Tag: angle theorems for a right angled triangle

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Multiple choice maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

P, Q, R are the points of intersection of a line 1 with sides BC, CA, AB of a $\Delta$ ABC 
respectively, then $\dfrac{BP}{PC} \dfrac{CQ}{QA} \dfrac{AR}{RB}$

  1. 1

  2. 2

  3. -1

  4. -2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a direct application of Menelaus' Theorem, which states that for a line intersecting the sides of a triangle, the product of the ratios of the segments is 1.

Multiple choice maths pythagoras theorem similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

Sides of triangle are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.

  1. 7 cm, 24 cm, 25 cmj

  2. 3 cm, 8 cm, 6 cm

  3. 50 cm, 80 cm, 100 cm

  4. 13 cm, 12 cm, 5 cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For option A: 7, 24, 25. Check if right triangle: 7² + 24² = 49 + 576 = 625 = 25². Also forms valid triangle (7+24 > 25). Option A is a right triangle with hypotenuse 25 cm. The question has typo ('25 cmj') but this doesn't affect the answer.

Multiple choice maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

The lengths of the medians through acute angles of a right-angled triangle are 3 and 4. Find the area of the triangle:

  1. $\displaystyle \frac{4}{3}\sqrt{11}$
  2. $\displaystyle \frac{2}{3}\sqrt{11}$
  3. $\displaystyle \frac{1}{3}\sqrt{11}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, $AD=3,CE=4$
Using Appaloneaus theorem for median $AD$
We have $\displaystyle{ c }^{ 2 }+{ b }^{ 2 }=2\left( \frac { { a }^{ 2 } }{ 4 } +9 \right) $   ...(1)
Using Appaloneaus theorem for median $CE$
We have $\displaystyle{ b }^{ 2 }+{ a }^{ 2 }=2\left( \frac { { c }^{ 2 } }{ 4 } +10 \right) $   ...(2)
Also, ${ a }^{ 2 }+{ c }^{ 2 }={ b }^{ 2 }$
Adding (1) and (2)
$\displaystyle 3{ b }^{ 2 }=2\left( \frac { { b }^{ 2 } }{ 4 } +25 \right) \Rightarrow { b }^{ 2 }=20$
Solving (1) and (2) we get,
$\displaystyle c=\frac { 4 }{ \sqrt { 3 }  }$ and $\displaystyle a=2\frac { 4 }{ \sqrt { 3 }  } $
Hence, area of triangle
$\displaystyle = \frac{1}{2}\left ( \frac{4}{\sqrt{3}} \right )\left ( 2\sqrt{\frac{11}{3}} \right )= \frac{4}{3}\sqrt{11}$.

Multiple choice maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

Let $ABC$ be a fixed triangle and $P$ be variable point in the plane of a triangle $ABC$. Suppose $a, b, c$ are lengths of sides $BC,  CA,AB$ opposite to angles $A, B, C $ respectively. If $a(PA)^{2} + b(PB)^{2} + c(PC)^{2}$ is minimum, then the point $P$ with respect to $\triangle{ABC}$ is

  1. Centroid

  2. Circumcenter

  3. Orthocenter

  4. Incenter
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The point P that minimizes the weighted sum of squared distances a(PA)^2 + b(PB)^2 + c(PC)^2 is the centroid of the triangle.

Multiple choice maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

If $AD,BE$ and $CF$ are the medians of a $\Delta ABC,$ then evaluate  $\displaystyle \left ( AD^{2}+BE^{2}+CF^{2} \right ):\left ( BC^{2}+CA^{2}+AB^{2} \right )=$

  1. $3:4$
  2. $4:3$
  3. $5:3$
  4. $4:1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, $AD,BE$ and $CF$ are the medians of a $\Delta ABC$.
$\Rightarrow AB^2+AC^2=2(AD^2+BD^2)$
$\Rightarrow AB^2+AC^2=2AD^2+\displaystyle\frac{BC^2}{2}$
$\Rightarrow 2AD^2=AB^2+AC^2-\displaystyle\frac{BC^2}{2}$ -----(1)
Similarly,
$2BE^2=BC^2+BA^2-\displaystyle\frac{AC^2}{2}$ -----(2)
$2CF^2=CA^2+CB^2-\displaystyle\frac{AB^2}{2}$ -----(3)
Adding equation 1,2 and 3, we get
$2(AD^2+BE^2+CF^2)=\displaystyle\frac{3}{2}(AB^2+BC^2+CA^2)$
$\therefore (AD^2+BE^2+CF^2):(AB^2+BC^2+CA^2)=3:4$