Tag: apollonius's theorem

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

The hypotenuse and the semi-perimeter of right triangle are 20 cm and 24 cm, respectively. The other two sides of the triangle are :

  1. 16 cm, 15 cm

  2. 14 cm, 16 cm

  3. 20 cm, 16 cm

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

Let the sides containing the right angle be a and b, with hypotenuse c = 20 cm. The semi-perimeter is given as 24 cm, meaning the perimeter is 48 cm and a + b + c = 48, so a + b = 28 cm. Using the Pythagorean theorem, a^2 + b^2 = 20^2 = 400. From (a + b)^2 = a^2 + b^2 + 2ab, we get 28^2 = 400 + 2ab, leading to ab = 192. Solving the quadratic equations or checking options, the sides are 12 cm and 16 cm, which are not listed in options A, B, or C. Therefore, None of these is correct.

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
$2(AD^2+BE^2+CF^2)=\displaystyle\frac{3}{2}(AB^2+BC^2+CA^2)$
$\therefore (AD^2+BE^2+CF^2):(AB^2+BC^2+CA^2)=3:4$