{{2.5}+{R}{T}}}$$
$$\displaystyle\Rightarrow\frac{{4}}{{6.9}}=\frac{{2.5}}{{{2.5}+{R}{T}}}$$
$$\displaystyle\Rightarrow{4}{\left({2.5}+{R}{T}\right)}={6.9}\times{2.5}$$
$$\displaystyle\Rightarrow{4}{R}{T}+{10}={17.25}$$
$$\displaystyle\Rightarrow{4}{R}{T}={17.25}-{10}$$
$$\displaystyle\Rightarrow{4}{R}{T}={7.25}$$
$$\displaystyle\Rightarrow{R}{T}=\frac{{7.25}}{{4}}$$
$$\displaystyle\Rightarrow{R}{T}={1.8125}$$
So,
The lenght of RT = 1.8cm

### Relevant Questions

To prove : The similarity of $$\displaystyle\triangle{N}{R}{T}$$ with respect to $$\displaystyle\triangle{N}{S}{P}$$.
Given information: Here, we have given that $$\displaystyle\overline{{{S}{P}}}$$ is altitude to $$\displaystyle\overline{{{N}{R}}}\ {\quad\text{and}\quad}\ \overline{{{R}{T}}}$$ is altitude to $$\displaystyle\overline{{{N}{S}}}$$.
The student engineer of a campus radio station wishes to verify the effectivencess of the lightning rod on the antenna mast. The unknown resistance $$\displaystyle{R}_{{x}}$$ is between points C and E. Point E is a "true ground", but is inaccessible for direct measurement because the stratum in which it is located is several meters below Earth's surface. Two identical rods are driven into the ground at A and B, introducing an unknown resistance $$\displaystyle{R}_{{y}}$$. The procedure for finding the unknown resistance $$\displaystyle{R}_{{x}}$$ is as follows. Measure resistance $$\displaystyle{R}_{{1}}$$ between points A and B. Then connect A and B with a heavy conducting wire and measure resistance $$\displaystyle{R}_{{2}}$$ between points A and C.Derive a formula for $$\displaystyle{R}_{{x}}$$ in terms of the observable resistances $$\displaystyle{R}_{{1}}$$ and $$\displaystyle{R}_{{2}}$$. A satisfactory ground resistance would be $$\displaystyle{R}_{{x}}{<}{2.0}$$ Ohms. Is the grounding of the station adequate if measurments give $$\displaystyle{R}_{{1}}={13}{O}{h}{m}{s}$$ and R_2=6.0 Ohms?
To prove: The similarity of $$\displaystyle\triangle{N}{W}{O}$$ with respect to $$\displaystyle\triangle{S}{W}{T}$$.
Given information: Here, we have given that NPRV is a parallelogram.
To prove: The similarity of $$\displaystyle\triangle{B}{C}{D}$$ with respect to $$\displaystyle\triangle{F}{E}{D}$$.
Given information: Here, we have given that $$\displaystyle\overline{{{A}{C}}}\stackrel{\sim}{=}\overline{{{A}{E}}}\ {\quad\text{and}\quad}\ \angle{C}{B}{D}\stackrel{\sim}{=}\angle{E}{F}{D}$$

A block of mass m=3.6 kg, moving on africtionless surface with a speed $$\displaystyle{v}_{{1}}={9.3}$$ m/s makes a perfectly elastic collision with a block of mass Mat rest. After the collision, the 3.6 kg block recoils with a speed of $$\displaystyle{v}_{{1}}={2.7}$$ m/s in figure, the speed of the vlock of mass M after the collision is closest to:
a. 9.3 m/s
b. 6.6 m/s
c. 8.0 m/s
d. 10.7 m/s
e. 12.0 m/s
Two oppositely charged but otherwise identical conducting plates of area 2.50 square centimeters are separated by a dielectric 1.80 millimeters thick, with a dielectric constant of K=3.60. The resultant electric field in the dielectric is $$\displaystyle{1.20}\times{10}^{{6}}$$ volts per meter.
Compute the magnitude of the charge per unit area $$\displaystyle\sigma$$ on the conducting plate.
$$\displaystyle\sigma={\frac{{{c}}}{{{m}^{{2}}}}}$$
Compute the magnitude of the charge per unit area $$\displaystyle\sigma_{{1}}$$ on the surfaces of the dielectric.
$$\displaystyle\sigma_{{1}}={\frac{{{c}}}{{{m}^{{2}}}}}$$
Find the total electric-field energy U stored in the capacitor.
u=J
To check: whether the additional information in the given option would be enough to prove the given similarity.
Given:
The given similarity is $$\displaystyle\triangle{A}{D}{C}\sim\triangle{B}{E}{C}$$

The given options are:
A.$$\displaystyle\angle{D}{A}{C}{\quad\text{and}\quad}\angle{E}{C}{B}$$ are congruent.
B.$$\displaystyle\overline{{{A}{C}}}{\quad\text{and}\quad}\overline{{{B}{C}}}$$ are congruent.
C.$$\displaystyle\overline{{{A}{D}}}{\quad\text{and}\quad}\overline{{{E}{B}}}$$ are parallel.
D.$$\displaystyle\angle{C}{E}{B}$$ is a right triangle.
Suppose the light bulb in the figure below is replaced with a short wire of zero resistance, and the resistance of the rails is negligible. The only resistance is from the moving rod, which is iron (resistivity $$\displaystyle={9.50}\times{10}^{{-{8}}}$$ ohm.m). The rod has a cross-sectional area of $$\displaystyle{3.50}\times{10}^{{-{6}}}$$m