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

Let $\Delta _1$ denotes the area of the triangle formed by the vertices $(a^3m^3 _1, am _1), (a^3m^3 _2am _2), (a^3m^3 _3, am _3)$ and $\Delta _2$ denotes the area of the triangle formed by the vertices $(2am _1m _2, a^2(m^2 _1+m^2 _2))$, $(2am _2m _3, a^2(m^2 _2+m^2 _3))$ and $(2am _3m _1, a^2(m^2 _3+m^2 _1))$. Then $\dfrac{\Delta _1}{\Delta _2}(a > 0)$ equals?

  1. $\dfrac{a}{2}$
  2. $2a$
  3. $\dfrac{a^3}{8}$
  4. $8a^3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Calculating the area of the triangles using the determinant formula for coordinates and simplifying the ratio yields a/2.

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
$\displaystyle = \frac{1}{2}\left ( \frac{4}{\sqrt{3}} \right )\left ( 2\sqrt{\frac{11}{3}} \right )= \frac{4}{3}\sqrt{11}$.